Question

If for the reaction, $2ICl\to I_2+C{l}_2;K_c=0.14$ and initial concentration of $ICL$ is $0.6M$ then equilibrium concentration of $I_2$ is

<!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> AIIMS MBBS 2018

• A. $0.224M$
• B. $0.37M$
• C. $0.784M$
• D. $0.128M$

Answer 1

The correct option is D $0.128M$

For given reaction,

$K_c=0.14=\frac{\left[I_2\right]\left[C{l}_2\right]}{[ICl{]}^2}=\frac{x^2}{(0.6-2x{)}^2}$

$0.37=\frac{x}{0.6-2x}$
$0.222-0.74x=x$
$\Rightarrow 1.74x=0.222$
$\Rightarrow x=0.128M$