Formation of $HCl$ is given by the equation $H_2$ $+$ $C{l}_2$ $=$ $2HCl$. The concentration of $H_2$, $C{l}_2$ and $HCl$ are $1M$, $2M$ and $3M$ respectively. Find the equilibrium constant.

The correct option is C $4.5$
For the reaction $H_2$ $+$ $C{l}_2$ $=$ $2HCl$, the equilibrium constant is expressed as $K$ $=$ $[HCl{]}^2$$/$$\left[H_2\right]$$\left[C{l}_2\right]$. The concentrations are $1M$, $2M$, $3M$ respectively. Thus, the value of $K$ $=$ $[3{]}^2$$/$$\left[1\right]$$\ast$$\left[2\right]$, that is $4.5$