chem1101 2014-j-9 june 2014 • use the following equilibria

CHEM1101 2014-J-9 June 2014

bull Use the following equilibria

2CH4(g) C2H6(g) + H2(g) K1 = 95 times 10ndash13

CH4(g) + H2O(g) CH3OH(g) + H2(g) K2 = 28 times 10ndash21

to calculate the equilibrium constant K3 for the following reaction

2CH3OH(g) + H2(g) C2H6(g) + 2H2O(g)

Show all working

Marks 2

The equilibrium constant expressions for reactions (1) and (2) are given by

K1 = 119810120784119815120788 119840 [119815120784 119840 ]

[119810119815120786 119840 ]120784 and K2 =

119810119815120785119822119815 119840 [119815120784 119840 ][119810119815120786 119840 ][119815120784119822 119840 ]

The equilibrium constant for reaction (3) is

K3 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

[119810119815120785119822119815 119840 ]120784[119815120784 119840 ]

K3 can be obtained by multiplying K1 by 1 K22

K1 K22 =

119810120784119815120788 119840 [119815120784 119840 ][119810119815120786 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]120784

[119810119815120786 119840 ]120784[119815120784119822 119840 ]120784 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

[119810119815120785119822119815 119840 ]120784[119815120784 119840 ] = K3

K3 = (95 times 10ndash13) (28 times 10ndash21)2 = 12 times 1029

Answer 12 times 1029

CHEM1101 2014-J-10 June 2014

bull Consider the following reaction

N2O4(g) 2NO2(g) Kc = 461 times 10ndash3 at 25 degC

A 0086 mol sample of NO2 is allowed to come to equilibrium with N2O4 in a 050 L container at 25 degC Calculate the amount (in mol) of NO2 and N2O4 present at equilibrium Show all working

Marks 4

The initial concentration of NO2(g) is [NO2(g)] = number of moles volume = 0086 mol 050 L = 0172 mol L-1

A reaction table can be used to calculate the equilibrium concentrations

N2O4(g) 2NO2(g) initial 0000 0172

change +x -2x

equilibrium x 0172 ndash 2x

Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120783120789120784120784119961)

120784

119961 = 461 times 10ndash3

So

(0172 ndash 2x)2 = 461 times 10ndash3 x 00296 ndash 0688x + 4x2 = 461 times 10ndash3 x 4x2 ndash 0693x + 00296 = 0

Solving the quadratic gives x = 00766 and x = 00965 The second route would give a negative value for [NO2(g)] Using x = 00766

[N2O4(g)] = 00766 M and [NO2(g)] = (0172 ndash 2 times 00766) M = 00188 M

As these concentrations are in a 050 L container

number of moles of N2O4(g) = concentration times volume

= 00766 mol L-1 times 050 L = 0038 mol number of moles of NO2(g) = 00188 mol L-1 times 050 L = 00094 mol

Amount of NO2 00094 mol Amount of N2O4 0038 mol

CHEM1101 2014-N-10 November 2014

bull Use the following equilibria 2CH4(g) C2H6(g) + H2(g) K1 = 95 times 10ndash13



2CH3OH(g) + H2(g) C2H6(g) + 2H2O(g)

Show all working

Marks 2

The equilibrium constant expressions for the three reactions are

K1 = 119810120784119815120788 119840 [119815120784 119840 ]

[119810119815120786 119840 ]120784 K2 =

119810119815120785119822119815 119840 [119815120784 119840 ]119810119815120786 119840 [119815120784119822 119840 ]

K3 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]

Hence K3 = K1 K2

2

K1 K22 =

119810120784119815120788 119840 [119815120784 119840 ][119810119815120786 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]120784

119810119815120786 119840 120784[119815120784119822 119840 ]120784 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]

= (95 times 10ndash13) (28 times 10ndash21)2 = 12 times 1029



bull The standard Gibbs free energy of the following reaction is +6973 kJ molndash1 COCl2(g) CO(g) + Cl2(g)

What is the expression for the equilibrium constant Kp for this reaction

Marks 5

Kp = 119927119810119822119927119810119845120784119927119810119822119810119845120784

Calculate the value of the equilibrium constant at 298 K

Using ΔGo = -RTlnKp +6973 times 103 J = -(8314 J K-1 mol-1) times (298 K) times lnKp lnKp = -2814 Kp = 598 times 10-13

Kp = 598 times 10-13

In which direction will this reaction proceed if a mixture of gases is made with PCOCl2 = 100 atm PCl2 = 001 atm PCO = 050 atm Show working

Using Qp = 119927119810119822119927119810119845120784119927119810119822119810119845120784

Qp = (120782120787120782)(120782120782120783)(120783120782120782)120784

= 0005 As Qp gt Kp the reaction will proceed towards reactants PCOCl2 will increase PCO and PCl2 will decrease

CHEM1101 2013-J-9 June 2013


N2O4(g) 2NO2(g)

An equilibrium mixture in a 100 L container is found to contain [N2O4] = 100 M and [NO2] = 046 M The vessel is then compressed to half its original volume while the temperature is kept constant Calculate the concentration [N2O4] when the compressed system has come to equilibrium Show all working

Marks 4

For this reaction the equilibrium constant expression is given by

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786]

As mixture is at equilibrium when [N2O4] = 100 M and [NO2] = 046 M

Kc = (120782120786120788)120784

(120783120782120782) = 021

If the volume of the vessel is halved the initial concentrations will double [N2O4] = 200 M and [NO2] = 092 M The reaction is no longer at equilibrium and Le Chatelierrsquos principle predicts it will shift towards the side with fewer moles it will shift towards reactants A reaction table needs to be used to calculate the new equilibrium concentrations

N2O4(g) 2NO2(g) initial 200 092

change +x -2x

equilibrium 200 + x 092 ndash 2x

Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120791120784120784119961)

120784

(120784120782120782119961) = 021

So

(092 ndash 2x)2 = 021 (200 + x) 08464 - 368x + 4x2 = 042 + 021x

4x2 ndash 389x + 043 = 0

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1101 2013-J-9 June 2013

With a = 4 b = -389 and c = 043 this quadratic equation has roots

x = 119939 plusmn 119939120784120786119938119940

120784119938 = 120785120790120791 plusmn (120785120790120791)120784120786times120786times120782120786120785

120784times120786

This gives x = 013 or 085 The latter makes no chemical sense as it gives a negative concentration for NO2

Hence using x = 013

[N2O4] = (200 + x) M = (200 + 013) M = 213 M

[NO2] = (092 - 2x) M = (092 ndash 2 times 013) M = 066 M

Answer [N2O4] = 213 M


bull The standard Gibbs free energy of the following reaction is +6973 kJ molndash1

COCl2(g) CO(g) + Cl2(g)


Marks 5

Kp = p(Cl2 )p(CO)p(COCl2 )


As ∆Gdeg = ndashRT lnK 6973 times 103 = -(8314 times 298) times lnK K = 598 times 10ndash13

Kp = 598 times 10ndash13


The reaction quotient Qp is

Qp =p(Cl2 )p(CO)p(COCl2 )

= (001)(050)(100)

= 0005

As Qp gt Kp the reaction proceeds to decrease Qp It moves to reduce the amount of produce and increase the amount of reactant it shifts to the left (towards reactants)

