A certain reaction is spontaneous at temperatures below 400. K but is not spontaneous at temperatures above 400. K. If \(\Delta H^\circ\) for the reaction is \(-20\). \(\pu{ kJ mol^{-1}}\) and it is assumed that \(\Delta H^\circ\) and \(\Delta S^\circ\) do not change appreciably with temperature, then the value of \(\Delta S^\circ\) for the reaction is

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$$ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ $$

Rearranging the equation and solving for \(\Delta S^\circ\)

$$ \Delta S^\circ = \frac{\Delta G^\circ - \Delta H^\circ}{-T} $$

We can set \(\Delta G^\circ\) to be zero when the reaction is in equilibrium. Make sure to convert units to Joules.

$$ \Delta S^\circ = \frac{0-(-20{,}000)}{-400} $$ $$ \Delta S^\circ = \frac{-20{,}000}{-400} $$ $$ \Delta S^\circ = -50 $$