Question

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is 1.5 atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=\) \(1.00 \mathrm{atm}) ?\)

Short answer

In short, the following results are obtained: a. \(K_{p} = 0.212\) at \(25^{\circ} C\) b. At the new equilibrium position (total pressure = 1.0 atm), the partial pressures of N2O4(g) and NO2(g) are \(P'_{N2O4} = 0.4 \ \text{atm}\) and \(P'_{NO2} = 0.6 \ \text{atm}\), respectively. c. At the new equilibrium position, 33.3% of the original N2O4(g) has dissociated.

Step-by-step solution

Calculate the initial moles of N2O4(g)

We are given that 16% (by moles) of N2O4(g) has dissociated. Since the initial moles of NO2(g) are zero, we can set up a simple proportion to calculate the initial moles of N2O4(g) before dissociation: initial moles of N2O4(g) = \(\frac{100-16}{16}\) × moles of NO2(g) We know that the total pressure of the system at equilibrium is 1.5 atm, and we can calculate the partial pressures of N2O4(g) and NO2(g) at equilibrium.

Calculate the partial pressures of N2O4(g) and NO2(g) at equilibrium

Let the partial pressure of NO2(g) be P_NO2 and the partial pressure of N2O4(g) be P_N2O4: P_NO2 = 2P_N2O4 (since 1 mole of N2O4 dissociates into 2 moles of NO2) The total pressure at equilibrium is given by: P_total = P_N2O4 + P_NO2 = 1.5 atm Now we have two equations with two unknowns: 1. P_NO2 = 2P_N2O4 2. P_N2O4 + P_NO2 = 1.5 atm We can solve for P_N2O4 and P_NO2.

Calculate the value of Kp

The equilibrium constant Kp in terms of partial pressures is given by: \[K_p = \frac{P_{NO2}^{2}}{P_{N2O4}}\] Using the values for P_N2O4 and P_NO2 calculated in Step 2, we can now calculate Kp at 25°C. #b. Calculate the equilibrium pressure of N2O4(g) and NO2(g) when the total pressure is 1.0 atm#

Calculate the new equilibrium pressures

When the volume of the cylinder is increased, the total pressure decreases to 1.0 atm. We can use the Kp value we calculated in Step 3 along with the new total pressure to calculate the new equilibrium pressures of N2O4(g) and NO2(g). Let the partial pressure of NO2(g) be P'_NO2 and the partial pressure of N2O4(g) be P'_N2O4: P'_NO2 = 2P'_N2O4 The new total pressure is given by: P'_total = P'_N2O4 + P'_NO2 = 1.0 atm Now we have two equations with two unknowns: 1. P'_NO2 = 2P'_N2O4 2. P'_N2O4 + P'_NO2 = 1.0 atm We can solve for P'_N2O4 and P'_NO2. #c. Calculate the percentage of dissociated N2O4(g) at the new equilibrium position#

Calculate the new percentage of dissociated N2O4(g)

Having found the new equilibrium pressures of N2O4(g) and NO2(g) in Step 4, we can calculate the new percentage of dissociated N2O4(g) at the equilibrium position with a total pressure of 1.00 atm: New percentage of N2O4 dissociated (by moles) = \(\frac{P'_{NO2}}{2 \times P'_{N2O4}} \times 100\) Using the values of P'_N2O4 and P'_NO2 calculated in Step 4, we can determine the percentage of dissociated N2O4(g) at the new equilibrium position.

Key concepts

Dissociation Reaction

In chemistry, a dissociation reaction is a process in which a compound is broken down into smaller molecules, atoms, or ions. For the exercise you are working on, the compound in question is dinitrogen tetroxide (\(\text{N}_2\text{O}_4\)), which dissociates into nitrogen dioxide (\(\text{NO}_2\)). This reaction is represented by:
\[\text{N}_2\text{O}_4 (g) \rightleftharpoons 2\text{NO}_2 (g)\]
Here, each molecule of \( \text{N}_2\text{O}_4\) produces two molecules of \(\text{NO}_2\). The double arrow indicates that the reaction is reversible and can reach equilibrium, where the forward and reverse reactions occur at the same rate, leading to constant concentrations of reactants and products.

• It's important to understand that a change in conditions such as pressure or temperature can shift the equilibrium, leading to more or fewer products being formed.
• In this exercise, 16% of the original \(\text{N}_2\text{O}_4\) initially dissociates, meaning that a portion of it is converted to \(\text{NO}_2\), while the rest remains as \(\text{N}_2\text{O}_4\).
• The percentage dissociation tells us how much of the \(\text{N}_2\text{O}_4\) shifted to the right side of the equation to form \(\text{NO}_2\).

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), is a special number that helps us understand the proportion of products to reactants in a chemical reaction at equilibrium. For gas-phase reactions like ours, \(K_p\) is calculated in terms of partial pressures.
For the reaction \(\text{N}_2\text{O}_4\ (g) \rightleftharpoons 2\text{NO}_2\ (g)\), the relationship is defined as:
\[K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}\]
where:

• \(P_{\text{NO}_2}\) is the partial pressure of nitrogen dioxide, and
• \(P_{\text{N}_2\text{O}_4}\) is the partial pressure of dinitrogen tetroxide.

A high \(K_p\) value suggests that at equilibrium, the reaction favors the formation of products, while a low \(K_p\) indicates a preference for reactants. In our task, once we determine the equilibrium pressures in the initial scenario, we can input those values into the \(K_p\) expression to calculate this constant at 25°C.
Calculating and understanding the \(K_p\) value gives us insight into how changes in conditions might affect the reaction.

Partial Pressure Calculation

Understanding partial pressures is crucial in solving gas-phase equilibria problems. The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied the entire volume alone.
In our dissociation reaction where \(\text{N}_2\text{O}_4\ (g) \rightleftharpoons 2\text{NO}_2 (g)\), calculating partial pressures help us find out how gases split up their domination of the pressure in the system.
For the given equilibrium problem:

• The expression \(P_{\text{NO}_2} = 2P_{\text{N}_2\text{O}_4}\) arises since each molecule of dinitrogen tetroxide yields two nitrogen dioxide molecules.
• The total pressure \(P_{\text{total}}\) given is the sum of both gas partial pressures: \(P_{\text{N}_2\text{O}_4} + P_{\text{NO}_2} = 1.5 \text{ atm}\).

By solving these equations, we can find the specific partial pressures of each gas.
When conditions change, like a reduction in total pressure to 1.0 atm, we redo these calculations using the fixed \(K_p\) value to look at new equilibriums, revealing how the gases shift again. Calculations like these give a great insight into the behavior of gases in equilibrium, applying the concept to practical chemical scenarios.