Question

For a particular reversible reaction at temperature T, Δ H and Δ S were found to be both positive. If Tₑ is the temperature at equilibrium, then the reaction would be spontaneous when:

• A. \(T_e > T\)
• B. \(T > T_e\)
• C. \(T_e = 5T\)
• D. T = Tₑ

Hint:

For Δ H>0 and Δ S>0: • Low T: non-spontaneous • High T: spontaneous

Answer 1

The Correct Option is B

Step 1: Gibbs free energy change: Δ G=Δ H-TΔ S
Step 2: For spontaneity: Δ G<0 ⟹ T>(Δ H)/(Δ S)
Step 3: At equilibrium: Tₑ=(Δ H)/(Δ S) ⟹ Reaction is spontaneous when T>Tₑ