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If an endothermic reaction occurs spontaneously at constant temperature $T$ and $P$, then which of the following is true ?
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Verified Answer
The correct answer is:
$\Delta \mathrm{S}>0$
For a reaction to take place spontaneously the value of $\Delta \mathrm{G}$ must be negative i.e. $\Delta \mathrm{G} < 0$. Now, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$.
As the reaction is endothermic, so value of $\Delta \mathrm{H}$ must be positive, i.e. $\Delta \mathrm{H}>0$.
Hence to have a negative $\Delta \mathrm{G}$.
$\Delta \mathrm{H} < \mathrm{T} \Delta \mathrm{S}$. As $\mathrm{T} \& \mathrm{P}$ are constant.
$\mathrm{T} \Delta \mathrm{S}$ must be positive to give the total value a negative sign.
Hence $\Delta \mathrm{S}>0$
As the reaction is endothermic, so value of $\Delta \mathrm{H}$ must be positive, i.e. $\Delta \mathrm{H}>0$.
Hence to have a negative $\Delta \mathrm{G}$.
$\Delta \mathrm{H} < \mathrm{T} \Delta \mathrm{S}$. As $\mathrm{T} \& \mathrm{P}$ are constant.
$\mathrm{T} \Delta \mathrm{S}$ must be positive to give the total value a negative sign.
Hence $\Delta \mathrm{S}>0$
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