At 1 atm pressure,if Delta S_{text{vap}} = 7 | vedclass

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(A) For a phase transition at equilibrium,the relationship between entropy change,enthalpy change,and temperature is given by: $\Delta S_{	ext{vap}}

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(A) For a phase transition at equilibrium,the relationship between entropy change,enthalpy change,and temperature is given by: $\Delta S_{	ext{vap}} = \frac{\Delta H_{	ext{vap}}}{T_b}$ Rearranging for the boiling point $(T_b)$: $T_b = \frac{\Delta H_{	ext{vap}}}{\Delta S_{	ext{vap}}}$ Given: $\Delta H_{	ext{vap}} = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$ and $\Delta S_{	ext{vap}} = 75 \ J \ K^{-1} \ mol^{-1}$ Substituting the values: $T_b = \frac{30000}{75} = 400 \ K$

At $1 \ atm$ pressure,if $\Delta S_{\text{vap}} = 75 \ J \ K^{-1} \ mol^{-1}$ and $\Delta H_{\text{vap}} = 30 \ kJ \ mol^{-1}$,what will be the boiling point of the liquid in $K$?

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