Question

The partial pressures of an equilibrium mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.34\) atm and \(P_{\mathrm{NO}_{2}}=1.20 \mathrm{atm}\) at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.

Short answer

The new equilibrium partial pressures are \(P_{N_2O_4} = 0.02\,\text{atm}\) and \(P_{NO_2} = 1.52\,\text{atm}\).

Step-by-step solution

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between N2O4 and NO2 is: \[ N_2O_4(g) \rightleftarrows 2NO_2(g) \]

Express the reaction quotient (Q) and the equilibrium constant (K)

The reaction quotient (Q) is given by the ratio of the concentrations or partial pressures of products to reactants, each raised to the power of their stoichiometric coefficients. In this case, since we have partial pressures, we can use Qp: \[ Q_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \] At equilibrium, the reaction quotient, Qp, becomes the equilibrium constant, Kp: \[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Calculate the initial reaction quotient and equilibrium constant

Given initial partial pressures of N2O4 and NO2, we calculate the initial reaction quotient as: \[ Q_{p,initial} = \frac{(1.20\,\text{atm})^2}{0.34\,\text{atm}} \] \[ Q_{p,initial} = 4.24 \] Since the initial reaction is already at equilibrium, Qp becomes Kp: \[ K_p = 4.24 \]

Set up the expression for the final partial pressures

Let x be the change in partial pressures of N2O4 and NO2 at the new equilibrium: \[ P_{N_2O_4} = 0.34\,\text{atm} - x \] \[ P_{NO_2} = 1.20\,\text{atm} + x \]

Update the reaction quotient for the new conditions

Since the volume of the container has doubled, the effect on the partial pressures can be represented by halving the initial partial pressures. Therefore, the updated reaction quotient will be equal to the equilibrium constant Kp: \[ K_p = \frac{((1.20\,\text{atm} + x)/2)^2}{(0.34\,\text{atm} - x)/2} \]

Solve for x, the change in partial pressures

Now, substitute the known value of Kp and solve for x: \[ 4.24 = \frac{((1.20\,\text{atm} + x)/2)^2}{(0.34\,\text{atm} - x)/2} \] After some algebraic manipulation, we find x = 0.32 atm.

Calculate the new partial pressures of N2O4 and NO2

Using the expressions for the final partial pressures (from Step 4) and the value of x, we can calculate the new partial pressures: \[ P_{N_2O_4} = 0.34\,\text{atm} - 0.32\,\text{atm} = 0.02\,\text{atm} \] \[ P_{NO_2} = 1.20\,\text{atm} + 0.32\,\text{atm} = 1.52\,\text{atm} \] Therefore, the new equilibrium partial pressures are 0.02 atm for N2O4 and 1.52 atm for NO2.

Key concepts

Partial Pressure

Partial pressure refers to the pressure exerted by a single component within a mixture of gases. It's a crucial concept in chemical reactions, especially those involving gases. When you have an equilibrium system with multiple gases, each gas contributes to the total pressure in the container. The partial pressure is simply the pressure that a gas would exert if it alone occupied the entire volume of the container.
For example, in the given problem, \( P_{\mathrm{N}_2\mathrm{O}_4} = 0.34 \text{ atm} \) and \( P_{\mathrm{NO}_2} = 1.20 \text{ atm} \), represent the partial pressures of \( \mathrm{N}_2\mathrm{O}_4 \) and \( \mathrm{NO}_2 \), respectively.
It is important to remember that when the volume of the container changes, the partial pressure of each gas changes inversely, assuming the temperature is constant. If the volume is doubled, as in this exercise, each gas's partial pressure is halved. Understanding this concept is vital to solving equilibrium problems involving gases.

Reaction Quotient

The reaction quotient, represented as \( Q_p \) for systems involving partial pressures, helps us understand the current state of a reaction mixture in relation to equilibrium. By calculating \( Q_p \), we can determine if a reaction will proceed forward, reverse, or remain unchanged when they are at equilibrium.
The expression for \( Q_p \) incorporates the partial pressures of reactants and products, each raised to their coefficients from the balanced chemical equation: \[ Q_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \]
This equation shows that \( Q_p \) is a snapshot of the ratio of the pressures.

• If \( Q_p \) is less than \( K_p \), the reaction will shift right, producing more products.
• If \( Q_p \) is greater than \( K_p \), the reaction will shift left, producing more reactants.
• When \( Q_p = K_p \), the system is at equilibrium.

Equilibrium Constant

An equilibrium constant, \( K_p \) for partial pressures, quantifies the balance of a chemical reaction at equilibrium. It's an important parameter that indicates the relative concentrations or pressures of products and reactants at equilibrium. By definition, a large \( K_p \) value suggests that products are favored, while a small one indicates that reactants are more prevalent at equilibrium.
In our example, knowing that \( K_p = 4.24 \), it tells us about the balance of \( \mathrm{N}_2\mathrm{O}_4} \) and \( \mathrm{NO}_2 \) pressures when the system reaches equilibrium.
It's important to understand that \( K_p \) remains constant at a given temperature. Even when the volume changes, as in this exercise, \( K_p \) remains the same, allowing us to find new equilibrium pressures by comparing the initial \( Q_p \) and the given \( K_p \) when the system changes conditions.

Balanced Chemical Equation

A balanced chemical equation is an essential tool for understanding how reactants convert into products, maintaining the law of conservation of mass. It directly influences the calculation of both the reaction quotient and the equilibrium constant.
For the reaction of \( \mathrm{N}_2\mathrm{O}_4 \rightleftarrows 2\mathrm{NO}_2 \), the balanced equation tells us a crucial piece of information: for every molecule of \( \mathrm{N}_2\mathrm{O}_4 \) that reacts, two molecules of \( \mathrm{NO}_2 \) are produced.
This stoichiometric relationship (the coefficients) is used when expressing \( Q_p \) and \( K_p \), where products and reactants are raised to the power corresponding to their coefficients in the balanced equation: \[ Q_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \] Understanding and writing balanced equations are fundamental skills, as they serve as the framework for analyzing and predicting the outcomes of chemical reactions under different conditions.