THIS QUESTION CONTINUES ON THE NEXT PAGE


This reaction mixture is now allowed to come to equilibrium at 298 K in a fixed volume container Calculate the equilibrium pressure of Cl2

Marks 4

As shown in 2013-N-12 the reaction will shift towards reactants to reach equilibrium The equilibrium constant for the forward reaction is very small K = 598 times 10ndash13 With PCl2 = 001 atm initially it is easiest to assume that (i) all of the Cl2 reacts and (ii) a little of the COCl2 then reacts to reach equilibrium

(i) All of the Cl2 reacts with CO to make COCl2

PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm

PCO = (050 - 001) atm = 049 atm (iii) This reaction will then shift to equilibrium


initial 101 049 000

change -x +x +x

equilibrium 101 ndash x 049 + x x


= x(049+ x)(101minus x)

As the equilibrium constant is so small x will be tiny In this case 049 + x ~ 049 and 101 ndash x ~ 101 Hence

Kp ~ x(049)(101)

= 598times10minus13

x = 598 times 10ndash13 times 101049

= 123 times 10ndash12

Answer 123 times 10ndash12

CHEM1101 2012-J-10 June 2012


SO2(g) + NO2(g) SO3(g) + NO(g)

An equilibrium mixture in a 100 L vessel was found to contain [SO2(g)] = 0800 M [NO2(g)] = 0100 M [SO3(g)] = 0600 M and [NO(g)] = 0400 M If the volume and temperature are kept constant what amount of NO(g) needs to be added to the reaction vessel to give an equilibrium concentration of NO2(g) of 0300 M

Marks 4

From the chemical equation

Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

As the original mixture is at equilibrium

Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300

This equilibrium is now disturbed by the addition of x M of NO(g) To re-establish equilibrium the reaction will shift to the left by an unknown amount y The reaction table for this is

SO2(g) NO2(g) SO3(g) NO(g)

initial 0800 0100 0600 0400 + x

change +y +y -y -y

equilibrium 0800 + y 0100 + y 0600 ndash y 0400 + x - y As [NO2(g)] = 0300 M at the new equilibrium y = (0300 ndash 0100) M = 0200 M Hence the new equilibrium concentrations are [SO2(g)] = (0800 + 0200) M = 1000 M [NO2(g)] = 0300 M [SO3(g)] = (0600 ndash 0200) M = 0400 M [NO(g)] = (0400 + x ndash 0200) M = (0200 + x) M As the system is at equilibrium

Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300

Solving this gives x = 205 M As the reaction is carried out in a 100 L container this is also the number of moles required

Answer 205 mol

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY

CHEM1101 2012-J-12 June 2012

bull The diagram below represents the Gibbs Free energy change associated with the formation of 4 different oxides

Using the free energy data above write down the equation and indicate with an arrow the direction of the expected spontaneous reaction under the following conditions If you think no reaction would occur write ldquono reactionrdquo

Marks 4

a) C and SnO are mixed at 400 degC

At 400 degC the Sn SnO line is below the C CO line Hence there is no reaction

b) C and SnO are mixed at 900 degC SnO + C agrave Sn + CO

c) SnO Sn Zn and ZnO are mixed at 900 degC Zn + SnO agrave ZnO + Sn

Of the 4 oxide formation reactions write down one for which the entropy change is negative Provide a brief explanation for your choice

The entropy change is likely to be negative in three of them

bull 2Sn + O2 agrave 2SnO bull 2Zn + O2 agrave 2ZnO bull 43Al+ O2 agrave 23AlO

Each of these involves a decrease in the number of moles of gas Gases have far higher entropy than solids The fourth reaction involves an increase in the number of moles of gas so is likely to produce an increase in entropy

ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)


bull The standard Gibbs free energy of formation for ammonia NH3(g) is ndash164 kJ molndash1 Consider the following reaction at 298 K

N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )


The reaction as written produces 2 mol of NH3 hence ΔrGo = 2 times ΔfGo(NH3(g)) = 2 times -164 kJ mol-1 = -328 kJ mol-1 As ΔGo = -RTlnKp -328 times 103 J mol-1 = -(8314 J K-1 mol-1) times (298 K) times lnKp lnKp = 1324 Kp = 56 times 105

Kp = 56 times 105

In which direction will this reaction proceed if a mixture of gases is made with PNH3 = 100 atm PH2 = 100 atm PN2 = 050 atm

The reaction quotient with this mixture is

Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200

As Qp lt Kp the reaction will proceed to products (ie to the right)





Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010

bull Ammonia NH3(g) has a standard Gibbs free energy of formation equal to ndash164 kJ molndash1 Consider the following reaction at 298 K

N2(g) + 3H2(g) 2NH3(g)


Marks 6

The reaction as written forms 2 moles of NH3(g) The standard Gibbs free energy of the reaction is therefore (2 times -164 kJ mol-1) = -328 kJ mol-1 Using ∆Gdeg = -RTlnKp the value of Kp can be determined lnKp = - ∆Gdeg RT = -(-328 times 103 J mol-1) (8314 J K-1 mol-1)(298 K) = 1324 Kp = 562 times 105 The direction of the reaction can be determined by calculating the reaction quotient

Q =

=

= 16

As Q lt Kp the reaction will proceed to the right This will increase the partial pressure of NH3(g) and decrease the partial pressures of N2(g) and H2(g)

Answer to the right

What pressure of hydrogen gas should be added to a mixture already containing 020 atm NH3 and 050 atm N2 so that the amounts of NH3 and N2 will not change

If the partial pressures of NH3(g) and N2(g) do not change these values must be the values at equilibrium Hence

Kp =

=

= 562 times 105

= 142 times 10-7 atm3 = 52 times 10-3 atm

Answer 52 times 10-3 atm

CHEM1101 2010-J-11 June 2010

bull Determine the value of the equilibrium constant (at 298 K) for the following reaction

CO2(g) + H2O(l) H2CO3(aq)

Marks 3

Substance ∆fHordm kJ molndash1 Sordm J Kndash1 molndash1

H2CO3(aq) ndash700 187

H2O(l) ndash286 70

CO2(g) ndash394 214

Using ∆rxnHdeg = Σm∆fHdeg(products) - Σn∆fHdeg(reactants) the enthalpy of this reaction is

∆Hdeg = (∆fHdeg(H2CO3(aq)) - (∆fHdeg(CO2(g) + ∆fHdeg(H2O(l))

= [(-700) ndash (-286 + -394)] kJ mol-1 = -20 kJ mol-1

Using ∆rxnSdeg = ΣmSdeg(products) - ΣnSdeg(reactants) the entropy change of this reaction is

∆Sdeg = (Sdeg(H2CO3(aq)) - (Sdeg(CO2(g) + Sdeg(H2O(l))

= [(187) ndash (214 + 70))] kJ mol-1 = -97 J K-1 mol-1

Using ∆Gdeg = ∆Hdeg - T∆Sdeg ∆Gdeg = (-20 times 103 J mol-1) ndash (298 K)(-97 J K-1 mol-1) = +8906 J mol-1 Using ∆Gdeg = -RTlnKp the value of Kp can be determined lnKp = - ∆Gdeg RT = -(+8906 J mol-1) (8314 J K-1 mol-1)(298 K) = -3595 Kp = 0027

