CH101 Spring 2012 Final Exam Review Questions Chapter 8
#1. When nickel (73.0 g) at 100 oC is added to ice (20.0 g) contained in an
insulated vessel, 10.0 g of ice is melted. Calculate the total entropy change for
the thermodynamic universe above (i.e. the nickel and the ice/water bath).
[Specific heat capacity of Ni at constant pressure is 0.46 J K -1 g-1. The enthalpy
of fusion of ice is 334 J K-1 g-1. Take 0 oC as 273 K]
Assume constant pressure throughout and perfect insulation for the ice/water
bath.
Here, the nickel is the system and the ice/water bath the surroundings and heat
flows from the nickel into the bath and melts some of the ice. Hence the entropy
of the nickel decreases, whilst that of the bath increases. The final equilibrium
temperature for both the system and surroundings is 0.0oC, as indicated by the
presence of ice.
Since the nickel is cooled at constant pressure, the entropy change for the nickel
is given by
S = m cp ln (TF/TI)
i.e. S = 73.0 (g) x 0.46 (J K-1 g-1) x ln 0.73
= - 10 J K-1
Since the ice bath has remained at constant temperature throughout the
experiment, it can be treated as a “large heat bath” and its entropy change is
given by
S = - H(system)/T (bath)
= - {- 334 (J g-1)} x {10.0 (g)}/273.15 (K)
= +12 J K-1
Now, S (total) = S (Ni) + S (bath)
= - 10 + 12
= + 2 J K-1
Thus the process is spontaneous, driven by the fact that the entropy gain of the
melting ice exceeds that of the entropy loss of the cooling nickel.
#2. If 4.00 mol of hydrogen is expanded reversibly at 400 K from an initial volume
of 12.0 L to a final volume of 30.0 L (also at 400 K), calculate U, q, w, H and
S for the gas.
[R = 8.315 J mol-1 K-1]
This is an isothermal expansion, as there is no temperature change.
Hence, U = H = 0
Also q = -w (from the first law of thermodynamics)
S = nR ln (V2/V1) for an isothermal expansion
= 4 (mol) x 8.315 (J K-1mol-1) x ln(30.0/12.0)
= +30.5 J K-1
S = +30.5 J K-1
For reversible isothermal process,
S = q(rev)/T
= q(rev) (J)/400 (K) = 30.5 (J K-1)
Hence, q(rev) = +12,200 J
And w = -12,200 J
#3. A sample of iron (55.9 g) initially at 100.0 oC was added to 90.1 g of water
initially at 10.0 oC and an equilibrium temperature of 15.8 oC was reached. If the
specific heat capacities (cp) of iron and water are 25.1 J K-1 mol-1 and 75.3 J K-1
mol-1, respectively, calculate S for the iron, S for the water and S (total) for
this process.
[molar masses (g mol-1): Fe = 55.9; H2O = 18.02. Take 0 K  -273 oC]
Number of moles of Fe = 55.9 (g)/55.9 (g mol-1) = 1.00
S(Fe) = ncp ln (T2/T1) = 1.00 (mol) x 25.1 (J K-1 mol-1) x ln(288.8/373.0)
= -6.42 J K-1
Number of moles of water = 90.1 (g)/18.02 (g mol-1) = 5.00
S(H2O) = ncp ln(T2/T1) = 5.00 (mol) x 75.3 (J K-1 mol-1) ln (288.8/283.0)
= +7.64 J K-1
S(total) = S(Fe) + S(H2O) = +1.22 J K-1 (spontaneous)
#4. Tungsten (VI) oxide (WO3) can be reduced to the metal by reaction with
hydrogen gas at high temperatures:
WO3(s) + 3H2(g)
W(s) + 3H2O(g)
Ho for this reaction is +117.4 kJ and So is +131.2 J K-1.
Determine the temperature range in which this reaction is spontaneous.
G = H - TS = 117400 (J) - T x 131.2 (J K-1)
At equilibrium, G = 0 and T = Ho/So
= 117400 (J)/131.2 (J K-1)
= 895 K
At temperatures above 895 K, TSo > Ho and G < 0: the reaction is
spontaneous above 895 K, but non-spontaneous below 895 K.
#5. At 1200 oC, the reduction of iron (III) oxide (Fe2O3) to elemental iron is not
spontaneous:
G = +840 kJ
2Fe2O3(s)
4Fe(s) + 3O2(g)
Show how this process can be made to proceed if all the oxygen generated
reacts with carbon as follows:
_
C(s) + O2(g)
CO2(g) G = 400 kJ
2Fe2O3(s)
4Fe(s) + 3O2(g)
G = +840 kJ
3C(s) + 3O2(g)  3CO2(g); G = 3 x (-400 J) = -1200 J
Coupling these two equations, where all the oxygen from the first reaction is used
up in combustion of carbon, gives
3C(s) + 2Fe2O3(s)  3CO2(g) + 4Fe(s); G = 840 - 1200 = -360 kJ
If ferrous oxide and carbon (ratio 2:3) are heated together, conversion of ferrous
oxide to iron is spontaneous.
#6. Use Boyle’s law to show that the entropy change for an isothermal change in
volume of an ideal gas can be written as S = nR ln (P1/P2). Use this formula to
calculate the entropy change, when 0.250 mol of N 2(g) is compressed from 1.00
atm to 6.00 atm.
[R = 8.315 J mol-1 K-1]
At constant temperature, Boyle’s law applies: P1V1 = P2V2 or P1/P2 = V2/V1.
S = nR ln (V2/V1) = nR ln (P1/P2)
= 0.250 (mol) x 8.315 (J K-1 mol-1) x ln (1.00/6.00)
= -3.72 J K-1 (As expected, S is negative for an isothermal compression)
#7. Deduce the sign of S for the following reactions.
(a) 2Na(s) + Cl2(g)  2NaCl(s) (b) Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) (c)
HCl(g) + NH3(g)  NH4Cl(s) (d) CaO(s) + SO2(g)  CaSO3(s) (e) H2(g) + Br2(g)
 2HBr(g) (f) H2O(l) + D2O(l)  2HOD (D = deuterium, 2H).
The question is best answered in tabular form.
Reaction
a
b
c
d
e
f
n(gas)
-1
+1
-2
-1
0
0
Sign of So(system)
Negative
Positive
Negative
Negative
Positive
Positive
Generally, if n(gas) < 0, So < 0 and if n(gas) >0, So > 0. Part (e) and (f) have
n(gas) = 0, but the products are more disordered than reactants (compounds,
versus elements; ‘scrambling’ of isotopes), hence we predict So > 0.
#8. Explain briefly whether the following processes have positive or negative
system entropy changes.
(a) Dissolution of glucose in water
(b) Separation of pure nitrogen from a sample of air at constant pressure and
temperature.
(c) Dimerization of acetic acid in an organic solvent.
(a) Positive. Disorder is increased: the molecular crystal lattice of glucose is
broken by water molecules and the glucose molecules are dispersed
throughout water.
(b) Negative. Separation and isolation of a pure substance (N2) from a solution
leads to decrease in disorder.
(c) Negative. The clustering together of acetic acid molecules into dimers is
accompanied by a decrease in disorder.