Question

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

Short answer

\(x = 1.5 \) 6. Calculate the equilibrium partial pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 4.5 - 1.5 = 3.0\ \mathrm{atm}\) \(P_{\mathrm{NO}_{2}} = 2 (1.5) = 3.0\ \mathrm{atm}\) #tag_title# Part B: Calculate equilibrium partial pressures when starting with NO2 #tag_content# 1. Let the change in pressure of \(NO_{2}\) be y at equilibrium. 2. Since two moles of \(NO_{2}\) yields one mole of \(N_{2}O_{4}\), the change in pressure of \(N_{2}O_{4}\) is y/2. 3. Write the expressions for the equilibrium pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = \frac{y}{2}\) \(P_{\mathrm{NO}_{2}} = 9.0-y\) 4. Substitute the equilibrium pressure values into the Kp expression: \(0.25 = \frac{(9.0-y)^{2}}{\frac{y}{2}}\) 5. Solve for y: \(y = 6.0\) 6. Calculate the equilibrium partial pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = \frac{6.0}{2} = 3.0\ \mathrm{atm}\) \(P_{\mathrm{NO}_{2}} = 9.0 - 6.0 = 3.0\ \mathrm{atm}\) #tag_title# Part C: Discuss the dependence of the equilibrium position on the initial conditions #tag_content# In both cases, the equilibrium partial pressures are the same: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 3.0\ \mathrm{atm}\) and \(P_{\mathrm{NO}_{2}} = 3.0\ \mathrm{atm}\). This indicates that, for this reaction, the equilibrium position does not depend on the initial conditions or the direction from which the equilibrium is approached.

Step-by-step solution

Write the Kp Expression for the Reaction

We're given the reaction: \[ N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g) \] The Kp expression for this reaction will look like: \[ K_{p} = \frac{(P_{\mathrm{NO}_{2}})^{2}}{P_{\mathrm{N}_{2}\mathrm{O}_{4}}} \] Where, \(P_{\mathrm{NO}_{2}}\) and \(P_{\mathrm{N}_{2}\mathrm{O}_{4}}\) represent the equilibrium partial pressures of the corresponding gases. In this problem, we are given that \(K_{\mathrm{p}}=0.25\).

Part A: Calculate equilibrium partial pressures when starting with N2O4

1. Let the change in pressure of \(N_{2}O_{4}\) be x at equilibrium. 2. Since one mole of \(N_{2}O_{4}\) yields two moles of \(NO_{2}\), the change in pressure of \(NO_{2}\) is 2x. 3. Write the expressions for the equilibrium pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 4.5-x\) \(P_{\mathrm{NO}_{2}} = 2x\) 4. Substitute the equilibrium pressure values into the Kp expression: \(0.25 = \frac{(2x)^{2}}{(4.5-x)}\) 5. Solve for x:

Key concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), is a unique value that quantifies the ratio of the concentrations of the products to the reactants at equilibrium for a gaseous reaction, with each being raised to the power of their stoichiometric coefficients.
In the context of the reaction \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \), \(K_p\) is expressed using the partial pressures of the gases involved:

• \(K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N_2O_4}}}\)

This formula indicates how the partial pressure of \(NO_2\) at equilibrium is squared due to its coefficient of 2 in the balanced chemical equation.
The constant \(K_p\) is only affected by changes in temperature. It does not change with variations in pressure or concentrations, as long as the system is at equilibrium.
Understanding \(K_p\) is crucial because it allows us to predict the extent of a reaction's progression and to determine whether more products or reactants are present at equilibrium.

Partial Pressures

Partial pressures are vital in reactions involving gases, as each gas contributes to the total pressure of a system.
According to Dalton's law, the total pressure is the sum of the partial pressures of all the gases present. For the equilibrium reaction \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \), calculating the equilibrium partial pressures involves understanding how molecular shifts affect system pressure.
Let's look at scenario (a), where only \(N_2O_4\) is initially present at 4.5 atm.

• The partial pressure of \(N_2O_4\) will decrease by an amount \(x\) as it forms \(NO_2\).
• Consequently, \(NO_2\)'s partial pressure increases by \(2x\), due to the stoichiometry of the reaction (two moles of \(NO_2\) form for every mole of \(N_2O_4\) consumed).

Understanding partial pressures helps us track the changes during reactions and calculate the final equilibrium conditions.
These pressures are key to applying \(K_p\) and solving for equilibrium states.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept that predicts how a system at equilibrium reacts to external changes. It states that if a change is made to a system at equilibrium, the system adjusts in a direction that counteracts the change.
In gaseous reactions like \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \), changes can include altering the concentration of reactants or products, pressure, or temperature.
For example, increasing the partial pressure of \(N_2O_4\) will shift the equilibrium toward the right, producing more \(NO_2\), as the system works to reduce the added pressure.

• If \(NO_2\) pressure is initially high, the system will shift leftward to form more \(N_2O_4\), counteracting the added gases.

Le Chatelier's Principle is especially helpful in industrial processes where optimizing product yield is crucial.
It provides insights into how we can shift the balance of reactions to favor the formation of products or reactants as desired.