Question

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 of $\left[HI\right]$ 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 $7$

$\left[H_2\right]=\frac{0.5}{10}=0.05$

$\left[I_2\right]=\frac{0.5}{10}=0.05$

$\Rightarrow H_2+I_2=2HI$

$t=0$; $0.05$ $0.05$ $0$

$(0.05-x)(0.05-x)$ $2x$ --- at eqbm

$\Rightarrow k_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{4x^2}{(0.05-x)(0.05-x)}$

$\Rightarrow 49=\frac{4x^2}{{(0.05-x)}^2}$

$\sqrt{49}=\sqrt{\frac{4x^2}{{(0.05-x)}^2}}$

$7=\frac{2x}{(0.05-x)}$

$\therefore 0.35-7x=2x$

$9x=0.35$

$x=\frac{0.35}{9}=0.039$

$\therefore \left[H_2\right]=0.05-0.039=0.011$

$\left[I_2\right]=0.05-0.039=0.011$

$\left[HI\right]=2\times 0.039=0.078$

$\therefore \frac{\left[HI\right]}{\left[I_2\right]}=\frac{0.078}{0.011}=7.09\simeq 7$

Hence, the answer is $7$.