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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

Option 1)
$T= T_{e}$

Option 2)
$T_{e}> T$

Option 3)
$T> T_{e}$

Option 4)
$T_{e}\: is\: 5 \: times\: T$

Answers (1)

best_answer

As we learnt in

Gibb's free energy (Δ G) -

$\Delta G= \Delta H-T \Delta S$

- wherein

$\Delta G=$ Gibb's free energy

$\Delta H=$ enthalpy of reaction

$\Delta S=$ entropy

$T=$ temperature

At equilibrium $\Delta G=0$

Hence, $\Delta G=\Delta H-T_{e}\Delta S=0$

$\therefore \: \Delta H=T_{e}.\Delta s\ \;or\ \; T_{e}=\frac{\Delta H}{\Delta S}$

For a spontaneous reaction $\Delta G$ must be negative which is possible only if $\Delta H< T\Delta S$

This will happen when T > T_e

Option 1)

$T= T_{e}$

This solution is incorrect

Option 2)

$T_{e}> T$

This solution is incorrect

Option 3)

$T> T_{e}$

This solution is correct

Option 4)

$T_{e}\: is\: 5 \: times\: T$

This solution is incorrect

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