09(2) physchem exam-answers

7/27/2019 09(2) PhysChem Exam-Answers

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** Insert the following equation into the formulae sheet:

1 0

i i

i

ma

m

=

Section A. Physical Chemistry. (Total of 60 marks)

Part 1: Kinetics (20 marks)

Question 1 (3 + 2 = 5 marks)

A set of experiments were performed to determine the rate law for the reaction of NO and O 2,according to the following equation:

2 NO(g) + O 2(g) 2 NO 2 (g)

Expt [NO](mol L -1)

[O 2](mol L -1)

Initial Rate(mol L -1 s-1)

1 0.020 0.010 0.028

2 0.020 0.020 0.057

3 0.020 0.040 0.114

4 0.040 0.020 0.227

5 0.010 0.020 0.014

a) What is the rate law for this reaction? (3 marks)

Look at each reactants effect on rate, in turn:

Comparison of Expts 1, 2 & 3 shows that as [O 2] doubles, the rate doubles, hence first order in O

Comparison of Expts 2, 4 & 5 shows that as [NO] doubles, rate increases by a factor of 4, hencesecond order in NO.

Hence rate law is Rate = k [O 2] [NO] 2

(1 mark for determining the reaction order of each component and 1 mark for the rate law or if justthe correct rate law given, all 3 marks)

b) What is the value of the rate constant, k? Ensure you include the appropriate unit for k. (2marks)

Choose one experiment and use the info to deduce the rate constant according to k =rate/([O 2] [NO] 2)e.g. from expt 1 k = 0.028 mol L -1 s-1 / (0.010 mol L -1 x (0.020 mol L -1)2) = 7.0 x 10 3 L2 mol -2 s-1.


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(1 mark for the correct numerical value, 1 mark for the correct units (standard units for 3rd order reactions)) Note that because these are real data, slight differences in the valueof the rate constant will be found (e.g. expt 4 gives 7.1 x 10 3 rather than 7.0 x 10 3)

Question 2 (3 marks)

The first-order rate constant for the decomposition of the insecticide, Killemquick, in the environment= 3.43 x 10 -2 day -1. Calculate how long it takes for the concentration of Killemquick to fall to 10% of itsinitial value.

Use ln[A] t = ln[A] o k t. (1 mark)

Rearrange this to ln[A] t - ln[A] o = k t. Or ln[A] 0 - ln[A] t = k t.

which equals ln ([A] 0 / ln[A]) = k t (1 mark)

Given that [A] t = [A] o x 0.1, we then have ln(10) = k t

So t = 2.303/3.43 x 10 -2 day -1 = 67.1 days (1 mark)


The rate constant for the reaction of iodide with methyl bromide is 7.70 x 10 -3 L mol -1 s-1 at 50C and4.25 x 10 -5 L mol -1 s-1 at 0C. Calculate the activation energy for this reaction.

Use the form of the Arrhenius equation with two rate constants and two temps:(1 mark)

So ln(7.70 x 10 -3/4.25 x 10 -5) = - E a/8.314 x (1/323 1/273) (1 mark for inserting values in right places)

So E a = 76237 J mol -1 = 76.2 kJ mol -1. (1 mark) (certainly accept 76.3 if temps given as 273.15 &323.15)

Question 4 (1 + 1 + 2 = 4 marks)

The catalysed oxidation of the thallium(I) ion by cerium(IV) in aqueous solution is described by theoverall equation:

2 Ce 4+ + Tl + 2 Ce 3+ + Tl 3+

The accepted mechanism for this reaction is:

Step 1: Ce 4+ + Mn 2+ Ce 3+ + Mn 3+

Step 2: Ce 4+ + Mn 3+ Ce 3+ + Mn 4+

Step 3: Mn 4+ + Tl + Mn 2+ + Tl 3+


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a) Circle all of the intermediates in this mechanism (1 mark)

Mn 3+ & Mn 4+ are the intermediates (1/2 mark each)

b) Identify the catalyst in this reaction (1 mark)

Mn 2+ (1 mark). No other option is correct.

c) Assuming that Step 1 is the rate determining step in the reaction, what is the rate law? (2 marks)

Rate = k [Mn 2+] [Ce 4+] (can omit the Rate =). (2 marks) No other option is correct.


