# 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?

May 18, 2015

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, $P C {l}_{5}$, at equilibrium. Moreover, since you only start with reactants, you can predict that the concentrations of both $P C {l}_{3}$, and of $C {l}_{2}$ will decrease.

At the same time, the concentration of $P C {l}_{5}$ will increase. Use an ICE table to help you determine the equilibrium concentrations for your reaction

$\text{ } P C {l}_{3 \left(s\right)} + C {l}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s P C {l}_{5 \left(g\right)}$
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

${K}_{c} = \frac{\left[P C {l}_{5}\right]}{\left[P C {l}_{3}\right] \cdot \left[C {l}_{2}\right]} = \frac{x}{\left(0.22 - x\right) \cdot \left(0.42 - x\right)} = 96.2$

Rearrange this equation to quadratic form

$96.2 \cdot \left(0.22 - x\right) \cdot \left(0.42 - x\right) = x$

$96.2 \cdot \left(0.0924 - 0.22 x - 0.42 x + {x}^{2}\right) = x$

$96.2 {x}^{2} - 62.568 x + 8.888 = 0$

This equation will produce two solutions for $x$

$\left\{\begin{matrix}\cancel{{x}_{1} = 0.4408} \\ {x}_{2} = 0.2096\end{matrix}\right.$

The first solution is eliminated because it will result in negative equilibrium concentrations for $P C {l}_{3}$ and $C {l}_{2}$, which means that you'll get

$\left[P C {l}_{3}\right] = 0.22 - 0.2096 = \textcolor{g r e e n}{\text{0.0104 M}}$
$\left[C {l}_{2}\right] = 0.42 - 0.2096 = \textcolor{g r e e n}{\text{0.210 M}}$
$\left[P C {l}_{5}\right] = 0 + 0.2096 = \textcolor{g r e e n}{\text{0.210 M}}$

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