Problem 91 - Chemistry The Molecular Science

Question

The total pressure for a mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) is 1.5 atm. If \(K_{\mathrm{p}}=7.0\) (at \(25^{\circ} \mathrm{C}\) ), calculate the partial pressure of each gas in the mixture. $$ 2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) $$

Step-by-Step Solution

Step 1: Write the Expression for Kp

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

Key Concepts

Partial Pressure

In a mixture of gases, each individual gas exerts its own pressure, which is called its partial pressure. Partial pressure is essential to understand because it is the contribution of a single gas to the total pressure of a gas mixture. The total pressure of the mixture is the sum of the individual partial pressures of all the gases present.

When calculating partial pressures, we use Dalton's Law of Partial Pressures. According to this law, the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas component.

So, if a gas mixture consists of gases A and B, the total pressure can be represented as:

• \[ P_{ ext{total}} = P_A + P_B \]

where \( P_A \) and \( P_B \) are the partial pressures of gases A and B, respectively.

Understanding partial pressures is crucial when dealing with Kp equations in chemical equilibrium scenarios.

Equilibrium Constant

The equilibrium constant, represented as \( K_{ ext{p}} \), is a vital concept in chemical equilibrium, especially when dealing with gas-phase reactions. It provides a quantitative measure of the ratio of product and reactant concentrations at equilibrium.

For reactions involving gases, \( K_{ ext{p}} \) is often used, where the p signifies that the constant is expressed in terms of partial pressures. The equilibrium constant is calculated using the equilibrium partial pressures of the products and reactants, raised to the power of their stoichiometric coefficients in the balanced chemical equation.

For example, considering the reaction: \[ 2 ext{NO}_2 (g) \rightleftharpoons ext{N}_2 ext{O}_4 (g) \] the expression for \( K_{ ext{p}} \) is given by:

• \[ K_{ ext{p}} = \frac{P_{ ext{N}_2 ext{O}_4}}{P_{ ext{NO}_2}^2} \]

Here:

• \( P_{ ext{N}_2 ext{O}_4} \) is the partial pressure of \( ext{N}_2 ext{O}_4 \)
• \( P_{ ext{NO}_2} \) is the partial pressure of \( ext{NO}_2 \)

In the context of equilibrium, knowing \( K_{ ext{p}} \) allows us to determine the extent to which a reaction will proceed and helps calculate unknown partial pressures in a gas mixture.

Gas Mixtures

A gas mixture is simply a combination of different gas particles existing together. Understanding gas mixtures is significant in chemistry because real-world gas samples often contain more than one type of gas.

Each gas in the mixture behaves independently according to its own set of equations, and together they form what is known as a gas mixture. We apply the concept of partial pressures to assess how each gas in the mixture contributes to the overall behavior of the mixture.

For a gas mixture in a closed system, the pressure exerted by the mixture depends not only on the amount and identity of its constituents but also on their ability to reach equilibrium in a particular reaction.

For a system like \( 2 \text{NO}_2 (g) \rightleftharpoons \text{N}_2\text{O}_4 (g) \) with total pressure \( 1.5 \) atm and equilibrium constant \( K_p = 7.0 \), understanding gas mixtures helps solve for the individual gas partial pressures.

In chemical reactions involving gas mixtures, knowing the total pressure, temperature, and the gas constants helps predict how the mixture will behave at equilibrium.