Question

The equilibrium constants for the stepwise formation of $MCl$, $MC{l}_2$ and $MC{l}_3$ are $a,b$ and $c$ respectively. If the equilibrium constant of formation of $MC{l}_3$ is K, which of the following is correct?

• A. $logK=loga+logb+logc$
• B. $K=a+b+c$
• C. $\frac{1}{K}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
• D. $\frac{1}{K}=\frac{1}{a}\times \frac{1}{b}\times \frac{1}{c}$

Answer 1

The correct option is A $logK=loga+logb+logc$

Given

(1) $M+1/2{Cl}_2\to MCl{K}_1=a$

(2) $MCl+1/2{Cl}_2\to {MCl}_2{K}_2=b$

(3) ${MCl}_2+1/2{Cl}_2\to {MCl}_3{K}_3=c$

(4) $M+3/2{Cl}_2\to {MCl}_3$ $Kntt$

Clearly solution (4) sum of (1) , (2) , (3)

So, $K_4=K_1\times K_2\times K_3$

$K=a\times b\times c$

$logK=loga+logb+logc$