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CHM1046 (Ref #625001) Chemical Equilibrium and Acid-Base Chemistry (solutions). Please let me know if you find errors/typos in this document.

1. Consider the following reaction: 2 NOCl (g) 2 NO (g) + Cl2 (g)

In one experiment 2.00 moles of NOCl is placed in a 1.00 L flask, and the concentration of NO after equilibrium is achieved is 0.66 M. Calculate the equilibrium constant Kc at 25°C for the reaction.

NOCl]0 = 2.00 mol/1.00 L = 2.00 M

[NO]eq = 0.66 M

2NOCL 2NO Cl2

I 2.00 0 0

C -2x 2x X

E (2.00 – 2x) 2x = 0.66 x

[NO]eq = 2x = 0.66 M x = 0.33 M [Cl2] = 0.33 M [NOCl] = (2.00 – 2x) = (2.00 – 2(0.33)) = 1.34 M

Kc = !"! [!"]!

[!"#$]! = !.!! [!.!!]

!

[!.!"]! = 0.080

2. When solid ammonium carbonate sublimes, it dissociates completely into ammonia and carbon dioxide according to the following equation:

N2H6CO2 (s) 2 NH3 (g) + CO2 (g)

At 25°C, an experiment shows that the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium constant, Kp?

N2H6CO2 (s) 2NH3 (g) CO2 (g)

I SOLID 0 0

C NA 2x X

E NA 2x = p(NH3) x = p(CO2)

Ptotal = 0.116 atm (at equilibrium)

ptotal = p(NH3) + p(CO2) 0.116 atm = 2x + x = 3x x =0.03867 atm

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therefore: p(NH3) = 2x = 0.07734 atm p(CO2) = x = 0.03867 atm Kp = (pNH3)2 (pCO2) = 2.31 x 10-4

3. A sample of 0.0020 moles of F2 was sealed into a 2.0 L reaction vessel and heated to 1000 K to study the dissociation into F atoms:

F2 (g) 2 F (g)

At this temperature, Kc = 1.2 x 10-4. What are [F2] and [F] at equilibrium? What is the percent dissociation of F2?

[F2]o = (0.0020 mol/2.0 L) = 0.0010 M

F2 2F

I 0.0010 0

C -x 2x

E 0.0010 – x 2x

Kc = 1.2 x 10-4 = ([F]2/[F2]) = (!")!

(!.!!"!!!)

4x2 +(1.2 x 10-4)x – (1.2 x 10-7) = 0

x = 1.59 x 10-4 and x = -1.89 x 10-7

[F2] = (0.0010 –x ) = 8.4 x 10-4 [F] = 2x = 3.2 x 10-4 % dissociation = x/initial x 100% = (1.59 x 10-4) / (0.0010) x 100% = 15.9 % = 16%

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4. Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO3 (g) 2 SO2 (g) + O2 (g)

Initially, the vessel is charged at 1000 K with SO3 (g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

pSO3 (initial) = 0.500 atm pSO3 (equilibrium) = 0.200 atm

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

I 0.500 0 0

C -2x 2x X

E 0.500 – 2x = 0.200 2x x

0.500 – 2x = 0.200 2x = 0.300 x = 0.150 atm at equilibrium: pSO3 = 0.200 atm pSO2 = 2x = 0.300 atm pO2 = 0.150 atm

Kp = !!! (!!!!)!

(!!"!)! = !.!"# [!.!""]

!

[!.!""]! = 0.338 atm

5. Consider the following reaction: 2 NO (g) + O2 2NO2 (g)

At 430°C, an equilibrium mixture consist of 0.020 mole of O2, 0.040 mole NO, and 0.96 moles NO2. Calculate Kp for the reaction, given that the total pressure is 0.20 atm.

Total # of moles: 0.020 mol + 0.040 mol + 0.90 mol = 1.02 moles

Partial pressure:

PNO = XNO PT = (0.040/1.02) x 0.20 atm = 0.0078 atm

PO2 = XO2 PT = (0.020/1.02) x 0.20 atm = 0.0039 atm

PNO2 = XNO2 PT = (0.96/1.02) x 0.20 atm = 0.19 atm

Kp = (!"!!)!

(!"#)!  !!! = (!.!")!

(!.!!"#)!  (!.!!"#) = 1.5 x 105 atm

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6. A mixture of 0.47 moles of H2 and 3.59 moles HCl is heated to 2800°C. Calculate the equilibrium partial pressures of H2, Cl2, and HCl if the total pressure is 2.00 atm. For the reaction:

H2 (g) + Cl2 (g) 2 HCl (g)

Kp is 193 at 2800°C.

