Final answer:
a) The reaction is exothermic with ΔH°rxn = -2220.6 kJ. b) The reaction is not spontaneous with respect to entropy with ΔS°rxn = -618.6 J/(mol*K). c) The reaction is spontaneous with respect to Gibbs Free energy with ΔG° = -2030.4 kJ.
Step-by-step explanation:
a) ΔH°rxn:
To calculate ΔH°rxn, we need to sum up the enthalpy changes of the products and subtract the enthalpy changes of the reactants. In this case, we have 3 moles of CO2 with an enthalpy change of -393.5 kJ/mol and 4 moles of H2O with an enthalpy change of -285.8 kJ/mol. On the reactant side, we have 1 mole of C3H8 with an enthalpy change of 0 kJ/mol and 5 moles of O2 with an enthalpy change of 0 kJ/mol. Therefore, ΔH°rxn = (3*(-393.5) + 4*(-285.8)) - (0 + 5*0) = -2220.6 kJ. Since ΔH°rxn is negative, the reaction is exothermic.
b) ΔS°rxn:
To calculate ΔS°rxn, we subtract the entropy changes of the reactants from the entropy changes of the products. In this case, we have 3 moles of CO2 with an entropy change of 213.6 J/(mol*K) and 4 moles of H2O with an entropy change of 188.7 J/(mol*K). On the reactant side, we have 1 mole of C3H8 with an entropy change of 270.7 J/(mol*K) and 5 moles of O2 with an entropy change of 205.0 J/(mol*K). Therefore, ΔS°rxn = (3*(213.6) + 4*(188.7)) - (270.7 + 5*(205.0)) = -618.6 J/(mol*K). Since ΔS°rxn is negative, the reaction is not spontaneous with respect to entropy.
c) ΔG° of the reaction:
ΔG° can be calculated using the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. Plugging in the values, we have ΔG° = (-2220.6 kJ) - (298 K * (-618.6 J/(mol*K)) / 1000 = -2030.4 kJ. Since ΔG° is negative, the reaction is spontaneous with respect to Gibbs Free energy.