CH 342 – Third Midterm
Tuesday, April 10, 2012
1) Calculate the entropy change when argon at 25°C and 1 atm pressure in a container of
volume 250 cm3 is allowed to expand to 500 cm3 and simultaneously is heated to a
temperature of 90°C.
30 points
The initial state is (250 cm3, 298 K); the final state is (500 cm3, 363 K). The sample can
be taken from the initial to the required final state in two steps. The first is an isothermal
expansion from Vi to Vf; this is followed by heating from Ti to Tf at constant volume. This
is because S is a state function.
The entropy change in the first step is S = nR ln (Vf / Vi).
The entropy change in the second step can be obtained as follows:
Tf
S 
 dq 
rev V
Ti
Tf
Tf
Ti
Ti
/ T   dU / T   CV dT / T  CV ln(Tf / Ti )
The total entropy change for the overall change is the sum
S = nR ln (Vf / Vi) + nCV,,m ln(Tf / Ti)
n = PiVi/RTi = (1 atm) x (0.25 dm3) /(0.082 atm dm3 K-1 mol-1) x (298 K) = 0.010 mol.
S = (0.010 mol) x (8.31 J K-1 mol-1) ln 2 + (0.010 mol) x (12.5 J K-1 mol-1) x
ln (363 K / 298 K)
= 0.058 J K-1 + 0.024 J K-1 = 0.083 J K-1
1
2) Calculate S, Stotal, and Ssurroundings when the volume of 85.0 g of CO initially at 298
K and 1.00 bar increases by a factor of three in (a) an adiabatic reversible expansion, (b)
an expansion against Pexternal = 0, and (c) an isothermal reversible expansion. Take CP,m to
be constant at the value 29.14 J mol–1K–1 and assume ideal gas behavior. State whether
each process is spontaneous.
40 points
a) An adiabatic reversible expansion
Ssurroundings = 0 because q = 0.S = 0 because the process is reversible.
Stotal= S + Ssurroundings = 0. The process is not spontaneous.
b) An expansion against Pexternal = 0
T and w = 0. Therefore U = q = 0.
Vf
85.0 g
S  nR ln

 8.314 J mol-1 K -1  ln 3
-1
Vi 28.01g mol
 3.03 mol  9.13 J mol -1 K -1 = 27.7 J K -1
Stotal=S + Ssurroundings = 27.7 J K-1 + 0 = 27. 7 J K-1. The process is spontaneous.
c) An isothermal reversible expansion
T = 0. Therefore U = 0.
V
w   q  nRT ln f  3.03 mol  8.314 J mol -1 K -1  298 K  ln 3  8.25  103 J
Vi
qreversible 8.25  103 J
S 

