For the equilibirum, `PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g)` at 298 K, `K = 1.8 xx 10^(-7)`. Calculate `Delta G^(@)` for the reaction `(R = 8.314 JK^(-1) mol^(-1))`.

Explanation

`PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g), " "K = 1.8 xx 10^(-7)`
`Delta G^(@) = -2.303 RT log K = -2.303 xx 8.314 JK^(-1) mol^(-1) xx 298 K xx log(1.8 xx 10^(-7))`
`= -2.303 xx 8.314 xx 298[log 1.8 + log10^(-7)] = -19.147 xx 298[0.2553 - 7.000]`
`Delta G^(@) = (-19.147 xx 298 xx (-6.7447))/(1000) = +34.48 kJ mol^(-1)`