Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(\mathrm{C} .\) In each case, explain the sign of \(\Delta S^{\circ} .\) $$ \begin{array}{l}{\text { (a) } \mathrm{HNO}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(s)} \\ {\text { (b) } 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g)} \\ {\text { (c) } \mathrm{CaCO}_{3}(s, \text { calcite })+2 \mathrm{HCl}(g) \rightarrow} \\\ {\mathrm{CaCl}_{2}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)}\\\ {\text { (d) } 3 \mathrm{C}_{2} \mathrm{H}_{6}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)+6 \mathrm{H}_{2}(g)}\end{array} $$

Step by step solution

01

(a) HNO3(g) + NH3(g) → NH4NO3(s)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - HNO3(g): 266.4 - NH3(g): 192.8 - NH4NO3(s): 151.1 Using the formula, we calculate ∆S° for the reaction: \(\Delta S^{\circ} = (1 \times 151.1) - (1 \times 266.4 + 1 \times 192.8) = -307.9 \: \text{J/mol K}\) The negative value of ∆S° indicates that the disorder is decreasing, mainly because one of the gaseous molecules in the reactants is condensed into a solid.

02

(b) 2Fe2O3(s) → 4Fe(s) + 3O2(g)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - Fe2O3(s): 87.4 - Fe(s): 27.3 - O2(g): 205.2 Calculating ∆S° for the reaction: \(\Delta S^{\circ} = (4 \times 27.3 + 3 \times 205.2) - (2 \times 87.4) = 414.8 \: \text{J/mol K}\) The positive ∆S° signifies that the disorder is increasing, largely due to the production of three moles of gaseous O2 molecules from two moles of solid Fe2O3.

03

(c) CaCO3(s, calcite) + 2HCl(g) → CaCl2(s) + CO2(g) + H2O(l)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - CaCO3(s, calcite): 92.9 - HCl(g): 186.9 - CaCl2(s): 55.5 - CO2(g): 213.8 - H2O(l): 69.95 Calculating ∆S° for the reaction: \(\Delta S^{\circ} = (1 \times 55.5 + 1 \times 213.8 + 1 \times 69.95) - (1 \times 92.9 + 2 \times 186.9) = -104.8 \: \text{J/mol K}\) The decreasing entropy signified by the negative ∆S° indicates disorder reduction, primarily due to the transformation of a gaseous molecule (HCl) into a solid species (CaCl2).

04

(d) 3C2H6(g) → C6H6(l) + 6H2(g)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - C2H6(g): 229.6 - C6H6(l): 173.4 - H2(g): 130.7 Calculating ∆S° for the reaction: \(\Delta S^{\circ} = (1 \times 173.4 + 6 \times 130.7) - (3 \times 229.6) = 50.4 \: \text{J/mol K}\) The positive ∆S° shows a disorder increase, primarily owing to the production of six moles of gaseous H2 molecules from three moles of gaseous C2H6 molecules. Although C6H6 forms as a liquid, the production of a higher number of gaseous molecules results in increased disorder.

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