Answer 0027

bull Consider the following equilibrium

CO(g) + H2O(g) CO2(g) + H2(g) Kc = 314 at 588 K

If a 1000 L vessel contains 250 mol CO(g) 250 mol H2O(g) 500 mol CO2(g) and 500 mol H2(g) at 588 K what are the concentrations of all species at equilibrium

2

The initial concentrations are

[CO(g)] = number of moles volume = 250 mol 1000 L = 025 M

[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010

The reaction table is therefore

CO(g) H2O(g) CO2(g) H2(g)

initial 0250 0250 0500 0500

change -x -x +x +x

equilibrium 0250 - x 0250 - x 0500 + x 0500 + x The equilibrium constant is therefore

Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560

0500 + x = 560 times (0250 ndash x) 0500 + x = 140 ndash 560x x = 0136 M

Finally [CO(g)] = (0250 ndash 0136) M = 0114 M

[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M


bull The reaction below is endothermic

N2O3(g) NO(g) + NO2(g)

Indicate whether the equilibrium will shift right shift left or remain unchanged when disturbed in the following ways

Marks3

adding more NO(g) left

increasing the pressure at constant temperature left

removing NO2(g) right

increasing the volume at constant temperature right

adding some Ar(g) no change

increasing the temperature at constant pressure right


bull At 1000 K a reaction mixture containing SO2(g) O2(g) and SO3(g) was allowed to come to equilibrium in a reaction vessel The reaction is

2SO2(g) + O2(g) 2SO3(g)

At equilibrium the system was found to contain the following concentrations [SO2] = 000377 M [O2] = 000430 M and [SO3] = 000185 M Calculate Kc for this reaction

Marks3

The equilibrium constant in terms of concentrations is given by

Kc = =

= 56

Kc = 56

If a mixture containing [SO2] = 00471 M [O2] = 00280 M and [SO3] = 000125 M is placed in the vessel is the reaction at equilibrium If not which way will it shift in order to achieve equilibrium right or left

With these initial concentrations

Q = =

= 0025

As Q lt Kc the reaction will proceed to the right to increase the concentration of the product and reduce the concentrations of the reactants


bull What does it mean to say that a reaction has reached equilibrium Marks1

A reaction is at equilibrium when the rate of the forward reaction equals the rate of the backward reaction


Br2(g) + Cl2(g) 2BrCl(g) Kc = 70 at 400 K

If 044 mol of Br2 and 044 mol of Cl2 are introduced into a 20 L container at 400 K what are the equilibrium concentrations of Br2(g) Cl2(g) and BrCl(g)

4

The initial concentrations are [Br2(g)] = number of moles volume = (044 mol) (20 L) = 022 M [Cl2(g)] = (044 mol) (20 L) = 022 M [BrCl(g)] = 0 M A reaction table can then be used

Br2(g) Cl2(g) 2BrCl(g) initial 022 022 0 change -x -x +2x

equilibrium 022 ndash x 022 ndash x 2x

The equilibrium constant in terms of concentrations is then

Kc = =

=

= 70

or

= (70)12

x = 0125 Hence [Br2(g)] = (022 ndash 0125) M = 0095 M [Cl2(g)] = (022 ndash 0125) M = 0095 M [BrCl(g)] = 025 M

[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009

Consider the following reaction at equilibrium

CH3OH(g) CO(g) + 2H2(g) Kc = 130 10ndash2

What is the concentration of CO(g) when [CH3OH(g)] = 349 10ndash1

M and

[H2(g)] = 176 10ndash1

M

Marks

2

The equilibrium constant for the reaction is given by

Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2

When [CH3OH(g)] = 349 times 10-1

M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =

120783120785120782 times 120783120782minus120784 (120785120786120791 120783120782ndash120783)

(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M

Explain briefly the chemical principles behind a) froth flotation or b) electrorefining 2

Froth flotation is a technique to separate a mineral from unwanted dirt and

rocks The crude ore is crushed to a fine powder and then treated with water to

produce a slurry A surfactant that selectively coats the mineral thus making it

more hydrophobic is added and the mixture agitated and aerated The mineral

attaches to the air bubbles and floats to the surface (as a froth) where it is

collected before undergoing further refining

Electrorefining is a technique for purifying a metal eg copper An electrolytic

cell consisting of a pure copper cathode and an impure copper anode is

constructed A voltage is selectively applied so that noble metals (less

electropositive than Cu) do not dissolve When operating the current causes the

impure copper anode to dissolve including metal impurities more electropositive

than copper The noble metals do not dissolve and form a sludge Only pure

copper is deposited at the cathode - the more electropositive metals stay in

solution as cations

CHEM1101 2009-J-10 June 2009

Solid NH4HS in placed in an evacuated container at 25 ordmC and the following

equilibrium is established

NH4HS(s) NH3(g) + H2S(g) Hordm = +93 kJ molndash1

At equilibrium some solid NH4HS remains in the container Predict and explain each

of the following

(a) The effect on the equilibrium partial pressure of NH3 gas when additional solid

NH4HS is introduced into the container

Marks

3

The equilibrium constant is given by Kp = [NH3(g)][H2S(g)] It does not include

the concentration of the solid as this is a constant during the reaction

If extra solid is introduced it does not change the concentration of the solid and

there is no effect on the equilibrium The solid is not included in the equilibrium

constant so changing the amount present has no effect on the equilibrium

(b) The effect on the amount of solid NH4HS present when the volume of the

container is decreased

Decreasing the volume of the container will increase the partial pressure of both

gases The reaction will move to the left to reduce the overall pressure

The amount on solid will therefore increase

(c) The effect on the amount of solid NH4HS present when the temperature is

increased

The reaction is endothermic as Hordm is positive If the temperature is increased

the reaction will move to the right to absorb the extra heat

The amount of solid will therefore decrease

CHEM1101 2009-J-13 June 2009

Fe2O3 can be reduced by carbon monoxide according to the following equation

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) Kp =199 at 1000 K

At 1000 K what are the equilibrium partial pressures of CO and CO2 if the only gas

initially present is CO at a partial pressure of 0978 atm

Marks

2

The reaction table is

Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)

initial atm - 0978 - 0

change atm -3x +3x

equilibrium atm 0978 ndash 3x 3x

The solids do not appear in the equilibrium constant expression and do not need

to be considered The equilibrium constant in terms of partial pressures Kp is

given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271

3x = (271)(0978 ndash 3x) = 265 ndash 813x or 111x = 265 or x = 0238

From the reaction table

p(CO) = (0978 ndash 3x) atm = 0264 atm

p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm


The value of the equilibrium constant Kc for the following reaction is 0118 mol Lndash1

2CO2(g) + N2(g) 2CO(g) + 2NO(g)

What is the equilibrium concentration of CO(g) if the equilibrium concentration of

[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118

When [CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008

Explain the difference between an equilibrium constant and a reaction quotient Marks

2

Both the equilibrium constant (K) and the reaction quotient (Q) show the

relationship between the amounts of product and reactant

K refers to equilibrium concentrations and so refers to a system at equilibrium

It is a constant for any system depending only on the temperature

Q refers to concentrations that are not necessarily at equilibrium

The following reactions have been demonstrated in mammalian liver at 37 degC and

pH 75

aspartate + citrulline argininosuccinate + H2O Keq = 13 (1)

argininosuccinate arginine + fumarate Keq = 45 (2)

fumarate + NH4+ aspartate Keq =

Calculate the equilibrium constant at 37 degC and pH 75 for the following reaction

arginine + H2O citrulline + NH4+ (4)

