For a particular reversible reaction at temperature T,ΔH and ΔS were found to be both positive. If Te is the temperature at equilibrium, the reaction would be spontaneous when

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

1. A

$T={T}_{\mathrm{e}}$

2. B

${T}_{\mathrm{e}}>T$

3. C

$T>{T}_{e}$

4. D

${T}_{\mathrm{e}}$ is 5 times $T$

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### Solution:

Since $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }G$, we have

Reactions at equilibrium:                               $\mathrm{\Delta }G=\mathrm{\Delta }H-{T}_{\mathrm{e}}\mathrm{\Delta }S=0$

For a reaction to be spontaneous:              $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S<0$

$\mathrm{\Delta }G$ will be negative provided     $T>{T}_{\mathrm{e}}$ as both $\mathrm{\Delta }H$ and $\mathrm{\Delta }S$ are positive.

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