Answer:
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
<h2>
XCOCl₂ = 0.25</h2>