2

If the first three reactions are combined by reversing each one and adding the

reactions together reaction (4) results

-(1) argininosuccinate + H2O aspartate + citrulline Keq = 113

-(2) arginine + fumarate argininosuccinate Keq = 145

-(3) aspartate fumarate + NH4+ Keq = 1017

(4) arginine + H2O citrulline + NH4+ Keq = (113)times(145)times(1017)

When a reaction is reversed the new equilibrium constant is the reciprocal of

the original value This has been used in the final column to obtain the

equilibrium constants for the reverse of reactions (1) (2) and (3)

When reactions are added together the new equilibrium constant is the product

of the equilibrium constants for the individual reactions

Overall this gives for reaction (4)

Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008

Methanol CH3OH is produced commercially by the catalysed reaction of carbon

monoxide and hydrogen gas Kp for this reaction at 600 K is 113 10ndash6

CO(g) + 2H2(g) CH3OH(g)

The reaction is exothermic yet the equilibrium favours the reactants Explain why

this is the case

Marks

5

For a process to be spontaneous the entropy of the universe must increase

Exothermic reactions lead to heat being given to the surroundings leading to an

increase in the entropy of the surroundings surroundingsS

Formation of reactants is endothermic so surroundingsS is negative However as

formation of reactants leads to the formation of 3 molecules from each CH3OH

it leads to an increase in the entropy of the system systemS is postivie

For reactants to be favoured requires systemS to be more positive than

surroundingsS is negative As surroundingsS = q T it becomes less negative as T

increases and so formation of reactants becomes more favourable at high T

s

The reaction vessel at 600 K is filled with 200 atm of CO(g) and 200 atm H2(g)

What is the final pressure of CH3OH(g) at equilibrium

The equilibrium constant in terms of partial pressures Kp is given by

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x

equilibrium 200 - x 200 ndash 2x x

Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6

As Kp is very small x will be tiny compared to 200 atm and so (200 ndash x) ~ 200

This approximation simplifies the expression for Kp

Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2

x = 3CH OHp = 904 times 10

-3 atm

Answer

3CH OHp = 904 times 10-3

atm


CHEM1101 2008-J-10 June 2008

Under what conditions of temperature and pressure do you think an industrial reactor

would function to optimise the production of methanol Explain

As formation of methanol in this reaction is exothermic its formation is

favoured by low temperature However as a low temperature would also lead to

a slow reaction the temperature used would be chosen as a compromise between

maximizing the yield whilst keeping the rate of its formation high

As the reaction leads to a decrease in the number of moles of gas (3 mol of gas

1 mol of gas) high pressure will favour product formation Thus the industrial

process uses high pressures


Consider the following reaction

N2O4(g) 2NO2(g)

An experiment was conducted in which 01000 mol of N2O4(g) was introduced into a

1000 L flask After equilibrium had been established at a particular temperature the

concentration of N2O4(g) was found to be 00491 M Calculate the equilibrium

constant Kc for the reaction as written at that temperature

Marks

4

As initially 01000 mol of N2O4(g) is present in 1000 L its concentration is 01000

M The reaction table is

N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x

equilibrium 01000 ndash x 2x

As [N2O4(g)]eq = 01000 ndash x M = 00491 M x = 00509 M

Hence [NO2(g)]eq = 2x = (2 times 00509) M = 0102 M

The equilibrium constant in terms of concentrations Kc is

Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211



Hydrogen cyanide HCN(g) is prepared commercially by the reaction of methane

CH4(g) ammonia NH3(g) and oxygen O2(g) at high temperature The other product

is gaseous water Write a chemical equation for the reaction

Marks

2

CH4(g) + NH3(g) + 32O2(g) HCN(g) + 3H2O(l)

What volume of HCN(g) can be obtained from 200 L of CH4(g) 200 L of NH3(g)

and 200 L of O2(g) The volumes of all gases are measured at the same temperature

and pressure

From the ideal gas law pV = nRT the number of moles is directly proportional to

the volume of the substance if the same temperature and pressure are used As

the volumes of CH4(g) NH3(g) and O2(g) are the same the number of moles of

each is also the same

From the chemical equation 15 mol of O2(g) is required for every 1 mol of

CH4(g) and every 1 mol of NH3(g) As the number of moles of O2(g) is equal to

the number of moles of CH4(g) and NH3(g) the limiting reagent is O2(g)

From the chemical equation 1 mol of HCN(g) is made from every 15 mol of

O2(g) Each mole of O2(g) will lead to 23 mol of HCN(g) Hence the volume of

HCN(g) = 23 times 200 L = 133 L

Answer 133 L

The reaction of carbon disulfide with chlorine is as follows

CS2(g) + 3Cl2(g) CCl4(g) + S2Cl2(g) H298 = ndash238 kJ molndash1

In which direction will the reaction move when the following changes are made to the

system initially at equilibrium

(a) The pressure on the system is doubled by halving the volume

3

As the reaction involves 4 mol of gas 2 mol of gas increasing the pressure

favours products (Le Chatelierrsquos principle) The reaction will shift to the right

(b) CCl4 is removed

Removing product will lead to the reaction shifting to produce more product (Le

Chatelierrsquos principle) The reaction will shift to the right

(c) The system is heated

The reaction is exothermic Increasing the temperature will cause the reaction to

reduce the amount of the exothermic reaction (Le Chatelierrsquos principle) The

reaction will shift to the left

CHEM1101 2007-J-8 June 2007

A major disadvantage of hydrogen as a fuel is that it is a gas and therefore hard to store There is an enormous world-wide effort including research performed in the University of Sydney to develop novel chemical structures in which H2 can be stored much more efficiently One of the structures being tested in the School of Chemistry is shown below

What type of intermolecular force (or forces) are responsible for the binding between the Cu2+ and the H2

Marks 5

There are ion-induced dipole forces between Cu2+ and the H2 molecules as well as weak dispersion (London or induced dipole-induced dipole) forces

In order that such a material be useful for fuel storage the binding of the H2 must be reversible

cage(s) + H2(g) cageH2(s)

One simple way to reverse the binding is to increase the temperature so that at low temperature the equilibrium lies to the right and at high temperature to the left Use this information plus any chemical knowledge or intuition to infer the sign of ∆G ∆H and ∆S at ldquolowrdquo and ldquohighrdquo temperatures (You may assume that ∆H and ∆S do not change greatly with temperature)

At low temperature the equilibrium lies to the right favouring products ∆G lt 0 At high temperature it lies to the left favouring reactants ∆G gt 0

The reaction involves formation of a solid from a solid and a gas There is therefore a decrease in the entropy ∆S lt 0 This is true at all temperatures

The forward reaction becomes less favourable as the temperature is increased Le Chatelierrsquos principle therefore suggests that the reaction is exothermic ∆H lt 0 This is true at all temperatures (If the temperature is increased the equilibrium shifts to remove heat by increasing the backward reaction)

Temperature ∆G ∆S ∆H

low lt 0 lt 0 lt 0 high gt 0 lt 0 lt 0

H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007

bull N2(g) and O2(g) react with each other to a small extent if a catalyst is present to form nitric oxide NO(g) according to the following equation

