Question

At $500K$, equilibrium constant, $K_p$ for the following reaction is 5.

$\frac{1}{2}{H}_{2\left(g\right)}+\frac{1}{2}{I}_{2\left(g\right)}\rightleftharpoons H{I}_{\left(g\right)}$

What would be the equilibrium constant $K_c$ for the reaction :

$2{HI}_{\left(g\right)}\rightleftharpoons H_{2\left(g\right)}+I_{2\left(g\right)}?$

• A. 0.04
• B. 0.4
• C. 25
• D. 2.5

Answer 1

The correct option is A 0.04

$\frac{1}{2}{H}_{2\left(g\right)}+\frac{1}{2}{I}_{2\left(g\right)}\rightleftharpoons H{I}_{\left(g\right)}:K_c=5\dots \left(1\right)$

Multiply eqn (i) by 2 ,

$H_{2\left(g\right)}+I_{2\left(g\right)}\rightleftharpoons 2{HI}_{\left(g\right)};K_c=(5{)}^2\dots$(2)

Now, reverse the reaction

$2{HI}_{\left(g\right)}\rightleftharpoons H_{2\left(g\right)}+I_{2\left(g\right)};K_c=\frac{1}{(5{)}^2}$

$\therefore K_c=\frac{1}{25}=0.04$