Question

For a particular reversible reaction at temperature $T, \,\Delta H$ and $\Delta S$ were found to be both $+ve$. If $T_{e}$ is the temperature at equilibrium, the reaction would be spontaneous when

• A. $T_e > T$
• B. $T > T_e$
• C. $T_e$ is $5$ times $T$
• D. $T = T_e$

Answer 1

Answer: (B)

$T > T_e$

Answer 2

For a particular reversible reaction at temperature $T , \Delta H$ and $\Delta S$ were found to be both +ve. If $T _{ e }$ is the temperature at equilibrium, the reaction would be spontaneous when $T >$ $T _{ e }$ At equilibrium $\Delta G =0$ $\Delta G =\Delta H - T _{ e } \Delta S$ $0 = \Delta H - T _{ e } \Delta S$ $\therefore \Delta H = T _{ e } \Delta s$ or $T _{ e }=\frac{\Delta H }{\Delta S } \ldots(1)$ For a spontaneous reaction $\Delta$ G must be negative which is possible only if $\Delta H - T \Delta S <0$ $\therefore \Delta H < T \Delta S$ or $T >\frac{\Delta H }{\Delta S } \ldots \ldots(2)$ From (1) and (2) $T > T _{ e }$