Consider the following reaction at some temperature: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \quad K=2.0 $$ Some molecules of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}\) are placed in a \(1.0\) -L container as shown below. When equilibrium is reached, how many molecules of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}\), \(\mathrm{H}_{2}\), and \(\mathrm{CO}_{2}\) are present? Do this problem by trial and errorthat is, if two molecules of \(\mathrm{CO}\) react, is this equilibrium; if three molecules of \(\mathrm{CO}\) react, is this equilibrium; and so on.

First, let's write the balanced equation given in the problem statement: \[ \mathrm{H_{2}O}(g) + \mathrm{CO}(g) \rightleftharpoons \mathrm{H_{2}}(g) + \mathrm{CO_{2}}(g) \] Equilibrium constant \(K=2.0.\) Now let's write the equilibrium expression for the reaction: \[ K = \frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} \]

Let's set up a table to track concentration changes for each substance involved in the reaction: | Substance | Initial | Change | At Equilibrium | |-----------|---------|----------|----------------| | H2O | x | -y | x-y | | CO | x | -y | x-y | | H2 | 0 | +y | y | | CO2 | 0 | +y | y | The table assumes y moles of H2O and CO are reacting. To reach equilibrium, we need to calculate the number of H2 and CO2 molecules formed. We do this by replacing the concentrations in the equilibrium expression with the concentration values from the table.

We will now use trial and error to find how many molecules of H2 and CO2 are formed at equilibrium. We will do this by first assuming the number of CO molecules that react (y) is 1, 2, 3, and so on. When the value of y gives an equilibrium state, we will find equilibrium concentrations. Trial 1: Assuming 1 molecule of CO reacts (y=1) In this case, substituting the concentrations in the equilibrium expression: \[ 2.0 = \frac{(1)(1)}{(x-1)(x-1)} \] As the equation isn't satisfied, try another value for y. Trial 2: Assuming 2 molecules of CO react (y=2) In this case, substituting the concentrations in the equilibrium expression: \[ 2.0 = \frac{(2)(2)}{(x-2)(x-2)} \] As the equation isn't satisfied, try another value for y. Trial 3: Assuming 3 molecules of CO react (y=3) In this case, substituting the concentrations in the equilibrium expression: \[ 2.0 = \frac{(3)(3)}{(x-3)(x-3)} \] In this case, the equation is satisfied. This means at equilibrium, three molecules of CO react.

As we have found that three molecules of CO react (y=3) at equilibrium, we will use those values to calculate the concentration of each substance. | Substance | Initial | Change | At Equilibrium | |-----------|---------|--------|----------------| | H2O | x | -3 | x-3 | | CO | x | -3 | x-3 | | H2 | 0 | +3 | 3 | | CO2 | 0 | +3 | 3 | The table shows that at equilibrium, there are 3 molecules of \(\mathrm{H_2}\) and \(\mathrm{CO_2}\), and (x-3) molecules of \(\mathrm{H_2O}\) and \(\mathrm{CO}\).