2013 Final Exam Answer

8/20/2019 2013 Final Exam Answer

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Question 1 (15 marks)

C01

a b

12 8

1 mol of gas B expands adiabatically against a constant external pressure that equals to the

final pressure, P f = 4 atm. The initial state of the gas is T i = 27 °C and P i = 10 atm. Assume the

gas to be ideal with heat capacity of:

C P ,m =28.58+1.76×10-2

T

with C P,m is in (J K -1 mol-1) and T is in Kelvin.

(a) Determine the final temperature of the system, T f in °C. (12 marks)

(b) Calculate the enthalpy, Δ H (in kJ) and entropy, ΔS (in J K -1) of the process. (8 marks)

SOLUTION:

Part (a):

Given: T i =27 °C = 300 K, P i = 10 atm, n = 1 mol, P external = P f = 4 atmdU = dq – P external dV; since dq = 0 → dU = – PdV ∫ = ∆ = ∫ = ( ) …………. (1)∫ = ∆ = ∫ , …………. (2)C P,m – C V,m = R → C V,m = C P,m – R

C V,m = [(28.58 + 1.76 x 10-2T ) – 8.3145] = (20.2655 + 1.76 x 10-2 T ) J K -1 mol-1

Ideal gas, PV = nRT

Initial: = ⁄ ; Final: = From (1): ∆ = ( ) = ⁄ = ⁄ ∆ = 18.31454 4⁄ 300 10⁄ = 18.31454 4⁄ 30 1⁄ ∆ = (8.3145 ) + 997.74 …….….. (3)

From (2): ∆ = 1 ∫ 20.2655 + 1.76 × 10− ∆ = [20.2655( ) + (1 2⁄ ) 1.76 × 10− ( )]∆ = [20.2655( 300) + 0.00898 ( 300)]


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∆ = [20.2655 6079.65 + 0.00898 808.2] ..……….. (4) Eqn (3) = Eqn (4),

[20.2655 6079.65 + 0.00898

808.2] = (8.3145 ) + 997.74

0.00898 + 28.58 7885.59 = 0 Solving quadratic equation:

= 28.58 ± √ 28.58 (4 × 0.00898 × 7885.59)2 × 0.00898 Taking the posit ive root: T f =255.42 K = -17.58 °C

Part (b):

(i) For ideal gas: ∫ = ∆ = ∫ , ∆ = [128.58( ) + 10.00898 ( )] ∆ = 28.58255.42 300 + 0.00898 255.42 300 ΔH = -1496.6 J = -1.5 kJ

(ii) For ideal gas: ∫ = ∆ = ∫ , ln ∆ = 1 28.58 + 0.0176 18.3145 ln

410 ∆ = 28.58ln + 0.0176( ) 7.618 ∆ = 28.58ln 255.42300 + 0.0176255.42 300 + 7.618 ΔS = 2.24 J K

-1


CO2

a b c d

12 3 2 3

The data below show the temperature variation of the equilibrium constant of the reaction


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Ag2CO

3(s) 5 Ag

2O(s) + CO

2(g)

a) Draw a graph to demonstrate the relationship between the temperature and equilibrium

constant. (Tips: use the integrated form of van’t Hoff equation;

(12 marks)

b) Based on the graph in part (a); determine the standard reaction enthalpy of the

decomposition, Δ H°rxn. (Tips: assume that Δ H°rxn is independent of temperature).

(3 marks)

c)

State either the above reaction is endothermic or exothermic? Why? (2 marks)

d) In general, list three factors that affect the position of chemical equilibrium? (3 marks)

SOLUTION:

a)

T [K] 350 400 450 500

K p 3.98E- 04 1.41E- 02 1.86E- 01 1.48

1/T 0.002857 0.002500 0.002222 0.002000

ln K p -7.8291 -4.2616 -1.6820 0.3920

298.15K1

T1

RH298.15KlnKTlnK

f

rxnPf P

T [K] 350 400 450 500

K p 3.98 × 10−

1.41 × 10−

1.86 × 10−

1.48

dTT

ΔH

R

1lnKd

f f P

P

T

298.15

2

rxn

TK

298.15K

P


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C03

a b

15 5

The data in the following table have been obtained for the potential of the cell

Pt(s) | H2(g, f = 1 atm) | HCl(aq, m) | AgCl(s) | Ag(s)

as function of m at 25C. Given the cell reaction,

2AgCl(s) + H2(g) 2Ag(s) + 2H+(aq) + 2Cl – (aq)


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m (mol kg –

) E (V) m (mol kg – ) E (V) m (mol kg – ) E (V)

