Question

For reaction ,$2A+B\rightleftharpoons 2C,K=x$
Equilibrium constant for $C\rightleftharpoons A+\frac{1}{2}B$ will be:

• A. $x$
• B. $\frac{x}{2}$
• C. $\frac{1}{\sqrt{x}}$
• D. $\sqrt{x}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{x}}$

$\frac{1}{\sqrt{x}}$

For reaction-

$2A+B\rightleftharpoons 2C$

$K=\frac{{\left[C\right]}^2}{{\left[A\right]}^2\left[B\right]}=x\left(\text{Given}\right)$

$\Rightarrow \frac{{\left[A\right]}^2\left[B\right]}{{\left[C\right]}^2}=\frac{1}{K}=\frac{1}{x}$

For the reaction-

$C\rightleftharpoons A+\frac{1}{2}B$

$K^{\prime }=\frac{\left[A\right]{\left[B\right]}^{\frac{1}{2}}}{\left[C\right]}$

Squaring both sides, we have

${\left(K^{\prime }\right)}^2=\frac{{\left[A\right]}^2\left[B\right]}{{\left[C\right]}^2}$

$\Rightarrow {\left(K^{\prime }\right)}^2=\frac{1}{K}=\frac{1}{x}$

$\Rightarrow K^{\prime }=\frac{1}{\sqrt{x}}$

Hence for the reaction- $C\rightleftharpoons A+\frac{1}{2}B$,

Equillibrium constant, $K=\frac{1}{\sqrt{x}}$