For CdO(s)+ SO₃(g)→ CdSO₄(s),Δ_rH° = -279.4 kJ/mol,and Δ_rS° = -118.4 J/mol K.What is Δ_rG° in kJ/mol at 127 K?

Name: For CdO(s)+ SO 3 (g)→ CdSO 4 (s),Δ r H° = -279.4 kJ/mol,and Δ r S° = -118.4 J/mol K.What is Δ r G° in kJ/mol at 127 K?
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Quizplus · Accepted Answer

Δ_rG° = Δ_rH° - TΔ_rS° Plugging in the given values: Δ_rG° = -279.4 kJ/mol - (127 K)(-118.4 J/mol K)/1000 = -263.6 kJ/mol ≈ -264 kJ/mol Therefore, the best choice is C.

For CdO(s)+ SO3(g)→ CdSO4(s),ΔrH° = -279

For CdO(s)+ SO₃(g)→ CdSO₄(s),Δ_rH° = -279