Carbon monoxide and chlorine gas react to form phosgene: CO(g)+Cl2(g)⇌COCl2(g); KP = 3.10 at 700 K.

If a reaction mixture initially contains 270 torr of CO and 257 torr of Cl2, what is the mole fraction of COCl2 when equilibrium is reached?

XCOCl₂ = 0.25

Explanation:

Let's write the reaction again:

CO(g) + Cl₂(g) > COCl₂(g)      Kp = 3.10

With this reaction, we can write the Kp expression:

Kp = PpCOCl₂ / PpCO * PpCl₂

As the pressure are in Torr, we can work these pressure in Atmosphere instead of Torr, and it would be easier cause it would be lower numbers. Remeber that 1 atm is 760 Torr. So Doing this we have:

PpCO = 270 * 1/760 = 0.36 atm

PpCl₂ = 257 / 760 = 0.34 atm

Now, we can write an ICE chart with pressures, so we can know the pressure of COCl₂

         CO(g) + Cl₂(g) > COCl₂(g)      Kp = 3.10

I:         0.36     0.34               0

C:          -x          -x                +x

E:        0.36-x   0.34-x           x

Replacing in the Kp expressio:

3.1 = x / (0.36-x)(0.34-x)

Solving for x we have:

3.1 = x / (0.1224 - 0.7x + x²)

3.1(0.1224 - 0.7x + x²) = x

0.3794 - 2.17x + 3.1x² = x

3.1x² - 3.17x + 0.3794 = 0

Now, using the quadratic formula to solve for x, we have:

x = 3.17 ±√(3.17)² - 4*3.1*0.3794 / 2 * 3.1

x = 3.17 ±√10.0489 - 4.7046 / 6.2

x = 3.17 ±√5.3443 / 6.2

x = 3.17 ± 2.3118 / 6.2

x₁ = 0.88

x₂ = 0.14

Using the lowest value of x, (because if we use the highest then the partial pressure of the reactants will give a negative value) we have that the partial pressure of COCl₂ would be 0.14 atm

Now that we have this value, we can calculate the partial pressure at equilibrium of the reactants:

PpCO = 0.36 - 0.14 = 0.22 atm

PpCl₂ = 0.34 - 0.14 = 0.20 atm

Now with these values, we can finally calculate the mole fraction of COCl₂ in equilibrium by using this expression:

XCOCl₂ = PCOCl₂ / Pt

And Pt is the total pressure of all species in reaction so:

Pt = 0.22 + 0.2 + 0.14 = 0.56 atm

Finally, the mole fraction:

XCOCl₂ = 0.14 / 0.56