Burdge/Overby, Chemistry: Atoms First, 2e Ch14 | Page 16

SEC TION 14.4 ? Entropy Changes in the Universe

585

?H° = [2?H°(HI)] ? [?H°(H2) + ?H°(I2)]
rxn
f
f
f
= (2)(25.9 kJ/mol) ? [0 kJ/mol + 62.25 kJ/mol] = –10.5 kJ/mol
–?H rxn
–(–10.5 kJ/mol)
??
? ?? ?? ???0.0385 kJ/K ? mol = 38.5 J/K ? mol
= _____________ =
?Ssurr = ______? ??
T
273 K
?Suniv = ?Ssys + ?Ssurr = 21.03 J/K ? mol + 38.5 J/K ? mol = 59.5 J/K ? mol
?Suniv is positive; therefore the reaction is spontaneous at 0°C.
(b), (c)? ?S° = 160.5 J/K ? mol
rxn
?H° = [?H°(CaO) + ?H°(CO2)] ? [?H°(CaCO3)]
rxn
f
f
f

(b)

= [–635.6 kJ/mol + (–393.5 kJ/mol)] ? (–1206.9 kJ/mol) = 177.8 kJ/mol
T = 200°C and

–?Hsys
–(177.8 kJ/mol)
? ? _____________ =
? = ??
?Ssurr = ??______? ?? ???–0.376 kJ/K ? mol = –376 J/K ? mol
T
473 K

?Suniv = ?Ssys + ?Ssurr = 160.5 J/K ? mol + (–376 J/K ? mol) = –216 J/K ? mol

?Suniv is negative, therefore the reaction is nonspontaneous at 200°C.
(c)

T = 1000°C and

–?Hsys
–(177.8 kJ/mol)
? ? _____________?
? = ??
?? = –0.1397 kJ/K ? mol = –139.7 J/K ? mol
?Ssurr = ??______? ??
T
1273 K

?Suniv = ?Ssys + ?Ssurr = 160.5 J/K ? mol + (–139.7 J/K ? mol) = 20.8 J/K ? mol

In this case, ?Suniv is positive; therefore, the reaction is spontaneous at 1000°C.
(d) ?S° = S°[Na(l)] ? S°[Na(s)] = 57.56 J/K ? mol ? 51.05 J/K ? mol = 6.51 J/K ? mol
rxn

?H° = ?H°[Na(l)] ? ?H°[Na(s)] = 2.41 kJ/mol ? 0 kJ/mol = 2.41 kJ/mol
rxn
f
f

–?Hrxn
–(2.41 kJ/mol)
??
? ? ____________ =
? = ??
?Ssurr = ______? ?? ??? –0.0650 kJ/K ? mol = –6.50 J/K ? mol
T
371 K

?Suniv = ?Ssys + ?Ssurr = 6.51 J/K ? mol + (–6.50 J/K ? mol) = 0.01 J/K ? mol ? 0

?Suniv is zero; therefore, the reaction is an equilibrium process at 98°C. In fact, this is the melting
point of sodium.

Think About It?

Remember that standard enthalpies of formation have units of kJ/mol, whereas standard absolute
entropies have units of J/K ? mol. Make sure that you convert kilojoules to joules, or vi