Calculating equilibrium pressures--can someone check my math as everything is laid out. I am getting it incorrect.

Question

The equilibrium constant, Kp, for the following reaction is 0.636 at 600 K:

COCl2(g) CO(g) + Cl2(g)

Calculate the equilibrium partial pressures of all species when COCl2(g) is introduced into an evacuated flask at a pressure of 1.73 atm at 600 K.

PCOCl2=
PCO=
PCl2=

setting up ICE, I know that Kp=x/(1.73-x)=.636

(.636)[1.73-x]^2=x
(.636)[1.73^2-2(1.73)x + x^2]=x
(.636)x^2-[2(1.73)(.636)+1]x + (.636)(1.73)^2

x^2-[2(1.73)+1/(.636)]x + 1.73^2= 0

a=1
b= -[2(1.73)+1/(.636)]= -5.032
c= (1.73)^2= 2.993

quadratic equation such that:
[(-5.032)^2-4(2.993)]^1/2 = [13.349]^1/2= 3.654

x=(-(-5.032)+-3.654)/2
x=4.343 or .689

Go with .689 since the other value of x is greater than the initial pressure of 1.73.
PCO=PCL2=(1.73-x)=(1.73-.689)=1.041
PCOCL2=x=.689

Answer 1

I think you've inverted the fraction in the definition of K_p. Since Kp = (P_products)/(P_reactant) I get

 

Kp = (P_CO)(P_Cl2)/P_COCl2)

 

If since P_CO = P_Cl2 and is zero at the start, let P_CO = P_Cl2 = x, and P_COCl2 = (1.73 - x)

 

0.636 = x²/(1.73 - x)

 

It looks like you got the math part down, so you can take it from there.

 

Jim S.