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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.

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.

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.

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.