At equilibrium in a rigid 2.0 L vessel, [PCl5] = 0.20 M, [PCl3] = 0.10 M and [Cl2] = 0.10 M for PCl5(g) PCl3(g) + Cl2(g). Some Cl2 is then removed instantly so that [Cl2] drops to 0.030 M, and the system re-equilibrates at the same temperature. Calculate the new [PCl3] in mol/L, to three decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3ce029

At equilibrium in a rigid 2.0 L vessel, $[PCl X_{5}] = 0.20 M$ , $[PCl X_{3}] = 0.10 M$ and $[Cl X_{2}] = 0.10 M$ for $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ . Some $Cl X_{2}$ is then removed instantly so that $[Cl X_{2}]$ drops to 0.030 M, and the system re-equilibrates at the same temperature. Calculate the new $[PCl X_{3}]$ in mol/L, to three decimal places.

Numerical answer type, answer in mol/L, tolerance 0.0005.

K_{c} = (0.10) (0.10) /0.20 = 0.050

. After removal, let

y

M of

PCl X_{5}

dissociate:

(0.10 + y) (0.030 + y) = 0.050 (0.20 - y)

, which gives

y = 0.0329

. New

[PCl X_{3}] = 0.10 + y = 0.1329

mol/L.