Question

If $K_1,K_2,K_3$ are equilibrium constant for formation of $A{D}_1,A{D}_2,A{D}_3$, respectively as follows $A+D\rightleftharpoons A{D}_1$, $A{D}_1+D\rightleftharpoons A{D}_2,A{D}_2+D\rightleftharpoons A{D}_3$. Then equilibrium constant 'K' for $A+3D\rightleftharpoons A{D}_3$ is related as:

• A. $K_1+K_2+K_3=K$
• B. $log{K}_1+log{K}_2+log{K}_3=logK$
• C. $K_1+K_2=K_3+K$
• D. $log{K}_1+log{K}_2=log{K}_3+logK$

Answer 1

The correct option is B $log{K}_1+log{K}_2+log{K}_3=logK$

Equilibrium constant is the ratio of concentration of product to that of concentration of reactant during on equilibrium.

$A+D\leftrightharpoons A{D}_1$ Equilibrium Constant = $K_1$

$A{D}_1+D\leftrightharpoons A{D}_2$ Equilibrium Constant = $K_2$

$A{D}_2+D\leftrightharpoons A{D}_3$ Equilibrium Constant = $K_3$

Adding the above equation

$A+3D\leftrightharpoons A{D}_3$ Equilibrium Constant = $K$

We know that when equations are added, their equilibrium constants are multiplied.

$K_1\times K_2\times K_3=K$

Taking log of both sides

$log{K}_1+log{K}_2+log{K}_3=logK$