1. science
2. physics
3. thermodynamics

Chemistry 360
Dr. Jean M. Standard
Spring 2015
Name ____________KEY________________
Exam 2 Solutions
1.) (14 points) Using the values in the table below reported at 25°C, determine ΔGR! at 300ºC for the
hydrogenation of acetylene to ethane,
C2 H 2 ( g) + 2 H 2 (g) → C2 H 6 (g) .
ΔH !f (kJ/mol)
C2H2 (g)
C2H6 (g)
H2 (g)
€
ΔG !f (kJ/mol)
227.4
209.2
–84.0
–32.0
0
0
The standard molar Gibbs free energy at another temperature may be estimated using the Gibbs-Helmholtz
equation,
$ ∂ ( ΔG /T ) '
&&
)) = ΔH .
% ∂ (1/T ) (P
If we approximate the partials by finite changes, the Gibbs-Helmholtz equation becomes
€
Δ ( ΔG /T )
≈ ΔH ,
Δ (1 /T )
$ ΔG '
$1'
or Δ &
) ≈ ΔH ⋅ Δ & ) .
% T (
%T (
The changes are defined as the differences in the higher temperature (573 K in this case) and 298 K,
€
#1&
1
1
Δ% ( =
−
$T '
Tf
Ti
# ΔG &
ΔG
and Δ %
( =
$ T '
T
−
T=T f
ΔG
T
.
T=Ti
Inserting these relations into the approximate Gibbs-Helmholtz equation,
€
# ΔG &
#1&
Δ%
( ≈ ΔH ⋅ Δ % (
$ T '
$T '
ΔG
T
€
−
T=T f
ΔG
T
T=Ti
#1
1&
(.
≈ ΔH %%
−
T i ('
$T f
2
1.
Continued
Substituting the temperatures,
$1
ΔG
ΔG
1'
− ≈ ΔH && − ))
T T =Tf
T T =Ti
Ti (
% Tf
$ 1
ΔG573
ΔG298
1 '
− ≈ ΔH 298 &
− ).
573K
298K
298K (
% 573K
Here, it has been assumed that the enthalpy of reaction is constant and the value at 298 K has been used.
Isolating the unknown Gibbs free energy change at 573 K, we have
$ 1
ΔG573
ΔG298
1 '
≈ + ΔH 298 &
− ),
573K
298K
298K (
% 573K
* ΔG
$ 1
1 'or ΔG573 ≈ ( 573K ), 298 + ΔH 298 &
− )/ .
298K (.
% 573K
+ 298K
In order to complete the determination of the Gibbs free energy change at 573 K, the values of the Gibbs free
energy and the enthalpy of reaction at 298 K are required. These can be obtained from the data given in the
table.
The standard molar enthalpy of reaction at 25°C is
ΔH R! = ΔH !f (C2 H 6 ) − ΔH !f (C2 H 2 ) − 2ΔH !f ( H 2 )
= (−84.0 kJ/mol) − ( 227.4 kJ/mol) − 2 ( 0 kJ/mol)
ΔH R!
= −311.4 kJ/mol.
The standard molar Gibbs free energy of reaction at 25°C is
ΔGR! = ΔG !f (C2 H 6 ) − ΔG !f (C2 H 2 ) − 2 ΔG !f ( H 2 )
= (−32.0 kJ/mol) − ( 209.2 kJ/mol) − 2 ( 0 kJ/mol)
ΔGR!
= −241.2 kJ/mol.
Now we can substitute these values into the expression above to get the Gibbs free energy of reaction at 573 K,
* ΔG
$ 1
1 'ΔG573 ≈ ( 573K ), 298 + ΔH 298 &
− )/
298K (.
% 573K
+ 298K
* −241.2 kJ/mol
$ 1
1 '≈ ( 573K ),
+ (−311.4 kJ/mol) &
− )/
298K
298K (.
% 573K
+
ΔG573 ≈ −176.4 kJ/mol.
This is a fairly large change from the value at 298 K and we see that as expected for an exothermic reaction, the
Gibbs free energy increased (became less negative).
3
2.) (14 points) Consider a classroom that is 5 m × 10 m × 3 m in size. Initially, the temperature in the room
is 20°C and the pressure is 1 bar. There are 20 people in the class. Each person gives off heat to the
room at an average rate of 150 Watts [1 Watt = 1 Joule/second]. Assume that the walls, ceiling, floor, and
furniture are perfectly insulating and do not absorb any heat; that is, all the heat from the people in the
classroom is absorbed by the air in the room.
a.) Determine the heat (in Joules) required to raise the room temperature to 37°C. Assume that the
pressure remains constant at 1 bar, the air behaves ideally, and the molar constant pressure heat
7
capacity of air is C p,m = R .
