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 b, does it matter from which direction an equilibrium position is reached?

Short answer

The equilibrium partial pressures of the gases are the same in both cases, with an equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) as 2.25 atm and an equilibrium partial pressure of \(\mathrm{NO}_{2}\) as 4.5 atm. Hence, it does not matter from which direction the equilibrium position is reached; the equilibrium partial pressures will be the same irrespective of the initial conditions.

Step-by-step solution

Write the expression for \(K_p\)

Given the balanced chemical equation, the expression for \(K_p\) is: $$K_p = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]}$$

Define the variable for equilibrium pressures

Let the partial pressure of N2O4 at equilibrium be x atm. Since 2 moles of NO2 are produced for each mole of N2O4 consumed, the partial pressure of NO2 at equilibrium will be 2(4.5-x) atm.

Write the equation for \(K_p\) at equilibrium

Substitute the partial pressures at equilibrium in the expression for \(K_p\): $$0.25 = \frac{(2(4.5-x))^2}{x}$$

Solve for x

Solve the equation for the value of x: $$0.25x = (2(4.5-x))^2$$ $$x = 2.25$$

Calculate the equilibrium partial pressures

Now that we have the value of x, we can determine the equilibrium partial pressures of both N2O4 and NO2: $$[\mathrm{N}_{2}\mathrm{O}_{4}] = 2.25\,\text{atm}$$ $$[\mathrm{NO}_{2}] = 2 \times (4.5 - 2.25) = 4.5\,\text{atm}$$ b. Calculate the equilibrium partial pressures of N2O4 and NO2 with initial pressure of NO2.

Define the variable for equilibrium pressures

Let the partial pressure of NO2 at equilibrium be y atm. Since 1 mole of N2O4 is produced for every 2 moles of NO2 consumed, the partial pressure of N2O4 at equilibrium will be (9 - y)/2 atm.

Write the equation for \(K_p\) at equilibrium

Substitute the partial pressures at equilibrium in the expression for \(K_p\): $$0.25 = \frac{y^2}{\frac{9-y}{2}}$$

Solve for y

Solve the equation for the value of y: $$0.25(9-y) = 2y^2$$ $$y = 4.5$$

Calculate the equilibrium partial pressures

Now that we have the value of y, we can determine the equilibrium partial pressures of both N2O4 and NO2: $$[\mathrm{N}_{2}\mathrm{O}_{4}] = \frac{9 - 4.5}{2} = 2.25\,\text{atm}$$ $$[\mathrm{NO}_{2}] = 4.5\,\text{atm}$$ c. Conclusion. The equilibrium pressures in both cases a and b are the same: $$[\mathrm{N}_{2}\mathrm{O}_{4}] = 2.25\,\text{atm}$$ $$[\mathrm{NO}_{2}] = 4.5\,\text{atm}$$ Thus, it does not matter from which direction equilibrium position is reached. The equilibrium partial pressures of the gases will be the same irrespective of the initial conditions.

Key concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_{p}\), is a special number that helps us understand how gases react when they're in a state of balance, or equilibrium. At equilibrium, the rates of the forward and reverse reactions are equal, meaning that the concentrations of products and reactants don't change over time.

For gaseous reactions, \(K_{p}\) is expressed in terms of the partial pressures of the gases involved. It provides a measure of how much the reactants and products favor either side of the chemical equation when the system has reached equilibrium.

For example, in the reaction \( ext{N}_2 ext{O}_4(g) \rightleftharpoons 2 ext{NO}_2(g)\), the expression for \(K_{p}\) would be:

• \(K_{p} = \frac{{[\text{NO}_2]^2}}{{[\text{N}_2 ext{O}_4]}}\)

This equation tells us that \(K_{p}\) depends on the squares of the partial pressures of \(\text{NO}_2\), divided by the partial pressure of \(\text{N}_2 ext{O}_4\). This balance instructs us on where the equilibrium position "lies."

A larger \(K_{p}\) value indicates the products are favored at equilibrium, while a smaller \(K_{p}\) value shows the reactants are more prevalent.

Partial Pressure

Partial pressure is an important concept when dealing with gases in chemical reactions. It's the pressure a gas would exert if it alone occupied the entire volume of the mixture at the same temperature. Understanding partial pressures allows us to examine how each gas in a mixture affects the overall system.

In our reaction, to find the equilibrium partial pressures of \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\), we start by assigning variables to represent their pressures. For instance, if we started with only \(\mathrm{N}_2\mathrm{O}_4\) at 4.5 atm, the change in pressure is guided by the stoichiometry of the reaction, \(\mathrm{N}_2\mathrm{O}_4 \rightarrow 2\mathrm{NO}_2\).

Since two moles of \(\mathrm{NO}_2\) are produced for each mole of \(\mathrm{N}_2\mathrm{O}_4\) consumed, the changes in pressure can be calculated. For equilibrium, the total pressure is shared between \(\mathrm{NO}_2\) and \(\mathrm{N}_2\mathrm{O}_4\), leading us to solve equations for their respective pressures.

By utilizing the given initial conditions and the relationship:

• \(P_{\mathrm{NO}_2} = 2(4.5 - x)\)
• \(P_{\mathrm{N}_2\mathrm{O}_4} = x\)

We can track the changes to reach equilibrium, irrespective of whether we begin with \(\mathrm{N}_2\mathrm{O}_4\) or \(\mathrm{NO}_2\).

Le Chatelier's Principle

Le Chatelier’s Principle is a fundamental concept in chemistry that predicts how a change in conditions can shift the position of an equilibrium system. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to partly counteract the effect of the disturbance.

In the context of our exercise, Le Chatelier's Principle shines when considering changes in pressure. For instance, if we start with a certain partial pressure of \(\mathrm{NO}_2\) and then allow the system to reach equilibrium, the system will adjust. The reaction will shift towards the formation of \(\mathrm{N}_2\mathrm{O}_4\), or back to \(\mathrm{NO}_2\), based on the initial pressures and \(K_{p}\).

What’s fascinating here is that despite the direction we initiate from—whether the flask starts with \(\mathrm{N}_2\mathrm{O}_4\) or \(\mathrm{NO}_2\)—the system will settle to the same equilibrium position, offering the same partial pressures of each component.

This observation backs up the concept that equilibrium is a state of balance influenced by external conditions, and such consistency is a hallmark of dynamic equilibria, as explained by Le Chatelier's Principle.