For CO(g)+ H₂(g)→ H₂CO(g),Δ_rH° = -5.36 kJ/mol,and Δ_rS° = -109.8 J/mol K.What is Δ_rG° in kJ/mol at 300 K? A)-38.3 kJ/mol B)57.6 kJ/mol C)38.3 kJ/mol D)27.6 kJ/mol E)-57.6 kJ/mol

Question

Quizplus · Accepted Answer

Δ_rG° can be calculated using the formula Δ_rG° = Δ_rH° - TΔ_rS°. Substituting the given values: Δ_rG° = -5.36 kJ/mol - (300 K)(-109.8 J/mol K)(1 kJ/1000 J) = 27.6 kJ/mol.

For CO(g)+ H2(g)→ H2CO(g),ΔrH° = -5

For CO(g)+ H₂(g)→ H₂CO(g),Δ_rH° = -5