Hence ${K_C} = \frac{{{{(2x)}^2} \times x}}{{(a - x){{(1.5a - 2x)}^2}}}$
Given, at equilibrium
$\therefore \,\,(a - x) = (1.5a - 2x)$
$\therefore \,\,a = 2x$
On solving ${K_C} = 4$
$\left[\mathrm{N}_2\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right]=3 \times 10^{-2} \mathrm{M}$ અને $\left[\mathrm{NH}_3\right]=1.5 \times 10^{-2} \mathrm{M}$ પ્રક્રિયા માટે સંતુલન અચળાંક_______છે.
નો $K_{sp}$ ........ થશે.
$(R = 8.314\, J\, K^{-1}\,mol^{-1})$
$Ag_2CO_{3(s)} \rightleftharpoons 2Ag^+_{(aq)} + CO^{2-}_{3(s)}$