Experiments to determine the rate law for the reaction below show that the kinetics are very complex.As a result, it has been suggested that the pseudo-first order approach should be used.

S2O82- + I - products with a rate constant, k

a) Explain what is meant by pseudo-first order. (1 mark)The concentration of one of the reactants is in vast excess (so its concentration remains effectively

constant) and the rate of reaction then depends just on the concentration of one reactant. Or

suitable paraphrasing of this.

b) Explain how you would manipulate the concentrations of S 2O82- and I - to work out the pseudo-

first order rate constant k. (2 marks)

Make [S 2O82-] to be in vast excess of [I -] (e.g. 100-fold), then measure the rate of the first-order

reaction (follow product form) as a function of [I -] to get the pseudo-first order rate constant, k. (as

rate = k [I -] ). 1 mark for having one named reactant in vast excess and the second mark for how to

determine k.

c) Then explain how you would again manipulate concentrations of S 2O82- and I - to determine the

rate constant k. (2 marks)

Repeat the determination of k at several different [S 2O82-] (1 mark). k = k[S 2O82-]. So plot k vs

[S2O82-]. (1 mark)

For the answers to b) and c), [S2O82-] could be the reactant in excess, so swap the reactants in these

answers

Part 2: Equilibria (20 marks)



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Solid PbCrO 4 is added to an aqueous sample which already contains solid PbCrO 4 and Pb 2+ and CrO 42- in equilibrium, according to the equation below. Briefly explain what will happen to the position of theequilibrium.

There will be no change in the position of the equilibrium (the solution is already saturated with Pb 2+ and CrO 42-) (2 marks). It is incorrect to say that the equilibrium will shift towards the soluble products(i.e. to the right)

Question 2 (5 marks )

The water-gas shift reaction:CO(g) + H 2O(g) CO 2(g) + H 2(g)

is an important industrial reaction used for producing high purity hydrogen used in the synthesis of ammonia.

If 0.250 mol of CO(g) and 0.250 mol of H 2O(g) are placed in a 0.125 L flask at 900 K, what is thecomposition of the equilibrium mixture, if K = 1.56.Show all calculations and justify any approximations:

Complete the table below:Concentration [CO(g)] [H 2O(g)] [CO 2(g)] [H 2(g)]Initially 0.250/0.125

=2.00

0.250/0.125

=2.00

0 0

Change -x -x +x +x

Equilibrium 2.00-x 2.00-x x x

K =CO

2[ ] H 2[ ]CO[ ] H 2 O[ ]

=x 2

2.00 x[ ]2

= 1.56

x 2

2.00

x[ ]2

= 1.56

x

2.00 x[ ]= 1.25

x=2.50 1.25 xan dhen ce 2.25 x =2.50

x= 1.11M

Hence at equilibrium:


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[CO(g)] = 2.00 x = 0.89 M[H 2O(g)] = 2.00 x = 0.89 M[CO 2(g)] = 1.11 M[H 2(g)] = 1.11 M


An aqueous suspension of Mg(OH) 2 is (Milk of Magnesia) has traditionally been used as a treatment of minor gastric disorders by neutralizing stomach acid. If the K sp of Mg(OH) 2 is 6.3 x 10 -10, calculate itssolubility (S) in pure water.

Concentration [M] Mg(OH) 2 (s) [Mg2+

] [OH-

]Initially

- 0 0

Change- +S +2S

Equilibrium- S 2S

K sp = Mg

2 +[ ] OH [ ]2

= S[] 2 S[]2

= 4 S 3 = 6.3 10 10

S =

6.3 1010

43 = 5.40 10 4

Solubility of Mg(OH) 2 = 5.40 x 10 -4 M

Question 4 (4 + 1 + 1 + 1 = 7 marks)


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(a) You are asked to prepare one (1.0) litre of a carbonate buffer solution with a pH of 10.0. Howmany grams of Na 2CO 3 must be dissolved in 1.0 L of 0.20 M NaHCO 3 to prepare a buffer of this pH.

