Question

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Short answer

The equilibrium partial pressure of \(\mathrm{N}_{2}\mathrm{O}_{4}\) is 1.256 atm. The value of \(K_p\) for the reaction is 0.209, and the value of \(K_c\) for the reaction is 0.00850.

Step-by-step solution

Determine the change in pressure for both N2O4 and NO2

From the initial pressures, we know that the pressure of NO2 has increased from 1.00 atm to 0.512 atm. Since the stoichiometry of the reaction is 1:2 between N2O4 and NO2, for every mole of N2O4 that decomposes, two moles of NO2 are formed. Thus, we can find the change in pressure for both N2O4 and NO2 as follows: Change in pressure of NO2 = Final pressure of NO2 - Initial pressure of NO2 = 0.512 atm - 1.00 atm = -0.488 atm Change in pressure of N2O4 = 1/2 * Change in pressure of NO2 = 1/2 * (-0.488 atm) = -0.244 atm

Calculate the equilibrium partial pressure of N2O4

Now we can find the equilibrium partial pressure of N2O4 by adding the initial partial pressure and the change in pressure of N2O4: Equilibrium partial pressure of N2O4 = Initial pressure of N2O4 + Change in pressure of N2O4 = 1.500 atm - 0.244 atm = 1.256 atm

Calculate the value of Kp for the reaction

The equilibrium constant expression for the given reaction in terms of partial pressures (Kp) is: \[K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}\] Using the equilibrium partial pressures calculated in Steps 1 and 2: \[K_p = \frac{(0.512 \, \text{atm})^2}{1.256 \, \text{atm}} = 0.209\] The value of Kp for the reaction is 0.209.

Calculate Kc for the reaction

To calculate Kc, we need to relate Kp to Kc using the ideal gas law: \[K_p = K_c(RT)^{\Delta n}\] where R is the gas constant (0.0821 L atm/mol K), T is the temperature in Kelvin (25°C = 298 K), and Δn is the change in the number of moles of gas (2 moles of NO2 - 1 mole of N2O4 = 1). Plugging in the values: \[0.209 = K_c(0.0821 \frac{\text{L atm}}{\text{mol K}})(298 \, K)^1\] Now, solve for Kc: \[K_c = \frac{0.209}{(0.0821)(298)} = 0.00850\] The value of Kc for the reaction is 0.00850.

Key concepts

Partial Pressure

Partial pressure refers to the pressure that an individual gas in a mixture of gases would exert if it were the only gas present in the container. In a chemical reaction involving gases, each gas contributes to the total pressure of the system.
For example, in the given reaction of \[\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2},\]the partial pressures of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) add up to the total pressure in the reaction flask.
Calculating partial pressures allows us to understand how each gas behaves individually, which is crucial for finding the equilibrium constant \(K_p\). Partial pressures are a key aspect of gas reactions because they help in quantifying the changes happening during the reaction's shift towards equilibrium.

Le Chatelier's Principle

Le Chatelier's principle is a fundamental rule used to predict the effect of a change in conditions on chemical equilibria. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
For instance, in our reaction, if the pressure of the system changes, the reaction will shift to adjust the amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) to re-establish equilibrium.
A practical implication for gaseous reactions is that increasing the pressure tends to favor the side of the equilibrium with fewer moles of gas because it reduces the system's volume. This principle provides a way to control the yield of a reaction by changing pressure, temperature, or concentration.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure of the relative amounts of reactants and products present during a reaction at a given point in time. It uses the same expression as the equilibrium constant but with the current partial pressures or concentrations.
For the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), the reaction quotient \(Q_p\) is given by:\[Q_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}\]When \(Q\) is compared to the equilibrium constant \(K\), we can determine the direction in which the reaction will proceed:

• If \(Q < K\), the reaction will move forward, producing more products.
• If \(Q > K\), the reaction will shift backward, forming more reactants.
• If \(Q = K\), the system is at equilibrium.

This concept is crucial for predicting how a reaction will respond to changes and helps us understand how far a reaction is from reaching equilibrium.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation relating the pressure, volume, temperature, and number of moles of a gas. It is expressed by the equation:\[PV = nRT\]where:

• \(P\) is pressure,
• \(V\) is volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (0.0821 L atm/mol K), and
• \(T\) is temperature in Kelvin.

In the context of equilibrium calculations, the ideal gas law helps convert between \(K_p\) and \(K_c\), the equilibrium constants expressed in terms of partial pressures and concentrations, respectively. It plays a crucial role when adjusting equations for conditions like temperature or when converting between different forms of equilibrium expressions. Understanding this law is essential for analyzing gaseous reactions and predicting their behavior under various conditions.