Question

A reversible reaction is at temperature $T$ $\left(K\right)$. Both enthalpy change and entropy change have positive values. Let $T^{\prime }$ represents the equilibrium temperature.

For the reaction to be spontaneous, which of the following conditions should be satisfied?

• A. $T^{\prime }$ $>$ $T$
• B. $T^{\prime }<T$
• C. $T=T^{\prime }$
• D. $T=2T^{\prime }$

Answer 1

Answer: (B)

$T^{\prime }<T$

When the reaction is at equilibrium, the change in Gibbs free energy is zero. $\Delta G=\Delta H-T\Delta S=0$
$T^{\prime }=\frac{\Delta H}{\Delta S}$
When reaction is spontaneous, the change in Gibbs free energy is negative. $\Delta G<0$
$\Delta H<T\Delta S$
$\frac{\Delta H}{\Delta S}<T$
$T^{\prime }<T$