Question

In the Haber process, nitrogen gas reacts with hydrogen gas to form gaseous ammonia (see Problem 14.76). (a) Write the equilibrium expression for the reaction. (b) The equilibrium constant at a certain temperature is \(1.5 \times 10^{-2}\). If the equilibrium concentrations of hydrogen and nitrogen are both \(0.20 \mathrm{M}\) at this temperature, what is the equilibrium concentration of ammonia?

Short answer

The equilibrium expression for the Haber process reaction is \(K_\mathrm{eq} = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\). Given the equilibrium constant, \(K_\mathrm{eq} = 1.5 \times 10^{-2}\) and equilibrium concentrations of \([\mathrm{N_2}] = [\mathrm{H_2}] = 0.20\ \mathrm{M}\), the equilibrium concentration of ammonia is approximately \([\mathrm{NH_3}] \approx 0.011\ \mathrm{M}\).

Step-by-step solution

Write the balanced chemical equation for the reaction.

In the Haber process, nitrogen gas (\(\mathrm{N_2}\)) reacts with hydrogen gas (\(\mathrm{H_2}\)) to form ammonia (\(\mathrm{NH_3}\)). The balanced equation for this reaction is: \[ \mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)} \]

Write the equilibrium expression.

The equilibrium expression for the reaction is the product of concentrations of products raised to their respective coefficients in the balanced equation divided by the product of concentrations of reactants raised to their respective coefficients. In this case, the equilibrium expression is: \[K_\mathrm{eq} = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\]

Use the given values to solve for the equilibrium concentration of ammonia.

We can now use the given values of equilibrium constant \((K_\mathrm{eq}\)) and the equilibrium concentrations of nitrogen and hydrogen to calculate the equilibrium concentration of ammonia. The equilibrium constant is given as \(1.5 \times 10^{-2}\), and the equilibrium concentrations of hydrogen and nitrogen are both \(0.20 \mathrm{M}\). We can rearrange the equilibrium expression to solve for \([\mathrm{NH_3}]\): \[[\mathrm{NH_3}]^2 = K_\mathrm{eq} [\mathrm{N_2}][\mathrm{H_2}]^3\] Now, plug the given values and concentrations into the equation: \[[\mathrm{NH_3}]^2 = (1.5 \times 10^{-2})(0.20)((0.20)^3)\] Now, solve for \([\mathrm{NH_3}]\): \[[\mathrm{NH_3}] = \sqrt{(1.5 \times 10^{-2})(0.20)((0.20)^3)}\] \[[\mathrm{NH_3}] = \sqrt{0.00012}\] \[[\mathrm{NH_3}] \approx 0.011\ \mathrm{M}\]

The equilibrium concentration of ammonia.

Thus, the equilibrium concentration of ammonia at this temperature is approximately \(0.011\ \mathrm{M}\).

Key concepts

Equilibrium Expression

The equilibrium expression is a mathematical representation of a chemical reaction's balance at equilibrium. It takes into account the concentrations of all reactants and products in the reaction mixture. For a general reaction of the type \[ aA + bB \rightleftharpoons cC + dD \]The equilibrium expression (\(K_{eq}\)) is written as:\[ K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]Here, concentrations of the products are raised to the power of their coefficients in the balanced equation and placed in the numerator, while concentrations of reactants are similarly treated in the denominator. The equilibrium expression is unique to each chemical reaction and helps predict the direction the reaction will proceed to reach equilibrium.Writing correct equilibrium expressions is essential for solving chemical equilibrium problems as it allows one to calculate unknown concentrations once equilibrium is achieved.

Equilibrium Constant

The equilibrium constant \(K_{eq}\) is a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium. It is specific to each reaction at a given temperature and is derived from the balanced chemical equation. For the Haber process:\[ \mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)} \] The equilibrium constant expression is\[K_{eq} = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\]A high \(K_{eq}\) value indicates the reaction strongly favors the production of products, while a low \(K_{eq}\) value indicates that reactants are favored. Sometimes, \(K_{eq}\) can be very small, as in the Haber process where \(K_{eq} = 1.5 \times 10^{-2}\), suggesting that at equilibrium, the mixture contains a small concentration of products relative to the reactants.

Balanced Chemical Equation

A balanced chemical equation is fundamental in describing a chemical reaction. It ensures that matter is conserved by having the same number of each type of atom on both sides of the equation. This balance is achieved by adjusting coefficients before reactants and products. For the Haber process, the balanced equation is:\[ \mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)} \]Here:

• The coefficient "1" before \(\mathrm{N_2}\) indicates one molecule of nitrogen gas.
• The coefficient "3" before \(\mathrm{H_2}\) indicates three molecules of hydrogen gas.
• The coefficient "2" before \(\mathrm{NH_3}\) indicates two molecules of ammonia are formed.

Balancing equations is crucial because it provides the stoichiometric ratios needed to correctly write the equilibrium expression and calculate the equilibrium concentrations of all species involved.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each reactant and product at equilibrium in a chemical reaction. To find a particular equilibrium concentration, we use the equilibrium expression and known values of other concentrations. Given \(K_{eq}\) for the Haber process, and concentrations of nitrogen and hydrogen, we can calculate the equilibrium concentration of ammonia. If \[ K_{eq} = 1.5 \times 10^{-2}, [\mathrm{N_2}] = 0.20 \ \mathrm{M}, \]\[ [\mathrm{H_2}] = 0.20 \ \mathrm{M} \]We rearrange the equilibrium expression:\[[\mathrm{NH_3}]^2 = K_{eq} \times [\mathrm{N_2}] \times [\mathrm{H_2}]^3 \]Plug the values:\[[\mathrm{NH_3}]^2 = (1.5 \times 10^{-2})(0.20)(0.20^3)\]\[[\mathrm{NH_3}] = \sqrt{0.00012}\]\[[\mathrm{NH_3}] \approx 0.011 \ \mathrm{M}\]Thus, at equilibrium, ammonia has a concentration of approximately \(0.011 \, \mathrm{M}\), creatively using the stoichiometry in the balanced equation and the given \(K_{eq}\).