Question

PCl3(g) + Cl2(g) <---> PCl5(g) Kc = 0.18 In an initial mixture of 0.300M PCl3 (g), 0.400M Cl2 (g), what are the equilibrium concentrations of all gases?

Answer 1

The equilibrium concentrations are `[PCl_3] = 0.280M`, `[Cl_2] = 0.380M`, `[PCl_5] = 0.02M`.

One can approach this problem by using the ICE method (more here: http://en.wikipedia.org/wiki/RICE_chart); even before doing any calculations, the value of the equilibrium constant can give us an idea on how the result will look like; since `K_c<1`, the reaction will favor the reactans, so we'd expect bigger equilibrium concentrations for `PCl_3` and `Cl_2`, than for `PCl_5`.

..`PCl_3(g) + Cl_2(g) rightleftharpoons PCl_5(g)`
I:. 0.300.............0.400..............0
C:... -x....................-x...................+x
E:..(0.300-x).....(0.400-x)...........x

So, `K_c = ([PCl_5])/([PCl_3]*[Cl_2]) = x/((0.300 - x) * (0.400-x)) = 0.18`

The equation `0.18x^2 - 1.126x + 0.0216 = 0` produces two values for `x`, `x_1` = `6.23` and `x_2` = `0.020`; since concentrations cannot be negative, the correct value will be `x = 0.020` (`x = 6.23` would have produced negative concentrations for `PCl_3` and `Cl_2`).

Therefore, the equilibrium concentrations for all the gases are

`[PCl_3] = 0.300 - 0.020 = 0.28M`
`[Cl_2] = 0.400 - 0.020 = 0.38M`
`[PCl_5] = 0.020 M`

Notice that the initial predictions on the relative concentration values is correct, the concentrations of the reactans are bigger than the concentration of the product.