Data:

H2CO3* + H 2O H 3O+ + HCO K 1 = 4.47 x 10-7

at 25CHCO+ H 2O H 3O+ + CO 3

2 K 2 = 4.68 x 10 -11 at 25C

Formula weight (Na 2CO 3) = 106.0 g/mol

Show all calculations

pH = 10.0.

pK a (HCO 3-) = -log(4.68 x 10 -11) = 10.32

pH = 10.33 + logCO 3

2 [ ]HCO 3[ ]

= 10.33 + logCO 3

2 [0.20

10.00=10.33+logCO

32[ ]

0.20 and hence CO 32 [ = 0.20 x10 (10.00-10.33)

CO 3

2 [ = 0.20 x10 (-0.33) = 0.20 x 0.468 = 9.36 x 10 -2 MTherefore

Mass Na 2CO 3 = 9.36 x 10 -2 x 106.0 = 9.92 g

Mass of Na 2CO 3 added per litre = 9.92 ..................... g

(4 marks)

(b) The diagram below shows the curve for titration of a 20 mL aliquot of 0.10 M propanoic acidwith 0.10 M NaOH.


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(i) Mark on the titration curve the position that corresponds to the value of pK a (Propanoicacid), and record this value below.

pK a (Propanoic acid) 4.8 - 4.9 .......................(1 mark)

(ii) Mark on the diagram the approximate shape of the titration curve if the original propanoic acid solution also contained 1 M propanoate ions.

(1 mark)

(c) Could a buffer solution be prepared using a mixture of 1 M lactic acid and 1 M sodium lactate?If so, what would be the pH of this buffer? pK a (Lactic acid) = 3.86Explain your answer (1 mark)

Yes, weak acid (pKa = 3.86). pH = pKa + log (1/1) = 3.86 + 0 = 3.86.


a) Briefly define the activity coefficient, . (1 mark)

Would accept phrases like : the activity coefficient allows for the no-ideality of the solution, or,

it accounts for the effect of additional ions on the reactivity (strictly speaking chemical potential

but we didnt cover this). Or similar paraphrasing. Would also accept that it is the

mathematical correction to relate activity to concentration.

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0

Volume of titrant (mL)

Ka

Titration mid- oint

Typical bufferedtitration curve


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b) Equilibrium constants are unitless. Briefly explain why this is so. (2 marks)

Equilibrium constants are made up of activities of the various species and activities are

unitless (1 mark).Activities are unitless as they are a ratio:

(The ratio is of the molality of the species in

solution and at infinite dilution)

(1 mark for showing how activity is unitless)Students are given this

equation in the formula sheet

Part 3: Thermodynamics (20 marks)


Answer True or False to the following statements.

a) An increase in atmospheric pressure compresses a gas in a frictionless piston container.

The work done on the system is positive.

TRUE

b) The only two means by which energy can be transferred are through heat or work.TRUE

c) If a reaction is very exothermic, it must be spontaneous.

FALSE


Choose the member with the higher entropy in each of the following pairs, and briefly justify your

choice (assume constant temperature):a. 1 mol of O 2 (g) or 1 mol of O 2 (aq)

1 mol of O 2(g) as gases dissolved in liquids are more constrained in their movement thanin the gaseous phase, 1 mark

b. 3 mol of CH 3CH 2CH 2CH 3 (g) or 3 mol of CH 3CH 3 (g)

3 mol of CH 3CH 2CH 2CH 3 (g) as larger, more complex molecules have higher entropy


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than smaller molecules, 1 mark (might mention more vibrational and rotational states, but this is not necessary)


Decomposition of solid silver oxide (Ag 2O) to silver metal and oxygen gas requires heating. Thenecessary thermodynamic information for this question is provided in the table below.