2
At constant pressure, q = ΔH = C p ΔT = nC p,m ΔT (this equation assumes that the heat capacity is
constant, as given). In order to determine the heat required to raise the temperature to 37°C, we need the
number of moles of air in the room. Assuming ideal behavior, the moles of air may be calculated from the
ideal gas equation using the volume, initial temperature, and pressure of the room. Calculating the room
volume,
V = length ⋅ width ⋅ height
= ( 5 m ) (10 m ) (3 m )
$
'
1 L
= 150 m 3 &
)
% 1×10 −3 m 3 (
V = 1.50 ×10 5 L
Then, the number of moles of air in the room is
n = = PV
RT
(1 bar ) (1.50 ×10 5 L )
(0.08314 L bar mol–1K-1 ) (293 K)
n = 6158 mol .
The heat required to raise the temperature to 37°C now may be calculated,
q = nC p,m ΔT
" 7%
= ( 6158 mol) $ ' 8.314 Jmol–1K -1 (310 K - 293 K )
#2&
(
q = 3.05 ×10 6 J (or 3050 kJ)
)
4
2.) continued
b.) How long in minutes will it take for the air in the classroom to reach 37°C?
We know that each person generates 150 Watts of heat and there are 20 people in the room. This allows us
to calculate how many Joules of heat per minute are generated,
! 150 Watt $! 1 Js –1 $! 60 s $
5
20 people #
&#
&#
& = 1.80 ×10 J/min .
" person %" 1 Watt %" 1 min %
Since 3.05 ×106 Joules are required to raise the temperature in the room to 37°C, next we can determine
how long this will take,
" 1 min %
3.05 ×10 6 J $
' = 16.9 min .
# 1.80 ×10 5 J &
c.)
Determine the change in entropy (in J/K) of the air in the classroom.
At constant pressure, the entropy change for heating a system is
ΔS = T final
∫T
Cp
T
init
dT = T final
∫T
init
nC p,m
T
dT .
Since the heat capacity in this case is constant we can pull it out and integrate,
ΔS = T final
∫T
nC p,m
init
T
dT
1
dT
T
#T &
= nC p,m ln % final (
$ Tinit '
= nC p,m ∫
T final
Tinit
# 310 K &
# 7&
= ( 6158 mol) % ( 8.314 Jmol–1K -1 ln %
(
$2'
$ 293K '
(
(
)
)
ΔS = 10110 J/K or 1.011×10 4 J/K .
5
3.) (14 points) Consider the following standard molar entropies at 298 K.
!
Sm
(J/molK)
Compound
CO (g)
CO2 (g)
C2H4 (g)
C2H6 (g)
197.7
213.8
€
219.3
229.2
Explain why we are able to obtain absolute molar entropies of substances; that is, what is the theoretical
foundation that allows the determination of absolute molar entropies? Discuss and explain any trends
observed in the molar entropies of the substances given in the table above.
The opportunity for an absolute entropy scale is provided by the Third Law of Thermodynamics, which sets the
absolute entropy of a perfect crystalline substance as 0 at 0 K.
The absolute entropies tabulated above are all for gases of varying structural complexity. As the substances
vary from a diatomic gas (CO) to a triatomic gas (CO2) to polyatomic gases (C2H4 and C2H6), the entropy is
observed to increase. This is a result of the increase in the number of degrees of freedom of the substances. As
we saw in our discussion of the Equipartition Theorem, the number of degrees of freedom increases as the
complexity of the molecule increases. For the complex molecules with more degrees of freedom, there are
more ways of distributing the available energy among the degrees of freedom, which increases the disorder in
the system. Thus, more complex substances have higher entropies than less complex substances.
6
4.) (15 points) True/false, short answer, multiple choice.
a.) True or False : The reaction 2 H 2 (g) + O 2 (g) → 2 H 2O ( ℓ) is an example of a formation reaction.
b.) True or False : In the statistical definition of entropy, S = kB lnW , the quantity W corresponds to the
energy.
c.) Short answer
The ______Gibbs free energy, G________
equals H − TS .
d.) Short answer
The ______Second Law of Thermodynamics_______ states that that ΔS ≥ 0 for an isolated system.
e.) Multiple Choice: A spontaneous process at constant temperature and pressure must have:
1) ΔU < 0 .
2) ΔH < 0 .
3) ΔA < 0 .
4) ΔG < 0 .
7
5.) (14 points) Determine ΔS for the isothermal compression at 300 K from 100 L to 10 L of two moles of a
real gas obeying the equation of state given by
€
Z = 1 + B
,
Vm
where Z is the compression factor and B is a constant. Assume that B = −0.122 L/mol .
The definition of the entropy is dS =
process,
dqrev
. From the first law, dU = dq + dw , or for a reversible
T
€
dU = dqrev€ − PdV .
€
Solving for dqrev yields
€
dqrev = dU + PdV .
