Problem 8 - Biomass As A Sustainable Energy Source For The Future

Question

A container with a capacity of \(1 \mathrm{~L}\) holds \(3 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at \(343 \mathrm{~K}\). At this temperature, the dissociation grade is \(65 \%\). Find \(K_{p}\) for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftarrows\) \(2 \mathrm{NO}_{2}\).

Step-by-Step Solution

Answer

Based on the given information and calculations, the equilibrium constant Kp for the reaction N2O4 ⇌ 2NO2 at the given conditions is 410.48.

Step 1: Determine the initial moles of N2O4 and NO2

Initially, the container has 100% N2O4, no dissociation has occurred yet. Therefore, the initial moles of N2O4 are 3 moles and the initial moles of NO2 are 0. N2O4: 3 moles NO2: 0 moles

Step 2: Determine the change in moles

Once the dissociation occurs, 65% of the N2O4 turns into NO2. For every mole of N2O4 that dissociates, we get 2 moles of NO2. We calculate the change in moles for both species as follows: Change in N2O4 moles: - 65% * 3 Change in NO2 moles: 2 * 65% * 3

Step 3: Determine the equilibrium moles

By adding the initial moles and the changes, we can find the equilibrium moles for both species: N2O4 equilibrium moles: 3 - 65% * 3 = 1.05 moles NO2 equilibrium moles: 0 + 2 * 65% * 3 = 3.9 moles

Step 4: Calculate partial pressures

Now that we have the equilibrium moles of the reactant and product, we can calculate the partial pressure for each species using the ideal gas law (\(PV=nRT\)). In this case, we know that the volume is 1 L and the temperature is 343 K. We can use the gas constant R = 0.0821 L atm / (mol K) to keep the units consistent. The ideal gas law has to be rearranged to calculate the partial pressures: P_N2O4 = (n_N2O4 × R × T) / V = (1.05 moles × 0.0821 atm L/(mol K) × 343 K) / 1 L = 29.56 atm P_NO2 = (n_NO2 × R × T) / V = (3.9 moles × 0.0821 atm L/(mol K) × 343 K) / 1 L = 110.45 atm

Step 5: Calculate Kp

Now that we have the partial pressures for both species, we can find the equilibrium constant Kp using the balanced chemical equation: \(K_p = \frac{P_{NO_2}^2}{P_{N_2O_4}}\) Substituting the partial pressures, we get: Kp = (110.45 atm)^2 / (29.56 atm) = 410.48 The equilibrium constant Kp for the given reaction is 410.48.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates pressure, volume, temperature, and the number of moles of a gas. It is given by the equation \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. In practice, this law helps us estimate the behavior of gases under different conditions. For example, when calculating how the pressure changes with a new temperature, the ideal gas law provides a basis for understanding these interactions.

• \(P\) = Pressure
• \(V\) = Volume
• \(n\) = Number of moles
• \(R\) = Ideal gas constant (0.0821 L·atm/mol·K)
• \(T\) = Temperature in Kelvin

For a dissociation reaction in a container, the ideal gas law is used to determine the partial pressures of the gases at equilibrium based on the moles of gas and the conditions given.

Partial Pressures

Partial pressure refers to the pressure contributed by a single type of gas in a mixture of gases. In a chemical reaction, each gas present in the mixture exerts its own pressure, called the partial pressure. The total pressure in the system is the sum of the partial pressures of all gases present.In the context of the provided reaction \( \text{N}_2\text{O}_4 \rightleftharpoons 2 \text{NO}_2 \), we determine the partial pressures using the ideal gas law. Given the equilibrium moles of each species and the conditions of volume and temperature, we calculate:

• Partial pressure of \(\text{N}_2\text{O}_4\) as \(29.56\) atm
• Partial pressure of \(\text{NO}_2\) as \(110.45\) atm

Understanding partial pressures is essential because they allow us to determine how each gas in a mixture behaves, especially when used to find the equilibrium constant \(K_p\).

Equilibrium Constant

The equilibrium constant, denoted as \(K_p\) for reactions involving gases, quantifies the ratio of product pressures to reactant pressures at equilibrium. For a balanced reaction, it accounts for the stoichiometry of reactants and products. It is expressed as:\[K_p = \frac{(P_{products})^{\text{coefficients}}}{(P_{reactants})^{\text{coefficients}}}\]In the reaction provided:\( \text{N}_2\text{O}_4 \rightleftharpoons 2 \text{NO}_2 \), the expression is:\[K_p = \frac{(P_{NO_2}^2)}{(P_{N_2O_4})}\]By substituting the equilibrium partial pressures:

• \(\text{NO}_2\): \(110.45\) atm
• \(\text{N}_2\text{O}_4\): \(29.56\) atm

We find \(K_p\) to be approximately \(410.48\), showing favor towards the products at equilibrium. Knowing \(K_p\) helps predict the direction and extent of the reaction.

Dissociation Reactions

Dissociation reactions involve the breaking apart of a compound into two or more components. In the reaction \( \text{N}_2\text{O}_4 \rightleftharpoons 2 \text{NO}_2 \), the compound \(\text{N}_2\text{O}_4\) dissociates into two molecules of \(\text{NO}_2\). Understanding the dissociation process requires acknowledging key terms like:

• Initial Moles: Amount of compound before dissociation begins.
• Dissociation Degree: The percentage of the initial compound that dissociates, here, 65%.
• Equilibrium Moles: The remaining moles of the original compound and the moles of the products after equilibrium is reached.

Dissociation reactions are important because they often influence the final composition of the chemical system. By calculating both the dissociation and equilibrium states, it becomes easier to apply related concepts such as the ideal gas law and partial pressures.