91Ó°ÊÓ

• According to the Third Law, "the entropy of a perfect crystal is zero when the crystal's temperature is equal to absolute zero (0 K)." 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)

\(\Delta S = S\left( {C{O_2},g} \right) - S(C,graphite) - S\left( {{O_2},g} \right)\)

\(\Delta S = (213.8 - 5.740 - 205.2)\frac{J}{{K \cdot mol}}\)

\(\Delta S = 2.9\frac{J}{{K \cdot mol}}\)

(b)

\(\Delta S = 2S(NO,g) - S\left( {{O_2},g} \right) - S\left( {{N_2},g} \right)\)

\(\Delta S = (2 \cdot 210.8 - 205.2 - 191.6)\frac{J}{{K \cdot mol}}\)

\(\Delta S = 24.8\frac{J}{{K \cdot mol}}\)

(c)

\(\Delta S = S\left( {C{u_2}S,s} \right) - 2S(Cu,s) - S(S,g)\)

\(\Delta S = (120.9 - 2 \cdot 33.15 - 167.82)\frac{J}{{K \cdot mol}}\)

\(\Delta S = - 113.2\frac{J}{{K \cdot mol}}\)

(d)

\(\Delta S = S\left( {Ca{{(OH)}_2},s} \right) - S({\rm{CaO}},s) - S\left( {{{\rm{H}}_2}{\rm{O}},l} \right)\)

\(\Delta S = (83.4 - 38.1 - 70.0)\frac{J}{{K \cdot mol}}\)

\(\Delta S = - 24.7\frac{J}{{K \cdot mol}}\)

(e)

\(\Delta S = 2S(Fe,s) + 3S\left( {{\rm{C}}{{\rm{O}}_2},g} \right) - S\left( {{\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3},s} \right) - 3S({\rm{CO}},g)\)

\(\Delta S = (2 \cdot 27.3 + 3 \cdot 213.8 - 87.40 - 3 \cdot 197.7)\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

\(\Delta S = 15.5\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

(f)

\(\Delta S = S({\rm{CaSO}},s) + 2S\left( {{{\rm{H}}_2}{\rm{O}},g} \right) - S\left( {{\rm{CaSO}} \cdot 2{{\rm{H}}_2}{\rm{O}},s} \right)\)

\(\Delta S = (106.5 + 2 \cdot 188.8 - 194.14)\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

\(\Delta S = 290.0\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\)

Hence the change in entropy are given as \(2.9\frac{J}{{K \cdot mol}}\),\(24.8\frac{J}{{K \cdot mol}}\),\( - 113.2\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\),\( - 24.7\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\),\(15.5\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\) and \(290.0\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}\).