For a reversible reaction A underset(K_(2))overset(K_(1))(hArr) B rate constant K_(1) (forward) = 10^(15)e^(-(200)/(T)) and K_(2) (backward) = 10^(12)e^(-(200)/(T)). What is the value of (-Delta G^(@))/(2.303 RT) ?

For a reversible reaction A underset(K_(2))overset(K_(1))(hArr) B rate

1.	For a reversible reaction A underset(K_(2))overset(K_(1))(hArr) B rate constant K_(1) (forward) = 10^(15)e^(-(200)/(T)) and K_(2) (backward) = 10^(12)e^(-(200)/(T)). What is the value of (-Delta G^(@))/(2.303 RT) ?
Answer» SOLUTION :`K=A.e^(-(E_(o))/(RT))` Equilibrium constant `(K_(P))=(K_(1))/(K_(2))` `K_(1)=10^(15)e^((-2000)/(T)), K_(2)=10^(12)e^((-2000)/(T))` Thus `(K_(1))/(K_(2))=10^(3), DELTAG^(@)= -2.303 RT log K_(P)` `-((DeltaG^(@))/(2.303 RT))=log KP=(K_(1))/(K_(2))=log 10_(10)^(3)=3`