Use the standard entropy data in Appendix \(G\) to determine the change in entropy for each of the reactions listed in Exercise 16.33. All are run under standard state conditions and\({\bf{2}}{{\bf{5}}^{\bf{^\circ }}}{\bf{C}}\).

• According to the Third Law, "the entropy of a perfect crystal is zero when the crystal's temperature is equal to absolute zero (0K)." The third law of thermodynamics mentions a state known as "absolute zero."
• For values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K, standard entropies are labelled S298 °. The standard entropy change (S°) of any process can be calculated using the standard entropies of its reactant and product species, as shown below:

ΔS° = ∑vS° ₂₉₈(products) - ∑vS° ₂₉₈(reactants)

In order to calculate the change in entropy, we use the formula

\(\Delta S = S({\rm{ products }}) - S({\rm{ reactants }})\)

Since all the reactions occur under standard conditions, we can look up the standard entropy values for each species in the table in Appendix \(G\)in the book.

a)

\(\Delta S = S(Mn,s) + S\left( {{O_2},g} \right) - S\left( {Mn{O_2},s} \right)\)

\(\Delta S = (32.0 + 205.2 - 53.05)\frac{J}{{K \cdot mol}}\)

\(\Delta S = 184.2\frac{J}{{K \cdot mol}}\)

b)

\(\Delta S = 2S(HBr,g) - S\left( {{H_2},g} \right) - S\left( {B{r_2},l} \right)\)

\(\Delta S = (2 \cdot 198.7 - 130.7 - 152.23)\frac{J}{{K \cdot mol}}\)

\(\Delta S = 114.5\frac{J}{{K \cdot mol}}\)

(c)

\(\Delta S = S({\rm{CuS}},{\rm{s}}) - S({\rm{Cu}},{\rm{s}}) - S(\;{\rm{S}},g)\)

\(\Delta S = (66.5 - 33.15 - 167.82)\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

\(\Delta S = - 134.5\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

(d)

\(\Delta S = S\left( {{\rm{L}}{{\rm{i}}_2}{\rm{C}}{{\rm{O}}_3},\;{\rm{s}}} \right) + S\left( {{{\rm{H}}_2}{\rm{O}},{\rm{g}}} \right) - 2\;{\rm{S}}({\rm{LiOH}},{\rm{s}}) - S\left( {{\rm{C}}{{\rm{O}}_2},g} \right)\)\(\Delta S = S\left( {{\rm{L}}{{\rm{i}}_2}{\rm{C}}{{\rm{O}}_3},\;{\rm{s}}} \right) + S\left( {{{\rm{H}}_2}{\rm{O}},{\rm{g}}} \right) - 2\;{\rm{S}}({\rm{LiOH}},{\rm{s}}) - S\left( {{\rm{C}}{{\rm{O}}_2},g} \right)\)

\(\Delta S = (90.17 + 188.8 - 2 \cdot 42.8 - 213.8)\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

\(\Delta S = - 20.4\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

(e)

\(\Delta S = S(C\), graphite \() + 2S\left( {{H_2}{\rm{O}},g} \right) - S\left( {{\rm{C}}{{\rm{H}}_4},g} \right) - S\left( {{{\rm{O}}_2},g} \right)\)

\(\Delta S = (5.740 + 2 \cdot 188.8 - 186.3 - 205.2)\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

\(\Delta S = - 8.2\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

(f)

\(\Delta S = S(C{\rm{Cl}},g) + S\left( {{S_2}{\rm{C}}{{\rm{l}}_2},g} \right) - S\left( {{\rm{C}}{{\rm{S}}_2},g} \right) - 3S\left( {{\rm{C}}{{\rm{l}}_2},g} \right)\)

\(\Delta S = (309.7 + 319.45 - 238.0 - 3 \cdot 223.1)\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

\(\Delta S = - 278.2\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

Hence the change in entropy are given as \(184.2\frac{J}{{K \cdot mol}}\),\(114.5\frac{J}{{K \cdot mol}}\),\( - 134.5\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\),\( - 20.4\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\),\( - 8.2\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)and\( - 278.2\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\).