For the equilibrium `PCl_(5)hArr PCl_(3)(g)+Cl_(2)(g)` at `298 K, K_(p)=1.8xx10^(-7)`. What is `Delta G^(@)` for the reaction ?

For the equilibrium `PCl_(5)hArr PCl_(3)(g)+Cl_(2)(g)` at `298 K, K_(p

1.	For the equilibrium `PCl_(5)hArr PCl_(3)(g)+Cl_(2)(g)` at `298 K, K_(p)=1.8xx10^(-7)`. What is `Delta G^(@)` for the reaction ?
Answer» `Delta G = -2.303 "RT log " K_(p)=-2.303xx8.314xx10^(-3)kJ mol^(-1)K^(-1)xx298 K xx log 1.8xx10^(-7)=38.484 kJ mol^(-1)`

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For the equilibrium `PCl_(5)hArr PCl_(3)(g)+Cl_(2)(g)` at `298 K, K_(p)=1.8xx10^(-7)`. What is `Delta G^(@)` for the reaction ?

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