N2(g) + O2(g) 2NO(g)

The equilibrium constant Kp for this reaction is 435 times 10minus35 at 298 K and 275 times 10minus20 at 500 K Is the reaction exothermic or endothermic Give reasons for your answer

Marks 5

The reaction is favoured at higher temperatures (ie increasing the temperature increases the amount of product) Le Chatelierrsquos principle therefore suggests that the reaction is endothermic ∆H gt 0

The partial pressures of O2(g) and N2(g) in air are 0210 atm and 0780 atm respectively If air at atmospheric pressure is sealed in a 100 L container containing the catalyst at 298 K what will be the partial pressure of NO(g) and the total pressure inside the container at equilibrium

Two moles of NO are produced for every mole of O2 and N2 that is lost

partial pressures O2 N2 NO

start 021 078 0

change -x -x +2x

equilibrium 021-x 078-x 2x

The equilibrium constant in terms of partial pressures is then

352

2N2O

2NO

p 10354)x780)(x210(

)x2())((

)( minusminusminusminustimestimestimestimes====minusminusminusminusminusminusminusminus

========pp

pK

As Kp is very small x will be tiny so that (021-x) ~ 021 and (078-x) ~ 078 to a very good approximation Substituting these approximations in gives

18352 10331x)780()210()10354(x4 minusminusminusminusminusminusminusminus timestimestimestimes====rArrrArrrArrrArrtimestimestimestimestimestimestimestimestimestimestimestimes====

The partial pressure of NO is twice this value = atm10672 18minusminusminusminustimestimestimestimes

The overall number of moles of gas does not change during the reaction so neither does the total pressure The pressure at equilibrium = 100 atm

Pressure of NO(g) 267 times 10-18 atm Total pressure 100 atm


CHEM1101 2007-J-10 June 2007

Oxidation of NO(g) to produce the pollutant NO2(g) is favoured at higher temperatures such as those in a car exhaust

2NO(g) + O2(g) 2NO2(g)

The equilibrium constant Kp for this reaction is 13 times 104 at 500 K What is the value of Kc at 500 K

Marks 6

Kp and Kc are related through the equation Kp = Kc (RT)∆n where ∆n is the change in the number of moles of gas during the reaction As three moles go to two moles in the reaction ∆n = minusminusminusminus1 and hence

1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10

minusminusminusminus====

= times = times times times = times= times = times times times = times= times = times times times = times= times = times times times = times

Kc = 53 times 105

Using this value and the equilibrium constant for the formation of NO(g) from the previous page calculate the value of Kc for the formation of NO2(g) from N2(g) and O2(g) at 500 K according to the following equation

N2(g) + 2O2(g) 2NO2(g)

From 2007-J-9 Kp = 275 times 10-20 for the reaction N2(g) + O2(g) 2NO(g) For this reaction ∆n = 0 and hence Kp = Kc (RT)0 or Kc = Kp This reaction and the one above can be combined

N2(g) + O2(g) 2NO(g) K c = 275 timestimestimestimes 10minus20minus20minus20minus20

2NO(g) + O2(g) 2NO2(g) K c = 53 timestimestimestimes 105555

N2(g) + 2O2(g) 2NO2(g) K c = (275 timestimestimestimes 10minus20minus20minus20minus20)))) timestimestimestimes ((((53 timestimestimestimes 105555) = 15) = 15) = 15) = 15 timestimestimestimes 10-14

Kc = 15 times 10-14


bull Water (100 mol) was placed in an otherwise empty container with a fixed volume of 100 L The container was heated to 1705 K at which temperature the following equilibrium was established

2H2O(g) 2H2(g) + O2(g) Kp = 189 times 10ndash9 atm

Calculate Kc for this reaction at 1705 K

Marks5

For a reaction involving a change in the number of moles of gas of ∆∆∆∆n Kp and Kc are related by

Kp = Kc(RT)∆∆∆∆n

In this reaction 2 mol of gas 3 mol of gas and hence ∆∆∆∆n = 1 Thus

Kp = Kc(RT)1 or Kc = 9

p 11189 10135 10

RT (008206) (1705)

K minusminusminusminusminusminusminusminustimestimestimestimes= = times= = times= = times= = times

timestimestimestimes

Kc = 135 times 10-11

Determine the amount of O2 (in mol) in the container at equilibrium at 1705 K

As 100 mol of H2O is initially present in a container with volume 100 L the initial concentration of H2O is

concentration = numberof moles 100mol

= =00100Mvolume 100L

The reaction table with [O2(g)]equilibrium = x is

2H2O(g) 2H2(g) + O2(g) initial 00100 0 0 change -2x +2x +x equilibrium 00100-2x 2x x

Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)

Although the amount of water initially present is quite small the value of Kc is so very small that 00100 ndash 2x ~ 00100 This approximation simplifies the expression so that

Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)

Hence x = [O2(g)]equilibrium = 696 times 10-6 M



As this concentration of O2 is present in a 100 L container the number of moles of O2 present is

number of moles = concentration times volume = (696 times 10-6 M) times (100 L) = 696 times 10-4 mol

Answer 696 times 10-4 mol

CHEM1101 2006-J-9 June 2006

bull The thermal decomposition of lithium peroxide produces oxygen

2Li2O2(s) 2Li2O(s) + O2(g)

A 10 g sample of Li2O2 was placed in a closed container and heated to a temperature where some but not all of the Li2O2 decomposes The experiment is then repeated using a 20 g sample heated to the same temperature in an identical container How does the pressure of O2(g) produced vary between these two experiments Explain

Marks 2

The equilibrium constant for this reaction is given by

Kp = 2O (g)p as all other reactants and products are solids

As solid is left at the equilibrium point in both reactions it does not enter the equilibrium expression and has no effect on the equilibrium constant The pressure of O2(g) will be a constant equal to Kp



N2O4(g) 2NO2(g)


100 L flask After equilibrium had been established at 100 C the concentration of

N2O4(g) was found to be 00491 M Calculate the equilibrium constant Kc for the

reaction as written at 100 C

Marks

5

The initial concentration of N2O4(g) is

[N2O4(g)]0 = number of moles 01000

= = 0100 Mvolume 100

At equilibrium [N2O4(g)] = 00491 M

The change in the concentration is (0100 ndash 00491) = 0051 M From the

chemical equation each mole of N2O4 that reacts leads to two moles of NO2

Hence [NO2(g)] = 2 times 0051 = 010 M

The equilibrium constant in terms of concentrations Kc is therefore

Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021

Use your calculated value for Kc to calculate whether a mixture of N2O4(g) (0120 M)

and NO2(g) (0550 M) is at equilibrium at 100 C If not in which direction will it

move

Using these concentrations the reaction quotient Q is

Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)

As Q gt Kp the reaction will proceed to reduce Q by increasing the amount of

reactant (N2O4(g)) and decreasing the amount of product (NO2(g)) It will

proceed towards the left

CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)

Therefore [CO(g)]2 = 0126 M

2 or [CO(g)] = 0354 M

Answer 0354 M

When hydrogen cyanide (HCN) is dissolved in water it dissociates into ions according

to the following equation

HCN(aq) H+(aq) + CN

ndash(aq)

