Search any question & find its solution
Question:
Answered & Verified by Expert
A certain reaction is non spontaneous at $298 \mathrm{~K}$. The entropy change during the reaction is 121 $\mathrm{J} \mathrm{K}^{-1}$. Is the reaction is endothermic or exothermic ? The minimum value of $\Delta \mathrm{H}$ for the reaction is
Options:
Solution:
1038 Upvotes
Verified Answer
The correct answer is:
endothermic, $\quad \Delta \mathrm{H}=36.06 \mathrm{~kJ}$
For non spontaneous reaction
$\Delta \mathrm{G}=+\mathrm{ve}$ $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ and $\Delta \mathrm{S}=121 \mathrm{~J} \mathrm{~K}^{-1}$ For $\Delta \mathrm{G}=+\mathrm{ve}$
$\Delta \mathrm{H}$ has to be positive. Hence the reaction is endothermic.
The minimum value of $\Delta \mathrm{H}$ can be obtained by putting $\Delta \mathrm{G}=0$
$\begin{aligned}
\Delta \mathrm{H} &=\mathrm{T} \Delta \mathrm{S}=298 \times 121 \mathrm{~J} \\
&=36.06 \mathrm{~kJ}
\end{aligned}$
$\Delta \mathrm{G}=+\mathrm{ve}$ $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ and $\Delta \mathrm{S}=121 \mathrm{~J} \mathrm{~K}^{-1}$ For $\Delta \mathrm{G}=+\mathrm{ve}$
$\Delta \mathrm{H}$ has to be positive. Hence the reaction is endothermic.
The minimum value of $\Delta \mathrm{H}$ can be obtained by putting $\Delta \mathrm{G}=0$
$\begin{aligned}
\Delta \mathrm{H} &=\mathrm{T} \Delta \mathrm{S}=298 \times 121 \mathrm{~J} \\
&=36.06 \mathrm{~kJ}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.