Question

For the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ the equilibrium constant, \(K,\) has the value \(5.21 \times 10^{-3}\) at a particular temperature. If the system is analyzed at equilibrium at this temperature, it is found that \([\mathrm{CO}(g)]=4.73 \times 10^{-3} M,\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=5.21 \times 10^{-3} M,\) and \(\left[\mathrm{CO}_{2}(g)\right]=3.99 \times 10^{-2} M .\) What is the equilibrium concentration of \(\mathrm{H}_{2}(g)\) in the system?

Short answer

The equilibrium concentration of H2(g) in the system is 0.0247 M.

Step-by-step solution

Write the equilibrium expression

According to the given reaction, the equilibrium expression for the reaction can be represented as: \[K = \frac{[\mathrm{CO}][\mathrm{H_{2}O}]}{[\mathrm{CO_{2}][\mathrm{H_{2}}]}\]

Insert the given values

We are given the values of K, [CO], [H2O], and [CO2]. Now, we will substitute these values into the equilibrium expression: \(5.21 \times 10^{-3} = \frac{(4.73 \times 10^{-3})(5.21 \times 10^{-3})}{(3.99 \times 10^{-2})([\mathrm{H_{2}}])}\)

Solve for [H2]

Now, we can solve the equation for [H2]: \[[\mathrm{H_{2}}] = \frac{(4.73 \times 10^{-3})(5.21 \times 10^{-3})}{(5.21 \times 10^{-3})(3.99 \times 10^{-2})}\]

Calculate the equilibrium concentration of H2

Calculate the value of [H2]: \[[\mathrm{H_{2}}] = \frac{(4.73 \times 10^{-3})(5.21 \times 10^{-3})}{(5.21 \times 10^{-3})(3.99 \times 10^{-2})} = 0.0247 \, M\] The equilibrium concentration of H2(g) in the system is 0.0247 M.

Key concepts

Equilibrium Constant

In any reversible chemical reaction, the equilibrium constant, denoted as \( K \), plays a crucial role in determining the balance between reactants and products. It is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced equation.

This constant provides insight into which direction the reaction favors:

• If \( K \) is much greater than 1, the reaction favors the formation of products.
• If \( K \) is much less than 1, the reaction favors the reactants.
• When \( K \) is close to 1, neither side is favored, and the quantities of reactants and products are similar.

Determining \( K \) is essential for predicting how changes in concentration, pressure, or temperature can affect the equilibrium state. In our exercise, a small \( K \) value indicates that the reactants are favored.

Equilibrium Expression

The equilibrium expression is developed from the balanced chemical equation of the reaction. For the reaction\[\mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2} \mathrm{O}(g)\]we write the expression as:\[K = \frac{[\mathrm{CO}][\mathrm{H_{2}O}]}{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}\]This formula highlights how concentrations of products \([\mathrm{CO} \text{ and } \mathrm{H}_{2} \mathrm{O}]\) are in the numerator, while the reactants \([\mathrm{CO_{2}} \text{ and } \mathrm{H}_{2}]\) are in the denominator. Each term represents the molar concentration at equilibrium.

Using this expression allows us to insert known concentrations and solve for unknowns, such as the equilibrium concentration of \([\mathrm{H}_{2}]\). In complex reactions, these expressions might include more terms, but their formulation follows the same principles:

• Balance the chemical equation.
• Apply the law of mass action.

Concentration Calculations

To find unknown concentrations at equilibrium, we use the equilibrium expression, inserting the known values and solving for the unknown variable. This process is methodical and relies on algebraic manipulation. Here’s how it's executed in our exercise:

• Start with the equilibrium expression: \[ K = \frac{[\mathrm{CO}][\mathrm{H_{2}O}]}{[\mathrm{CO_{2}}][\mathrm{H_{2}}]} \]
• Substitute the known quantities into the expression given: \( K = 5.21 \times 10^{-3} \), \([\mathrm{CO}] = 4.73 \times 10^{-3}\, M\), \([\mathrm{H_{2}O}] = 5.21 \times 10^{-3}\, M\), and \([\mathrm{CO_{2}}] = 3.99 \times 10^{-2}\, M\).
• Set up the equation to solve for \([\mathrm{H_{2}]\). Replace the known values, rearrange the equation to isolate \([\mathrm{H_{2}}]\).

This method systematically reveals unknown concentrations by manipulating the equation. Using the given example, we calculated the \([\mathrm{H_{2}}] = \frac{(4.73 \times 10^{-3})(5.21 \times 10^{-3})}{(5.21 \times 10^{-3})(3.99 \times 10^{-2})} = 0.0247 \, M \)Understanding these calculations is pivotal for analyzing and predicting the behavior of chemical systems at equilibrium.