Pressure equilibrium -- I am getting my math wrong somewhere

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SUMMARY

The equilibrium constant, Kp, for the reaction COCl2(g) ⇌ CO(g) + Cl2(g) is 0.636 at 600 K. When COCl2(g) is introduced into an evacuated flask at a pressure of 1.73 atm, the equilibrium partial pressures can be calculated using the ICE table method. The calculation leads to the quadratic equation x^2 - 5.032x + 2.993 = 0, where x represents the partial pressure of CO and Cl2. The correct value for x is determined to be 0.689 atm, resulting in equilibrium partial pressures of PCO = PCl2 = 1.041 atm and PCOCl2 = 0.689 atm.

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maylandchastity
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Homework Statement


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.

Homework Equations


Included below

The Attempt at a Solution


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

Already doesn't look correct to me.
 

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