The Equilibrium Law

\[\text{a}\text{A}(g) + \text{b}\text{B}(g) \rightleftharpoons \text{c}\text{C}(g) + \text{d}\text{D}(g)\]

The reaction quotient, \(Q\) is written as:

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

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

Worked Example

For example, in the equilibrium system:

\[\text{2SO}_2(g) + \text{O}_2(g) \rightleftharpoons \text{2SO}_3(g) \quad\quad \Delta H = \text{-ve}\]

the equilibrium expression would be:

\[K_c = \frac{[\text{SO}_3]^2}{[\text{O}_2][\text{SO}_2]^2}\]

For a given equilibrium reaction, when the concentrations of each reactant and product are measured at equilibrium, the equilibrium constant can be calculated.

Worked example

For example, in the equilibrium system:

\[\text{2SO}_2(g) + \text{O}_2(g) \rightleftharpoons \text{2SO}_3(g) \quad\quad \Delta \text{H} = \text{-ve}\]

The units for \(\text{K}_c\) would resolve as follows:

\[\begin{align} \text{K}_c &= \frac{[\text{SO}_3]^2}{[\text{O}_2][\text{SO}_2]^2}\\ &=\frac{[\text{M}]^{2}}{[\text{M}][\text{M}]^{2}}\\  &=\frac{1}{\text{M}}\\ &=\text{M}^{-1} \end{align}\]

Worked example

For example, for the reaction:

\[\text{2SO}_2(g) + \text{O}_2(g) \rightleftharpoons \text{2SO}_3(g)\]

A chemist measures the concentrations at equilibrium of each reactant and product inside a sealed reaction vessel at 400℃, and finds the following:

\[\begin{align} &[\text{SO}_2] = \text{0.20 M} \\ &[\text{O}_2] = \text{0.20 M} \\ &[\text{SO}_3] = \text{0.02 M} \end{align}\]

To solve for the value of \(K_{c}\) we first need to write the equilibrium expression.

\[\text{K}_c = \frac{[\text{SO}_3]^2}{[\text{O}_2][\text{SO}_2]^2}\]

Then we substitute in the values for concentration, evaluate and include the units for \(K_c\):

\[K_c=\frac{[0.02]^2}{[0.20][0.2]^2}\]

\[K_c=0.05 \text{ M}^{-1}\]

Worked example

For example, for the reaction:

\[\text{3H}_2(g) + \text{N}_2(g) \rightleftharpoons \text{2NH}_3(g)\]

A chemist measures the amounts of each reactant and product inside a 2.0 L sealed reaction vessel at equilibrium at 500℃ and finds the following:

\[\begin{align} \text{H}_2 &= 0.60\ \text{mol} \\ \text{N}_2 &= 0.30\ \text{mol} \\ \text{NH}_3 &= 0.20\ \text{mol} \end{align}\]

To calculate the value of \(K_c\) we first need to write the equilibrium expression:

\( \text{K}_c =\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\)

Then solve for concentration:

\[[\text{N}_2] = \frac{0.30 \text{mol}}{2.0 \text{L}}=0.15 \text{M}\]

\[[\text{H}_2] = \frac{0.60 \text{mol}}{2.0 \text{L}}=0.30 \text{M}\]

\[[\text{NH}_3] = \frac{0.20 \text{mol}}{2.0 \text{L}}=0.10 \text{M}\]

Then we substitute in the values for concentration, evaluate and include the units for \(K_c\):

\[\text{K}_c =\frac{[0.10]^2}{[0.15][0.30]^3}\]

\[\text{K}_c = 2.47 \text{M}^{-2}\]

Worked example

For example, for the reaction:

\[\text{3H}_2(g) + \text{N}_2(g) \rightleftharpoons \text{2NH}_3(g) \text{ ; } \text{K} = 1.45 \times 10^{-5}\ \text{M}^{-2} \text{ at } 500^{\circ}\text{C}\]

