1. Tardigrade
  2. Question
  3. Chemistry
  4. A certain reaction is non-spontaneous at 300 K. The entropy change during the reaction is 120 JK -1. Then the minimum value of Δ H for the reaction and the nature of the reaction will be:

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Q. A certain reaction is non-spontaneous at $300\, K$. The entropy change during the reaction is $120\, JK ^{-1}$. Then the minimum value of $\Delta H$ for the reaction and the nature of the reaction will be:

VITEEEVITEEE 2017

Solution:

For non-spontaneous reactions, $\Delta G$ is $+ve$.
$\Delta G =\Delta H - T \Delta S ; \Delta S \text { is }+120\, JK ^{-1}$
To make $\Delta G$ as tve, $\Delta H$ should be $+ve,$
i.e., endothermic For minimum value of entropy,
$\Delta H = T \Delta S$
$=300 \times 120=36000 J$
$=36.0\, kJ$