[H2] = 0.47 M [HCl] = 3.59 M Reaction lies to the left:

H2 (g) Cl2 (g) 2HCl(g)

I 0.47 0 3.59

C X X -2x

E 0.47 + x X (3.59 – 2x)

Kp = Kc because Δn = 0

Kc = 193 = [!"#]!

!! [!"!] = (!.!"!!")

!

!.!"!! (!)

Solve using the quadratic formula:

x = 0.103

at equilibrium:

[H2] = 0.573 M (in 1 L ) = 0.573 moles [Cl2] = 0.103 M (in 1L) = 0.103 moles [HCl] = 3.384 M ( in L) = 3.384 moles total moles = 4.06 moles partial pressures at equilibrium:

PH2 = XH2 PT = (0.573/4.06) x 2.00 atm = 0.28 atm

PCl2 = XCl2 PT = (0.103/4.06) x 2.00 atm 0.051 atm

PHCl = XHCl PT = (3.384/4.06) x 2.00 atm = 1.67 atm

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Acid-Base Problems:

1. Complete the following table:

pH [H+] pOH [OH-] Acidic, Basic or neutral?

A 3.27 5.4 x 10-4 10.73 1.85 x 10-11 Acidic

B 4.89 1.35 x 10-5 9.11 7.8 x 10-10 Acidic

C 10.75 1.8 x 10-11 3.25 5.6 x 10-4 Basic

D 9.00 1.0 x 10-9 5.00 1.0 x 10-5 Basic

2. Calculate the pH and the percent ionization of a 0.250 M HC2H3O2 solution.

HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq) Ka (H C2H3O2) = 1.8 x 10-5.

HC2H3O2 (aq) H2O (l) H3O+ (aq) C2H3O2- (aq

I 0.250 NA 0 0

C -x NA +x +x

E 0.250 – x NA x x Ka = 1.8 x 10-5 = ! (!)

(!.!"#!!) = !!

(!.!"#)

x = 0.0022 = [H3O+] pH = - log [H3O+] = - log (0.0022) pH = 2.67 % ionization = (x/[HA]o) x 100% = (0.0022/0.250) x 100% = 0.85 %

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3. A solution of methylamine (CH3NH2) has a pH of 10.64. How many grams of methylamine are there in 100.0 mL of the solution? pH = 10.64 pOH = 14 – 10.64 = 3.36 [OH-] = 10-3.36 = 4.4 x 10-4 M CH3NH2 (aq) + H2O (l) CH3NH3

+ (aq) + OH- (aq) At equil: x – 4.4 x 10-4 4.4 x 10-4 4.4 x 10-4

Kb = !.!  !  !"!!  (!.!  !  !"!!)

!!(!.!  !  !"!!) = 4.4 x 10-4

x = 8.6 x 10-4 M the molar mass of CH3NH2 is 31.06 g/mol

(100.0 mL) ( !.!  !  !"!!  !"#  !!!!!!!"""  !"

) (!".!"  !  !!!!!!!  !"#  !!!!!!)

) = 2.7 x 10-3 g CH3NH2

4. Calculate the pH and the concentration of OH- in a solution of 0.15 M NH3.

(Kb for ammonia is 1.8 x 10-5) NH3 (aq) + H2O (l) NH4

+ (aq) + OH- (aq) Kb = 1.8 x 10-5

NH3 (aq) H2O (l) NH4+ (aq) OH- (aq

I 0.15 NA 0 0

C -x NA +x +x

E 0.15 – x NA x x Kb = 1.8 x 10-5 = ! (!)

(!.!"#!!) = !!

(!.!"#)

x = 1.6 x 10-3 M = [OH-] pOH = - log [OH-]= - log (1.6 x 10-3) = 2.80 pH = 11.20

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5. Calculate the percentage of HF molecules ionized in 0.010 M HF solution.

HF (aq) H+ (aq) + F- (aq) Ka = 6.8 x 10-4

HF (aq) H+ (aq) F- (aq )

I 0.010 0 3.59

C - X X x

E 0.010 –x X X

Ka = 6.8 x 10-4 = ! (!)(!.!"!!!)

Need to use the quadratic formula (% ionization is greater than 5%) x = [H+] = 2.3 x 10-3 % ionized: (x/[HA]o) x 100% = 23 %