Calculate ΔG° at 298 K for the reaction below. I₂(g)+ Br₂(g)→ 2IBr(g) Given: Δ_rH° = -11.6 kJ/mol-rxn; Δ_rS° = 12 J/K⋅mol-rxn at 298 K. A) -8.02 kJ/mol-rxn B) 15.2 kJ/mol-rxn C) -15.2 kJ/mol-rxn D) 3.59 × 10³ kJ/mol-rxn E) -3.59 × 10³ kJ/mol-rxn

Question

Quizplus · Accepted Answer

ΔG° = ΔH° - TΔS°

ΔS° is given as 12 J/K⋅mol-rxn, which needs to be converted to kJ/K⋅mol-rxn.
ΔS° = 12 J/K⋅mol-rxn × (1 kJ/1000 J) = 0.012 kJ/K⋅mol-rxn

Substituting the values given:
ΔG° = (-11.6 kJ/mol-rxn) - (298 K)(0.012 kJ/K⋅mol-rxn) = -15.2 kJ/mol-rxn

Therefore, the answer is C, -15.2 kJ/mol-rxn.

Calculate ΔG° at 298 K for the Reaction Below