The equilibrium constant for this reaction is Kc = 62 10ndash10

mol Lndash1

If 100 mol of

HCN is dissolved to make 100 L of solution calculate the percentage of HCN that

will be dissociated

3

The initial concentration of HCN is 100 M The reaction table is

HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x

[equilibrium] 100-x x x

Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10

As Kc is very small the amount of dissociation will be tiny and 100-x ~ 100

Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]

As [HCN(aq)] = 100 ndash x [HCN(aq)] = 100 and the percentage dissociation is

percentage dissociation = 5

25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004

A saturated solution of iodine in water contains 0330 g I2 per litre but more than this

amount can dissolve in a potassium iodide solution because of the following

equilibrium

I

(aq) + I2(aq) I3(aq)

A 0100 M KI solution dissolves 125 g of I2 per litre most of which is converted

to I3(aq) Assuming that the concentration of I2(aq) in all saturated solutions is the

same calculate the equilibrium constant for the above reaction

Marks

4

The molar mass of I2 is (2 times 12690) g mol-1

= 2538 g mol-1

As 0330 g of I2 dissolves in a litre of water the concentration of I2 in the

saturated solution of iodine in water is therefore

[I2(aq)] = 0330

00013M2538

125 g of I2 corresponds to 1

125g00493mol

2538gmol

and when this dissolves in

a litre of KI solution the concentration is 00493 M

For the equilibrium the reaction table is therefore

I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x

[equilibrium] 0100-x 00493-x x

Assuming that [I2(aq)] is the same as in the saturated solution (as stated in the

question) 00493 ndash x = 00013 so x = 0048 giving

[I-(aq)] = 0100 ndash 0048 = 0052 M [I2(aq)] = 00013 M and [I3

-(aq)] = 0048 M

The equilibrium constant is therefore

Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710


bull When 10 mol of acetic acid and 10 mol of ethanol are mixed they react according to the following equation

C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

At equilibrium the mixture contains 067 mol of the ester (CH3COOC2H5) What is the equilibrium constant for the reaction

Marks 3

At equilibrium there are (10 ndash 067) mol of C2H5OH and CH3COOH together with 067 mol of CH3COOC2H5 and of H2O(l) The equilibrium constant is therefore

Kc = 3 2 5 2

2 5 3

[CH COOC H (l)][H O(l)] (067)(067)[C H OH(l)][CH COOH(l)] (10 067)(10 067)

====minus minusminus minusminus minusminus minus

= 41

Note that although amounts (mol) rather than concentrations are known the volume is the same for each species and cancels through in the equation The reaction involves mixtures of liquids and not aqueous solution so the concentration of water is included in the expression

ANSWER 41

bull Bromine-containing compounds are even more ozone depleting than the analogous chlorine-containing compounds In the stratosphere an equilibrium exists between bromine and the NOx species One of these equilibrium reactions is

2NOBr(g) 2NO(g) + Br2(g)

To study this reaction an atmospheric chemist places a known amount of NOBr in a sealed container at 25 degC to a pressure of 0250 atm and observes that 34 of it decomposes into NO and Br2 What is Kp for this reaction

3

The initial partial pressure of NOBr is 0250 atm At equilibrium 34 of it has decomposed so its partial pressure at equilibrium is 066 times 025 = 0165 atm The reaction stoichiometry means that 2 moles of NO(g) are produced for every 2 moles of NOBr(g) that decompose The equilibrium partial pressure of NO is therefore 034 times 025 = 0085 atm 1 mole of Br2(g) is produced for every 2 moles of NOBr(g) that decompose The equilibrium partial pressure of Br 2(g) is therefore frac12 times 034 times 025 = 00425 atm

The equilibrium constant in terms of partial pressures Kp is therefore

Kp =

1313

13= 00113

ANSWER 00113


The bacterium Azotobacter chroococcum growing aerobically in a medium free of

nitrogen containing compounds obtains all of its nitrogen by the fixation of

atmospheric N2 The solubility of N2 in water is governed by the following

equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1

What is the concentration of dissolved N2 available to the bacterium at 10 atm and

30 C (Air is 78 N2)

Marks

4

The equilibrium constant is given by K = N (g)2

2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)

The atmospheric pressure is the sum of the partial pressures of the constituent

gases in the air As N2 represents 78 of the air and the total pressure is 10

atm the partial pressure of N2 is given by N (g)2p = 078 times 10 atm = 078 atm

Hence the concentration of dissolved N2 is

[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846

120783120788 times 120783120782120785 119834119853119846 119819 119846119848119845minus120783 = 49 times 10

-4 M

Answer 49 times 10

-4 M

A culture of these bacteria (10 L) grows to a density of 084 mg dry weight per mL of

culture and has a nitrogen content of 70 of the dry weight What volume of air at

10 atm and 30 C would supply this nitrogen requirement

As the density of the culture is 084 mg mL-1

the mass of 10 L (1000 mL) is

1000 mL times (086 times 10-3

g mL-1

) = 086 g The nitrogen content is 70 so the

mass of nitrogen in the culture is 007 times 086 g = 0060 g

As nitrogen has a atomic mass of 1401 g mol-1

this mass corresponds to

moles of nitrogen atoms = -1

0060gmass of nitrogen= = 00043mol

atomicmass 1401g mol

Nitrogen is present in the air as N2 so the number of moles of N2 required is frac12 times

00043 mol = 00021 mol

The partial pressure due to N2 is 078 atm if the atmospheric pressure is 10 atm

Using PV = nRT the volume of air containing 00021 mol of N2(g) is

-1 -1(0021mol)times(008206L atm K mol )times((30+273)K)

= =(078atm)

= 0068 L = 68 mL

nRTV

P

Answer 0068 L or 68 mL

CHEM1101 2014-J-10 June 2014


N2O4(g) 2NO2(g) Kc = 461 times 10ndash3 at 25 degC

A 0086 mol sample of NO2 is allowed to come to equilibrium with N2O4 in a 050 L container at 25 degC Calculate the amount (in mol) of NO2 and N2O4 present at equilibrium Show all working

Marks 4

The initial concentration of NO2(g) is [NO2(g)] = number of moles volume = 0086 mol 050 L = 0172 mol L-1

A reaction table can be used to calculate the equilibrium concentrations

N2O4(g) 2NO2(g) initial 0000 0172

change +x -2x

equilibrium x 0172 ndash 2x

Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120783120789120784120784119961)

120784

119961 = 461 times 10ndash3

So

(0172 ndash 2x)2 = 461 times 10ndash3 x 00296 ndash 0688x + 4x2 = 461 times 10ndash3 x 4x2 ndash 0693x + 00296 = 0

Solving the quadratic gives x = 00766 and x = 00965 The second route would give a negative value for [NO2(g)] Using x = 00766

[N2O4(g)] = 00766 M and [NO2(g)] = (0172 ndash 2 times 00766) M = 00188 M

As these concentrations are in a 050 L container

number of moles of N2O4(g) = concentration times volume

= 00766 mol L-1 times 050 L = 0038 mol number of moles of NO2(g) = 00188 mol L-1 times 050 L = 00094 mol

Amount of NO2 00094 mol Amount of N2O4 0038 mol





2CH3OH(g) + H2(g) C2H6(g) + 2H2O(g)

Show all working

Marks 2


K1 = 119810120784119815120788 119840 [119815120784 119840 ]

