For a chemical reaction A + B ⇌ C + D
(Δ ᵣH^Θ = 80 kJ mol⁻¹) the entropy change Δ ᵣS^Θ depends on the temperature T (in K) as Δ ᵣS^Θ = 2T (J K⁻¹mol⁻¹).
Minimum temperature at which it will become spontaneous is ___________ K. (Integer)

Solution

Chemistry ma’am · 4 steps

Here Δ S itself grows with T, so TΔ S becomes 2T². Set Δ G = 0 and solve.

1. Spontaneity condition. A reaction is spontaneous when Δ G < 0. The minimum temperature is the point where Δ G just reaches zero.

   Δᵣ G^Θ = Δᵣ H^Θ - T Δᵣ S^Θ = 80000 - T(2T) = 80000 - 2T²

2. Match the units. Δᵣ H^Θ = 80 kJ mol⁻¹ = 80000 J mol⁻¹, because Δᵣ S^Θ is given in joules.
3. Set it to zero. 80000 - 2T² = 0 gives T² = 40000, so T = 200 K. Above this, 2T² > 80000 and Δ G turns negative.
4. So. The reaction becomes spontaneous from 200 K onwards.

The common slip: 40000. Stopping at T² = 40000 gives 40000. Take the square root to get T = 200 K.

The idea to keep. For the minimum temperature of spontaneity, put Δ G = 0. If Δ S depends on T, substitute that expression before solving, and keep Δ H and Δ S in the same energy unit.

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