1.0 mol of $SO X_{2} Cl X_{2} (g)$ is heated in a 5.0 L vessel; at equilibrium 0.30 mol has dissociated, $SO X_{2} Cl X_{2} (g) SO X_{2} (g) + Cl X_{2} (g)$ . The mixture is then transferred to a 2.0 L vessel at the same temperature and starting from the same 1.0 mol of $SO X_{2} Cl X_{2}$ it re-equilibrates there. Calculate the amount of $SO X_{2}$ at equilibrium in the 2.0 L vessel, in mol, to two decimal places.

Numerical answer type, answer in mol, tolerance 0.005.

Answer: 0.2 mol

K_{c} = (0.30/5)^{2} / (0.70/5) = 0.02571 mol L^{- 1}

. In 2.0 L, with

x

mol dissociated,

(x /2)^{2} = 0.02571 (1 - x) /2

, i.e.

x^{2} + 0.0514 x - 0.0514 = 0

where

0.0514 = 2 K_{c}

. The positive root is

x = 0.2025

mol of

SO X_{2}