[119810119815120786 119840 ]120784 K2 =

119810119815120785119822119815 119840 [119815120784 119840 ]119810119815120786 119840 [119815120784119822 119840 ]

K3 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]

Hence K3 = K1 K2

2

K1 K22 =

119810120784119815120788 119840 [119815120784 119840 ][119810119815120786 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]120784

119810119815120786 119840 120784[119815120784119822 119840 ]120784 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]






Marks 5

Kp = 119927119810119822119927119810119845120784119927119810119822119810119845120784



Kp = 598 times 10-13


Using Qp = 119927119810119822119927119810119845120784119927119810119822119810119845120784

Qp = (120782120787120782)(120782120782120783)(120783120782120782)120784


CHEM1101 2013-J-9 June 2013


N2O4(g) 2NO2(g)


Marks 4


Kc = [119821119822120784 119840 ]120784

[119821120784119822120786]


Kc = (120782120786120788)120784

(120783120782120782) = 021



change +x -2x


Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120791120784120784119961)

120784

(120784120782120782119961) = 021

So

(092 ndash 2x)2 = 021 (200 + x) 08464 - 368x + 4x2 = 042 + 021x

4x2 ndash 389x + 043 = 0


CHEM1101 2013-J-9 June 2013


x = 119939 plusmn 119939120784120786119938119940


120784times120786


Hence using x = 013

[N2O4] = (200 + x) M = (200 + 013) M = 213 M







Marks 5








= (001)(050)(100)

= 0005





Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






2CH3OH(g) + H2(g) C2H6(g) + 2H2O(g)

Show all working

Marks 2


K1 = 119810120784119815120788 119840 [119815120784 119840 ]

[119810119815120786 119840 ]120784 K2 =

119810119815120785119822119815 119840 [119815120784 119840 ]119810119815120786 119840 [119815120784119822 119840 ]

K3 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]

Hence K3 = K1 K2

2

K1 K22 =

119810120784119815120788 119840 [119815120784 119840 ][119810119815120786 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]120784

119810119815120786 119840 120784[119815120784119822 119840 ]120784 = 119810120784119815120788 119840 [119815120784119822 119840 ]120784

119810119815120785119822119815 119840 120784[119815120784 119840 ]






Marks 5

Kp = 119927119810119822119927119810119845120784119927119810119822119810119845120784



Kp = 598 times 10-13


Using Qp = 119927119810119822119927119810119845120784119927119810119822119810119845120784

Qp = (120782120787120782)(120782120782120783)(120783120782120782)120784


CHEM1101 2013-J-9 June 2013


N2O4(g) 2NO2(g)


Marks 4


Kc = [119821119822120784 119840 ]120784

[119821120784119822120786]


Kc = (120782120786120788)120784

(120783120782120782) = 021



change +x -2x


Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120791120784120784119961)

120784

(120784120782120782119961) = 021

So

(092 ndash 2x)2 = 021 (200 + x) 08464 - 368x + 4x2 = 042 + 021x

4x2 ndash 389x + 043 = 0


CHEM1101 2013-J-9 June 2013


x = 119939 plusmn 119939120784120786119938119940


120784times120786


Hence using x = 013

[N2O4] = (200 + x) M = (200 + 013) M = 213 M







Marks 5








= (001)(050)(100)

= 0005





Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P





Marks 5

Kp = 119927119810119822119927119810119845120784119927119810119822119810119845120784



Kp = 598 times 10-13


Using Qp = 119927119810119822119927119810119845120784119927119810119822119810119845120784

Qp = (120782120787120782)(120782120782120783)(120783120782120782)120784


CHEM1101 2013-J-9 June 2013


N2O4(g) 2NO2(g)


Marks 4


Kc = [119821119822120784 119840 ]120784

[119821120784119822120786]


Kc = (120782120786120788)120784

(120783120782120782) = 021



change +x -2x


Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120791120784120784119961)

120784

(120784120782120782119961) = 021

So

(092 ndash 2x)2 = 021 (200 + x) 08464 - 368x + 4x2 = 042 + 021x

4x2 ndash 389x + 043 = 0


CHEM1101 2013-J-9 June 2013


x = 119939 plusmn 119939120784120786119938119940


120784times120786


Hence using x = 013

[N2O4] = (200 + x) M = (200 + 013) M = 213 M







Marks 5








= (001)(050)(100)

= 0005





Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2013-J-9 June 2013


N2O4(g) 2NO2(g)


Marks 4


Kc = [119821119822120784 119840 ]120784

[119821120784119822120786]


Kc = (120782120786120788)120784

(120783120782120782) = 021



change +x -2x


Hence

Kc = [119821119822120784 119840 ]120784

[119821120784119822120786] = (120782120791120784120784119961)

120784

(120784120782120782119961) = 021

So

(092 ndash 2x)2 = 021 (200 + x) 08464 - 368x + 4x2 = 042 + 021x

4x2 ndash 389x + 043 = 0


CHEM1101 2013-J-9 June 2013


x = 119939 plusmn 119939120784120786119938119940


120784times120786


Hence using x = 013

[N2O4] = (200 + x) M = (200 + 013) M = 213 M







Marks 5








= (001)(050)(100)

= 0005





Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2013-J-9 June 2013


x = 119939 plusmn 119939120784120786119938119940


120784times120786


Hence using x = 013

[N2O4] = (200 + x) M = (200 + 013) M = 213 M







Marks 5








= (001)(050)(100)

= 0005





Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






Marks 5








= (001)(050)(100)

= 0005





Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P




Marks 4



PCOCl2 = (100 + 001) atm = 101 atm

PCl2 = (001 - 001) atm = 000 atm



initial 101 049 000

change -x +x +x



= x(049+ x)(101minus x)


Kp ~ x(049)(101)

= 598times10minus13




CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2012-J-10 June 2012




Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2012-J-12 June 2012



Marks 4









ndash500 -

ndash1000 -

ΔG

(kJ

mol

ndash1)

Temperature (degC)



N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P




N2(g) + 3H2(g) 2NH3(g)


Marks 5

Kp = p2(NH3 )p(N2 ) p3(H2 )



Kp = 56 times 105



Qp = p2(NH3 )p(N2 ) p3(H2 )

= (100)2

(050)(100)3 = 200






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






Marks 4


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120788120782120782)(120782120786120782120782)(120782120790120782120782)(120782120783120782120782)

= 300



initial 0800 0100 0600 0400 + x

change +y +y -y -y


Keq = [119826119822120785 119840 ][119821119822 119840 ][119826119822120784 119840 ][119821119822120784 119840 ]

= (120782120786120782120782)(120782120784120782120782119961)(120783120782120782120782)(120782120785120782120782)

= 300


Answer 205 mol

CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2010-J-10 June 2010


N2(g) + 3H2(g) 2NH3(g)


Marks 6


Q =

=

= 16


Answer to the right



Kp =

=

= 562 times 105



CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2010-J-11 June 2010



Marks 3



H2O(l) ndash286 70

CO2(g) ndash394 214








Answer 0027




2



[H2O(g)] = 250 mol 1000 L = 0250 M

[CO2(g)] = 500 mol 1000 L = 0500 M

[H2(g)] = 500 mol 1000 L = 0500 M


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2010-J-11 June 2010



initial 0250 0250 0500 0500

change -x -x +x +x


Kc =

= 1313

= =

13

= 314

Hence

13

= (314)12 = 560



[H2O(g)] = (0250 ndash 0136) M = 0114 M

[CO2(g)] = (0500 + 0136) M = 0636 M

[H2(g)] = = (0500 + 0136) M = 0636 M

[CO] = 0114 M [H2O] = 0114 M [CO2] = 0636 M [H2] = 0636 M





Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






Marks3









2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P




2SO2(g) + O2(g) 2SO3(g)


