Question

The equilibrium constant for the following reaction are given at $25{}^{\circ }C$
$2A\leftrightharpoons B+C,K_1=1.0$
$2B\leftrightharpoons C+D,K_2=16$
$2C+D\leftrightharpoons 2P,K_3=25$
The equilibrium constant for the reaction $P\leftrightharpoons A+\frac{1}{2}B$ at $25{}^{\circ }C$ is

• A. $\frac{1}{20}$
• B. $20$
• C. $\frac{1}{42}$
• D. $21$

Answer 1

The correct option is A $\frac{1}{20}$

Given,
$2A\leftrightharpoons B+C,K_1$
$2B\leftrightharpoons C+D,K_2$
$2C+D\leftrightharpoons 2P,K_3$

$\therefore K_1=\frac{\left[B\right]\left[C\right]}{[A{]}^2}$
$K_2=\frac{\left[C\right]\left[D\right]}{[B{]}^2}$
$K_3=\frac{[P{]}^2}{\left[C{]}^2\right[D]}$
Required reaction is
$P\rightleftharpoons A+\frac{1}{2}B$
$K_4=\frac{\left[A\right][B{]}^{\frac{1}{2}}}{\left[P\right]}$
$\therefore K_4=\sqrt{\frac{1}{K_3}\times \frac{1}{K_2}\times \frac{1}{K_1}}$
$\Rightarrow K_4=\sqrt{\frac{1}{25}\times \frac{1}{16}\times \frac{1}{1.0}}$
Solving these we get $K_4=\frac{1}{20}$