QWAQWA CAMPUS / QWAQWA KAMPUS
CEM 214
UNIVERSITEIT VAN DIE VRYSTAAT
UNIVERSITY OF THE FREE STATE
SEMESTER TEST 1
ASSESSOR:
Sat, 10 March 2012
Ms. C.E. Clarke
MODERATOR:
Prof. A.S. Luyt
TIME : 1 Hour
MARKS: 50
ANSWER ALL QUESTIONS AND SHOW ALL CALCULATIONS WITH AN INK PEN -- NO
PENCILS OR "TIPPEX" ALLOWED / PAPER CONSISTS OF PAGE(S) AND QUESTIONS.
RECORD ALL ANSWERS UP TO TWO (2) SIGNIFICANT FIGURES.
THERMODYNAMICS / 40 marks:
1. Estimate the molar combustion enthalpy of ethanol, CH3CH2OH(l), by using the appropriate average
bond energies data. Assume that the evaporation enthalpy of CH3CH2OH(l) = 47 kJ mol-1 and that of
H2O(l) = 31 kJ mol-1. Write the equation of the combustion of ethanol. Use T = 298 K and show all
steps.
[10]
Bond Energy /kJ mol-1
C-C
343
O-H
464
C=C
615
O-O
144
CC
812
O=O (in O2)
498
C-O
351
H-H
436
C=O
724
C-H
416
C=O (in CO2)
799
2. Derive from the Second Law of Thermodynamics an equation in terms of a, b, c and T to calculate the
entropy change, S, of water vapour as a function of temperature in a closed system during a reversible,
isobaric temperature change.
Given Cp(H2O) (g) = a + bT + cT-2
[7]
3. One mole ideal gas undergoes a reversible isothermal expansion until its volume is doubled. If the gas
performed 1 kJ work on the surroundings, what is the temperature?
[4]
4. The densities of ice and water at 0C are 0.9168 g.cm-3 and 0.9998 g.cm3, respectively. If H for the
fusion process at atmospheric pressure is 6.025 kJ.mol-1, what is U? How much work is done on the
system?
[15]
5. One mole of an ideal gas, with CV,m = 3/2R, is heated from 298 K to 353 K, at constant pressure.
Calculate S for the system.
[4]
MEMORANDUM:
1. 10 marks
CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) 
∆Ug  [BE(C-C) + 5xBE(C-H) + BE(C-O) + BE(O-H) + 3xBE(O=O)] – [4xBE(C=O) + 6xBE(H-O)] 
∆Ug  [343 + 5x416 + 351 + 464 + 3x498] kJ mol-1 - [4x799 + 6x464] kJ mol-1
∆Ug  [4732 – 5980] kJ mol-1 = -1248 kJ mol-1

∆Hg = ∆Ug + ∆(PV) where ∆(PV) is small
∆Hg  ∆Ug = -1248 kJ mol-1
-1248 kJ mol-1
CH3CH2OH(g) + 3O2(g)  2CO2(g) + 3H2O(g)
47 kJ mol-1 
CH3CH2OH(l)  CH3CH2OH(g)
3H2O(g)  3H2O(l)
-93 kJ mol-1 
CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
-1294 kJ mol-1
2. 7 marks
 S  
dq p

T
T2
C p dT
 S  
T
T1
dH
T
 and dH  C p dT
a  bT  cT  2
dT
T
T1
T2




T2
T2
c 
a


 S     b  cT 3 dT  S  a ln T  bT 
T
2T 2  T1


T1 
 S  a ln
T2
 bT2  T1  
T1
c  1
1 
 2 
2

2  T2
T1 

3. 4 marks

dw  - PdV
dw  - (nRT/V)dV
 dw  - nRT
- w  -nRTln
V2
1
 VdV
V1
V2
V1

- 1000J  - (1 mol)(8.314 J.K - 1.mol - 1)T ln
 1000 J
- (1 mol)(8.314 J.K - 1.mol - 1)ln2
 173.5K 
T
2
1



4. 15 marks
Mr(H2O) = 2(1.00) + (16.0)g.mol-1 = 18.01 g.mol-1 
1 mol of ice has a MASS (m) = Mr(H2O) x n = (18.01 g.mol-1) x (1 mol) = 18.01 g 
1 mol of water has a MASS (m) = Mr(H2O) x n = (18.01 g.mol-1) x (1 mol) = 18.01 g 
1 mol of ICE has a VOLUME (Vice) = m/
= (18.01 g) / (0.9168 g.cm-3) 
= 19.64 cm3 
1 mol of WATER has a VOLUME (Vwater) = m/
= (18.01 g) / (0.9998 g.cm-3) 
= 18.01 cm3 
V(water  ice) = (18.01 – 19.64) cm3mol-1 
= -1.63 cm3.mol-1 = -0.00163 dm3.mol-1 
(PV) = -0.00163 dm3.mol-1 x 1 atm
= -0.00163 dm3.mol-1 x 101.325 J 
= - 0.165 J.mol-1 
H = U + PV = 6.025kJ = 6025 J 
6025 J = U + (-0.165 J.mol-1) 
U = 6024.84 J.mol-1 = 6025 J.mol-1 = 6.025 kJ.mol-1 
Work done on the system = 0.165 J.mol-1 
5. 4 marks
CV,m = 3/2R
 CP,m = 3/2R + R = 5/2R 
353
Sm = ∫298 (CP,m)dT/T
= 5/2R ln(T2/T1) 
= 5/2 x (8.3145 J.K-1.mol-1) x ln(353/298) 
= 3.52 J.K-1.mol-1 