Question

For the reaction $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$,
If $0.5$ mole $H_2$ is reacted with $0.5$ mole $I_2$ in a ten-litre container at ${444}^{o}C$ and at same temperature value of equilibrium constant $K_c$ is $49$, the ratio for $\left[HI\right]$ and $\left[I_2\right]$ will be:

• A. $7$
• B. $\frac{1}{7}$
• C. $\sqrt{\frac{1}{7}}$
• D. $49$

Answer 1

The correct option is A $7$

The reaction is $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
Equilibrium constant $K_c$ for this reaction is given by
$K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
if $\left[H_2\right]=\left[I_2\right]=\frac{0.5}{10}$
Then $K_c=\frac{[HI{]}^2}{[I_2{]}^2}$
$or\frac{[HI{]}^2}{[I_2{]}^2}=K_c$
$\Rightarrow \frac{\left[HI\right]}{\left[I_2\right]}=\sqrt{K_C}$
$\Rightarrow \frac{\left[HI\right]}{\left[I_2\right]}=\sqrt{49}=7$