Question

The total pressure for a mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) is 0.15 atm. If \(K_{p}=7.1\) (at \(25^{\circ} \mathrm{C}\) ), calculate the partial pressure of each gas in the mixture. $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

Short answer

\( P_{\text{NO}_2} = 0.066 \text{ atm}, \) \( P_{\text{N}_2\text{O}_4} = 0.084 \text{ atm} \).

Step-by-step solution

Write the Expression for Kp

For the equilibrium reaction \(2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)\), the equilibrium constant expression in terms of partial pressures is: \[ K_p = \frac{P_{\text{N}_2\text{O}_4}}{(P_{\text{NO}_2})^2} \] where \(P_{\text{N}_2\text{O}_4}\) and \(P_{\text{NO}_2}\) are the partial pressures of \(\text{N}_2\text{O}_4\) and \(\text{NO}_2\) respectively.

Express Total Pressure

The given total pressure of the gas mixture is the sum of the partial pressures of \(\text{N}_2\text{O}_4\) and \(\text{NO}_2\). Therefore, we have: \[ P_{\text{N}_2\text{O}_4} + P_{\text{NO}_2} = 0.15 \text{ atm} \]

Substitute Expressions into Kp Equation

We can use the substitution method to find either \( P_{\text{N}_2\text{O}_4} \) or \( P_{\text{NO}_2} \). Let’s express \( P_{\text{N}_2\text{O}_4} \) in terms of \( P_{\text{NO}_2} \) using the total pressure equation: \[ P_{\text{N}_2\text{O}_4} = 0.15 - P_{\text{NO}_2} \] Substituting into the Kp equation gives: \[ K_p = \frac{0.15 - P_{\text{NO}_2}}{(P_{\text{NO}_2})^2} = 7.1 \]

Solve for Partial Pressure of NO2

Rearrange the equation to express in terms of \( P_{\text{NO}_2} \): \[ 7.1(P_{\text{NO}_2})^2 = 0.15 - P_{\text{NO}_2} \] Solving this quadratic equation: \[ 7.1x^2 + x - 0.15 = 0 \] where \( x = P_{\text{NO}_2} \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 7.1 \), \( b = 1 \), \( c = -0.15 \).

Calculate Roots Using Quadratic Formula

Compute the discriminant \( b^2 - 4ac = 1^2 - 4 \times 7.1 \times (-0.15) \). Calculate: \[ x = \frac{-1 \pm \sqrt{1 + 4.26}}{2 \times 7.1} = \frac{-1 \pm \sqrt{5.26}}{14.2} \] Solve for the positive root (since pressure cannot be negative).

Determine Partial Pressures

Calculate the positive root from the earlier step to find \( P_{\text{NO}_2} \). After computation, \( P_{\text{NO}_2} \approx 0.066 \text{ atm} \). Use the total pressure to find \( P_{\text{N}_2\text{O}_4} = 0.15 - 0.066 = 0.084 \text{ atm} \).

Key concepts

Partial Pressure

When dealing with gas mixtures, understanding partial pressure is essential. Each gas in a mixture exerts pressure independently of the others. This pressure, known as partial pressure, represents how much pressure a particular gas contributes to the total pressure of the mixture.

In the context of chemical equilibria, it's important to remember that the total pressure of the system is the sum of the partial pressures of all gases present. For our exercise, the total pressure is given as 0.15 atm, comprising the partial pressures of \( ext{N}_2 ext{O}_4\) and \( ext{NO}_2\). Hence, you set up the equation: \[ P_{\text{N}_2\text{O}_4} + P_{\text{NO}_2} = 0.15 \text{ atm} \] This forms the basis for calculating the individual partial pressures of the gases.

Equilibrium Constant

The concept of the equilibrium constant \(K_p\) is central in chemical equilibria, especially for gaseous systems. It provides insight into the position of equilibrium at a given temperature.

For the reaction \(2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)\), \(K_p\) is expressed in terms of the partial pressures of the reactants and products as follows: \[ K_p = \frac{P_{\text{N}_2\text{O}_4}}{(P_{\text{NO}_2})^2} \]

Here, \(K_p\) value of 7.1 indicates that at equilibrium, the pressure of \( ext{N}_2 ext{O}_4\) relative to \(\text{NO}_2\) squared is consistent, as dictated by this constant. The equilibrium constant helps predict how the reaction shifts in response to changes in conditions, vital for understanding the dynamic nature of equilibrium.

Quadratic Formula

To determine the individual partial pressures of gases in our equilibrium system, the quadratic formula is employed. Quadratic equations often arise in chemical equilibrium problems when expressing \(K_p\).

In this case, substituting into the \(K_p\) equation gives a quadratic equation that can be expressed as: \[ 7.1(P_{\text{NO}_2})^2 = 0.15 - P_{\text{NO}_2} \] Rearranging gives the quadratic form: \[ 7.1x^2 + x - 0.15 = 0 \] where \( x = P_{\text{NO}_2} \).

• \(a = 7.1\)
• \(b = 1\)
• \(c = -0.15\)

The quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] is then used to find the possible values of \(x\). Solving this gives the partial pressures required for the exercise.

Gaseous Equilibria

Gaseous equilibria involve reactions in which all reactants and products are gases. These systems are influenced by changes in pressure, volume, and temperature. When a system reaches equilibrium, the forward and reverse reactions occur at equal rates, stabilizing concentrations or pressures over time.

In our reaction \(2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)\), the system reacts to establish a dynamic balance, characterized by the equilibrium constant \(K_p\). It's crucial to consider that changes in conditions such as pressure will shift the equilibrium position—a concept explained by Le Chatelier’s Principle.

Understanding these principles allows chemists to manipulate conditions to favor the formation of desired products, an application valuable in industrial and laboratory settings.