Let us consider a reversible reaction at temperature, T. In this reaction, both Δ H and Δ S were observed to have positive values. If the equilibrium temperature is Te , then the reaction becomes spontaneous at:

Solution

Chemistry ma’am · 4 steps

With both Δ H and Δ S positive, the TΔ S term wins only once T climbs above Tₑ.

1. Spontaneity condition. A reaction is spontaneous when Δ G = Δ H - TΔ S is negative. Here Δ H > 0 works against it, and -TΔ S works for it.
2. Equilibrium temperature. At Tₑ the two terms balance, so Δ G = 0. That gives Tₑ = (Δ H)/(Δ S).

   Δ G < 0 ⇔ TΔ S > Δ H ⇔ T > (Δ H)/(Δ S) = Tₑ

3. Check both sides. Below Tₑ, TΔ S is smaller than Δ H, so Δ G > 0 and the reaction is non-spontaneous. Above Tₑ, TΔ S is larger, so Δ G < 0.
4. So. The reaction becomes spontaneous when T > Tₑ, which is option (b).

The idea to keep. When Δ H and Δ S are both positive, the reaction is entropy-driven: it turns spontaneous only above Tₑ = Δ H/Δ S. Both negative means the opposite, spontaneous only below Tₑ.

Answer from the NTA final answer key. Solution by PrepWiser.