Question

A sample of air with a mole ratio of \(\mathrm{N}_{2}\) to \(\mathrm{O}_{2}\) of 79: 21 is heated to 2500 K. When equilibrium is established in a closed container with air initially at 1.00 atm, the mole percent of \(\mathrm{NO}\) is found to be \(1.8 \% .\) Calculate \(K_{\mathrm{p}}\) for the reaction. $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$$

Short answer

The calculation gives \(K_{\mathrm{p}} = 0.00171\).

Step-by-step solution

Calculate the initial mole fractions

The mole fractions of nitrogen and oxygen given by the ratio 79: 21 sum up to 100. Thus the mole fractions are \(0.79\) for \(\mathrm{N}_{2}\) and \(0.21\) for \(\mathrm{O}_{2}\). Remember that the mole fraction is defined as the amount of a specific component divided by the total amount.

Calculate the partial pressures at equilibrium

The total pressure at equilibrium is the sum of the partial pressures of each gas. Since the initial pressure is 1.00 atm and no gas was added or removed (only the proportions changed), the total pressure at equilibrium is still 1.00 atm. Use the mole percent of nitric oxide at equilibrium to find its partial pressure. A \(1.8 \%\) mole percent corresponds to a partial pressure of \(0.018 \times 1.00 \, \mathrm{atm} = 0.018 \, \mathrm{atm}\). The reaction consumes equal moles of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to produce \(\mathrm{NO}\), so the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) decrease by the same amount, \(0.018 \, \mathrm{atm}\). The final partial pressures are therefore \(0.79 \times 1.00 \, \mathrm{atm} - 0.018 \, \mathrm{atm} = 0.772 \, \mathrm{atm}\) for \(\mathrm{N}_{2}\) and \(0.21 \times 1.00 \, \mathrm{atm} - 0.018 \, \mathrm{atm} = 0.192 \, \mathrm{atm}\) for \(\mathrm{O}_{2}\).

Use the equilibrium equation to find \(K_{\mathrm{p}}\)

The equilibrium constant, \(K_{\mathrm{p}}\), for the reaction is defined as \[K_{\mathrm{p}}= \frac{\left(\mathrm{P}_{\mathrm{NO}}\right)^2}{\mathrm{P}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{O}_{2}}}\] where \(\mathrm{P}_{\mathrm{NO}}\), \(\mathrm{P}_{\mathrm{N}_{2}}\), and \(\mathrm{P}_{\mathrm{O}_{2}}\) are the partial pressures of \(\mathrm{NO}\), \(\mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) at equilibrium. Substituting the calculated pressures, we get \[K_{\mathrm{p}}= \frac{(0.018 \, \mathrm{atm})^2}{0.772 \, \mathrm{atm} \cdot 0.192 \, \mathrm{atm}}\]

Calculate \(K_{\mathrm{p}}\)

Finally, perform the multiplication and division to obtain the value of \(K_{\mathrm{p}}\).

Key concepts

Partial Pressure

In the context of gases, the concept of partial pressure is crucial for understanding how different gases contribute to the total pressure of a mixture. Partial pressure is defined as the pressure that a gas in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature. To find the partial pressure of a particular gas, you multiply its mole fraction by the total pressure of the gas mixture. For example, in our problem, nitrogen (\( \mathrm{N}_2 \)) and oxygen (\( \mathrm{O}_2 \)) have mole fractions of \(0.79\) and \(0.21\), respectively. If the total pressure is \(1.00 \, \text{atm}\), the partial pressures of \(\mathrm{N}_2\) and \(\mathrm{O}_2\) initially are \(0.79 \, \text{atm}\) and \(0.21 \, \text{atm}\).
This concept helps us in many calculations, including those involving chemical equilibria, as the change in partial pressures during reactions gives us the information needed to find equilibrium constants.

Equilibrium Constant

The equilibrium constant, \(K_p\), for a gaseous reaction is calculated using the partial pressures of the reactants and products at equilibrium. It reflects the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. For the reaction \(\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\), this means \(K_p\) is calculated by the formula:

• \(K_p = \frac{(P_{\mathrm{NO}})^2}{P_{\mathrm{N}_2} \cdot P_{\mathrm{O}_2}}\)

Knowing the equilibrium constant for a reaction is valuable because it indicates the extent to which reactants convert to products under given conditions.
In the example provided, substituting the equilibrium partial pressures of \(P_{\mathrm{NO}} = 0.018 \, \text{atm}\), \(P_{\mathrm{N}_2} = 0.772 \, \text{atm}\), and \(P_{\mathrm{O}_2} = 0.192 \, \text{atm}\) into the formula gives us the equilibrium constant for the formation of NO at 2500 K.

Gaseous Reactions

Gaseous reactions often involve multiple reactants and products that exist in a gas phase. They have notable features and behaviors, primarily dictated by the principles of gas laws and equilibrium. In a closed system, they reach a state where the rate of the forward reaction equals the rate of the reverse reaction, creating what is known as a dynamic equilibrium. When dealing with gaseous reactions, partial pressures are used as a measure of the concentration of gases, and they play a vital role in calculating the equilibrium state, given by \(K_p\).
Gaseous reactions are highly influenced by changes in temperature and pressure. At higher temperatures, such as 2500 K in our example, molecules move more rapidly, potentially affecting the extent of the reaction. The equilibrium achieved is specific to the conditions present, and shifts can occur if temperature or pressure are altered. Understanding these reactions through the concept of equilibrium and the associated mathematics enables predictions of how systems react to changes.