To find the equilibrium composition of PCl5, we need to use the given equilibrium constant (Kc) and the initial amounts of PCl3 and Cl2.
The balanced equation for the reaction is:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Given:
Kc = 33.3
Initial amount of PCl3 = 0.31 mol
Initial amount of Cl2 = 0.22 mol
Volume of reaction vessel = 8 L
Let's assume the equilibrium amount of PCl5 in the reaction vessel is x mol.
The equilibrium amounts of PCl3 and Cl2 can be calculated using the initial amounts and the changes in moles during the reaction:
PCl3 = 0.31 mol - x mol
Cl2 = 0.22 mol - x mol
Now we can use the equilibrium constant expression to set up an equation:
Kc = [PCl3][Cl2] / [PCl5]
33.3 = (0.31 - x)(0.22 - x) / x
Expanding the equation:
33.3x = (0.31 - x)(0.22 - x)
Solving for x:
33.3x = 0.0682 - 0.53x + x^2
x^2 - 0.53x + 0.0682 - 33.3x = 0
Rearranging the equation:
x^2 - 33.83x + 0.0682 = 0
Using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 1, b = -33.83, c = 0.0682
Calculating the values:
x = (33.83 ± sqrt((-33.83)^2 - 4(1)(0.0682))) / (2(1))
x ≈ 33.75 mol/L (approximately)
Therefore, the equilibrium composition of PCl5 is approximately 33.75 mol/L.