€
#∂ U &
Substituting dU = C v dT makes the assumption that %
( = 0 , which it is. This can be shown from the
$ ∂ V 'T
€
Maxwell relations. The expression for dqrev then becomes
€
dqrev = C v dT + PdV .
€
€
Finally, the entropy is given by
€
dS =
C v dT
P
+
dV .
T
T
For an isothermal process, dT = 0 , and so the entropy becomes
€
dS =
€
P
dV .
T
This expression can be integrated if P/T for this gas can be evaluated. From the equation of state,
€
B
,
V
PVm
B
or
= 1 + .
RT
Vm
Z = 1 + Solving for P/T, we get
P
R
BR
= + 2 ,
T
Vm
Vm
or P
nR
n 2 BR
= + 2 .
T
V
V
8
5.
Continued
Substituting into the expression for entropy leads to
" nR
n 2 BR %
'' dV .
dS = $$
+
V2 &
#V
Breaking this up into two terms and integrating gives
€
2
V2 n BR
nR
dV + ∫
dV
V1
1 V
V2
#V &
# 1
1&
= nR ln % 2 ( + n 2 BR % − + (
V2 '
$ V1 '
$ V2
# 10 L &
= ( 2 mol) 8.314 J mol−1K −1 ln %
(
$ 100 L '
ΔS = V2
∫V
(
)
# 1
1 &
2
+ ( 2 mol) −0.122 L mol−1 8.314 J mol−1K −1 % −
+ (
100 L '
$ 10 L
= −38.29 J/K + 0.37 J/K
(
)(
)
ΔS = −37.92 J/K.
Note that the entropy change is negative because the process involves a compression.
9
6.) (14 points) The standard molar enthalpy of combustion of graphite is
ΔH R!
= − 393.51 kJ/mol and the
ΔH R!
= −1299.58 kJ/mol , both at 298 K. Use this
information, along with the standard molar enthalpy changes for the following reactions at 298 K,
standard molar enthalpy of combustion of C2H2 (g) is
(1)
CaC 2 (s) +
(2)
Ca (s)
(3)
CaO (s)
2 H 2O (ℓ)
1
2
+
+
O 2 (g)
H 2O (ℓ)
Ca(OH) 2 (s)
→
→
→
ΔH 1" = − 127.9 kJ/mol
+ C 2 H 2 (g)
ΔH 2" = − 635.1 kJ/mol
CaO (s)
ΔH 3" =
Ca(OH) 2 (s)
− 65.2 kJ/mol ,
to determine the standard molar enthalpy of formation of CaC2 (s) at 298 K.
€
The formation reaction of CaC2 is
Ca (s) + 2 C (s) → CaC2 (s).
Note that reaction (1) above contains CaC2 (s) as a reactant, so it will need to be flipped in order to get it as a
product in the formation reaction shown above. Reaction (2) contains Ca (s) as a reactant, and the combustion
reaction for graphite will include C (s) as a reactant. The remainder of the reactions will be needed to cancel out
the other products and reactants:
Ca (OH )2 (s) + C2 H 2 (g)
Ca (s) + 1
2
→ CaC2 (s) + 2 H 2O (ℓ)
O 2 (g) → CaO(s)
CaO (s) + H 2O (ℓ) → Ca (OH )2 (s)
2 [ C (s) + O 2 (g)
2 CO 2 (g) + H 2O (ℓ)
→ CO 2 (g) ]
→ C2 H 2 (g) + 25 O 2 (g)
Ca (s) + 2 C (s) → CaC2 (s)
From the combination of reactions, the enthalpy of formation of CaC2 can be calculated from Hess' Law,
!
!
ΔH !f = −ΔH1! + ΔH 2! + ΔH 3! + 2 ΔH comb
(C) − ΔH comb
(C2 H 2 ) = − (−127.9 kJ/mol) + (−635.1) + (−65.2 ) + 2 (−393.51) − (−1299.58 kJ/mol)
ΔH !f
= −59.8 kJ/mol.
10
7.) (15 points) True/false, short answer, multiple choice.
a.) True or False: For the reaction 2 CO (g) + O2 (g) → 2 CO2 (g), the entropy change is expected to be
negative.
!∂ G $
b.) True or False : The expression – #
& is equal to the volume V.
" ∂ P %T
c.) Short answer
The ______Helmholtz energy A______ is equivalent to the isothermal work under reversible conditions.
d.) Short answer
The differential dH = T dS + V dP corresponds to one of four ______Fundamental Equations_______ .
e.) Multiple Choice: The molar constant pressure heat capacity of a crystalline solid at very low temperatures
is given by the expression:
1) C p,m = aT 2 .
T1
2) C p,m = ∫ 0
aT 2
dT .
T
3) C p,m = aT 3.
T1
4) C p,m = ∫ 0
aT 3
dT .
T