Problem 38 - Chemistry: The Central Science in SI Units, Global Edition

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

The equilibrium partial pressure of N2O4 is 176.7 kPa. The value of Kp for the reaction is approximately 15.19, and the value of Kc for the reaction is approximately 0.61.

Step 1: Write down the given initial partial pressures

The initial partial pressures of the gases are: N2O4(g): 152.0 kPa NO2(g): 101.3 kPa

Step 2: Calculate the change in partial pressures at equilibrium

At equilibrium, the partial pressure of NO2 is 51.9 kPa. Therefore, the change in the partial pressure of NO2 is: ΔP_NO2 = final P_NO2 - initial P_NO2 = 51.9 kPa - 101.3 kPa = -49.4 kPa Since the stoichiometry of the reaction is 1 N2O4(g) to 2 NO2(g), the change in the partial pressure of N2O4 is half the change in the partial pressure of NO2: ΔP_N2O4 =1/2 * ΔP_NO2 = -49.4 kPa * (1/2) = 24.7 kPa

Step 3: Calculate the equilibrium partial pressure of N2O4 and NO2

Equilibrium partial pressures: P_N2O4(eq) = initial P_N2O4 + ΔP_N2O4 = 152.0 kPa + 24.7 kPa = 176.7 kPa P_NO2(eq) = initial P_NO2 + ΔP_NO2 = 101.3 kPa - 49.4 kPa = 51.9 kPa

Step 4: Calculate Kp for the reaction

Now that we have the equilibrium partial pressures, we can calculate the Kp of the reaction: Kp = (P_NO2(eq))^2 / P_N2O4(eq) = (51.9 kPa)² / 176.7 kPa ≈ 15.19

Step 5: Calculate Kc for the reaction

To calculate Kc, we must relate Kp to Kc using the ideal gas law and the stoichiometry of the reaction: Kp = Kc * (RT)^(Δn) Where R is the ideal gas constant (R = 0.0821 L atm/mol K), T is the temperature in Kelvin, and Δn is the difference in moles of gaseous products and reactants in balanced chemical equation. For this reaction, T = 25°C = 298 K and Δn = 2 - 1 = 1 Kc = Kp / (RT)^(Δn) = 15.19 / ((0.0821 L atm/mol K) * (298 K))^(1) ≈ 0.61 Kc for the reaction is approximately 0.61 (rounded to two decimal places).

Key Concepts

Equilibrium Constant (Kp)

In chemical equilibrium, the Equilibrium Constant denoted as \(K_p\) is crucial when dealing with gases. It links the partial pressures of reactants and products at equilibrium. To find \(K_p\), first determine the equilibrium partial pressures of the involved gases. In the given exercise, it’s established from the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\).
In this example, the equilibrium partial pressure for \(\mathrm{NO}_{2}\) is 51.9 kPa and for \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 176.7 kPa. The formula for \(K_p\) is:

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

Substitute the known values and solve it to find \(K_p = 15.19\). This reveals the equilibrium ratio of pressures. The squared term is due to the stoichiometric coefficient of \(\mathrm{NO}_2\), which is 2. Always remember that \(K_p\) depends on the reaction temperature as well.

Equilibrium Constant (Kc)

The Equilibrium Constant in terms of concentration, \(K_c\), is useful for reactions in solution but applicable in gas reactions through conversion. Knowing \(K_p\), you can find \(K_c\) using the formula:

• \[ K_p = K_c \times (RT)^{\Delta n} \]

Here, \(R\) is the ideal gas constant (0.0821 L atm/mol K), \(T\) is the temperature in Kelvin, and \(\Delta n\) represents the change in moles of gas (products minus reactants). For the given reaction, \(\Delta n = 1\) because 2 moles of \(\mathrm{NO}_2\) come from 1 mole of \(\mathrm{N}_2 \mathrm{O}_4\).
Substitute the values into the equation, you find \(K_c\) to be approximately 0.61, showing how interactions are measured in terms of concentration vs. pressure. This reflects the same equilibrium state in a different unit.

Partial Pressure

Partial pressure is the individual pressure exerted by a gas in a mixture of gases. When gases are mixed, each gas contributes to the total pressure based on its proportion. In equilibrium contexts, partial pressures are key to calculating \(K_p\).
In the exercise, partial pressure of \(\mathrm{N}_2 \mathrm{O}_4\) and \(\mathrm{NO}_2\) was critical. Initially, \(\mathrm{N}_2 \mathrm{O}_4\) had 152.0 kPa, and \(\mathrm{NO}_2\) had 101.3 kPa. Understanding changes in these pressures helps in solving equilibrium problems. Upon reaching equilibrium, pressures shift according to stoichiometry. This type of calculation reveals how much of each gas is present in the balance, influencing the reaction's direction and rate.

Stoichiometry

Stoichiometry involves the quantitative relationships of reactants and products in a chemical reaction. It's the backbone for determining how species change during reactions. In equilibrium problems, stoichiometry helps in converting changes in concentration or pressure into a useful form, such as calculating equilibrium pressures.
Within the equilibrium reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), the stoichiometric ratio is 1:2. This means for every mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociating, 2 moles of \(\mathrm{NO}_{2}\) are produced. The exercise used these ratios to adjust the partial pressures when equilibrium was reached. For instance, knowing the change in \(\mathrm{NO}_{2}\) helped to find the shift in \(\mathrm{N}_{2} \mathrm{O}_{4}\), illustrating how stoichiometry directly affects equilibrium computations. Always consider stoichiometric coefficients for accurate calculations in any chemical reaction.