A certain reaction is non spontaneous at $298 \mathrm{~K}$. The entropy change during the reaction is 121…

A certain reaction is non spontaneous at $298 \mathrm{~K}$. The entropy change during the reaction is 121 $\mathrm{JK}^{-1} .$ Is the reaction is endothermic or exothermic ? The minimum value of $\Delta \mathrm{H}$ for the reaction is
  1. endothermic, $\Delta \mathrm{H}=36.06 \mathrm{~kJ}$
  2. exothermic, $\Delta \mathrm{H}=-36.06 \mathrm{~kJ}$
  3. endothermic, $\Delta \mathrm{H}=60.12 \mathrm{~kJ}$
  4. exothermic, $\Delta \mathrm{H}=-60.12 \mathrm{~kJ}$

Solution

For non spontaneous reaction
$\Delta \mathrm{G}=+\mathrm{ve}$
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ and
$\Delta \mathrm{S}=121 \mathrm{JK}^{-1}$
For $\Delta \mathrm{G}=+$ ve
$\Delta \mathrm{H}$ has to be positive. Hence the reaction is endothermic.

The minimum value of $\Delta \mathrm{H}$ can be obtained
by putting $\Delta \mathrm{G}=0$
$\Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S}=298 \times 121 \mathrm{~J}$
$=36.06 \mathrm{~kJ}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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