Question

Let us consider a reversible reaction at temperature, T . In this reaction, both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ were observed to have positive values. If the equilibrium temperature is $\mathrm{T}_{\mathrm{e}}$, then the reaction becomes spontaneous at:

• A. $\mathrm{T}=\mathrm{T}_{\mathrm{e}}$
• B. $\mathrm{T}_{\mathrm{e}}>\mathrm{T}$
• C. $\mathrm{T}>\mathrm{T}_{\mathrm{e}}$
• D. $\mathrm{T}_{\mathrm{e}}=5 \mathrm{~T}$

Hint:

A reaction is spontaneous when the Gibbs free energy change is negative.

Answer 1

The Correct Option is C

1. For a reaction to be spontaneous, $\Delta \mathrm{G}<0$. \[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S} \]
2. Given that both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are positive: \[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S}<0 \] \[ \mathrm{T}>\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} = \mathrm{T}_{\mathrm{e}} \] Therefore, the correct answer is (3) $\mathrm{T}>\mathrm{T}_{\mathrm{e}}$.