Question

Two parallel first-order liquid phase reactions \( A \xrightarrow{k_1} B \) and \( A \xrightarrow{k_2} C \) are carried out in a well-mixed isothermal batch reactor. The initial concentration of \( A \) in the reactor is \( 1 \, \text{kmol} \, \text{m}^{-3} \), while that of \( B \) and \( C \) is zero. After \( 2 \, \text{hours} \), the concentration of \( A \) reduces to half its initial value, and the concentration of \( B \) is twice that of \( C \). The rate constants \( k_1 \) and \( k_2 \), in \( \text{h}^{-1} \), are respectively:

• A. 0.40, 0.20
• B. 0.23, 0.12
• C. 0.50, 0.25
• D. 0.36, 0.18

Hint:

For parallel reactions, use the overall rate constant as the sum of individual rate constants and relate the product concentrations to their respective rate constants.

Answer 1

The Correct Option is B

Step 1: Use the first-order decay law for \( A \). The decay of \( A \) is governed by the overall rate constant \( k \): \[ \ln \left( \frac{C_A}{C_{A0}} \right) = -kt. \] Here: - \( C_A = \frac{C_{A0}}{2} = \frac{1}{2} \, \text{kmol} \, \text{m}^{-3} \), - \( C_{A0} = 1 \, \text{kmol} \, \text{m}^{-3} \), - \( t = 2 \, \text{hours} \). Substitute into the equation: \[ \ln \left( \frac{\frac{1}{2}}{1} \right) = -k \cdot 2. \] Simplify: \[ \ln \left( \frac{1}{2} \right) = -2k \quad \Rightarrow \quad k = \frac{\ln(2)}{2} \approx 0.3466 \, \text{h}^{-1}. \] Step 2: Relate \( k_1 \) and \( k_2 \). The overall rate constant is the sum of the individual rate constants for parallel reactions: \[ k = k_1 + k_2. \] Thus: \[ k_1 + k_2 = 0.3466. \] Step 3: Use the concentration ratio of \( B \) and \( C \). The concentration of \( B \) is twice that of \( C \): \[ C_B = 2C_C. \] The concentrations are related to the rate constants: \[ \frac{C_B}{C_C} = \frac{k_1}{k_2}. \] Substitute \( \frac{C_B}{C_C} = 2 \): \[ \frac{k_1}{k_2} = 2 \quad \Rightarrow \quad k_1 = 2k_2. \] Step 4: Solve for \( k_1 \) and \( k_2 \). Substitute \( k_1 = 2k_2 \) into \( k_1 + k_2 = 0.3466 \): \[ 2k_2 + k_2 = 0.3466 \quad \Rightarrow \quad 3k_2 = 0.3466 \quad \Rightarrow \quad k_2 = 0.1155 \, \text{h}^{-1}. \] Then: \[ k_1 = 2k_2 = 2 \cdot 0.1155 = 0.231 \, \text{h}^{-1}. \] Step 5: Conclusion. The rate constants are \( k_1 = 0.23 \, \text{h}^{-1} \) and \( k_2 = 0.12 \, \text{h}^{-1} \).