4.5 moles each of hydrogen and iodine heated in a sealed 10 litre vessel. At equilibrium, 3 moles of HI was found. The equilibrium constant for 

${\mathrm{H}}_{2\left(\mathrm{g}\right)} + {\mathrm{I}}_{2\left(\mathrm{g}\right)} \rightleftharpoons 2\mathrm{HI}\left(\mathrm{g}\right)$ is

                   ${\mathrm{H}}_2 + {\mathrm{I}}_2 \rightleftharpoons 2\mathrm{HI}$

Initial conc. 4.5      4.5          0

                     x         x           2x

From question 2x = 3

$\mathrm{x}=\frac{3}{2}=1.5$

So conc. at equilibrium 4.5 - 1.5 of  ${\mathrm{H}}_2$

= 4.5 - 1.5 of ${\mathrm{I}}_2$ and 3 of HI

$\mathrm{K}=\frac{{\left[\mathrm{HI}\right]}^2}{\left[{\mathrm{I}}_2\right] \left[{\mathrm{H}}_2\right]}= \frac{3 \times 3}{3 \times 3}=1$