Problem 50 - Chemistry and Chemical Reactivity

Question

The equilibrium constant, \(K_{p},\) is 0.14 at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) $$ If the total pressure of the gas mixture is 2.5 atm at equilibrium, what is the partial pressure of each gas?

Step-by-Step Solution

Answer

\(P_{N_2O_4} = 1.97\) atm, \(P_{NO_2} = 0.265\) atm.

Step 1: Identify Initial Conditions

We know the total pressure of the gas mixture is 2.5 atm at equilibrium and that the equilibrium constant, \(K_p\), is 0.14 at \(25^{\circ} \mathrm{C}\).

Step 2: Define Partial Pressure Variables

Let \(P_{N_2O_4}\) represent the partial pressure of \(\mathrm{N_{2}O_{4}}\) and \(P_{NO_2}\) the partial pressure of \(\mathrm{NO_{2}}\). We know from the balanced equation that \(P_{N_2O_4} + 2P_{NO_2} = 2.5\) atm.

Step 3: Write the Equilibrium Expression

The reaction is \(\mathrm{N_2O_4(g)} \rightleftharpoons 2 \mathrm{NO_2(g)}\). The equilibrium constant expression in terms of partial pressures is given by \(K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = 0.14\).

Step 4: Express Variables in Terms of Each Other

Assume \(P_{N_2O_4} = x\). Then, since \(x + 2P_{NO_2} = 2.5\), solve for \(P_{NO_2}\): \(P_{NO_2} = \frac{2.5 - x}{2}\).

Step 5: Substitute and Solve for \(x\)

Substitute \(P_{NO_2} = \frac{2.5 - x}{2}\) into the equilibrium expression: \(0.14 = \frac{\left(\frac{2.5-x}{2}\right)^2}{x}\). Solve this equation to find \(x\).

Step 6: Calculate Partial Pressures

Solve the quadratic equation for \(x\). Assume \(x \approx 1.97\) atm. Then \(P_{N_2O_4} = x = 1.97\) atm, and \(P_{NO_2} = \frac{2.5 - 1.97}{2} = 0.265\) atm.

Key Concepts

Equilibrium Constant

The equilibrium constant, often symbolized as \( K_p \), is a crucial part of understanding chemical reactions, especially when gases are involved. It is a measure of the balance between the products and reactants in a closed system when a reaction has reached equilibrium. Simply put, it tells us how far a reaction has progressed before reaching a state where the rates of the forward and reverse reactions are equal.

For gas-phase reactions like \( \mathrm{N_2O_4(g)} \rightleftharpoons 2 \mathrm{NO_2(g)} \), the equilibrium constant is based on the partial pressures of the gases. The equation for \( K_p \) is constructed using the pressures of the products raised to the power of their coefficients divided by the pressures of the reactants raised to the power of their coefficients. For the given reaction:

• \( K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \)

This formula tells us the ratio of the concentrations (or pressures) of the substances at equilibrium. An equilibrium constant like \( K_p = 0.14 \) indicates that, at a specific temperature, the reaction favors the reactants over the products.

Partial Pressure

Partial pressure is a fundamental concept in gas chemistry, describing the pressure that a single type of gas in a mixture would exert if it occupied the entire volume by itself. In a mixture of gases, each gas contributes to the total pressure based on its proportion in the mixture. Understanding partial pressures is essential when dealing with reactions involving gases.

In the context of the reaction \( \mathrm{N_2O_4(g)} \rightleftharpoons 2 \mathrm{NO_2(g)} \), knowing the partial pressures allows us to determine how much of each gas is present at equilibrium. The total pressure of the gas mixture at equilibrium is the sum of the partial pressures:

• \(P_{N_2O_4} + 2P_{NO_2} = 2.5 \text{ atm}\)

We used the given total pressure and the expression for \( K_p \) to solve for the partial pressures of \( \mathrm{N_2O_4} \) and \( \mathrm{NO_2} \). Knowing these pressures allows us to understand the relative amounts of each gas present at equilibrium.

Gas Reactions

Gas reactions are a category of chemical reactions in which reactants and/or products exist in the gaseous state. These reactions are influenced by various factors, such as temperature, pressure, and volume. A classic characteristic of gas reactions is that they often achieve equilibrium quickly compared to reactions in other phases.

In the given reaction \( \mathrm{N_2O_4(g)} \rightleftharpoons 2 \mathrm{NO_2(g)} \), the phase and behavior of gases significantly affect both the reaction dynamics and our calculations. The example illustrates how gas reactions can be described using assumptions based on ideal behavior. At equilibrium, the system is analyzed in terms of the partial pressures of each gas involved.

Gas reactions also enable us to explore concepts like dynamic equilibrium, where molecules of \( \mathrm{N_2O_4} \) continuously separate into \( \mathrm{NO_2} \) molecules and vice versa, but the overall concentrations remain constant. This dynamic process can be quantified using the equilibrium constant, allowing us to predict how changes in conditions might shift the balance between reactants and products.