Hydrogen gas is being considered as a fuel for automobiles. There are many chemical means for producing hydrogen gas from water. One of these reactions is $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ In this case the form of carbon used is graphite. a. Calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction using data from Appendix \(4 .\) b. At what temperatures is this reaction spontaneous? Assume $\Delta H^{\circ}\( and \)\Delta S^{\circ}$ do not depend on temperature.

We have the following reaction: \(C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)\) Using the thermodynamic data in Appendix 4, we can calculate ΔH° and ΔS° based on the values for each of the constituent species. The standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) can be calculated using the following formulae: ΔH° = Σ (ΔH_product°) - Σ (ΔH_reactant°) ΔS° = Σ (ΔS_product°) - Σ (ΔS_reactant°) From Appendix 4, we have: ΔH° and ΔS° for C(s): 0 kJ/mol, 5.74 J/(mol * K) ΔH° and ΔS° for H2O(g): -241.826 kJ/mol, 188.84 J/(mol * K) ΔH° and ΔS° for CO(g): -110.530 kJ/mol, 197.66 J/(mol * K) ΔH° and ΔS° for H2(g): 0 kJ/mol, 130.68 J/(mol * K) Now, we can plug these values into the respective formulae and calculate ΔH° and ΔS° for the overall reaction. ΔH° = [(-110.530) + (0)] - [(0) + (-241.826)] = 131.296 kJ/mol ΔS° = [(197.66) + (130.68)] - [(5.74) + (188.84)] = 133.760 J/(mol * K)

A reaction is spontaneous if ΔG° < 0, where ΔG° is the standard Gibbs free energy change. We can use the following formula to find ΔG°: ΔG° = ΔH° - T * ΔS° Assuming ΔH° and ΔS° do not depend on temperature, to find the temperature range where the reaction is spontaneous, we solve for T: ΔG° < 0 => ΔH° - T * ΔS° < 0 => T > (ΔH° / ΔS°) T > (131.296 kJ/mol) / (133.760 J/(mol * K)) => T > (131296 J/mol) / (133.760 J/(mol * K)) T > 981.74 K The reaction is spontaneous for temperatures greater than 981.74 K.