Burdge/Overby, Chemistry: Atoms First, 2e Ch14 | Page 32

A n swe r s to I n - C h a pte r Mate r i a ls

14.65 Hydrogenation reactions (for example, the process
of converting C C bonds to C C bonds in the food
industry) are facilitated by the use of a transition
metal catalyst, such as Ni or Pt. The initial step is the
adsorption, or binding, of hydrogen gas onto the metal
surface. Predict the signs of ?H, ?S, and ?G when
hydrogen gas is adsorbed onto the surface of Ni metal.
14.66 At 0 K, the entropy of carbon monoxide crystal is not
zero but has a value of 4.2 J/K ? mol, called the residual
entropy. According to the third law of thermodynamics,

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this means that the crystal does not have a perfect
arrangement of the CO molecules. (a) What would be
the residual entropy if the arrangement were totally
random? (b) Comment on the difference between the
result in part (a) and 4.2 J/K ? mol. (Hint: Assume that
each CO molecule has two choices for orientation, and
use Equation 14.1 to calculate the residual entropy.)
14.67 Which of the following thermodynamic functions are
associated only with the first law of thermodynamics:
S, U, G, and H?

Answers to In-Chapter Materials
Practice Problems

14.1A 0.34 J/K. 14.1B __??. 14.2A (a) 173.6 J/K ? mol, (b) –139.8 J/K ? mol,
?? 1
5
(c) 215.3 J/K ? mol. 14.2B (a) S°[K(l)] = 71.5 J/K ? mol, (b) S°[S2Cl2(g)]
= 331.5 J/K ? mol, (c) S°[MgF2(s)] = 57.24 J/K ? mol. 14.3A (a) negative,
(b) negative, (c) positive. 14.3B The sign of ?H° for both dissolution
processes is negative. Something must favor spontaneity; if not entropy
change, then enthalpy change. Because these processes both involve
decreases in the system’s entropy, they must be exothermic, or they could
not be spontaneous. 14.4A (a) ?Suniv = –27.2 J/K ? mol, nonspontaneous,
(b) ?Suniv = –28.1 J/K ? mol, nonspontaneous, (c) ?Suniv = 0 J/K ? mol,
equilibrium. 14.4B (a) ?Suniv = 5.2 J/K ? mol, spontaneous, (b) 346°C,
(c) 58°C. 14.5A 3728°C. 14.5B 108 J/K ? mol. 14.6A (a) –106 kJ/mol,
(b) –2935 kJ/mol. 14.6B (a) ?G°[Li2O(s)] = –561.2 kJ/mol,
f
(b) ?G°[NaI(s)] = –286.1 kJ/mol. 14.7A ?Sfus = 16 J/K ? mol,
f
?Svap = 72 J/K ? mol. 14.7B (a) ?H° = 2.41 kJ/mol,
fus
?S° = 6.51 J/K ? mol, Tmelting = 97°C. (b) ?H° = 105.3 kJ/mol,
fus
vap
?S° = 96.1 J/K ? mol, Tboiling = 823°C.
vap

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Section Review

14.3.1 (a) negative, (b) positive, (c) positive, (d) positive, (e) negative.
14.3.2 (a) positive, (b) negative, (c) positive, (d) positive, (e) positive.
14.3.3 A2 + 3B2
2AB3, negative. 14.4.1 –145.3 J/K ? mol.
14.4.2 –242.8 J/K ? mol. 14.4.3 (a) negative, (b) positive, (c) not enough
information to determine. 14.5.1 –800.8 kJ/mol. 14.5.2 196 J/K ? mol.
14.5.3 790°C.

9/10/13 12:01 PM