E - Calculating Equilibrium Concentrations (Lesson)

Calculating Equilibrium Concentrations

The previous lesson described how ICE tables can be used as a technique to determine the concentrations of all substances in a reaction at equilibrium when given the initial concentration of a reactant. Perhaps more importantly, ICE tables can also be used to determine the equilibrium concentrations of all substances when provided with the equilibrium constant.

Consider the equilibrium involving the decomposition of phosphorous pentachloride (PCl₅) at 271 ^{$^\circ$}C :

PCl₃ (g) + Cl₂ (g) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ PCl₅ (g) K_c = 1.60

Suppose that the initial concentrations of PCl₃ and Cl₂ are 0.200 M and 0.100 M, respectively. To calculate the equilibrium concentrations of all substances involved, an ICE table can also be employed. The ICE table for this scenario would look as follows:

ICE TABLE
	PCl₃	Cl₂	PCl₅
Initial	0.200	0.100	0
Change	-x	-x	+x
Equilibrium	0.200 - x	0.100 - x	x

Notice that in this table, it is not possible to calculate the specific amount of change for any substance because there is no information on the concentration of any of the substances at equilibrium. Therefore, it is necessary to simply assume that PCl₅ will increase by some amount "x" causing PCl₃ and Cl₂ to both decrease by that same amount "x" based on the stoichiometry of each substance involved. Using the expressions to describe the equilibrium concentrations on the final "equilibrium" line, it is possible to substitute these into the equilibrium expression to solve for K:

$LaTeX: K_c=\frac{[PCl_{5}]}{[PCl_{3}][Cl_{2}]}$ $K_c=\frac{[PCl_{5}]}{[PCl_{3}][Cl_{2}]}$

$LaTeX: 1.60=\frac{x}{(0.200 - x)(0.100 - x)}$ $1.60=\frac{x}{(0.200 - x)(0.100 - x)}$

Rearranging to combine like terms will give the following expression:

1.60x² - 1.48x + 0.0320 = 0

This results in a quadratic equation in the form of ax² + bx + c = 0 that can be solved using the quadratic equation shown below:

$LaTeX: x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Using this equation to solve for x results in two values: x = 0.903 or 0.0222. Because this is a quadratic equation, which always results in two solutions, we have two different possible answers; however, only one of these values can be physically possible. When examining the equilibrium line of the ICE table, it can be seen that subtracting 0.903 from 0.200 or 0.100 would result in negative answers. Because this is impossible, the 0.903 answer is tossed out and the value of 0.0222 is accepted. The concentrations of all substances present at equilibrium would therefore be:

[PCl₅] = x = 0.0222 M

[Cl₂] = 0.100 - 0.0222 = 0.078 M

[PCl₃] = 0.200 - 0.0222 = 0.178 M

Another example can be found in the video shown below:

You Try It!

In the following self-assessment activity, calculate the equilibrium concentrations for the situation provided. Click on the plus sign to check your answer!

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