Question

If $K_1,K_{2}and{K}_3$ are equilibrium constant for formation of $AD,A{D}_{2}andA{D}_3$ respectively as follow $A+D\rightleftharpoons AD,AD+D\rightleftharpoons A{D}_2,A{D}_2+D\rightleftharpoons A{D}_3$ then equilibrium constant 'K" for $A+3D\rightleftharpoons A{D}_3$ is related as

• A. $K_1+K_2+K_3=K$
• B. $\log K_1+\log K_2+\log K_3=\log K$
• C. $K_1+K_2=K_3+K$
• D. $\log K_1+\log K_2=\log K_3+\log K$

Answer 1

The correct option is B $\log K_1+\log K_2+\log K_3=\log K$

$A+D\rightleftharpoons AD;K_{1}AD+D\rightleftharpoons A{D}_2;K_{2}A{D}_2+D\rightleftharpoons A{D}_3;K_{3}A+3D\rightleftharpoons A{D}_3;K$
As we have that $K=K_1.K_2.K_3$
Taking log on both sides,
$\log K=\log K_1+\log K_2+{\log }_3$