Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

Solution

Chemistry ma’am · 3 steps

Endothermic means Δ H > 0; if heating makes it go, Δ S must be positive too.

1. Sign of ΔH. The reaction is endothermic, so Δ H is positive. That already rules out options (a) and (c).

   Δ G = Δ H - TΔ S

2. Sign of ΔS. At 273 K the reaction is non-spontaneous, so Δ G > 0. At 373 K it is spontaneous, so Δ G < 0. With Δ H fixed and positive, Δ G can only fall as T rises if the term -TΔ S grows more negative, and that needs Δ S > 0.

   Δ H	Δ S	When is Δ G < 0?
   -	+	at all temperatures
   -	-	only at low T
   +	-	never
   +	+	only at high T, above T = Δ H/Δ S

3. So. Both Δ H and Δ S are positive, and the crossover temperature Δ H/Δ S lies between 273 K and 373 K. The answer is (d).

The idea to keep. Write Δ G = Δ H - TΔ S and watch the signs. When temperature decides spontaneity, Δ H and Δ S share a sign. Both positive means it goes at high T; both negative means it goes at low T.

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