JEE Main 2023 — Chemical Equilibrium Question with Solution

The equilibrium moles of each reactant and product can be calculated as follows,

 $$\begin{gathered} {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right) \rightleftharpoons 2{\mathrm{HI}}_{\left(\mathrm{g}\right)} \\ \mathrm{initial} 4.5 4.5 0 \\ \mathrm{At} \mathrm{equilibrium} 4.5-1.5 4.5-1.5 3 \end{gathered}$$

Now, using equilibrium moles, equilibrium constant calculated as shown below.
${\mathrm{K}}_{\mathrm{eq}}=\frac{[\mathrm{HI}{]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}=\frac{(3{)}^2}{3\times 3}=1$