Question

The \(\mathrm{N}_{2} \mathrm{O}_{4}-\mathrm{NO}_{2}\) reversible reaction is found to have the following equilibrium partial pressures at \(100^{\circ} \mathrm{C}\). Calculate \(K_{p}\) for the reaction. $$ \begin{array}{rl} \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows & 2 \mathrm{NO}_{2}(g) \\ 0.0014 \mathrm{~atm} & 0.092 \mathrm{~atm} \end{array} $$

Short answer

The equilibrium constant \( K_{p} \) is approximately 6.046.

Step-by-step solution

Identify Reaction Equation and Partial Pressures

The reaction given is \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \). The partial pressure of \( \mathrm{N}_{2} \mathrm{O}_{4} \) at equilibrium is \( 0.0014 \mathrm{~atm} \), and the partial pressure of \( \mathrm{NO}_{2} \) at equilibrium is \( 0.092 \mathrm{~atm} \).

Write the Expression for \( K_{p} \)

For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \), the equilibrium constant \( K_{p} \) is given by the formula: \[ K_{p} = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \] where \( P_{\mathrm{NO}_2} \) is the partial pressure of \( \mathrm{NO}_{2} \) and \( P_{\mathrm{N}_2\mathrm{O}_4} \) is the partial pressure of \( \mathrm{N}_{2} \mathrm{O}_{4} \).

Substitute the Values into the Expression

Substitute the given equilibrium partial pressures into the expression for \( K_{p} \): \[ K_{p} = \frac{(0.092)^2}{0.0014} \].

Calculate \( K_{p} \)

Perform the calculation: \[ K_{p} = \frac{0.008464}{0.0014} \]. Compute the final result: \[ K_{p} \approx 6.046 \].

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, represented as \( K \), is a crucial value that helps us understand the balance between reactants and products in a reversible reaction at equilibrium. For gaseous reactions, this constant is often expressed in terms of partial pressures and is denoted as \( K_{p} \). It provides insights into the extent of the reaction and whether the equilibrium position lies towards the products or the reactants.

For our specific reaction \( \text{N}_{2} \text{O}_{4}(g) \rightleftarrows 2 \text{NO}_{2}(g) \), \( K_{p} \) is calculated using the formula:

• \( K_{p} = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} \)

This equation shows that \( K_{p} \) is the ratio of the partial pressure of the products raised to the power of their stoichiometric coefficients to the partial pressure of the reactants.

With the calculated \( K_{p} \approx 6.046 \), it suggests this equilibrium favors the formation of \( \text{NO}_{2} \), as the constant is greater than 1, indicating more products at equilibrium.

Partial Pressure

Partial pressure is a concept that describes the pressure exerted by a single type of gas in a mixture of gases. It can be thought of as the contribution of that particular gas to the total pressure of the gas mixture in equilibrium. In reactions involving gases, each gas component has its own partial pressure, which plays a critical role in determining the equilibrium constant \( K_{p} \).

For our example of the \( \text{N}_{2} \text{O}_{4}-\text{NO}_{2} \) reaction, the given partial pressures were:

• Partial pressure of \( \text{N}_{2} \text{O}_{4} \) = 0.0014 atm
• Partial pressure of \( \text{NO}_{2} \) = 0.092 atm

By knowing these partial pressures, we can utilize them to determine the equilibrium constant \( K_{p} \). The calculation highlights how each molecule's pressure influences the overall equilibrium state of the system.

Understanding partial pressure is fundamental to studying gas-phase equilibria, as it helps chemists control and predict how gases will behave under changing conditions.

Reversible Reaction

A reversible reaction is a chemical reaction where the reactants form products, which can themselves react to reform the original reactants. Essentially, it's a process in which the conversion happens in both directions until it reaches a state of equilibrium. During this state, the rate of the forward reaction (reactants to products) is equal to the rate of the reverse reaction (products to reactants).

The example reaction \( \text{N}_{2} \text{O}_{4}(g) \rightleftarrows 2 \text{NO}_{2}(g) \) perfectly illustrates a reversible reaction. Here, the decomposition of \( \text{N}_{2} \text{O}_{4} \) to form \( \text{NO}_{2} \) can reverse, allowing \( \text{NO}_{2} \) molecules to combine back into \( \text{N}_{2} \text{O}_{4} \).

Understanding reversible reactions is vital since it introduces the concept of dynamic equilibrium, where the concentration of reactants and products remains constant over time, even though the reactions are still occurring. This balance is a fundamental concept in chemical equilibrium, allowing chemists to predict how changes in conditions can affect the composition of the reaction mixture at equilibrium.