A chemist measures the following concentrations of each reactant inside a 2.0 \(\text{L}\) sealed reaction vessel at equilibrium at 500℃ and finds the following:

\(\text{H}_2 = 0.60\text{ M}\)
\(\text{N}_2 = 0.30 \text{ M}\)

To find the concentration at equilibrium of the product, \(\text{NH}_3\), we first need to write the equilibrium expression. Let \(\text{NH}_3\) be represented by \(X\):

\[\begin{align}
1.45 \times 10^{-5}\ \text{M}^{-2} &= \frac{[X]^2}{[\text{N}_2][\text{H}_2]^3} \\
[X]^2 &= 1.45 \times 10^{-5}\ \text{M}^{-2} \times [0.30\ \text{M}][0.60\ \text{M}]^3 \\
[X]^2 &= 9.40 \times 10^{-7}\ \text{M}^2 \\
[X] &= 9.69 \times 10^{-4}\ \text{M}
\end{align}\]

Worked example

For example, for the reaction:
\[\text{3H}_2(g) + \text{N}_2(g) \rightleftharpoons \text{2NH}_3(g)\]

An industrial chemist adds \(1.0\ \text{mol}\) of \(\text{H}_2(g)\) and \(0.5\ \text{mol}\) of \(\text{N}_2(g)\) to a \(2.0\ \text{L}\) sealed reaction vessel at equilibrium at \(500^{\circ}\text{C}\) and finds there is \(0.2\ \text{mol}\) of \(\text{NH}_3\) at equilibrium.

To solve for the value of the unknown concentrations, we first need to set up an ICE table.

\[\text{Initial + Change = Equilibrium}\]

\[0+2x = 0.2 \text{ mol}\] \[x=0.1 \text{ mol}\]

With the value of \(x\), we can complete the ICE table to find all equilibrium values.

We can then write the equilibrium expression.

\[\text{K}_c = \frac{[\text{NH}_3]^2}{[\text{H}_2]^3[\text{N}_2]}\]

Then substitute in the known values.

\[\begin{align}
\text{K}_c &= \frac{[0.1]^2}{[0.35]^3[0.2]} \\
\text{K} &= 1.17\ \text{M}^{-2}
\end{align}\]

The same calculation can occur and result in a quadratic equation that needs to be solved:

\[\text{I}_2(g) + \text{Cl}_2(g) \rightleftharpoons \text{2ICl}(g) \quad \text{K}_c = 82 \text{ at } 30^{\circ}\text{C}\]

An industrial chemist adds \(1.0\ \text{mol}\) of \(\text{I}_2(g)\) and \(1.0\ \text{mol}\) of \(\text{Cl}_2(g)\) to a \(12.0\ \text{L}\) sealed reaction vessel at equilibrium at \(30^{\circ}\text{C}\). What are the unknown values at equilibrium?

To solve for the value of the unknown concentrations, we first need to set up an ICE table.

Note: we can interchange between mole or concentration values in the ICE table. For this one we will use concentration.

\text{K}_c = \frac{[\text{ICl}]^2}{[\text{I}_2][\text{Cl}_2]}

Then substitute in the known values:
\[
\begin{align}
\text{K}_c &= \frac{[\text{ICl}]^2}{[\text{I}_2][\text{Cl}_2]} \\
82 &= \frac{[2x]^2}{[0.083 - x][0.083 - x]} \\
82 &= \frac{[2x]^2}{[0.083 - x]^2} \\
\sqrt{82} &= \frac{2x}{0.083 - x} \\
9.05 &= \frac{2x}{0.083 - x} \\
0.751 - 9.05x &= 2x \\
0.751 &= 11.05x \\
x &= 0.06798 \\
x &= 0.068\ \text{M} \\
[\text{I}_2] &= 0.083 - 0.068 = 0.015\ \text{M} \\
[\text{Cl}_2] &= 0.083 - 0.068 = 0.015\ \text{M} \\
[\text{ICl}] &= 0 + 2 \times 0.068 = 0.136\ \text{M}
\end{align}
\]