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)

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 $T_e=\frac{\Delta H}{\Delta S}\dots (1)$

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 $T>\frac{\Delta H}{\Delta S}\dots \dots (2)$

From (1) and (2)

$T>T_e$