Problem 45 - Chemistry and Chemical Reactivity

Question

The equilibrium reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) has been thoroughly studied (Figure 16.8 ). (a) If the total pressure in a flask containing \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas at \(25^{\circ} \mathrm{C}\) is 1.50 atm and the value of \(K_{\mathrm{p}}\) at this temperature is \(0.148,\) what fraction of the \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to \(\mathrm{NO}_{2} ?\) (b) What happens to the fraction dissociated if the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm?

Step-by-Step Solution

Answer

N₂O₄ dissociates more when pressure decreases, as equilibrium shifts to increase NO₂, the side with more moles.

Step 1: Write the equilibrium expression for Kp

The equilibrium expression for the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \) can be written as: \[ K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2O_4}} \]. The given \( K_p \) at 25°C is 0.148.

Step 2: Define partial pressures and total pressure

Let \( x \) be the partial pressure of \( \mathrm{N}_2O_4 \) dissociated. Then, \( P_{\mathrm{N}_2O_4} = 1.50 - x \) and \( P_{\mathrm{NO}_2} = 2x \). The total pressure is \( 1.50 \) atm.

Step 3: Substitute partial pressures into Kp equation

Substitute the partial pressures into the \( K_p \) expression: \[ K_p = \frac{(2x)^2}{(1.50-x)} \]. Since \( K_p = 0.148 \), set the equation: \[ 0.148 = \frac{4x^2}{1.50-x} \].

Step 4: Solve the equation for x

Rearrange and solve the quadratic equation: \[ 0.148(1.50 - x) = 4x^2 \]giving \[ 0.222 = 4x^2 + 0.148x \].Use the quadratic formula to find \( x \).

Step 5: Calculate the fraction dissociated

The fraction of \( \mathrm{N}_2O_4 \) dissociated is \( \frac{x}{\text{initial partial pressure of } \mathrm{N}_2O_4} \). Initial \( \mathrm{N}_2O_4 \) is 1.50 atm, so the fraction is \( \frac{x}{1.50} \). Calculate this using the value of \( x \) found in the previous step.

Step 6: Consider effect of pressure decrease

When the total pressure decreases to 1.00 atm, establish new expressions for partial pressures: \( y \) is the new dissociation, giving \( P_{\mathrm{N}_2O_4} = 1.00 - y \) and \( P_{\mathrm{NO}_2} = 2y \). Find \( y \) using the same steps to calculate \( x \) with total pressure 1.00 atm.

Step 7: Calculate new fraction dissociated

Compute the new dissociation fraction \( \frac{y}{1.00} \) using the value of \( y \) found after pressure change. Compare this with the previous fraction to check effect.

Key Concepts

Partial Pressure

Partial pressure refers to the pressure exerted by a single component of a gas mixture in a container. It's a crucial concept in chemical equilibrium, especially when dealing with gaseous reactions.
In a closed system with multiple gases, the total pressure is composed of the sum of the individual partial pressures of each gas present. For example, in the reaction \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\), the partial pressure of \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\) are key to understanding the equilibrium state.
When calculating partial pressures in equilibrium problems, you may often define a variable such as \(x\), representing the quantity of gas that has reacted or dissociated. This helps in making a clear equation using the equilibrium expression.

• The partial pressure for each gas is calculated by subtracting the dissociated part from the total pressure for the original, and multiplying the dissociated part for the product.
• Thus, for a system at equilibrium with total pressure 1.5 atm and some \(\mathrm{N}_2\mathrm{O}_4\) dissociated, you'd find \(P_{\mathrm{N}_2\mathrm{O}_4} = 1.5 - x\).

Equilibrium Constant

The equilibrium constant, often denoted as \(K\), is a value that expresses the ratio of the concentrations or partial pressures of the reactants and products at equilibrium. For reaction calculations involving gases, this is referred to as \(K_p\), where "p" stands for pressure.
In this context, the reaction \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\) has a given \(K_p\) value of 0.148 at 25°C. The equation for \(K_p\) is given by \[ K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2O_4}} \].
Understanding this equilibrium concept allows us to predict the positions of equilibrium under certain conditions.

• A smaller \(K_p\) value suggests the equilibrium position favors the reactants more than the products.
• By substituting the partial pressures into the \(K_p\) expression, we can solve for the unknown quantities within the reaction.

Le Chatelier's Principle

Le Chatelier’s Principle provides insight into how a system at equilibrium reacts to disturbances. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system shifts in such a way as to counteract that change.
In the exercise, the total pressure in the flask changes from 1.50 atm to 1.00 atm, indicating a shift in equilibrium. This principle helps us understand what happens when you alter conditions such as pressure, temperature, or concentration.

• If the pressure is reduced, the system will shift towards the side of the reaction with more gaseous molecules to increase the pressure.
• For the reaction \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\), a decrease in pressure would cause the equilibrium to shift in the direction that results in more moles of gas, favoring the production of \(\mathrm{NO}_2\).

Dissociation Reaction

A dissociation reaction involves the breaking down of a compound into two or more simpler substances. In chemical equilibria involving gases, dissociation plays a pivotal role in understanding how equilibrium is achieved.
For example, \(\mathrm{N}_2\mathrm{O}_4\) dissociates into \(2\mathrm{NO}_2\) as shown in the equilibrium equation \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\). This is a classic case of a dissociation reaction.

• The degree of dissociation is expressed as the fraction of the original substance that converts into products during the reaction.
• In our scenario, calculating the fraction dissociated helps us ascertain how much \(\mathrm{N}_2\mathrm{O}_4\) turned into \(\mathrm{NO}_2\) under different pressure conditions.
• Finding this fraction involves solving for \(x\) or \(y\) (the amount of substance dissociated) and comparing it against the initial total pressure.

Understanding dissociation is vital for grasping more complex interactions in chemical equilibrium.