Question

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{B}(g) . \mathrm{A}\) flask is charged with 0.75 atm of pure \(\mathrm{A},\) after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\) . At equilibrium, the partial pressure of \(\mathrm{A}\) is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?(\mathbf{c})\) What could we do to maximize the yield of B?

Short answer

The total pressure at equilibrium is 1.14 atm. The value of \(K_p\) is approximately 1.69. To maximize the yield of B, we can increase the pressure of the system, remove B from the system as it is formed, and use a catalyst to reach equilibrium faster.

Step-by-step solution

Establish the reaction and set up the table

The reaction is given by: \[A(g) \rightleftharpoons 2B(g)\] Set up the initial, change, and equilibrium pressures in relation to the reaction: For A: Initial Pressure: 0.75 atm Change: -x Equilibrium Pressure: 0.36 atm For B: Initial Pressure: 0 atm Change: 2x Equilibrium Pressure: 2x

Determine x

As we know the equilibrium pressure of A is 0.36 atm, we can find the change in pressure (x): \[0.75-x=0.36 \Rightarrow x=0.39\]

Calculate the equilibrium pressure of B

Now that we have the value of x, we can find the equilibrium pressure of B: Equilibrium Pressure of B = 2x = 2(0.39) = 0.78 atm

Calculate total pressure at equilibrium

Finally, the total pressure at equilibrium is the sum of the equilibrium pressures of A and B Total Pressure at equilibrium = 0.36 atm (A) + 0.78 atm (B) = 1.14 atm (b) Finding the value of \(K_p\)

Write the equilibrium expression

Since the reaction is given by A(g) ⟷ 2B(g), the expression for the equilibrium constant will be: \[K_p=\frac{P(B)^2}{P(A)}\]

Substitute the equilibrium pressures in the expression

Substitute the equilibrium pressures of A and B in the expression: \[K_p=\frac{(0.78)^2}{0.36}\]

Calculate the value of \(K_p\)

Now, find the value of \(K_p\): \[K_p=\frac{0.6084}{0.36} \approx 1.69\] (c) Maximizing the yield of B

Apply Le Chatelier's principle

Le Chatelier’s principle states that if an equilibrium system is subjected to a change in temperature, volume, or pressure, the system will adjust the equilibrium position to counteract the change. To maximize the yield of B, we can: 1. Increase the pressure of the system: Since there are fewer moles of gas on the right side of the reaction (2 moles of B) compared to the left side of the reaction (1 mole of A), increasing the pressure will shift the equilibrium in the direction of B, increasing its formation. 2. Remove B from the system as it is formed: Continuously removing B from the system will shift the equilibrium towards the right side, thus producing more B. 3. Use a catalyst: A catalyst can increase the rate of the reaction without affecting equilibrium; however, it can help in reaching equilibrium faster. Note, changes in temperature will not maximize the yield of B since no information is given about the heat associated with the reaction.

Key concepts

Le Chatelier's Principle

Le Chatelier's principle is a fundamental concept in chemistry that explains how a system at equilibrium responds to external changes. This principle is crucial for predicting how changes in concentration, pressure, or temperature can affect the position of equilibrium.
When a change is imposed on a system at equilibrium, the system will react in a way that tends to counteract that change and restore a new balance.
For example, in the reaction \( A(g) \rightleftharpoons 2B(g) \), if you increase the pressure, the system will shift to the side with fewer moles of gas to reduce the pressure.
In this case, it shifts to the right since converting 1 mole of \( A \) into 2 moles of \( B \) results in an increase in the total number of moles, which is counteracted by this shift. This is also applicable if you continuously remove \( B \) from the system, which will prompt the system to produce more \( B \).
While catalysts can help the system reach equilibrium faster, they do not change the position of the equilibrium.

Equilibrium Constant (\(K_p\))

The equilibrium constant, \( K_p \), is a number that reflects the ratio of the concentrations of products to reactants at equilibrium for a given reaction. It tells us which side of the reaction is favored under a set of conditions.
For gas-phase reactions, \( K_p \) is calculated using the partial pressures of the gases. For reactions involving gases, such as \( A(g) \rightleftharpoons 2B(g) \), the expression for \( K_p \) is given by
\[ K_p = \frac{P(B)^2}{P(A)} \]
where \( P(B) \) is the partial pressure of B and \( P(A) \) is the partial pressure of A.
In our example, substituting the equilibrium pressures, \( K_p \) is \( \frac{0.78^2}{0.36} \), which calculates to approximately 1.69. This value indicates the tendency of the reaction to shift towards the formation of products (B in this case) under equilibrium conditions.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It's based on the idea that each gas in a mixture behaves independently.
In a chemical reaction involving gases, knowing the partial pressures helps in understanding the behavior of the gas reaction mixture at equilibrium.
For the reaction \( A(g) \rightleftharpoons 2B(g) \), initially, only \( A \) was present with a partial pressure of 0.75 atm. As the reaction progresses toward equilibrium, the partial pressure of \( A \) decreases, while that of \( B \) increases as it is being produced.
The total pressure at equilibrium is the sum of the partial pressures of \( A \) and \( B \), which at equilibrium became 0.36 atm and 0.78 atm, respectively, adding up to a total of 1.14 atm.
Understanding partial pressures is critical for calculating \( K_p \) and applying Le Chatelier's principle to predict how the system will react to changes in pressure or other conditions.