Problem 58 - Chemistry: An Atoms-Focused Approach

Question

In which direction will the following hypothetical reaction proceed to reach equilibrium under the conditions given? \(A(g)+B(g) \rightleftharpoons C(g) \quad K_{p}=1.00\) at \(300 K\) a. \(P_{\mathrm{A}}=P_{\mathrm{C}}=1.0 \mathrm{atm}, P_{\mathrm{B}}=0.50 \mathrm{atm}\) b. \([\mathrm{A}]=[\mathrm{B}]=[\mathrm{C}]=1.0 \mathrm{M}\)

Step-by-Step Solution

Answer

a. Partial pressures: \(P_\mathrm{A} = 1.0 \mathrm{atm}, P_\mathrm{C} = 1.0 \mathrm{atm}, P_\mathrm{B} = 0.50 \mathrm{atm}\) b. Concentrations: \([\mathrm{A}] = [\mathrm{B}] = [\mathrm{C}] = 1.0 \mathrm{M}\) Answer: a. The reaction will proceed to reach equilibrium in the reverse (left) direction. b. The reaction is already at equilibrium.

Step 1: Write down the reaction and the given parameters

The reaction is given as: \(A(g) + B(g) \rightleftharpoons C(g)\) with \(K_p = 1.00\) at \(300 K\). Part A: \(P_\mathrm{A} = 1.0 \mathrm{atm}, P_\mathrm{C} = 1.0 \mathrm{atm}, P_\mathrm{B} = 0.50 \mathrm{atm}\) Part B: \([\mathrm{A}] = [\mathrm{B}] = [\mathrm{C}] = 1.0 \mathrm{M}\)

Step 2: Calculate the reaction quotient Q_p for Part A

The reaction quotient \(Q_p\) in terms of partial pressures in part A can be calculated as follows: \(Q_p = \frac{P_\mathrm{C}}{P_\mathrm{A} \cdot P_\mathrm{B}}\) Substitute given values of partial pressures: \(Q_p = \frac{1.0 \cdot \mathrm{atm}}{1.0 \cdot \mathrm{atm} \cdot 0.50 \cdot \mathrm{atm}} = 2\)

Step 3: Compare Q_p and K_p for Part A

Now, we can compare \(Q_p\) and \(K_p\): \(Q_p = 2 > K_p = 1\) Since \(Q_p > K_p\), the reaction will proceed in the reverse direction (from right to left) to reach equilibrium.

Step 4: Calculate the reaction quotient Q_c for Part B

The reaction quotient \(Q_c\) in terms of concentrations in part B can be calculated as follows: \(Q_c = \frac{[\mathrm{C}]}{[\mathrm{A}] \cdot [\mathrm{B}]}\) Substitute given values of concentrations: \(Q_c = \frac{1.0 \cdot \mathrm{M}}{1.0 \cdot \mathrm{M} \cdot 1.0 \cdot \mathrm{M}} = 1\)

Step 5: Compare Q_c and K_p for Part B

Now, we can compare \(Q_c\) and \(K_p\): \(Q_c = K_p = 1\) Since \(Q_c = K_p\), the reaction is already at equilibrium in part B. In summary: a. The reaction will proceed to reach equilibrium in the reverse (left) direction. b. The reaction is already at equilibrium.

Key Concepts

Reaction Quotient

The reaction quotient, represented as \( Q \), is a valuable tool when analyzing chemical reactions and determining their direction under non-equilibrium conditions. It's similar to the equilibrium constant, but it applies to a reaction that hasn't yet reached equilibrium. To find the reaction quotient, you use the same formula as for the equilibrium constant, substituting the current concentration or partial pressures of the reactants and products.

For gases, this takes the form \( Q_p \) to imply relationships written in terms of partial pressures. For example, in the reaction \( A(g) + B(g) \rightarrow C(g) \), \( Q_p \) would be defined as:

• \( Q_p = \frac{P_C}{P_A \cdot P_B} \)

In Part A of the problem, given the values \( P_A = 1.0\, \text{atm} \), \( P_B = 0.50\, \text{atm} \), and \( P_C = 1.0\, \text{atm} \), we calculate \( Q_p = 2 \).

This result is then compared with the equilibrium constant \( K_p \) to forecast which way the reaction will shift. If \( Q > K \), it suggests the reaction will go in reverse to achieve equilibrium. If \( Q < K \), it will go forward. If \( Q = K \), it's already at equilibrium.

Equilibrium Constant

The equilibrium constant, typically denoted as \( K \), is a numerical value that expresses the ratio of products to reactants at equilibrium for any given reaction at a specific temperature. It's crucial because it tells us about the extent to which a reaction will proceed.

In gases, this constant is known as \( K_p \) and it relies on partial pressures rather than concentrations. For the example reaction \( A(g) + B(g) \rightarrow C(g) \) at 300 K, \( K_p \) is given as 1.00. This value means that when the system reaches equilibrium, the ratio of the pressures (products over reactants) will be equal to 1.

Understanding \( K \) values helps to predict the direction in which a reaction mixture under non-equilibrium conditions will move to achieve equilibrium. If the reaction quotient \( Q \) is different from \( K \), the system will adjust by shifting in a direction that pushes \( Q \) towards \( K \).

• If \( Q > K \): Reaction shifts to the left (forms reactants)
• If \( Q < K \): Reaction shifts to the right (forms products)
• If \( Q = K \): Reaction is at equilibrium

Partial Pressure

Partial pressure refers to the pressure that a single gas component in a mixture contributes to the total pressure. It's crucial when dealing with gases in chemical equilibrium, as it helps calculate reaction quotients and equilibrium constants for gaseous reactions.

The total pressure of a gas mixture is the sum of the partial pressures of the individual gases present. If you know the partial pressures, you can substitute these into the expression for \( Q_p \) or \( K_p \):

• For \( A(g) + B(g) \rightarrow C(g) \): \( Q_p = \frac{P_C}{P_A \cdot P_B} \)

This process involves using the actual measured partial pressures of each gas in the expression. In the problem provided, \( P_A = 1.0 \text{ atm} \), \( P_B = 0.50 \text{ atm} \), and \( P_C = 1.0 \text{ atm} \), making \( Q_p = 2 \).

Understanding partial pressures is essential for predicting the course of a reaction based on changes in pressure, particularly when dealing with reactions involving gases. This concept ties back to fundamental principles of gas laws, where each gas in a mixture behaves independently and contributes to the total atmospheric pressure. Knowing these pressures allows chemists to determine how a reaction will shift to find equilibrium effectively.