Problem 38 - Chemistry The Central Science In Si Units

Question

A flask is charged with \(152.0 \mathrm{kPa}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(101.3 \mathrm{kPa}\) \(\mathrm{NO}_{2}(g)\) 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 \(51.9 \mathrm{kPa}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?(\mathbf{b})\) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Step-by-Step Solution

Answer

(a) 127.3 kPa; (b) \( K_p \approx 21.2 \); (c) \( K_c \approx 0.861 \).

Step 1: Identify Initial Pressures

Initially, we have the pressures of \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \) as \( 152.0 \ \mathrm{kPa} \) and \( \mathrm{NO}_{2}(g) \) as \( 101.3 \ \mathrm{kPa} \).

Step 2: Determine Change in Pressure

At equilibrium, the partial pressure of \( \mathrm{NO}_{2}(g) \) is \( 51.9 \ \mathrm{kPa} \). Since equilibrium decreases pressure from dissociation, the change in \( \mathrm{NO}_{2}(g) \) is \( 101.3 - 51.9 = 49.4 \ \mathrm{kPa} \).

Step 3: Calculate Change in \( \mathrm{N}_{2} \mathrm{O}_{4} \) Pressure

Since \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g) \), for every mole of \( \mathrm{N}_{2} \mathrm{O}_{4} \) dissociated, two moles of \( \mathrm{NO}_{2} \) are formed. Hence, for a \( 49.4 \ \mathrm{kPa} \) change in \( \mathrm{NO}_{2} \), \( 49.4 / 2 = 24.7 \ \mathrm{kPa} \) of \( \mathrm{N}_{2} \mathrm{O}_{4} \) is lost.

Step 4: Calculate Equilibrium Pressure of \( \mathrm{N}_{2} \mathrm{O}_{4} \)

Subtract the change from the initial pressure: \( 152.0 - 24.7 = 127.3 \ \mathrm{kPa} \). Therefore, the equilibrium partial pressure of \( \mathrm{N}_{2} \mathrm{O}_{4} \) is \( 127.3 \ \mathrm{kPa} \).

Step 5: Calculate \( K_p \)

Utilize the equilibrium expression \[ K_p = \frac{{P_{\mathrm{NO}_2}^2}}{{P_{\mathrm{N}_2\mathrm{O}_4}}} \]Substitute \[ K_p = \frac{{(51.9)^2}}{{127.3}} \approx 21.2 \].

Step 6: Find \( K_c \)

Use the relationship \[ K_p = K_c (RT)^{\Delta n} \]. Assume \( R = 0.0821 \ \mathrm{L} \bullet \mathrm{atm} \bullet \mathrm{mol}^{-1} \bullet \mathrm{K}^{-1} \), \( T = 298 \ \mathrm{K} \), \( \Delta n = 1 \). Convert pressures to atm: \[ K_c = \frac{{K_p}}{{(0.0821 \times 298)}} \approx 0.861 \].

Key Concepts

Equilibrium Constant

An equilibrium constant, denoted as either \( K_p \) for gas-phase reactions or \( K_c \) for reactions in solution, quantifies the balance between reactants and products in a chemical reaction at equilibrium. For the reaction \( \mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \), the expression for the equilibrium constant in terms of partial pressure is given by:

• \( K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \)

Here, \( P_{\mathrm{NO}_2} \) is the partial pressure of \( \mathrm{NO}_2 \) and \( P_{\mathrm{N}_2\mathrm{O}_4} \) is the partial pressure of \( \mathrm{N}_2\mathrm{O}_4 \) at equilibrium.
The importance of the equilibrium constant lies in its ability to convey the extent of a reaction. A large \( K \) value means a higher concentration of products, suggesting that the reaction favors products, while a small \( K \) value indicates a reaction favoring reactants.
In the given problem, once equilibrium pressures are determined, we calculate \( K_p \) using the partial pressures. Knowing \( K_p \) allows us to calculate \( K_c \) through the relationship \( K_p = K_c(RT)^{\Delta n} \), where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas.

Partial Pressure

Partial pressure is the individual pressure contributed by each gas in a mixture. In terms of chemical equilibrium, understanding partial pressures is crucial for determining the position of the equilibrium. Each gas in the mixture contributes to the total pressure based on its molar concentration and temperature.
In the problem, we started with initial pressures for \( \mathrm{N}_2\mathrm{O}_4 \) and \( \mathrm{NO}_2 \). At equilibrium, the partial pressure of \( \mathrm{NO}_2 \) changed to \( 51.9 \ \mathrm{kPa} \).
The principle of partial pressures dictates that the sum of individual partial pressures equals the total pressure in the system. For a reaction such as \( \mathrm{N}_2\mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2 \), the change in the partial pressure of \( \mathrm{NO}_2 \) directly affects \( \mathrm{N}_2\mathrm{O}_4 \), with the corresponding reaction stoichiometry determining how one pressure changes relative to the other. By calculating the initial and final pressures, we can deduce unknown pressures at equilibrium.

Reaction Kinetics

Reaction kinetics explores the rates of chemical processes and the factors affecting them, such as concentration, temperature, and catalysts.
In the context of chemical equilibrium, reaction kinetics is essential for understanding how quickly a system reaches equilibrium. Although this exercise primarily focuses on equilibrium, kinetics provide insights like how initial fluctuations in pressure might occur before equilibrium is achieved.
For example, the dissociation of \( \mathrm{N}_2\mathrm{O}_4 \) into \( \mathrm{NO}_2 \) involves kinetic processes:

• The rate of dissociation: how fast \( \mathrm{N}_2\mathrm{O}_4 \) molecules split into \( 2 \mathrm{NO}_2 \).
• The reverse reaction rate: how quickly \( 2 \mathrm{NO}_2 \) molecules recombine into \( \mathrm{N}_2\mathrm{O}_4 \).

The equilibrium point of interest emerges when these rates become equal. At this juncture, the reaction ceases to show macroscopic change, signifying that the dynamic balance of forward and reverse reactions has been achieved, thereby linking concepts of equilibrium with kinetic factors.