If Delta_fG^o [X_{(l)}] = -65 , kcal , mol^{- | vedclass

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(A) The process of vaporization is $X_{(l)} \rightleftharpoons X_{(g)}$.For this process,the change in standard Gibbs free energy is $\Delta_rG^o = \D

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(A) The process of vaporization is $X_{(l)} ightleftharpoons X_{(g)}$.For this process,the change in standard Gibbs free energy is $\Delta_rG^o = \Delta_fG^o [X_{(g)}] - \Delta_fG^o [X_{(l)}]$.$\Delta_rG^o = -60.4 - (-65) = 4.6 \, kcal \, mol^{-1} = 4600 \, cal \, mol^{-1}$.We know that $\Delta_rG^o = -RT \ln P$,where $P$ is the vapour pressure in $atm$.$4600 = - (2 \, cal \, K^{-1} \, mol^{-1}) 	imes (500 \, K) 	imes \ln P$.$4600 = -1000 \ln P$.$\ln P = -4.6$.Using $\ln P = 2.3 \log P$,we get $2.3 \log P = -4.6$.$\log P = -2$.$P = 10^{-2} \, atm = 0.01 \, atm$.

If $\Delta_fG^o [X_{(l)}] = -65 \, kcal \, mol^{-1}$ and $\Delta_fG^o [X_{(g)}] = -60.4 \, kcal \, mol^{-1},$ the vapour pressure of $X$ at $500 \, K$ would be about ...... $atm$.
Given: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.

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If $\Delta_fG^o [X_{(l)}] = -65 \, kcal \, mol^{-1}$ and $\Delta_fG^o [X_{(g)}] = -60.4 \, kcal \, mol^{-1},$ the vapour pressure of $X$ at $500 \, K$ would be about ...... $atm$.Given: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.