0.00100 0.59715 0.0200 0.43024 0.500 0.27231

0.00200 0.54425 0.0500 0.38588 1.000 0.23328

0.00500 0.49846 0.100 0.35241 1.500 0.20719

0.0100 0.46417 0.200 0.31874 2.000 0.18631

a) Determine E° using a graphical method. (15 marks)

b)

Calculate γ± for HCl at m = 0.00100, 0.0100, and 0.100. (5 marks)

SOLUTION:


m/m° (m/m°)1/2 E E+(2RT/F)ln(m/m°)

0.001 0.031623 0.579150 0.224212

0.002 0.044721 0.544250 0.224909

0.005 0.070711 0.498460 0.226203

0.010 0.100000 0.464170 0.227531

0.020 0.141421 0.430240 0.229218

0.050 0.223607 0.385880 0.231943

0.100 0.316228 0.352410 0.234090

m/m ln 0.001 -0.010654 0.989

0.010 -0.075511 0.927

0.100 -0.203036 0.816

a) Cell reaction: 2AgCl(s) + H2(g) 2Ag(s) + H+ (aq) + 2Cl – (aq)

lnγF

2RTE

m

mln

F

2RTEE

mγaaa

aalnF

RTEaaln

2F

RTEE

o

222

ClH

ClH

2

ClH

Low concentration limit use the Debye-Hückel result

oo

10m

m1.172614ln

m

m0.50926loglnγ

For dilute solutions

oo m

mln1.172614

F

2RTE

m

mln

F

2RTE

Plot, E+(2RT/F) ln(m/mo) (y-axis) vs (m/mo) (x-axis), intercept E

Use data up to m = 0.1 Debye-Hückel model is no valid for more concentrated solutions

b) Given E we can now find from

y = 0.035x + 0.2237

0.2200

0.2240

0.2280

0.2320

0.2360

0.0000 0.1000 0.2000 0.3000 0.4000

E + ( 2 R T / F ) l n ( m

/ m )

(m/m)^1/2

om

mlnEE

RT

Flnγ

From graph, E = 0.2237

E = 0.2237


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C04

a b c

10 5 5

Consider the schematic reaction,

B A k

a) If the reaction is 3.5-th order with respect to [A], solve the derivative of integrated rate

law expression for this reaction? (10 marks)

b)

Explain what k is, and how to determine its value from the above reaction? (5 marks)

c) Derive and evaluate the half-life expression for this reaction. (5 marks)

SOLUTION:


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C05

a b c

4 3 13

a. What is the difference between a homogeneous and a heterogeneous catalyst? (4 marks)

b.

What are the inherent assumptions in the Langmuir model of surface adsorption?

(3 marks)c. The adsorption of nitrogen on mica measured at different pressures is as follows

Vads (cm3g

-1) P (Torr)

0.494 2.110-3

0.782 4.610-3

1.16 1.310-3

Langmuir equation can be written as

1

Vads=

1

KVm1P

+ 1Vm


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where Vm is the maximum adsorption and K is equilibrium constant.

Using Langmuir isotherm, determine the

i. Langmuir parameters (7 marks)

ii.

Fractional coverage, at each pressure. (6 marks)

Fractional coverage, is defined as ratio of adsorbed volume to the volume of

maximum absorption

SOLUTION:

a. For a homogeneous catalysis both the catalyst and the substrate exist in the same phase.

1.5 marks

Heterogeneous catalysis occurs when the catalyst and the substrate molecules exist in

different phases. 1.5 marks

b.

Langmuir assumptions:

1. Adsorption finishes once one monolayer of coverage results.

2.

The surface is uniform and all adsorption sites are equivalent.3. The occupancy of a site will not affect the adsorption or desorption processes in

adjacent sites.

c.

i. Plot 1/Vads versus 1/P as follows: 4marks

Vads (cm g- ) 1/Vads P (Torr) 1/P

0.494 2.024 2.110-3 476.19

0.782 1.2794.6

10

-3

217.39

1.16 0.862 1.310-2 76.92


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Best fit to the data by a straight line yields:

1

Vads=

0.0029

1

P

+

0.642

The maximum adsorption volume is,Vm= 1intercept of y-axis Vm= 10.642

Vm=1.56 cm3g-1 2marks

The equilibrium constant, K = 1 = 10.00291.56 K=2.21 torr -1 2marks

ii. =Vads/Vm

Vads (cm g- ) P (Torr)

0.494 0.317 2.110-3

0.782 0.501 4.610-3

1.16 0.744 1.310-3

2013 Final Exam Answer

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