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

The equilibrium moles can be calculated as follows,

$A\left(g\right)\overset{}{⇌}P\left(g\right)initial60Atequilibrium6 - yy$

at ${T}_{1}y = 4$ at ${T}_{2}y = 2$

$∴ {\left({K}_{eq}\right)}_{1} = \dfrac{4}{2} = 2 ∴ {\left({K}_{eq}\right)}_{2} = \dfrac{2}{4} = \dfrac{1}{2}$

The relation between Gibbs free energy and equilibrium constant,

$∆ {G}^{o} = - {nRTlnK}_{eq}$

${\Delta G}_{1}^{o} = - {RT}_{1}ln{\left({K}_{eq}\right)}_{1}$

${\Delta G}_{1}^{o} = - 2{RT}_{2}ln{\left({K}_{eq}\right)}_{1}$ [Given: ${T}_{1} = 2{T}_{2}$]

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

$∴ {\Delta G}_{2}^{o} - {\Delta G}_{1}^{o} = {RT}_{2}ln\dfrac{{\left({K}_{{eq}_{1}}\right)}^{2}}{\left({K}_{{eq}_{2}}\right)}$

$= {RT}_{2}ln\dfrac{{2}^{2}}{\dfrac{1}{2}} = {RT}_{2}\ln 8$

$∴ {\Delta G}_{2}^{o} - {\Delta G}_{1}^{o} = RTlnx$ has $x = 8$