Is the correct answer(s) PCl3 = 2.50 M, Cl2 = 1.40 M, and PCl5 = 1.51 x 10^-5 M?

Question

Consider the following reaction: 

 PCl3 (g)+ Cl2 (g) PCl5 (g) K = 4.30 x 10-6 

Initially, 2.50 M PCl3 and 1.40 M Cl2 are placed into a container and allowed to reach equilibrium. Determine the equilibrium concentrations of all the species.

Answer 1

Hello, Andrew,

You got the same answer I got, so either it is correct or we are both wrong.

PCl₃(g) + Cl₂(g) <---> PCl₅(g)

Let's start with the equilibrium expression:

K_c = [PCl₅]/[PCl₃]*[Cl₂]

We are given starting concentrations of both reactants, so I'll assign x to be the concentration of the product, PCL₅. We note that the equilibrium coefficent is rater low at 1.51x10^-5. I will assume for now that the change in starting concentrations will be low relative to the initial concentrations of the starting reagents. We can check that assumption later.

The ICE table allows us to keep the data organized. The change in PCl₅, x, will mean that an equal amount will be removed from both PCl₃ and Cl₂. That leads to the (2.5-x) and (1.4-x) expressions for the change, C. Assuming x is small, we can simplify the final (equilibrium) concentrations to just the starting values.

4.30x10^-6 = K_c

4.30x10^-6 = [PCl₅]/[PCl₃]*[Cl₂]

4.30x10^-6 = [x]/[2.5]*[1.4]

x = 1.51x10^-5

x is indeed much smaller that the 2.5 and 1.4 starting concentrations, so our assumption is valid and the ending concentrations of both reactants are essentially the same as the starting values. A quadratric equation is avoided, and that is a nice start to my day.

Bob