Question

At $500K$, equilibrium constant, $K_c$ for the following reaction is $5$.
$\frac{1}{2}{H}_{2\left(g\right)}+\frac{1}{2}{I}_{2\left(g\right)}\rightleftharpoons H{I}_{(g)}$
What would be the equilibrium constant $K_c$ for the reaction :
$2H{I}_{(g)}\rightleftharpoons H_{2(g)}+I_{2(g)}$ ?

• A. $0.04$
• B. $0.4$
• C. $25$
• D. $2.5$

Answer 1

Answer: (A)

$\frac{1}{2}{H}_{2\left(g\right)}+\frac{1}{2}{I}_{2\left(g\right)}\rightleftharpoons H{I}_{(g)}$ ; $K_c=5\dots (i)$
Multiply eqn $(i)$ by $2$,
$H_{2(g)}+I_{2(g)}\rightleftharpoons 2H{I}_{(g)}$; $K_c=(5{)}^2\dots (ii)$
Now, reverse the reaction
$2H{I}_{(g)}\rightleftharpoons H_{2(g)}+I_{2(g)}$; $K_c=\frac{1}{{\left(5\right)}^2}$
$\therefore K_c=\frac{1}{25}=0.04$