Question

Nitrogen dioxide decomposes according to the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ where \(K_{\mathrm{p}}=4.48 \times 10^{-13}\) at a certain temperature. If \(0.75 \mathrm{~atm}\) of \(\mathrm{NO}_{2}\) is added to a container and allowed to come to equilibrium, what are the equilibrium partial pressures of \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g) ?\)

Short answer

Equilibrium partial pressures: \( P_{\text{NO}} = 1.0 \times 10^{-4} \text{ atm} \), \( P_{\text{O}_{2}} = 5.0 \times 10^{-5} \text{ atm} \).

Step-by-step solution

Write the equilibrium expression

The equilibrium constant expression for the given reaction is: \[ K_{\text{p}} = \frac{P_{\text{NO}}^2 \times P_{\text{O}_{2}}}{P_{\text{NO}_2}^2} = 4.48 \times 10^{-13} \] where \(P\) denotes partial pressures of the gases.

Set up initial conditions

Initially, only \( \mathrm{NO}_{2} \) is present with a partial pressure of 0.75 atm. So we have: \[ P_{\text{NO}_2,\text{initial}} = 0.75 \text{ atm} \] \[ P_{\text{NO},\text{initial}} = 0 \text{ atm} \] \[ P_{\text{O}_{2},\text{initial}} = 0 \text{ atm} \]

Define change in partial pressures

As the system reaches equilibrium, \( \mathrm{NO}_{2} \) will dissociate into \( \mathrm{NO} \) and \( \mathrm{O}_{2} \). Define the change in partial pressure of \( \mathrm{NO}_{2} \) as \( -2x \), resulting in: \[ P_{\text{NO}_2,\text{equilibrium}} = 0.75 - 2x \] \[ P_{\text{NO},\text{equilibrium}} = 2x \] \[ P_{\text{O}_{2},\text{equilibrium}} = x \]

Substitute equilibrium partial pressures into the expression

Substitute the equilibrium partial pressures into the equilibrium expression: \[ 4.48 \times 10^{-13} = \frac{(2x)^2 \times (x)}{(0.75 - 2x)^2} \] Simplifying this yields: \[ 4.48 \times 10^{-13} = \frac{4x^3}{(0.75 - 2x)^2} \]

Solve for x

Solve the equation for \( x \). Due to the small value of \( K_{\text{p}} \), we can assume that \( 0.75 - 2x \approx 0.75 \). Thus: \[ 4.48 \times 10^{-13} \approx \frac{4x^3}{0.75^2} \] \[ x^3 \approx \frac{4.48 \times 10^{-13} \times 0.75^2}{4} \] \[ x^3 \approx 1.26 \times 10^{-13} \] \[ x \approx (1.26 \times 10^{-13})^{1/3} \approx 5.0 \times 10^{-5} \text{ atm} \]

Calculate equilibrium partial pressures

Calculate the equilibrium partial pressures using the value of \( x \): \[ P_{\text{NO}_{2},\text{equilibrium}} = 0.75 - 2x \approx 0.75 \text{ atm} \] \[ P_{\text{NO},\text{equilibrium}} = 2x = 1.0 \times 10^{-4} \text{ atm} \] \[ P_{\text{O}_{2},\text{equilibrium}} = x = 5.0 \times 10^{-5} \text{ atm} \]

Key concepts

Partial Pressure

In a chemical equilibrium context, partial pressure refers to the pressure exerted by a single gas in a mixture of gases. Each gas in the mixture behaves independently, and its partial pressure is proportional to its concentration. This means that if you know the concentration of a gas, you can calculate its partial pressure using the ideal gas law as follows: \( P = \frac{nRT}{V} \), where \( P \) is the partial pressure, \( n \) is the number of moles, \( R \) is the gas constant, \( T \) is the temperature, and \( V \) is the volume of the container.
In our given problem, partial pressures play a crucial role in determining the equilibrium state of the reaction. Initially, we only have nitrogen dioxide (\(NO_2\)) with a partial pressure of 0.75 atm. As the reaction reaches equilibrium, this \(NO_2\) will decompose into nitrogen monoxide (\(NO\)) and oxygen (\(O_2\)), each establishing their own partial pressures according to the stoichiometry of the reaction.
By understanding how partial pressures change as the reaction proceeds, we can set up the equilibrium expression and calculate the equilibrium concentrations of all involved gases.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), is a measure of the ratio of the partial pressures of the products to the reactants at equilibrium for a given reaction. It specifically applies to reactions involving gases. For our exercise, the equilibrium expression using \(K_p\) is given by: \[ K_p = \frac{P_{NO}^2 \times P_{O_2}}{P_{NO_2}^2} \]
This equation tells us that the equilibrium constant depends on the partial pressures of \(NO\), \(O_2\), and \(NO_2\). The value of \(K_p\) is 4.48 x 10^-13, which is very small.
Because \(K_p\) is so small, it indicates that at equilibrium, the concentration of nitrogen dioxide (\(NO_2\)) will be much higher than the concentrations of nitrogen monoxide (\(NO\)) and oxygen (\(O_2\)). This is because the reaction heavily favors the reactants over the products. By inserting the expressions for partial pressures at equilibrium into the \(K_p\) equation, we can solve for the unknown quantities, giving us a clearer understanding of the equilibrium state.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure of the ratio of the concentrations or partial pressures of the products to the reactants at any point in time, not necessarily at equilibrium. It uses the same expression as \(K_p\): \[ Q = \frac{P_{NO}^2 \times P_{O_2}}{P_{NO_2}^2} \]
By comparing \(Q\) to \(K_p\), we can predict which direction the reaction will proceed to reach equilibrium:

• If \(Q = K_p\), the system is at equilibrium.
• If \(Q > K_p\), the reaction will shift towards the reactants.
• If \(Q < K_p\), the reaction will shift towards the products.

In our problem, since we start with only \(NO_2\), the initial reaction quotient \(Q\) is zero because the numerator of the equation (products part) is zero. This means \(Q < K_p\), so the reaction will proceed in the forward direction, forming more \(NO\) and \(O_2\) until equilibrium is achieved.
Understanding the reaction quotient helps us anticipate how the system evolves over time, giving us a way to track the progression of a reaction towards its equilibrium state.