12.
Consider a reversible reaction 𝐴 ⇌ 𝐵 with forward and backward rate constants
𝑘! = 𝑘! = 1 sec!!. Suppose we start with a 1 molar solution of A. How long will the
concentration of A take to reach 0.75 molar?
a. 0.25 sec
b. ln (4/3) ~ 0.29 sec
c. ln 2 ~ 0.35 sec
d. It will never reach that concentration
Reversible Reaction A ⇌ B: Time for [A] to Reach 0.75 M from 1 M
The correct answer is c. ln 2 ~ 0.35 sec, as the concentration of A reaches 0.75 M in approximately 0.347 seconds for this first-order reversible reaction with equal rate constants.
Reaction Kinetics Analysis
For the reversible first-order reaction A ⇌ B where k_f = k_b = 1 s⁻¹, starting from [A]_0 = 1 M and [B]_0 = 0 M, the concentration of A follows:
[A](t) = (k_b + k_f e^{-(k_f + k_b)t}) / (k_f + k_b)
Substituting values gives [A](t) = (1 + e^{-2t})/2. At equilibrium, [A]_∞ = 0.5 M, so A decreases from 1 M toward 0.5 M and will pass through 0.75 M.
Solving [A](t) = 0.75 yields e^{-2t} = 0.5, so -2t = ln(0.5) = -ln 2, thus t = (ln 2)/2 ≈ 0.693/2 ≈ 0.3466 seconds, matching numerical simulation results.
Option Explanations
| Option | Time | Explanation |
|---|---|---|
| a. 0.25 sec | 0.25 s | This assumes an irreversible reaction rate where [A] = e^{-t}, so e^{-0.25} ≈ 0.7788 (not 0.75). Too fast for the reversible case. |
| b. ln(4/3) ~ 0.29 sec | 0.2877 s | ln(4/3) ≈ 0.2877 s gives [A] ≈ 0.75 only for irreversible kinetics ([A] = e^{-t}). Reversibility slows the approach. |
| c. ln 2 ~ 0.35 sec | 0.3466 s | Correct, as derived analytically and verified numerically at t ≈ 0.3466 s. |
| d. It will never reach | – | Incorrect, since 0.75 M lies between initial 1 M and equilibrium 0.5 M. |
SEO-Friendly Article: Reversible Reaction Kinetics Time Calculation
The reversible first-order reaction A ⇌ B with equal rate constants k_f = k_b = 1 s⁻¹ represents a classic kinetics problem for CSIR NET Life Sciences preparation, testing analytical derivation of time-dependent concentrations. Starting from 1 molar A, determine how long until [A] reaches 0.75 molar.
Analytical Derivation
The rate equations d[A]/dt = -k_f[A] + k_b[B] and conservation [A] + [B] = 1 yield the solution [A](t) = 0.5 + 0.5 e^{-2t}. Setting [A](t) = 0.75 gives 0.75 = 0.5(1 + e^{-2t}), so e^{-2t} = 0.5, t = ln(2)/2 ≈ 0.3466 s, closest to option c.
CSIR NET Exam Insights
This problem distinguishes reversible vs. irreversible kinetics:
- Irreversible: t = ln(1/0.75) = ln(4/3) ≈ 0.29 s (option b)
- Reversible: Accounts for back reaction, doubling the decay constant to 2 s⁻¹, halving the time constant.
Keywords: reversible reaction kinetics, concentration time calculation, first-order reversible reaction, CSIR NET biochemistry, A ⇌ B rate constants.
