Question

4.5 moles each of hydrogen and iodine heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were 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:-
(A) 1 (B) 10 (C) 5 (D) 0.33

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{H}}_2\left(\mathrm{g}\right)\hspace{0.17em}+{\mathrm{I}}_2\left(\mathrm{g}\right)\leftrightarrow 2\mathrm{HI}\left(\mathrm{g}\right) \\ \mathrm{Initial}4.5\mathrm{mol}4.5\mathrm{mol}0 \\ \mathrm{At}\mathrm{equil}.4.5-\mathrm{x}4.5-\mathrm{x}2\mathrm{x} \\ \mathrm{Now},2\mathrm{x}=3 \\ \mathrm{So},\mathrm{x}=1.5 \\ \mathrm{Thus},\mathrm{at}\mathrm{equiibrium},\left[{\mathrm{H}}_2\right]=4.5-1.5=3 \\ \left[{\mathrm{I}}_2\right]=4.5-1.5=3 \\ {\mathrm{K}}_{\mathrm{c}}=\frac{[\mathrm{HI}{]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]} \\ \mathrm{or},{\mathrm{K}}_{\mathrm{c}}=\frac{({\displaystyle \frac{3}{10}}{)}^2}{\left({\displaystyle \frac{3}{10}}\right)\left({\displaystyle \frac{3}{10}}\right)}=1 \end{gathered}$$