Compound H f o, kJ mol -1 So, J mol -1 K -1

Ag 2O -31.05 121.3O2 0 205.1Ag 0 42.6

a) Calculate whether this reaction occurs spontaneously at 200C. (4 marks)

2Ag 2O 4 Ag + O 2 (1 mark)

For a reaction to be spontaneous, Gr must be < 0.Gr = H r - TSr

Hr o = 4 x H f o(Ag) + H f o(O 2) - 2 H f o(Ag 2O) = 4 x 0 + 2 x 0 - 2 x (-31.05) = 62.10 kJ mol -1

Sr o = 4 x S o(Ag) + S o(O 2) - 2 S o(Ag 2O) = 4 x 42.6 + 205.1 - 2 x 121.3 = 132.9 J K -1 mol -1

So Gr o = 62.1 ((200+273) x 132.9/1000) = -0.8 kJ mol -1 (2 marks)

Hence it is spontaneous (just!) (1 mark)

If the stoichiometry of the equation is wrong, but the rest of the answer is consistent, then up to3 of the 4 marks can still be given.

Also, subtract a mark if they forget to convert T to Kelvin, but the rest of the answer isconsistent - will end up with a non-spontaneous reaction in this case (i.e. 35 kJ mol -1 ).

b) Calculate the Gibbs free energy, G r , for this reaction at equilibrium. (1 mark)At equilibrium, G r = 0 (1 mark)


Ethene (CH 2=CH 2) reacts with hydrogen (H 2) at 25C to produce ethane (CH 3-CH 3) in a reversiblereaction. Use the thermodynamic data provided to calculate the equilibrium constant for this reaction.

Compound H f o, kJ mol -1 So, J mol -1 K -1

CH 2=CH 2 (g) 52.26 219.6H2 (g) 0 130.7


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CH 3-CH 3 (g) -84.68 229.6

CH 2=CH 2 + H 2 CH 3-CH 3 (1 mark)

G r = Hr - T Sr Hr o = H f o(CH 3-CH 3) - H f o(H 2) - H f o(CH 2=CH 2) = -84.68 - 0 - 52.26 = -136.94 kJ mol -1

Sr o = S o(CH 3-CH 3) - S o(H 2) - S o(CH 2=CH 2) = 229.6 130.7 219.6 = -120.7 J K -1 mol -1

So Gr o = -136.94 ((25+273) x (-120.7)/1000) = -100.97 kJ mol -1 (2 marks)

Gr o = -RTln(K)so K = e^(- G r o/RT) = e^(100970/(8.314 x 298)) = 5.0 x 10 17 (1 mark)


For the ammonia synthesis reaction at 25C, calculate the change in Gibbs Free Energy ( G r ) if 1.00mol N 2(g) at 0.23 atm pressure and 3.00 mol H 2(g) at 0.42 atm are converted to 2.00 mol NH 3 (g) at1.45 atm. G f (NH 3)(g) = -16.45 kJ mol -1.

N 2(g) + 3H 2(g) 2NH 3(g)

G r = 2 x G f o(NH 3) - Gf o(N 2) 3 x G f o(H 2)= 2 x (-16.45) 0 3 x 0 = -32.90 kJ mol -1. (1 mark)

Not at equilibrium so calculate Q. (1 mark)Q = p(NH 3)2/(p(N 2) x p(H 2)3) = 1.45 2/(0.23 x 0.42 3) = 123.4 (2 marks)

Gr = G r + RT ln Q = -32.90 kJ mol -1 + 8.314 x 298 x ln (123.4) J mol -1 (1 mark)

=-32.90 + 11.94 kJ mol -1 = -20.96 kJ mol -1 (1 mark)

Note: conditional marks can be given if an earlier stage is wrong but the answers are consistent fromthat point on.

09(2) physchem exam-answers

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