Question

A first order reversible reaction A to B occurs in a batch reactor. The exponential decay of the concentration of A has the time constant

• A. \( \frac{1}{k_1} \)
• B. \( \frac{1}{k_2} \)
• C. \( \frac{1}{(k_1 - k_2)} \)
• D. \( \frac{1}{(k_1 + k_2)} \)

Hint:

For reversible reactions, remember that the time constant depends on both the forward and reverse rate constants, and the total effect is their sum.

Answer 1

The Correct Option is D

Step 1: Understanding the reaction.
For a first-order reversible reaction, the rate of decay of the concentration of A can be described by the rate constant, which depends on both the forward rate constant (\(k_1\)) and the reverse rate constant (\(k_2\)). The time constant for the exponential decay is influenced by both \(k_1\) and \(k_2\) combined, as the reaction proceeds in both directions.
Step 2: Analyzing the options.
(1) \( \frac{1}{k_1} \): This is incorrect because it only considers the forward reaction and ignores the reverse.
(2) \( \frac{1}{k_2} \): This is incorrect as it only considers the reverse reaction.
(3) \( \frac{1}{(k_1 - k_2) \):} This is incorrect as the time constant is a sum, not a difference of the rate constants.
(4) \( \frac{1}{(k_1 + k_2) \):} Correct — The time constant is derived by combining the effects of both the forward and reverse rate constants, making this the correct choice.
Step 3: Conclusion.
The correct answer is \(\frac{1}{(k_1 + k_2)}\), which correctly describes the time constant for the first-order reversible reaction.