Question

Consider the following reaction at some temperature:

Some molecules of H₂O and CO are placed in a 1.0-L container as shown below.

When equilibrium is reached, how many molecules of H₂O, CO, H₂, and CO₂ are present? Do this problem by trial and error—that is, if two molecules of CO react, is this equilibrium; if three molecules of CO react, is this equilibrium; and so on.

Short answer

The number of molecules of H₂O, CO, H₂, and CO₂ present at equilibrium are 4, 2, 4 and 4 respectively.

Step-by-step solution

Give the equilibrium constant expression.

The expression for the equilibrium constant, K in terms of concentrations for the given reaction is written below:

$K=\frac{\left[H_2\right]\left[C{O}_2\right]}{\left[H_{2}O\right]\left[CO\right]}$ …(1)

According to the balanced chemical equation in the problem, one equivalent of water reacts with one equivalent of carbon monoxide to form one equivalent of hydrogen and one equivalent of carbon dioxide. So, the stoichiometric factor between any species (reactants/products) is 1.

Find the number of molecules of each gas for trial 1.

For Trail 1: Two molecules of carbon monoxide reacted.

The number of molecules of each gas present at equilibrium can be calculated as follows:

Setup the ICE table:

$$\begin{gathered} H_{2}O\left(g\right)+CO\left(g\right)\to H_2\left(g\right)+C{O}_2\left(g\right) \\ Initial\left(molecules\right):8600 \\ Change\left(molecules\right):-2-2+2+2 \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ Equilibrium\left(molecules\right):6422 \end{gathered}$$

Substitute these values into Equation (1).

$$\begin{gathered} K_{trial 1}=\frac{2\times 2}{6\times 4} \\ =0.17<K \end{gathered}$$

This is not equilibrium because the calculated equilibrium constant value is less than the actual value that is 2.0.

Find the number of molecules of each gas for trial 2.

For Trail 2: Three molecules of carbon monoxide reacted.

The number of molecules of each gas present at equilibrium can be calculated as follows:

Setup the ICE table:

$$\begin{gathered} H_{2}O\left(g\right)+CO\left(g\right)\to H_2\left(g\right)+C{O}_2\left(g\right) \\ Initial\left(molecules\right):8600 \\ Change\left(molecules\right):-3-3+3+3 \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ Equilibrium\left(molecules\right):5333 \end{gathered}$$

Substitute these values into Equation (1).

$$\begin{gathered} K_{trial 2}=\frac{3\times 3}{5\times 3} \\ =0.60<K \end{gathered}$$

This is not equilibrium because $K_{trial 2}<K$.

Find the number of molecules of each gas for trial 3.

For Trail 3: Four molecules of carbon monoxide reacted.

The number of molecules of each gas present at equilibrium can be calculated as follows:

Setup the ICE table:

$$\begin{gathered} H_{2}O\left(g\right)+CO\left(g\right)\to H_2\left(g\right)+C{O}_2\left(g\right) \\ Initial\left(molecules\right):8600 \\ Change\left(molecules\right):-4-4+4+4 \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ Equilibrium\left(molecules\right):4244 \end{gathered}$$

Substitute these values into Equation (1).

$$\begin{gathered} K_{trial 3}=\frac{4\times 4}{4\times 2} \\ =2.0=K \end{gathered}$$

The $K_{trial 3}$value is equal to the K value. Hence, this is equilibrium.

Therefore, the number of molecules of H₂O, CO, H₂, and CO₂ present at equilibrium are 4, 2, 4 and 4 respectively.