CO (g) + Cl 2(g) ↔ COCl 2(g) K = 8.2 x 10^-2 at 626 ̊C

Question

If 2.5 mol of carbonyl chloride gas, COCl 2 is initially introduced into a 10.0 L reaction vessel at 626 ̊C, calculate the equilibrium concentrations of carbon monoxide and chlorine.

Answer 1

Volume is fixed, so we can use molarity as a surrogate for moles. I assume that the K is a K_C The original concentration of COCl₂ is 2.5 moles/10 liters or .25 M. If we let x be the extent of reaction to the left in order to move the system to equilibrium, the equilibrium concentrations of the species will be:

CO: x Cl₂: x and COCl₂ (.25 - x)

The equilibrium expression is K = [COCl₂]/([CO][Cl₂]) or .082 = (.25-x)/x²

This can be multiplied out and solved using the quadratic formula for x which are the answers you are looking for. (.245 M or so)