Question

1. For the reaction, `PCl_(3(g))+Cl_(2(g)) rightleftharpoons PCl_(5(g))`, `K_c = 96.2` at 400 K. If the initial concentrations are 0.22 mol/L of `PCl_3` and 0.42 mol/L of `Cl_2`, what are the equilibrium concentrations of all species?

Answer 1

So, you know that you're dealing with an equilibrium reaction that has its equilibrium constant equal to 96.2 at a certain temperature.

The fact that `K_c` is larger than one tells you that the reaction will favor the product, `PCl_5`, at equilibrium. Moreover, since you only start with reactants, you can predict that the concentrations of both `PCl_3`, and of `Cl_2` will decrease.

At the same time, the concentration of `PCl_5` will increase. Use an ICE table to help you determine the equilibrium concentrations for your reaction

`" "PCl_(3(s)) + Cl_(2(g)) rightleftharpoons PCl_(5(g))`
I.....0.22...........0.42.................0
C.....(-x).............(-x)..................(+x)
E...0.22-x.......0.42-x................x

By definition, the equilibrium constant will be equal to

`K_c = ([PCl_5])/([PCl_3] * [Cl_2]) = x/((0.22-x) * (0.42 - x)) = 96.2`

Rearrange this equation to quadratic form

`96.2 * (0.22-x) * (0.42-x) = x`

`96.2 * (0.0924 - 0.22x - 0.42x + x^2) = x`

`96.2x^2 - 62.568x + 8.888 = 0`

This equation will produce two solutions for `x`

`{ (cancel(x_1 = 0.4408)), (x_2 = 0.2096) :}`

The first solution is eliminated because it will result in negative equilibrium concentrations for `PCl_3` and `Cl_2`, which means that you'll get

`[PCl_3] = 0.22 - 0.2096 = color(green)("0.0104 M")`
`[Cl_2] = 0.42 - 0.2096 = color(green)("0.210 M")`
`[PCl_5] = 0 + 0.2096 = color(green)("0.210 M")`

SIDE NOTE I've left the answers with three sig figs.