Calculate the standard free energy change (R = 8.314 J/K·mol).

Question

Debate continues on the practicality of H₂ gas as a fuel. The equilibrium constant for the reaction below is 1.0 × 10⁵ at 25°C.

CO₂+H₂O --> CO₂+H₂

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A. Calculate the standard free energy change (R = 8.314 J/K·mol).

B. Without doing any calculations, estimate ΔS°_rxn.

C. Without doing any calculations, estimate ΔH°_rxn.

Answer 1

Assuming this is the water-gas shift reaction CO + H₂O --> CO₂ + H₂

A. ΔG°_rxn = –RTln K_eq = –(8.314 J/mol-K)(298 K)ln(10⁵) = –28,524 J/mol = –28.524 kJ/mol at 25 °C,

B. Since the number of product and reactant molecules are the same, each side of the equation has the same number of translational degrees of freedom. I'd expect ΔS°_rxn to be low.

CO, CO₂ and H₂ each have two rotational degrees of freedom, but H₂O has three rotational degrees of freedom. So there is a small loss in entropy in going from reactants to products.

I would predict ΔS°_rxn < 0, but small in magnitude.

C. ΔG°_rxn = ΔH°_rxn – TΔS°_rxn which means that ΔH°_rxn = ΔG°_rxn + TΔS°_rxn

If ΔS°_rxn is slightly negative, then ΔH°_rxn will be a little bit more negative than ΔG°_rxn

I might be overthinking parts B and C, though. Simpler answers are

B. ΔS°_rxn ≈ 0 and C. ΔH°_rxn ≈ ΔG°_rxn