Use data in Appendix C to calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each of the following reactions. $$ \begin{array}{l}{\text { (a) } 4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)} \\ {\text { (b) } \mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)} \\\ {\text { (c) } 2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)} \\ {\text { (d) } \mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)}\end{array} $$

(a) \(4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\)

For the reactants, we have: Cr(s): \(\Delta H_f^{\circ} = 0 \ kJ/mol\), \(S^{\circ} = 7.19 \ J/(mol \cdot K)\), \(G^{\circ} = 0 \ kJ/mol\) O2(g): \(\Delta H_f^{\circ} = 0 \ kJ/mol\), \(S^{\circ} = 205.14 \ J/(mol \cdot K)\), \(G^{\circ} = 0 \ kJ/mol\) For the product, we have: Cr2O3(s): \(\Delta H_f^{\circ} = -1134.20 \ kJ/mol\), \(S^{\circ} = 81.40 \ J/(mol \cdot K)\), \(G^{\circ} = -1046.50 \ kJ/mol\)

For the reaction, we can write: \(\Delta H^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) = 2(-1134.20) - (4(0) + 3(0)) = -2268.40 \ kJ/mol\) \(\Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants}) = 2(81.40) - (4(7.19) + 3(205.14)) = -543.84 \ J/(mol \cdot K)\) Now, we can calculate \(\Delta G^{\circ}\) using the formula: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), where \(T\) is the temperature in Kelvin: \(T = 25^{\circ} C + 273.15 = 298.15 \ K\) \(\Delta G^{\circ} = -2268.40- 298.15(-543.84/1000) = -2104.29 \ kJ/mol\) For reaction (a), we have: \(\Delta H^{\circ} = -2268.40 \ kJ/mol\) \(\Delta S^{\circ} = -543.84 \ J/(mol \cdot K)\) \(\Delta G^{\circ} = -2104.29 \ kJ/mol\) Now, you should follow the same steps for reactions (b), (c), and (d) using the corresponding data from Appendix C and the given chemical equations.