One mole of NaCl (s) on melting absorbed 30.5 | vedclass

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(A) At the melting point,the process of melting is in equilibrium,so $\Delta G = 0$. From the relation $\Delta G = \Delta H - T\Delta S$,we get $T = \

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$1059$

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(A) At the melting point,the process of melting is in equilibrium,so $\Delta G = 0$. From the relation $\Delta G = \Delta H - T\Delta S$,we get $T = \frac{\Delta H}{\Delta S}$. Given: $\Delta H = 30.5 \ kJ \ mol^{-1} = 30500 \ J \ mol^{-1}$ and $\Delta S = 28.8 \ J \ K^{-1} \ mol^{-1}$. Substituting the values: $T = \frac{30500 \ J \ mol^{-1}}{28.8 \ J \ K^{-1} \ mol^{-1}} \approx 1059 \ K$.

One mole of $NaCl$ $(s)$ on melting absorbed $30.5 \ kJ$ of heat and its entropy increased by $28.8 \ J \ K^{-1}$. The melting point of $NaCl$ is ............. $K$.

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