Calculate the equilibrium constant (in multiples of 10^(-4)) for the reaction PCl_(5(g)) rarr PCl_(3(g)) + Cl_(2(g)) at 400K, if Delta H^(0) = 77.2KJ "mole"^(-1) and Delta S^(0) = 122JK^(-1) "mole"^(-1)

Calculate the equilibrium constant (in multiples of 10^(-4)) for the r

1.	Calculate the equilibrium constant (in multiples of 10^(-4)) for the reaction PCl_(5(g)) rarr PCl_(3(g)) + Cl_(2(g)) at 400K, if Delta H^(0) = 77.2KJ "mole"^(-1) and Delta S^(0) = 122JK^(-1) "mole"^(-1)
Answer» SOLUTION :`Delta G^(0) = Delta H^(0) - T Delta S^(0)` `= 77.2 - (400 xx 122)/(1000) = 28.4 KJ` `Delta G^(0) = -RT LN K rArr K = 2 xx 10^(-4)`