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

The value of \(K_p\) for the dissociation reaction under initial conditions is 0.70. After increasing the volume of the cylinder and reaching a new equilibrium at a total pressure of 1.0 atm, the equilibrium pressure of \(N_{2}O_{4}(g)\) is 0.35 atm, and that of \(NO_{2}(g)\) is 0.65 atm. The percentage of original \(N_{2}O_{4}(g)\) dissociated at the new equilibrium position is 43%.

Step-by-step solution

Write the Reaction Equilibrium Expression

The chemical reaction can be written as follows: \( N_2O_4(g) \leftrightarrow 2NO_2(g) \) The equilibrium constant expression (\(K_p\)) for this reaction can be written as: \( K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \), where \(P_{NO_2}\) is the partial pressure of NO2 and \(P_{N_2O_4}\) is the partial pressure of N2O4.

Calculate the Partial Pressures under Initial Conditions

Given 16% of N2O4 dissociates, so in every 100 moles of N2O4 initially present, 16 moles dissociate into 32 moles of NO2. Hence when equilibrium is established: For \(N2O4: P_{N2O_4} = 1.5 \times \frac{100 - 16}{100 + 32}\) atm = 0.765 atm For \(NO2: P_{NO_2} = 1.5 \times \frac{32}{100 + 32}\) atm = 0.735 atm

Calculate the Value of \(K_p\) under Initial Conditions

Substitute \(P_{N2O_4}\) and \(P_{NO_2}\) values into the \(K_p\) equation : \( K_p = \frac{(0.735)^2}{0.765} = 0.70 \)

Calculate the Partial Pressures after Volume Increase

The total pressure is given to have dropped to 1 atm due to the increase in volume. According to Le Chatelier's principle, the reaction will shift in the direction in which more gaseous moles are present. Thus, volume increase will cause more dissociation of N2O4. Let the extra percentage of N2O4 dissociated be x. So, for \(N2O4: P_{N2O_4} = 1 \times \frac{100 - (16 + x)}{100 + 32 + 2x}\) For \(NO2: P_{NO_2} = 1 \times \frac{32 + 2x}{100 + 32 + 2x}\)

Evaluate Percentage of Additional N2O4 Dissociated

Apply \(K_p\) value to the new equilibrium equation: \( 0.70 = \frac{(\frac{32 + 2x}{132 + 2x})^2}{\frac{84 - x}{132 + 2x}}\) Solving this yields x = 27. Therefore, the new pressures when equilibrium is reestablished: \( P_{N2O4} = 0.35 \) atm, and \( P_{NO2} = 0.65 \) atm

Calculate the Percentage of Original N2O4 Dissociated

Percentage dissociated = (Initial moles - Remaining moles) / Initial moles * 100 = [(100 - (100 - (16 + 27))] / 100 * 100 = 43\% of the original N2O4 has dissociated at the new equilibrium position.

Key concepts

Equilibrium Constant (Kp)

The equilibrium constant, represented as \( K_p \), is a crucial concept in understanding chemical equilibrium, especially for reactions involving gases. It describes the ratio of the concentrations of products to reactants at equilibrium. However, instead of concentration, \( K_p \) uses the partial pressures of gases. This is particularly helpful for gas-phase reactions where pressure changes affect equilibrium.
In the given problem, the reaction is \( \text{N}_2\text{O}_4(g) \leftrightarrow 2\text{NO}_2(g) \). Here, \( K_p \) is calculated using:

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

The partial pressures of \( \text{NO}_2 \) and \( \text{N}_2\text{O}_4 \) at equilibrium are plugged into this expression to find \( K_p \). Understanding this process helps in predicting how the equilibrium position shifts under various conditions, such as changes in pressure or volume.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental guideline in predicting the behavior of a system at equilibrium when it encounters a disturbance. It states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change and re-establish equilibrium.
In the problem, specifically when the volume is increased causing the pressure to drop from 1.5 atm to 1 atm, Le Chatelier's Principle suggests that the reaction will shift to produce more molecules to counter the pressure change. Since more gaseous moles are on the product side (\( \text{2NO}_2(g) \)), the dissociation of \( \text{N}_2\text{O}_4 \to \text{NO}_2 \) increases. This shift alters the partial pressures of both \( \text{N}_2\text{O}_4 \) and \( \text{NO}_2 \) and consequently affects the extent of reaction dissociation.

Partial Pressure Calculation

Calculating partial pressures is essential in deriving the equilibrium constant and understanding the behavior of gas reactions. Partial pressure refers to the pressure that each gas in a mixture would exert if it alone occupied the entire volume.
For this exercise, you start with knowing that 16% of \( \text{N}_2\text{O}_4 \) dissociates into \( \text{NO}_2 \) at equilibrium with a total pressure of 1.5 atm. To calculate partial pressures:

• \( P_{\text{N}_2\text{O}_4} = 1.5 \times \frac{100 - 16}{100 + 32} \)
• \( P_{\text{NO}_2} = 1.5 \times \frac{32}{100 + 32} \)

These calculations give \( P_{\text{N}_2\text{O}_4} = 0.765 \) atm and \( P_{\text{NO}_2} = 0.735 \) atm initially. Similar calculations are used after adjusting the volume to find the new partial pressures at 1 atm total pressure. Understanding how to work with partial pressures is central to mastering gas phase equilibria.