Thermodynamics:
Free Energy

In this module:

Free Energy and Temperature Just as with Entropy, you can use the definition of Gibbs Free Energy to find at what temperature a reactant-favored process will become product-favored, and vice versa. The assumption that is made is that both the enthalpy change and the entropy change of the reaction are constant at different temperatures. As long as none of the reactants or products undergoes a phase change over the temperature range, this is true, at least to a rough approximation. At what temperature do the following reactions switch from product-favored to reactant-favored?
N₂(g) + 3H₂(g) 2NH₃(g)
CaCO₃(s) CaO(s) + CO₂(g)
Correct! Remember to include units with your answer. That is incorrect. Try again. The solution is shown below. DHº = 2 mol•DH_fº(NH₃(g)) - 1 mol•DH_fº(N₂(g)) - 3 mol•DH_fº(H₂(g)) DHº = 2 mol•(-46.11 kJ/mol) - 1 mol•(0) - 3 mol•(0) = -92.22 kJ DSº = 2 mol•Sº(NH₃(g)) - 1 mol•Sº(N₂(g)) - 3 mol•Sº(H₂(g)) DSº = 2 mol•(192.45 J/mol•K) - 1 mol•(191.61 J/mol•K) - 3 mol•(130.684 J/mol•K) = -198.762 J/K DGº = 0 = DHº - TDSº T = DHº/DSº = -92.22 kJ/-198.762 J/K • 1000 J/1 kJ = 464.0 K = 191 ºC The solution is shown below. DHº = 1 mol•DH_fº(CaO(s)) + 1 mol•DH_fº(CO₂(g)) - 1 mol•DH_fº(CaCO₃(s)) DHº = 1 mol•(-635.09 kJ/mol) + 1 mol•(-393.509 kJ/mol) - 1 mol•(-1206.92 kJ/mol) = 178.321 kJ DSº = 1 mol•Sº(CaO(s)) + 1 mol•Sº(CO₂(g)) - 1 mol•Sº(CaCO₃(s)) DSº = 1 mol•(39.75 J/mol•K) + 1 mol•(213.74 J/mol•K) - 1 mol•(92.9 J/mol•K) = 160.59 J/K DGº = 0 = DHº - TDSº T = DHº/DSº = 178.321 kJ/160.59 J/K • 1000 J/1 kJ = 1110 K = 837.3 ºC

DG and Temperature