= 27.7 J K -1
T
298 K
 q 8.25  103J
S surroundings 

 27.7 J K -1
T
298 K
Stotal=S + Ssurroundings = 27.7 J K-1 – 27.7 J K-1 = 0. The system and surroundings are
at equilibrium.
2
3) (a) A container is divided into two equal compartments. One contains 4.0 mol H2 at
25°C; the other contains 1.0 mol N2 at 25°C. Calculate the Gibbs energy of mixing when
the partition is removed. Assume ideal gas behavior.
20 points
(b) What would be the Gibbs energy of mixing when 4.0 mol H2 mixes with 1.0 mol of
N2 at the same pressure, with the volumes of the compartments adjusted accordingly?
What is the entropy of mixing in this case? Assume ideal gas behavior.
20 points
40 points total
(a) Given that the pressure of nitrogen is P, the pressure of hydrogen is 4P; therefore,
the initial Gibbs energy is
Gi = (4.0 mol){µ°(H2) + RTln(4P/P°)} + (1.0 mol){µ°(N2) + RTln(P/P°)}
When the partition is removed and each gas occupies twice the original volume, the
partial pressure of nitrogen falls to ½P and that of hydrogen falls to 4/2P. Therefore, the
Gibbs energy changes to
Gf = (4.0 mol){µ°(H2) + RTln(4P/2P°)} + (1.0 mol){µ°(N2) + RTln(P/2P°)}
The Gibbs energy of mixing is the difference of these two quantities:
ΔGmixing = (4.0 mol)RT ln[(4/2P)/(4P)] + (1.0 mol)RT ln[(1/2P)/(P)]
= -(4.0 mol)RT ln2 –(1.0)RT ln2 = -(5.0 mol)RT ln2 = -8.58 kJ
(b) If pressure is constant, then
ΔGmixing = nRT (xH2lnxH2 + xN2lnxN2)
n = nH2 + nN2 = 5 mol; xH2 = 4/5; xN2 = 1/5
ΔGmixing = 5.0RT (4/5 ln4/5 + 1/5 ln1/5)
= - 5.0x8.314x298x10-3(0.178 + 0.322) = -6.2 kJ.
ΔSmixing = -nR (xH2lnxH2 + xN2lnxN2) = 20.8 JK-1
3
4)
(a) Explain the significance of the word “free” in the terms Gibbs free energy (G) and
Helmholtz free energy (A)?
5 points
(b) A process involving an ideal gas is carried out in which the temperature changes at
constant volume. For a fixed value T, the mass of the gas is doubled. The process is
repeated with the same initial mass and T is doubled. For which of these processes is S
greater? Why?
5 points
(c) Qualitatively plot the temperature and pressure dependence of G for a gas, liquid and
solid.
10 points
(d) What is the relationship between the KP for the two reactions 3/2H2 + 1/2N2 → NH3
and 3H2 + N2 → 2NH3?
5 points
(e) You are told that ΔS = 0 for a process in which the system is coupled to its
surroundings. Can you conclude that the process is reversible? Justify your answer.
5 points
30 points total
(a) The word “free” refers to energy available in the form of useful work. For G it is the
maximum nonexpansion work, whereas for A it is maximum expansion plus
nonexpansion work.
(b) S is greater if the mass is doubled, because S increases linearly with the amount of
material. By contrast, S only increases as the logarithm of the temperature. This
increase is much slower than a linear increase.
(c)
 G 
 G 

   S and 
 V
 T P
 P T
(d) KP for the second reaction is the square of KP for the first reaction as can be seen the
2
 PNH 3 
 PNH 3 
 P 
 P 




reactions quotients
and
.
32
12
3
 PH 2   PN 2 
 PH 2   PN 2 
 P   P 
 P   P 

 


 

(e) No. The criterion for reversibility is S + S surroundings = 0. To decide if this criterion
is satisfied, S surroundings must be known.
4
5) The pressure at the center of the earth is probably greater than 3x106 atm, and the
temperature there is about 4000°C. What is the change in the Gibbs function of reaction
on going from crust to core for a reaction in which Vm = 1 cm3 mol-1 and Sm = 2.1 J K-1
mol-1? Does the pressure or temperature effect dominate in determining the change?
30 points
Since (∂Gm/∂p)T = Vm and (∂Gm/∂T)P = -Sm we can make a rough estimate from
Gm (core) - Gm (crust)  Vm (pcore – pcrust) - Sm(Tcore – Tcrust)
Vm (pcore – pcrust)  (1 x 10-6 m3 mol-1) x (3 x 106 x 1.013 x 105 N m-2)
 3.04 x 105 J mol-1,
Sm(Tcore – Tcrust)  (2.1 J K-1 mol-1) x (4273 - 298 K)
 8300 J mol-1.
Then,
Gm (core) - Gm (crust)  300 kJ mol-1 – 8.3 kJ mol-1
 296 kJ mol-1.
The pressure effect dominates, and even for such small volume changes as used in the
example, the change in Gibbs function is dramatic. This is the thermodynamic basis of
why materials change their structure at great depths in the earth’s interior.
5
6) NO2 (g) and N2O4 (g) are in a reaction vessel with partial pressures of 0.350 and 0.650
bar, respectively, at 298 K. Is this system at equilibrium? If not, must the system move
toward reactants or products to reach equilibrium?
30 points
The reaction of interest is 2NO2 (g) ↔ N2O4 (g). We calculate ΔGR using
ΔGR = ΔGoR + RT lnQP
ΔGR = 99.8 kJ mol-1 – 2x51.3 kJ mol-1 + 8.314 J mol-1 K-1 x 298 K ln (0.650 bar/1
bar)/(0.350 bar/ 1 bar)2
= -2.80 x 103 J mol-1 + 4.13 x 103 J mol-1 = 1.33 x 103 J mol-1
= 1.33 x 103 J mol-1
Because ΔGR ≠ 0, the system is not at equilibrium. Because ΔGR > 0, the system moves
towards reactants.
6