Problem 75 - Chemistry The Central Science

Question

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) .\) A flask is charged with 0.75 atm of pure \(A\), 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 \(K_{p} ?\) (c) What could we do to maximize the yield of \(\mathrm{B}\) ?

Step-by-Step Solution

Answer

The total pressure in the flask at equilibrium is 1.14 atm. The value of \(K_p\) for the reaction is 1.68. To maximize the yield of B, we should increase the pressure of the system, as it will shift the equilibrium towards the production of B according to Le Chatelier's Principle.

Step 1: Write the reaction and the equilibrium constant expression

The given reaction is: \( A(g) \rightleftharpoons 2B(g) \) We can now write the equilibrium constant expression for this reaction in terms of partial pressures, Kp, as follows: \( K_p = \frac{P_B^2}{P_A} \) where \(P_A\) and \(P_B\) represent the partial pressures of A and B gases, respectively, at equilibrium.

Step 2: Calculate the change in partial pressures at equilibrium

Initially, the flask has a pressure of 0.75 atm for A and 0 atm for B. Let x be the drop in pressure of A at equilibrium. Therefore, the increase in pressure of B will be twice this change (since there are 2 moles of B produced for each mole of A that reacts). So, \( P_A = 0.75 - x \) \( P_B = 0 \Rightarrow P_B = 2x \) At equilibrium, we are given that the pressure of A is 0.36 atm. Therefore, we can now calculate the change in pressure (x) and the pressure of B at equilibrium.

Step 3: Calculate equilibrium pressures of A and B

Since we have the equilibrium pressure of A, \( P_A = 0.36 \, atm \) We can calculate the change in pressure (x): \( x = 0.75 - P_A = 0.75 - 0.36 = 0.39 \, atm \) Now, we can calculate the equilibrium pressure of B: \( P_B = 2x = 2 * 0.39 = 0.78 \, atm \)

Step 4: Calculate the total pressure at equilibrium

To find the total pressure at equilibrium, we simply add the partial pressures of A and B: \( P_{total} = P_A + P_B = 0.36 + 0.78 = 1.14 \, atm \) So, the total pressure in the flask at equilibrium is 1.14 atm.

Step 5: Calculate the value of Kp

Now, we can substitute the equilibrium pressures of A and B into the equilibrium constant expression: \( K_p = \frac{P_B^2}{P_A} = \frac{(0.78)^2}{0.36} = 1.68 \) Hence, the value of Kp for this reaction is 1.68.

Step 6: Suggest a method to maximize the yield of B

To maximize the yield of B, we can use Le Chatelier's Principle, which states that a change in external conditions (pressure, temperature, and concentration) will cause a shift in the equilibrium to counteract the change. In this case, since the formation of B is associated with a decrease in the number of moles of gas (1 mole of A produces 2 moles of B), increasing the pressure of the system will cause the equilibrium to shift towards the side with fewer moles of gas, i.e., towards the production of B. Therefore, to maximize the yield of B, we should increase the pressure of the system.

Key Concepts

Equilibrium Constant

Chemical reactions often don't go to completion; instead, they reach a state of balance called equilibrium, where the rates of the forward and reverse reactions are equal. In this state, the ratio of the concentrations of products to reactants is constant for a given temperature and is known as the equilibrium constant, denoted as \( K \).
For reactions involving gases, we use partial pressures instead of concentrations, and the equilibrium constant is then represented as \( K_p \). In our example of the reaction \( A(g) \rightleftharpoons 2B(g) \), the expression for \( K_p \) is:

• \( K_p = \frac{P_B^2}{P_A} \)

where \( P_A \) and \( P_B \) are the partial pressures of \( A \) and \( B \) at equilibrium.
Knowing \( K_p \) allows us to predict the ratio of products to reactants under certain conditions, and it remains constant unless the temperature changes. If \( K_p > 1 \), the reaction favors the formation of products; if \( K_p < 1 \), it favors reactants.

Le Chatelier's Principle

Le Chatelier's Principle offers insight on how a change in conditions affects a reaction at equilibrium. It states that if an external change such as pressure, temperature, or concentration occurs, the system shifts in the direction that counteracts the change.
Let's apply this to our chemical reaction \( A(g) \rightleftharpoons 2B(g) \). This principle helps us understand how to maximize the production of \( B \). When dealing with pressure changes, remember:

• An increase in pressure shifts equilibrium towards the side with fewer moles of gas.
• A decrease in pressure shifts equilibrium towards the side with more moles of gas.

In our case, since 1 mole of \( A \) turns into 2 moles of \( B \), increasing the pressure will favor the side with fewer initial moles of gas, encouraging a shift that maximizes \( B \). By understanding Le Chatelier's Principle, we can effectively manipulate conditions to achieve the desired yield in a reaction.

Partial Pressure

Partial pressure is a critical concept in chemical equilibrium involving gases. It refers to the pressure that a gas in a mixture would exert if it occupied the entire volume by itself.
In equilibrium scenarios, individual gases contribute to the total pressure based on their mole fractions and the total number of gas moles present. For the process \( A(g) \rightleftharpoons 2B(g) \), partial pressures help us understand and calculate the equilibrium state:

• Initial partial pressure of \( A \) is given, and it decreases as the reaction proceeds.
• Partial pressure of \( B \) increases, being twice the change of \( A \) because 2 moles of \( B \) are produced for each mole \( A \) consumed.
• Total pressure at equilibrium is the sum of the partial pressures of \( A \) and \( B \).

By calculating these pressures, we gain insight into the dynamics of the reaction under specific conditions. Partial pressure, therefore, plays a vital role, along with \( K_p \) and Le Chatelier's Principle, in predicting and understanding the behavior of gaseous equilibria.