Question

PCl3 + Cl2 ⇌ PCl5, calculate all three equilibrium concentrations when Kc = 16.0 and [PCl5] i = 1.00 M.

Answer 1

For the reaction:

PCl₃ + Cl₂ ⇌ PCl₅

the equilibrium concentration constant $K_c$ has the following expression:

$K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]}$ ,

where $[PCl_5], [PCl_3]$ and $[Cl_2]$ are the equilibrium concentrations of PCl₅, PCl₃ and Cl₂, respectively.

The equilibrium concentration of PCl₅ is its initial concentration, 1.00 M minus the quantity of PCl₅ dissociated $x$ :

$[PCl_5] = 1.00 - x$ .

Therefore, the equilibrium concentrations of PCl₃ and Cl₂ are $x$ :

$[PCl_3] = [Cl_2] = x$ .

Using these equations, let's find $x$ :

$16.0 = \frac{1-x}{x^2}$

$16x^2 + x - 1 =0$

$x = 0.221$ .

Finally, the equilibrium concentrations are:

$[PCl_5] = 1 - 0.221 = 0.779$ M

$[PCl_3] = [Cl_2] = 0.221$ M.

Answer: [PCl₅] = 0.779 M, [PCl₃] = [Cl₂] = 0.221 M.