0.50 mol of PCl5 is placed in a 5.0 L closed vessel and it dissociates 40% at equilibrium, PCl5(g) PCl3(g) + Cl2(g). Calculate K_c in mol L^-1 and report K_c x 10^3 to the nearest integer.

JEE Main levelNumerical answerDifficulty 2 of 3ce019

0.50 mol of $PCl X_{5}$ is placed in a 5.0 L closed vessel and it dissociates 40% at equilibrium, $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ . Calculate $K_{c}$ in $mol L^{- 1}$ and report $K_{c} \times 1 0^{3}$ to the nearest integer.

Numerical answer type, tolerance 0.5.

Dissociated

= 0.50 \times 0.40 = 0.20

mol, so

PCl X_{3}

and

Cl X_{2}

are 0.20 mol each and

PCl X_{5}

is 0.30 mol. Concentrations are 0.040 M, 0.040 M and 0.060 M.

K_{c} = (0.040)^{2} /0.060 = 0.02667 mol L^{- 1}

, so

K_{c} \times 1 0^{3} \approx 26.67