0.40 mol of PCl5 is heated in a 2.0 L closed vessel at 500 K and it dissociates to the extent of 35% according to PCl5(g) PCl3(g) + Cl2(g). Using R = 0.0821 L atm mol^-1 K^-1, calculate K_p in atm, correct to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3ce074

0.40 mol of $PCl X_{5}$ is heated in a 2.0 L closed vessel at 500 K and it dissociates to the extent of 35% according to $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ . Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , calculate $K_{p}$ in atm, correct to two decimal places.

Numerical answer type, answer in atm, tolerance 0.01.

At equilibrium:

PCl X_{5}

0.40 \times 0.65 = 0.26

mol,

PCl X_{3}

Cl X_{2}

0.14

mol.

K_{c} = (0.07)^{2} /0.13 = 0.03769 mol L^{- 1}

(concentrations are moles divided by 2.0 L).

K_{p} = K_{c} RT = 0.03769 \times 0.0821 \times 500 = 1.547

atm.