Question

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.

Short answer

To summarize, when equilibrium is reached in the given reaction, there will be 3 molecules of H2 and CO2, while the number of H2O and CO molecules will be (x-3) each, where x represents the initial number of H2O and CO molecules. The equilibrium is achieved by the reaction of 3 molecules of CO.

Step-by-step solution

Write the balanced equation and equilibrium expression

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: \[ K = \frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} \]

Set up a table to track concentration changes

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.

Use trial and error to find the number of H2 and CO2 molecules at equilibrium

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.

Calculate the concentrations of each substance at equilibrium

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 \(\mathrm{CO}\).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \(K\), is a crucial part of understanding chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium for a given reaction. In our case, the reaction between water vapor \(\mathrm{H}_2\mathrm{O}(g)\) and carbon monoxide \(\mathrm{CO}(g)\) forming hydrogen \(\mathrm{H}_2(g)\) and carbon dioxide \(\mathrm{CO}_2(g)\) gives us an equilibrium constant \(K\) of 2.0. This constant is determined by the equation:\[ K = \frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} \]At equilibrium, the concentrations of each substance are consistent, allowing \(K\) to serve as a snapshot of how far forward or backward the reaction favors under specific conditions. A higher \(K\) value suggests more products at equilibrium, whereas a lower \(K\) indicates more reactants.

Concentration Changes

When a reaction reaches equilibrium, the concentration of reactants and products changes until stability is achieved. For the given reaction, as some molecules of \(\mathrm{CO}\) and \(\mathrm{H_2O}\) react to form \(\mathrm{H_2}\) and \(\mathrm{CO_2}\), changes occur. We represent these with a set of concentration changes for each substance, usually in a table format.For example, starting with \(x\) initial molecules of \(\mathrm{H_2O}\) and \(\mathrm{CO}\), if some amount \(y\) is consumed, then \(\mathrm{H_2O}\) and \(\mathrm{CO}\) will each decrease by \(y\), while \(\mathrm{H_2}\) and \(\mathrm{CO_2}\) each increase by \(y\). At equilibrium, these changes are represented as:

• \(\mathrm{H_2O}(x-y)\)
• \(\mathrm{CO}(x-y)\)
• \(\mathrm{H_2}(y)\)
• \(\mathrm{CO_2}(y)\)

The changes in concentration stop when the equilibrium constant equation is satisfied, meaning equilibrium is reached.

Balanced Chemical Equation

A balanced chemical equation is essential to describe the specific ratio of molecules involved in the reaction and ensure the law of conservation of mass is not violated. In our chemical reaction:\[ \mathrm{H_2O}(g) + \mathrm{CO}(g) \rightleftharpoons \mathrm{H_2}(g) + \mathrm{CO_2}(g) \]Every reactant and product has a coefficient of 1, indicating each molecule reacts in a 1:1 molar ratio. Balancing the equation ensures an accurate calculation of initial concentrations and changes during the reaction.Having the balanced equation allows us to apply the equilibrium constant formula to determine the relationship between concentrations of the reactants and products accurately. The balance in the equation reflects that no atoms are lost or gained, only rearranged, during the chemical process.

Trial and Error Method

The trial and error method is a straightforward approach used to discover the equilibrium position of a reaction by testing various possible scenarios. It involves making a series of assumptions about how many molecules might react and then calculating whether those assumptions fit the equilibrium constant equation.In our given problem, we start by assuming different values of \(y\), the number of molecules reacting, such as 1, 2, or 3, and assess if these assumptions lead the concentration changes to satisfy the equilibrium expression. For example, by substituting \(y = 3\) into the equilibrium expression:\[ 2.0 = \frac{(3)(3)}{(x-3)(x-3)} \]We confirm that this condition satisfies the equation, indicating equilibrium is reached. This method teaches students the iterative process of hypothesizing and testing until the correct balance required by \(K\) is found, reinforcing the understanding of chemical equilibrium.