In a one-litre flask, 6 moles of A undergoes the reaction A (   g ) ⇌ P (   g ) . The…

The equilibrium moles can be calculated as follows,

$$\begin{gathered} \mathrm{A}\left(\mathrm{g}\right)\stackrel{}{\rightleftharpoons }\mathrm{P}\left(\mathrm{g}\right) \\ \mathrm{initial} 6 0 \\ \mathrm{At} \mathrm{equilibrium} 6-\mathrm{y} \mathrm{y} \end{gathered}$$

at ${\mathrm{T}}_1 \mathrm{y}=4$                               at  ${\mathrm{T}}_2 \mathrm{y}=2$

$\therefore {\left({\mathrm{K}}_{\mathrm{eq}}\right)}_1=\frac{4}{2}=2 \therefore {\left({\mathrm{K}}_{\mathrm{eq}}\right)}_2=\frac{2}{4}=\frac{1}{2}$

The relation between Gibbs free energy and equilibrium constant,

$∆{\mathrm{G}}^{\mathrm{o}}=-{\mathrm{nRTlnK}}_{\mathrm{eq}}$

${\mathrm{\Delta G}}_1^{\mathrm{o}}=-{\mathrm{RT}}_1\ln {\left({\mathrm{K}}_{\mathrm{eq}}\right)}_1$

${\mathrm{\Delta G}}_1^{\mathrm{o}}=-2{\mathrm{RT}}_2\ln {\left({\mathrm{K}}_{\mathrm{eq}}\right)}_1$                   [Given: ${\mathrm{T}}_1=2{\mathrm{T}}_2$]

${\mathrm{\Delta G}}_2^{\mathrm{o}}=-{\mathrm{RT}}_2\ln {\left({\mathrm{K}}_{\mathrm{eq}}\right)}_2$

$\therefore {\mathrm{\Delta G}}_2^{\mathrm{o}}-{\mathrm{\Delta G}}_1^{\mathrm{o}}={\mathrm{RT}}_2\ln \frac{{\left({\mathrm{K}}_{{\mathrm{eq}}_1}\right)}^2}{\left({\mathrm{K}}_{{\mathrm{eq}}_2}\right)}$

$={\mathrm{RT}}_2\ln \frac{2^2}{\frac{1}{2}}={\mathrm{RT}}_2\ln 8$

$\therefore {\mathrm{\Delta G}}_2^{\mathrm{o}}-{\mathrm{\Delta G}}_1^{\mathrm{o}}=\mathrm{RTlnx}$  has $\mathrm{x}=8$