If the equilibrium constant for A leftharpoons B + C is K eq ( 1) and that of B + C leftharpoons P is K eq (2), the equilibrium constant for A leftharpoons P is :-

Question

If the equilibrium constant for A leftharpoons B + C is K eq ( 1) and that of B + C leftharpoons P is K eq (2), the equilibrium constant for A leftharpoons P is :- (A) K eq (2)- K eq ( 1) (B) K eq (1) K eq (2) (C) K eq ( 1 ) / K eq (2) (D) Ke q(1)+Ke q(2)

Tardigrade Admin · Accepted Answer

A leftharpoons B + C K eq ( 1 )=([ B ][ C ]/[ A ]) ....(1) B + C leftharpoons P eq (2)=([ P ]/[ B ][ C ]) ...(2) For A leftharpoons P K eq =([ P ]/[ A ]) Multiplying equation (1) (2) K eq (1) × K eq (2)=([ P ]/[ A ])= K eq

Q. If the equilibrium constant for $A ⇌ B + C$ is $K_{e q}^{(1)}$ and that of $B + C ⇌ P$ is $K_{e q}^{(2)},$ the equilibrium constant for $A ⇌ P$ is :-

$K_{e q}^{(2)} - K_{e q}^{(1)}$

$K_{e q}^{(1)} K_{e q}^{(2)}$

$K_{e q}^{(1)} / K_{e q}^{(2)}$

$K_{e q}^{(1)} + K_{e q}^{(2)}$

$A ⇌ B + C$ $K_{e q}^{(1)} = \frac{[ B ] [ C ]}{[ A ]} .... (1)$

$B + C ⇌ P_{e q}^{(2)} = \frac{[ P ]}{[ B ] [ C ]} ... (2)$

For

$A ⇌ P K_{e q} = \frac{[ P ]}{[ A ]}$

Multiplying equation $(1)$ & $(2)$

$K_{e q}^{(1)} \times K_{e q}^{(2)} = \frac{[ P ]}{[ A ]} = K_{e q}$

Q. If the equilibrium constant for A⇌B+C is Keq(1)​ and that of B+C⇌P is Keq(2)​, the equilibrium constant for A⇌P is :-

Keq(2)​−Keq(1)​

Keq(1)​Keq(2)​

Keq(1)​/Keq(2)​

Keq(1)​+Keq(2)​