For a given reaction, Δ_rH = +35.5 kJ mol^-1 and Δ_rS = +83.6 J K^-1mol^-1. The reaction is spontaneous ________. Assume that Δ_rH and Δ_rS do not vary with temperature.

Question

Quizplus · Accepted Answer

The spontaneity of a reaction is determined by the sign of the Gibbs free energy change, ΔG = ΔH - TΔS. From the given information, we know that Δ_rH = +35.5 kJ mol^-1 and Δ_rS = +83.6 J K^-1mol^-1. To use these values, we need to convert the units of ΔS to match the units of ΔH: Δ_rH = 35.5 J K^-1 mol^-1 Δ_rS = 0.0836 J K^-1 mol^-1 Substituting these values into the equation for ΔG: ΔG = ΔH - TΔS ΔG = +35.5 kJ mol^-1 - T(0.0836 J K^-1 mol^-1) We can see that ΔG will be negative (and the reaction will be spontaneous) when T > 425 K. Therefore, the answer is (D) at T > 425 K.

For a Given Reaction, ΔrH = +35

For a Given Reaction, Δ_rH = +35