At 373 , K,steam and water are in equilibrium | vedclass

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(A) For a phase transition at equilibrium,the entropy change is given by the formula:$\Delta S = \frac{\Delta_{vap} H}{T_{BP}}$Given $\Delta_{vap} H =

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$105$

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(A) For a phase transition at equilibrium,the entropy change is given by the formula:$\Delta S = \frac{\Delta_{vap} H}{T_{BP}}$Given $\Delta_{vap} H = 39.2 \, kJ \, mol^{-1} = 39.2 	imes 10^3 \, J \, mol^{-1}$ and $T = 373 \, K$.Substituting the values:$\Delta S = \frac{39.2 	imes 10^3}{373} \approx 105.09 \, J \, K^{-1} \, mol^{-1}$.Rounding to the nearest integer,we get $105 \, J \, K^{-1} \, mol^{-1}$.

At $373 \, K$,steam and water are in equilibrium and $\Delta H = 39.2 \, kJ \, mol^{-1}$. What will be $\Delta S$ for the conversion of $1 \, mole$ of water into steam?
$H_2O_{(l)} \to H_2O_{(g)}$ ... $J \, K^{-1} \, mol^{-1}$

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At $373 \, K$,steam and water are in equilibrium and $\Delta H = 39.2 \, kJ \, mol^{-1}$. What will be $\Delta S$ for the conversion of $1 \, mole$ of water into steam?$H_2O_{(l)} \to H_2O_{(g)}$ ... $J \, K^{-1} \, mol^{-1}$