1.
Answer ALL parts
(a)
For an ideal monatomic gas:
CP = (5/2)R ; CV = (3/2)R ; and
S
 C P ln
T
T0

R ln
P
P0
An ideal monatomic gas is stored in a vessel with volume 0.0500 m3 at
2.00 bar and 2.50 × 102 K. The gas is heated, transferring 1.000 × 104 J of heat
(Q) to the gas without changing the volume of the vessel, and with no heat
being lost to the vessel. Calculate the temperature of the gas after heating, and
hence calculate the change in entropy that results.
(b)
One mole of an ideal diatomic gas [CP = (7/2)R and CV = (5/2)R] is initially
stored at P1 = 12.0 bar and T1 = 500 K. It is then depressurised to P2 = 1.0 bar
by an isothermal pathway. Assuming mechanical reversibility, and using the
ideal gas law, calculate W, Q, ΔU and ΔH.
Note, for this system:
5
R T
2
7
H  R T
2
W    PdV
U 
U  Q  W
(c) Thermodynamic data for the reversible hydrogenation reaction of gaseous
benzene is given below
C6H6 (g) + 6H2(g)
benzene

C6H12(g)
cyclohexane
Chemical Species
ΔHºf 298/ kJ mol–1
ΔGºf 298 / kJ mol–1
benzene (gas)
+83
+130
styrene (gas)
–123
+32
(i)
Calculate ΔHº298, ΔGº298 and ΔSº298for this reaction, and hence
determine the equilibrium constant Kp at 298 K.
 G 0 
K p  exp 

 RT 
(ii)
Would it be easier or harder to hydrogenate benzene at a higher
temperature? Why?
2.
Answer ALL parts.
(a)
Balance the following reaction and hence evaluate the standard enthalpy of
reaction if the reaction is carried out at 500 ºC?
C2H6 (g) + aO2 (g) → bCO2(g) + cH2O(g)
The enthalpy change for this equation is
H T01
C P0
dT
R
T0
T1
 R
 H T00
The enthalpy of formations of these compounds at 298 K are
compund
C2H6(g)
CO2(g)
H2O(g)
Hfo298 / kJ mol–1
-85
-394
-242
The heat capacity of each substance is well described by the equation
C P0
R

A 
DT 2
where the constants A and D are:
A
D / (105 K2)
C2H6
3.438
-0.177
O2
3.639
-0.227
CO2
3.217
0.114
H2O
3.470
0.121
Chemical species
Why is no value given for the enthalpy of formation of O2?