The molar heat of vaporisation of a liquid is | vedclass

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(B) At the boiling point,the process of vaporisation is in equilibrium,so $\Delta G = 0$. Using the relation $\Delta G = \Delta H - T\Delta S$,we get

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$375 \ K$

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(B) At the boiling point,the process of vaporisation is in equilibrium,so $\Delta G = 0$. Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta H = T\Delta S$. Therefore,$T = \frac{\Delta H}{\Delta S}$. Given: $\Delta H = 6 \ kJ \ mol^{-1} = 6000 \ J \ mol^{-1}$ and $\Delta S = 16 \ J \ mol^{-1} \ K^{-1}$. $T = \frac{6000 \ J \ mol^{-1}}{16 \ J \ mol^{-1} \ K^{-1}} = 375 \ K$.

The molar heat of vaporisation of a liquid is $6 \ kJ \ mol^{-1}$. If the entropy change is $16 \ J \ mol^{-1} \ K^{-1}$,the boiling point of the liquid is:

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