Marks3


Kc = =

= 56

Kc = 56



Q = =

= 0025








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P








4





Kc = =

=

= 70

or

= (70)12


[Br2(g)] = 0095 M [Cl2(g)] = 0095 M [BrCl(g)] = 025 M

CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2009-J-9 June 2009




M and

[H2(g)] = 176 10ndash1

M

Marks

2


Keq = 119810119822 119840 [119815120784 119840 ]120784

[119810119815120785119822119815 119840 ] = 130 times 10

-2


M and [H2(g)] = 176 times 10-1

M

[CO(g)] = 119922119942119954[119810119815120785119822119815 119840 ]

[119815120784 119840 ]120784 =


(120783120789120788 times120783120782minus120783)120784 M = 0146 M

Answer 0146 M















solution as cations

CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2009-J-10 June 2009





of the following



Marks

3












increased




CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2009-J-13 June 2009





Marks

2


Fe2O3(s) 3CO(g)

2Fe(s) 3CO2(g)


change atm -3x +3x




given by

Kp = 119953(119810119822120784)

120785

119953(119810119822)120785 =

(120785119961)120785

(120782120791120789120790minus120785119857)120785 = 199

Hence

(120785119961)

(120782120791120789120790minus120785119857) = (199)

13 = 271




p(CO2) = 3x atm = 0714 atm

p(CO) = 0264 atm p(CO2) = 0714 atm



2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P




2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0492 M [N2(g)] = 0319 M and [NO(g)] = 0350 M

Marks

2


Keq = 119810119822 119840 120784[119821119822 119840 ]120784

[119810119822120784 119840 ]120784[119821120784 119840 ] = 0118


[CO(g)]2 =

119922119942119954[119810119822120784 119840 ]120784[119821120784 119840 ]

[119821119822(119840)]120784 =

120782120783120783120790 (120782120786120791120784)120784(120782120785120783120791)

(120782120785120787120782)120784 M = 0146 M

[CO(g)] = 0273 M

Answer 0273 M

CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2008-J-9 June 2008


2







pH 75






2













Keq = (113)times(145)times(1017) = 010

Answer Keq = 010

CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2008-J-10 June 2008





this is the case

Marks

5










s




Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p = 113 times 10

-6


initial 200 200 000

change -x -2x +x


Thus at equilibrium

Kp = )

3

2

CH OH

2CO H

(

( )( )

p

p p =

2(200 - x)(200 - 2 )

x

x = 113 times 10

-6



Kp ~ 2

(200)(200)

x = 113 times 10

-6 so x = (113 times 10

-6) times (200) times (200)

2


-3 atm

Answer


atm


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2008-J-10 June 2008












N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P




N2O4(g) 2NO2(g)





Marks

4



N2O4(g)

2NO2(g)

initial 01000 0

change -x +2x





Kc = [119925119926120784 119944 ]120784

[119925120784119926120786 119944 ] =

120782120783120782120784120784

120782120782120786120791120783 = 0211

Answer 0211






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






Marks

2




and pressure










HCN(g) = 23 times 200 L = 133 L

Answer 133 L






3



(b) CCl4 is removed







CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2007-J-8 June 2007



Marks 5










H2

H2

Cu2+

Cu2+

O

C

O

CH2 ~

239 Aring

CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2007-J-9 June 2007


N2(g) + O2(g) 2NO(g)


Marks 5





start 021 078 0

change -x -x +2x



352

2N2O

2NO

p 10354)x780)(x210(

)x2())((


========pp

pK







CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2007-J-10 June 2007


2NO(g) + O2(g) 2NO2(g)


Marks 6


1p c

4 5c p

K K (RT)

K K (RT) (13 10 ) (008206 500) 53 10



Kc = 53 times 105


N2(g) + 2O2(g) 2NO2(g)





Kc = 15 times 10-14





Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






Marks5





p 11189 10135 10

RT (008206) (1705)



Kc = 135 times 10-11







Hence Kc =2 2

-112 22 2

2

[H (g)] [O (g)] (2x) (x)= =135times10

[H O(g)] (00100-2x)


Kc =3

-112

4x~ =135times10

(00100) or 4x3 = (00100)2 times (135 times 10-11)







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P







CHEM1101 2006-J-9 June 2006


2Li2O2(s) 2Li2O(s) + O2(g)


Marks 2






N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P




N2O4(g) 2NO2(g)





Marks

5



= = 0100 Mvolume 100




Hence [NO2(g)] = 2 times 0051 = 010 M


Kc =

2 2

2

2 4

[NO (g)] (010)= = 021

[N O (g)] (00491)

Kc = 021



move


Q =

2 2

2

2 4

[NO (g)] (0550)= = 252

[N O (g)] (0120)




CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2005-J-9 June 2005


2CO2(g) + N2(g) 2CO(g) + 2NO(g)


[CO2(g)] = 0392 M [N2(g)] = 0419 M and [NO(g)] = 0246 M

Marks

2


Kc = 2 2 2 2

2 22 2

[CO(g)] [NO(g)] [CO(g) ](0246)0118

[CO (g)] [N (g)] (0392) (0419)


2 or [CO(g)] = 0354 M

Answer 0354 M



HCN(aq) H+(aq) + CN

ndash(aq)


mol Lndash1

If 100 mol of


will be dissociated

3


HCN(aq) H+(aq) CN

-(aq)

[initial] 100 0 0

change -x +x +x


Kc = 2

[H (aq)][CN (aq)] (x)(x) x

[HCN(aq)] (100 x) (100 x)

= 62 times 10-10


Hence

x2 ~ 62 times 10

-10x so x = 25 times 10

-5 = [H

+(aq)] = [CN

-(aq)]



25 10100

100

25 times 10

-3

Answer 25 times 10

-3

CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


CHEM1101 2004-J-12 June 2004



equilibrium

I





Marks

4


= 2538 g mol-1



[I2(aq)] = 0330

00013M2538


125g00493mol

2538gmol




I-(aq) I2(aq) I3

-(aq)

[initial] 0100 00493 0

change -x -x +x





-(aq)] = 0048 M


Kc = 3

2

[I (aq)] (0048)710

(0052)(00013)[I (aq)][I (aq)]

Answer 710





Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






Marks 3


Kc = 3 2 5 2

2 5 3



= 41


ANSWER 41




3



Kp =

1313

13= 00113

ANSWER 00113





equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P






equilibrium

N2(aq) N2(g) K = 16 103 atm L mol

ndash1


30 C (Air is 78 N2)

Marks

4


2

( )

[N (aq)]

p where N (g)2

p is the partial

pressure of N2(g)





[N2(aq)] = 119927119821120784(119944)

119922 =

120782120789120790 119834119853119846


-4 M

Answer 49 times 10

-4 M







g mL-1









00043 mol = 00021 mol




= =(078atm)

= 0068 L = 68 mL

nRTV

P


chem1101 2014-j-9 june 2014 • use the following equilibria

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