Question

The system described by the reaction
CO(g)+Cl2(g)⇌COCl2(g)
is at equilibrium at a given temperature when PCO=0.31 atm , PCl2= 0.10 atm , and PCOCl2= 0.62 atm . An additional pressure of Cl2(g)= 0.42 atmis added.

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

Answer 1

first determine the intial pressures

P(CO) = 0.31 atm

P(Cl2) = 0.10 + 0.42 atm = 0.52 atm

P(OCCl2) = 0.62 atm

when reaction is in equilibrium you have added Cl2 so reaction will shift to wards product side according to lechatlear principal

so at equilibrium

P(CO) = 0.31 - x
P(Cl2) = 0.52 - x
P(OCCl2) = 0.62 + x

equlibrium expression

K = [P(OCCl2)]/[P(CO)][P(Cl2)]

find out the K before addiding the Cl2 since you know the pressures at equilibrium

K = 0.62 / 0.31 x 0.1

= 19.37

Using this value and our expressions for the new pressures at equilibrium gives:

19.37 = [0.62 + x] / [0.31 -x] [0.52 - x]

19.37 x2 - 16.077x + 3.122 = 0.62 +x

19.37 x2 - 17.077x +2.5 = 0

x = 0.6962 or = 0.185

since you can't possibly have negative reagents

P(CO) = 0.31 - x = 0.31 - 0.185 = 0.125 atm
P(Cl2) = 0.52 - x = 0.52 - 0.185 = 0.335 atm
P(OCCl2) = 0.62 + x = 0.62 + 0.185 = 0.805 atm