JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Equilibrium

Starting from only ${\mathrm{H}}_2$ and ${\mathrm{I}}_2$, as the forward reaction proceeds $[{\mathrm{H}}_2]$ and $[{\mathrm{I}}_2]$ fall while $[\mathrm{HI}]$ rises; the opposing rates become equal at equilibrium, after which all three concentrations stay constant. Because 2 mol HI form per mol ${\mathrm{H}}_2$ (or ${\mathrm{I}}_2$) consumed, the HI curve rises about twice as steeply as the reactant curves fall. Graph (B) shows both reactant curves decreasing and the product curve increasing, all levelling off at constant equilibrium values.