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.

Answer 2

For the reversible reaction \( A \rightleftharpoons B \) with forward rate constant \( k_1 \) and reverse rate constant \( k_2 \), the rate of change of the concentration of A is governed by both the forward loss and the reverse gain:

\[ -\frac{d[A]}{dt} = k_1[A] - k_2[B] \]

Using mass balance \( [B] = [A]_0 - [A] \) (starting from pure A) gives a linear first-order differential equation in \( [A] \) whose solution decays exponentially toward the equilibrium concentration, with the exponential rate constant equal to \( (k_1 + k_2) \). Testing each option against this exponent:

1. \( \frac{1}{k_1} \): This only accounts for the forward reaction and would apply only if the reverse reaction were absent, which is not the case here.
2. \( \frac{1}{k_2} \): This only accounts for the reverse reaction, ignoring the forward rate constant entirely, so it cannot represent a system where both directions actively proceed.
3. \( \frac{1}{(k_1 - k_2)} \): Subtracting the two rate constants does not correspond to any term that appears in the differential equation governing the approach to equilibrium, and it could even become undefined or negative if \( k_2 > k_1 \), which is not physically meaningful for a time constant.
4. \( \frac{1}{(k_1 + k_2)} \): This matches exactly the combined exponential decay constant that emerges when the forward and reverse rate expressions are combined into a single differential equation for \( [A] \).

The relaxation toward equilibrium in a reversible first-order reaction is governed jointly by both rate constants, and their sum is what sets the exponential time constant.

Therefore, the correct answer is \( \dfrac{1}{(k_1+k_2)} \).