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A reaction is spontaneous at all temperature, when
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Verified Answer
The correct answer is:
$\Delta H$ is -ve and $\Delta S$ is + ve
For a reaction to be spontaneous change in Gibbs free energy should be negative.
i.e. $\Delta G=\Delta H-T \Delta S < 0$
$$
=(-v e)-(+v e)=-\text { e }-\text { ve }=-2 \mathrm{ve}
$$
$\Delta H < 0$ and $\Delta S>0$
So, $\Delta H=-$ ve and $\Delta S=+$ ve
i.e. $\Delta G=\Delta H-T \Delta S < 0$
$$
=(-v e)-(+v e)=-\text { e }-\text { ve }=-2 \mathrm{ve}
$$
$\Delta H < 0$ and $\Delta S>0$
So, $\Delta H=-$ ve and $\Delta S=+$ ve
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