Question

In the gas phase, nitrogen dioxide is actually a mixture of nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right) .\) If the density of such a mixture is \(2.3 \mathrm{~g} / \mathrm{L}\) at \(74^{\circ} \mathrm{C}\) and 1.3 atm, calculate the partial pressures of the gases and \(K_{P}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\).

Short answer

The partial pressures are 0.337 atm ( \(\mathrm{N}_2\mathrm{O}_4\right) ) and 0.963 atm ( \(\mathrm{NO}_2\right) ) with \(K_P\right) \approx 2.75 .

Step-by-step solution

Understanding the Dissociation Reaction

The reaction in this problem is the dissociation of dinitrogen tetroxide (\(\mathrm{N}_2\mathrm{O}_4\right)\) into nitrogen dioxide (\(\mathrm{NO}_2\right)\) in the gas phase:\[\mathrm{N}_2\mathrm{O}_4\right\rightleftharpoons 2\mathrm{NO}_2\right\]We need to relate the partial pressures of \(\mathrm{N}_2\mathrm{O}_4\right\) and \(\mathrm{NO}_2\right\) to find the equilibrium constant \(K_P\).

Calculate the Average Molar Mass

Use the ideal gas law to calculate the average molar mass of the gas mixture. The gas density formula is:\[\text{Density} = \frac{PM}{RT}\]where \(P\) is pressure (1.3 atm), \(M\) is molar mass, \(R\) is the ideal gas constant (0.0821 L atm/mol K), and \(T\) is temperature in Kelvin (\(74^\circ\text{C} = 347\ ext{K})\). Calculate \(M:\)\[M = \frac{\text{Density} \times R \times T}{P} = \frac{2.3 \times 0.0821 \times 347}{1.3} \approx 57.95 \text{ g/mol}\]

Determine the Composition of the Mixture

Set up a system of equations based on the molar masses:- \(M_{\mathrm{N}_2\mathrm{O}_4} = 92.02 \text{ g/mol}\)- \(M_{\mathrm{NO}_2} = 46.01 \text{ g/mol}\)Let \(x\) be the mole fraction of \(\mathrm{N}_2\mathrm{O}_4\) and \(1-x\) be the mole fraction of \(\mathrm{NO}_2\). The average molar mass can be expressed as:\[M = x(92.02) + (1-x)(46.01)\]Using \(M = 57.95 \text{ g/mol}\), solve for \(x\):\[57.95 = 92.02x + 46.01(1-x)\]\[57.95 = 92.02x + 46.01 - 46.01x\]\[57.95 - 46.01 = 46.01x\]\[11.94 = 46.01x\]\[x = 0.2594\]

Calculate Partial Pressures

The total pressure is the sum of the partial pressures:- \(P_{\mathrm{N}_2\mathrm{O}_4} = x \times P_{\text{total}} = 0.2594 \times 1.3 \approx 0.337 \text{ atm}\)- \(P_{\mathrm{NO}_2} = (1-x) \times P_{\text{total}} = 0.7406 \times 1.3 \approx 0.963 \text{ atm}\)

Calculate the Equilibrium Constant \(K_P\)

For the reaction \(\mathrm{N}_2\mathrm{O}_4\rightleftharpoons 2\mathrm{NO}_2\right\) , the equilibrium expression is:\[K_P = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}}\]Substitute the partial pressures:\[K_P = \frac{(0.963)^2}{0.337} \approx 2.75\]

Key concepts

Dissociation Reaction

In the context of chemical equilibrium, a dissociation reaction involves breaking down a compound into simpler substances. A key example in our problem is the dissociation of dinitrogen tetroxide (\[\text{N}_2\text{O}_4\rightleftharpoons 2\text{NO}_2\]). Here, \(\text{N}_2\text{O}_4\) dissociates into two \(\text{NO}_2\) molecules. During this process, the original compound is in equilibrium with its products. For equilibrium, the rate of the forward reaction (dissociation) equals the rate of the reverse reaction (recombination).
This balance determines the concentration of reactants and products at equilibrium. Understanding dissociation reactions is crucial for calculating equilibrium constants, like \(K_P\), which reflect how favorably a reaction proceeds under certain conditions. These reactions play significant roles in fields like atmospheric chemistry and industrial syntheses.

Partial Pressure Calculation

Partial pressure refers to the pressure exerted by a single component in a mixture of gases. In a non-reacting gas mixture, it's essential to understand how each gas contributes to overall pressure. To find the partial pressures of \(\text{N}_2\text{O}_4\) and \(\text{NO}_2\), we use mole fractions and total pressure.
The total pressure equation is:

• \(P_{\text{total}} = P_{\text{N}_2\text{O}_4} + P_{\text{NO}_2}\)

Mole fraction (\(x\)) represents the ratio of moles of each component:

• \(P_{\text{N}_2\text{O}_4} = x \times P_{\text{total}}\)
• \(P_{\text{NO}_2} = (1-x) \times P_{\text{total}}\)

By using these equations, we better understand how gas mixtures behave and correlate them with chemical equilibria.

Equilibrium Constant Calculation

The equilibrium constant (\(K_P\)) quantifies the ratio of concentrations or pressures of products to reactants at equilibrium. For gaseous reactions, \(K_P\) involves partial pressures:\[K_P = \frac{(P_{\text{NO}_2}^2)}{P_{\text{N}_2\text{O}_4}}\]This formula arises because two \(\text{NO}_2\) molecules form from each \(\text{N}_2\text{O}_4\) molecule.
High \(K_P\) values indicate that at equilibrium, products are favored, as seen in our example. Understanding \(K_P\) helps predict the direction of chemical reactions when conditions change, assisting in applications ranging from industrial to laboratory settings.
By mastering \(K_P\) calculations, students gain insights into how molecular interactions dictate macroscopic chemical behaviors.

Ideal Gas Law

The Ideal Gas Law is vital for calculating various parameters of gases under specific conditions. It's expressed as:\[PV = nRT\]Here, \(P\) is pressure, \(V\) is volume, \(n\) is moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
In our problem, the Ideal Gas Law aids in estimating the average molar mass of the gas mixture using:\[\text{Density} = \frac{PM}{RT}\]By transforming this equation based on known density, pressure, and temperature, we calculated the molar mass, essential for analyzing gas mixtures. This law is fundamental in both theoretical and practical chemistry for evaluating conditions of gases in different systems.