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

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

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Answer 1 · 2015-05-18T21:14:37

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 $K_{c}$ $K_c$ is larger than one tells you that the reaction will favor the product, $P C l_{5}$ $PCl_5$ , at equilibrium. Moreover, since you only start with reactants, you can predict that the concentrations of both $P C l_{3}$ $PCl_3$ , and of $C l_{2}$ $Cl_2$ will decrease.

At the same time, the concentration of $P C l_{5}$ $PCl_5$ will increase. Use an ICE table to help you determine the equilibrium concentrations for your reaction

$P C l_{3 (s)} + C l_{2 (g)} ⇌ P C l_{5 (g)}$ $" "PCl_(3(s)) + Cl_(2(g)) rightleftharpoons PCl_(5(g))$
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

$K_{c} = \frac{[P C l_{5}]}{[P C l_{3}] \cdot [C l_{2}]} = \frac{x}{(0.22 - x) \cdot (0.42 - x)} = 96.2$ $K_c = ([PCl_5])/([PCl_3] * [Cl_2]) = x/((0.22-x) * (0.42 - x)) = 96.2$

Rearrange this equation to quadratic form

$96.2 \cdot (0.22 - x) \cdot (0.42 - x) = x$ $96.2 * (0.22-x) * (0.42-x) = x$

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

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

This equation will produce two solutions for $x$ $x$

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

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.

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