Question

The system described by the reaction
CO(g)+Cl2(g)?COCl2(g)
is at equilibrium at a given temperature when PCO= 0.32atm , PCl2= 0.10atm , and PCOCl2= 0.58atm . An additional pressure of Cl2(g)= 0.40atm is added.

Part A

Find the pressure of CO when the system returns to equilibrium.

Express your answer using two significant figures.

Answer 1

Hi,

CO(g) + Cl₂(g) => COCl₂(g)

K = p(COCl₂) / (p(CO) X p(Cl₂))

Equilibrium constant K can be compute from the given equilibrium pressures:
K = 0.58 / (0.32 X 0.10) = 18.125

Assuming ideal gas behavior partial pressure and number of moles are proportional:
p(X) X V = n(X) X R X T
<=>
p(X) = (R X T/V) X n(X)

That means a certain change in amount correspond to a certain change in partial pressure irrespective which compound you consider. So you can set up ICE table in terms of partial pressures

............... CO............... Cl?.............. COCl?
I.............. 0.32............ 0.50............. 0.58
C.............. -x................. -x................. +x
E...........0.32-x.......... 0.50-x............0.58+x

K = p(COCl₂) / (p(CO) X p(Cl₂))
<=>
18.125 = (0.58 + x)/( (0.50 - x) X (0.32 - x) )
<=>
(0.50 - x) X (0.32 - x) = (0.58 + x)/18.125
<=>
0.1600 - 0.8200x + x