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For a reversible reaction A underset(K(2...

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)` ?

Text Solution

Verified by Experts

The correct Answer is:
3
`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`
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