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When heated PCl5, phosphorus pentachloride, forms PCl3 and Cl2 as follows: PCl5(g)  PCl3(g) + Cl2(g) When 1.00 mol PCl5 in a 1.00-l container is allowed to come to equilibrium at a given temperature, the mixture is found to contain 0.135 mol PCl3. What is the molar composition of the mixture?

Accepted Answer

The molar composition of the mixture can be determined using the equilibrium constant expression for the reaction:

PCl5(g)  PCl3(g) + Cl2(g)

At equilibrium, the reaction quotient Q is equal to the equilibrium constant Kc:

Kc = [PCl3][Cl2]/[PCl5]

We can use the initial concentration of PCl5 and the equilibrium concentration of PCl3 to calculate the equilibrium concentration of Cl2:

Initial [PCl5] = 1.00 mol/L
Equilibrium [PCl3] = 0.135 mol/L

Let x be the equilibrium concentration of Cl2. Then, the equilibrium concentration of PCl5 is (1.00 - x) mol/L.

Substituting these values into the equilibrium constant expression, we get:

Kc = [0.135][x]/[1.00 - x]

Solving for x, we get:

x = 0.113 mol/L

Therefore, the equilibrium concentrations of PCl5, PCl3, and Cl2 are:

[PCl5] = 1.00 - x = 0.887 mol/L
[PCl3] = 0.135 mol/L
[Cl2] = x = 0.113 mol/L

The molar composition of the mixture is:

%PCl5 = ([PCl5]/[Total moles]) x 100%
= (0.887 mol / (0.887 mol + 0.135 mol + 0.113 mol)) x 100%
= 68.5%

%PCl3 = ([PCl3]/[Total moles]) x 100%
= (0.135 mol / (0.887 mol + 0.135 mol + 0.113 mol)) x 100%
= 10.5%

%Cl2 = ([Cl2]/[Total moles]) x 100%
= (0.113 mol / (0.887 mol + 0.135 mol + 0.113 mol)) x 100%
= 21.0%

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