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Let's learn to calculate the equilibrium constant ($K_{c}$) from concentration!

The subscript assigned to "K" indicates what is being used to determine the equilibrium constant. In the case of $K_{c}$, the "c" indicates that concentration is used to determine $K$.

Remember: the equilibrium constant of concentration uses the molarities of solutes and gases in dilute solutions to approximate their activities at equilibrium

The first step is to write the equilibrium expression for the reaction. Solid species or species that are pure liquid or solvent are NOT shown in the expression.

The general equilibrium expression for a reaction:

$a A + bB \Leftrightarrow c C + d D$

is written as:

$K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$

• The "[ ]'s" represent the concentration of the species they contain in molarity ($\frac{mol}{L}$)

• The "little letters" (a, b, c, & d) represent the coefficients used to balance the equation

Let's first practice writing some equilibrium expressions:

1. The production of ammonia from nitrogen and hydrogen gases

Formula: $N_{2(g)} + 3 H_{2(g)} \Leftrightarrow 2 NH_{3(g)}$

Equilibrium expression: $K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$

2. The use of copper to replace silver ions

Formula: $Cu_{(s)} + 2 Ag^{+}_{(aq)} \Leftrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$

Equilibrium expression: $K_{c} = \frac{[Cu^{2+}]}{[Ag^{+}]^{2}}$

Remember: solid species are not shown

Now you try!

Okay, now that we know how to write an equilibrium expression, let's work on solving a problem where all the equilibrium amounts are known!

Say you have a 3.0 L reaction vessel containing 0.1870 moles of $CO_{2(g)}$, 0.4872 moles of $H_{2(g)}$, 0.0056 moles of $CO_{(g)}$, and 0.0573 moles of $H_{2}O_{(g)}$ present at equilibrium. Calculate the value of the equilibrium constant ($K_{c}$).

$CO_{2(g)} + H_{2(g)} \Leftrightarrow CO_{(g)} + H_{2}O_{(g)}$

First, write the equilibrium expression:

$K_{c} = \frac{[CO][H_{2}O]}{[CO_{2}][H_{2}]}$

Next, we are given the amounts in terms of moles, but since we are told that all is contained within a 3.0 reaction vessel, we can convert to concentration in terms of moles per liter:

$[CO_{2}] = \frac{0.1870 mol CO_{2}}{3.00 L} = 0.0623 M$

$[H_{2}] = \frac{0.4872 mol H_{2}}{3.00 L} = 0.1624 M$

$[CO] = \frac{0.0056 mol CO}{3.00 L} = 0.0019 M$

$[H_{2}O] = \frac{0.0573 mol H_{2}O}{3.00 L} = 0.0191 M$

If the equilibrium amounts are already expressed in molarity, conversion is NOT required

Last, substitute the concentrations into the equilibrium expression and solve for the value of $K_{c}$!

$K_{c} = \frac{[0.0019][0.0191]}{[0.0623][0.1624]} = 0.0000946$ or $9.46 \times 10^{-3}$

Sometimes, you will only be given a single equilibrium concentration and will be asked to calculate the equilibrium constant. Let's see how solving that would look like:

Suppose a mixture of 0.100 M $FeCl_{3}$, 0.050 M $SnCl_{2}$, 0.2 M $FeCl_{2}$, and 0.100 M $SnCl_{4}$ was allowed to reach equilibrium. At equilibrium, the concentration of $FeCl_{3}$ was found to be 0.058 M. Determine the value of the equilibrium constant for this reaction.

$2FeCl_{3(aq)} + SnCl_{2(aq)} \Leftrightarrow 2FeCl_{2(aq)} + SnCl_{4(aq)}$

Start by writing the equilibrium expression

$K_{c} = \frac{[FeCl_{2}]^{2}[SnCl_{4}]}{[FeCl_{3}]^{2}[SnCl_{2}]}$

Since the amounts are expressed in terms of molarity, no conversion is necessary

For this kind of problem, you will want to create an ICE chart that expresses the initial concentration, the change in concentration, and the equilibrium concentration for each species in the reaction.

$$\begin{array}{|r|c|c|c|c|} \hline \text{}&FeCl_{3}&SnCl_{2}&FeCl_{3}&SnCl_{4}\\ \hline \text{Initial Concentration (M)}&0.100&0.050&0.200&0.100\\ \hline \text{Change in Concentration (M)}&-2x&-x&+2x&+x\\ \hline \text{Equilibrium Concentration (M)}&0.058\\ \hline \end{array}$$

From this point, you can use the change in concentration of the $FeCl_{3}$ to solve for 'x' and complete the ice chart.

$0.058 M - 0.100 M = -0.042 M$

Change in equilibrium = -2x

Thus, -2x = -0.042

x = 0.021

Now that you have the value for 'x', fill in the remainder of the chart!

$$\begin{array}{|r|c|c|c|c|} \hline \text{}&FeCl_{3}&SnCl_{2}&FeCl_{3}&SnCl_{4}\\ \hline \text{Initial Concentration (M)}&0.100&0.050&0.200&0.100\\ \hline \text{Change in Concentration (M)}&-0.042&-0.021&+0.042&+0.021\\ \hline \text{Equilibrium Concentration (M)}&0.058&0.029&0.242&0.121\\ \hline \end{array}$$

And finally, substitute the equilibrium constants into the equilibrium expression and solve for $K_{c}$

$K_{c} = \frac{[0.024]^{2}[0.121]}{[0.058]^{2}[0.029]} = 0.714$

Whew! Now you know how to calculate the equilibrium constant from concentration! Give it a try on your own now.

(Video) Equilibrium Calculations (ICE Tables) | General Chemistry

By Chad's Prep

In this section of the video, Chad walks you through an example of calculating the equilibrium constant when you're given the equilibrium concentrations.