Chemistry - Thermodynamics - II Question with Solution | TestHub

$\text{Δ}{\text{H}}^{\text{o}}$for neutralization of strong acid and base is$-13.7$kcal/equivalent

${\text{H}}^{+}+\text{O}{\text{H}}^{-}\to {\text{H}}_2\text{O}$

Hence, for dissociation, enthalpy change will be$+13.7$kcal/equivalent

${\text{H}}_2\text{O}\to {\text{H}}^{+}+\text{O}{\text{H}}^{-}$

$\text{Δ}{\text{G}}^{\text{o}}$for dissociation is given as:$-19.14$kcal

$\text{Δ}{\text{G}}^{\text{o}}=\text{Δ}{\text{H}}^{\text{o}}-\text{T}\text{Δ}{\text{S}}^{\text{o}}$

$\Rightarrow$$\text{Δ}{\text{S}}^{\text{o}}=\frac{\left(\text{Δ}{\text{H}}^{\text{o}}-\text{Δ}{\text{G}}^{\text{o}}\right)}{\text{T}}=\frac{13.7-\left(-19.14\right)}{298}$

$\Rightarrow$$\text{Δ}{\text{S}}^{\text{o}}=\left(\frac{32.84\text{k}\text{c}\text{a}\text{l}}{298\text{K}}\right)=110.2\text{c}\text{a}\text{l}{\text{K}}^{-1}\text{m}\text{o}{\text{l}}^{-1}$