Phosphorus pentachloride decomposes according to the chemical equation
PCl5(g) <-> PCl3(g)+Cl2(g)
Kc=1.80 at 250 degrees Celsius
A 0.222 mol sample of PCl5(g) is injected into an empty 3.25 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
PCl5=________M
PCl3=________M
To solve this problem, we need to use the equation provided and the given value of the equilibrium constant (Kc) to determine the equilibrium concentrations of PCl5 and PCl3.
Let's assume that at equilibrium, the concentration of PCl5 is "x" M, the concentration of PCl3 is "y" M, and the concentration of Cl2 is also "y" M.
Now, let's set up an ICE (Initial, Change, Equilibrium) table:
Initial:
PCl5: 0.222 M
PCl3: 0 M (not present initially)
Cl2: 0 M (not present initially)
Change:
PCl5: -x
PCl3: +y
Cl2: +y
Equilibrium:
PCl5: 0.222 - x
PCl3: y
Cl2: y
According to the given equilibrium equation, the expression for Kc is:
Kc = [PCl3] * [Cl2] / [PCl5]
Substituting the equilibrium concentrations:
1.80 = (y * y) / (0.222 - x)
Now, we need to determine the value of "x" at equilibrium. Since PCl5 decomposes to form both PCl3 and Cl2 in a 1:1 ratio, the change in concentration of PCl5 is the same as the change in concentration of PCl3 and Cl2, which is "y" M.
Since the initial concentration of PCl5 is 0.222 M, the equilibrium concentration of PCl5 is 0.222 - y M.
Substituting these values into the expression for Kc:
1.80 = (y * y) / (0.222 - (0.222 - y))
Simplifying:
1.80 = y^2 / y
1.80 = y
Therefore, at equilibrium, the concentration of PCl3 is 1.80 M.
Also, since PCl5 decomposes in a 1:1 ratio with PCl3, the equilibrium concentration of PCl5 is equal to the initial concentration minus the concentration of PCl3:
PCl5 = 0.222 - 1.80 = -1.578 M
Note: Since concentration cannot be negative, it indicates that the reaction has gone to completion, and there is no PCl5 remaining at equilibrium. The negative value of PCl5 suggests that the reaction has exceeded equilibrium.