Calculating Kc Equilibrium for PCl3 + Cl2 = PCL5 at 230°C

[kittymaniac84]

The equilibrium constant Kc for the reaction, PCl3 (g) + Cl2 (g) = PCL5 (g),
equals 49 at 230°C. If 0.400 mol each of phosphorus trichloride and chlorine are
added to a 4.0 Liter reaction vessel, what is the equilibrium composition of the
mixture at 230°C?

 

[gravenewworld]

Do your own hw

 

[Blueshift5]

Based on the given information, we can use the equilibrium constant (Kc) equation to calculate the equilibrium composition of the mixture at 230°C:

Kc = [PCL5]/[PCl3][Cl2] = 49

Since the initial amounts of PCl3 and Cl2 added to the reaction vessel are equal at 0.400 mol each, we can assume that at equilibrium, the amounts of both substances will decrease by the same amount (x) and the amount of PCL5 formed will be equal to the amount of PCl3 and Cl2 reacted. Therefore, the equilibrium concentrations can be represented as:

[PCl3] = 0.400 - x
[Cl2] = 0.400 - x
[PCL5] = x

Substituting these values into the Kc equation, we get:

49 = x/(0.400 - x)^2

Solving for x, we get x = 0.119 mol. This means that at equilibrium, 0.119 mol of PCL5 will be formed and the concentrations of PCl3 and Cl2 will be 0.281 mol each.

In summary, at 230°C, the equilibrium composition of the mixture will be 0.119 mol of PCL5, 0.281 mol of PCl3, and 0.281 mol of Cl2 in a 4.0 Liter reaction vessel. This calculation assumes that the reaction reaches equilibrium and no other factors, such as pressure or volume changes, affect the equilibrium composition.