Reversible Reaction Kinetics

Introduction

Simple integrated rate laws assume a reaction runs to completion. Many reactions do not: as products accumulate, the reverse reaction becomes significant, and the composition levels off at equilibrium. Kinetics and thermodynamics meet here. The rate constants of the forward and reverse steps determine both how quickly equilibrium is reached and where it lies, providing a direct bridge between rate data and equilibrium constants.

Core explanation

The model. Consider the elementary first-order interconversion

A ⇌ B, forward rate constant k f, reverse rate constant k r

The net rate of loss of A is

−d[A]/dt = k f[A] − k r[B]

At the start, when [B] = 0, only the forward term matters and the reaction looks like simple first-order decay. As B builds up, the reverse term grows until the two terms are equal and the net rate is zero.

Equilibrium. Setting the net rate to zero gives k f[A]eq = k r[B]eq, so

K = [B]eq/[A]eq = k f/k r

The equilibrium constant is the ratio of forward and reverse rate constants. This is the principle of detailed balance applied to a single step: equilibrium is dynamic, with both directions running at equal and opposite rates.

The integrated form. Using mass balance, [A] + [B] = [A]₀ when only A is present initially. Substituting and integrating gives

[A] − [A]eq = ([A]₀ − [A]eq) exp[−(k f + k r)t]

The displacement from equilibrium decays exponentially with an observed rate constant

k obs = k f + k r

This is a striking result. The approach to equilibrium is faster than either the forward or the reverse reaction alone, because both processes act to remove any deviation from equilibrium. The relaxation time is τ = 1/(k f + k r).

Extracting both constants. One experiment measures k obs from the exponential approach and K from the final composition. Two equations, k obs = k f + k r and K = k f/k r, then give

k f = k obs K/(1 + K) and k r = k obs/(1 + K)

Analysis pitfall. Plotting ln[A] against t for a reversible reaction gives a curve that flattens at ln[A]eq, not a straight line. The correct linear plot is ln([A] − [A]eq) against t, whose gradient is −(k f + k r).

Higher orders. For reversible steps of other orders, such as A + B ⇌ C, the full integrated solution is more complicated, but near equilibrium any system relaxes exponentially, which is the basis of relaxation methods.

Formulae

−d[A]/dt = k f[A] − k r[B]; K = k f/k r; k obs = k f + k r; τ = 1/(k f + k r).

Step-by-step reasoning

1. Measure [A] over time until it stops changing, and record [A]eq. 2. Calculate K from [B]eq/[A]eq using mass balance. 3. Plot ln([A] − [A]eq) against t and take k obs from the gradient. 4. Solve the two simultaneous equations for k f and k r.

Visual explanation

Plot [A] and [B] against time on the same axes. [A] falls and [B] rises, each curving smoothly towards a horizontal plateau. On a second graph, plot forward and reverse rates: the forward rate falls, the reverse rate rises, and they meet at equilibrium.

Real-world analogy

Two connected water tanks with pipes flowing both ways reach equal flow in each direction. How level the final water sits depends on the pipe sizes' ratio, while how fast it settles depends on the total flow capacity of both pipes together.

Real-world example

Glucose in water exists as α- and β-anomers that interconvert through the open-chain form. Freshly dissolved α-glucose changes its optical rotation over tens of minutes until an equilibrium mixture of roughly 36% α and 64% β forms. The observed rotation change follows a single exponential with rate constant k f + k r.

Why?

Why is k obs the sum of the constants? If the system is displaced above equilibrium in A, the forward reaction removes the excess A while the reverse reaction also slows its production. Both effects push back towards equilibrium, so their rate constants add.

Common misconception

"At equilibrium the reaction stops." Molecules keep reacting in both directions; the forward and reverse rates are simply equal, so there is no net change in composition.

Worked example

Question: For A ⇌ B starting with pure A, the approach to equilibrium has k obs = 0.030 s⁻¹ and at equilibrium [B]/[A] = 2.0. Find k f and k r.

Reasoning: k r = k obs/(1 + K) = 0.030/3.0 = 0.010 s⁻¹. k f = K × k r = 2.0 × 0.010 = 0.020 s⁻¹. Check: sum = 0.030 s⁻¹.

Answer: k f = 0.020 s⁻¹; k r = 0.010 s⁻¹.

Quick check

1. For a reversible first-order reaction A ⇌ B, what is the observed rate constant for relaxation towards equilibrium? Answer: It is the sum of the forward and reverse rate constants, k obs = k f + k r.

Exam focus

State the link K = k f/k r and justify it by equating forward and reverse rates at equilibrium. Show the plot of ln([A] − [A]eq) against time when determining k obs. Examiners reward the explicit explanation that equilibrium is dynamic.

Advanced insight

Temperature-jump and pressure-jump experiments disturb an equilibrium in microseconds and follow the exponential return. Because k obs = k f + k r regardless of how big K is, these relaxation methods can measure extremely fast reactions, such as proton transfer in water with rate constants near 10¹¹ mol⁻¹ dm³ s⁻¹, that are impossible to mix manually.

Summary

For A ⇌ B, the net rate is k f[A] − k r[B]. At equilibrium the rates balance, so K = k f/k r. The deviation from equilibrium decays exponentially with rate constant k f + k r. Combining k obs with the equilibrium composition gives both individual rate constants, linking kinetics directly to thermodynamics.

Practice questions

1. Why does ln[A] against t not give a straight line for a reversible reaction? Answer: [A] approaches a non-zero equilibrium value, so the plot flattens; ln([A] − [A]eq) should be plotted instead. 2. If k f = 0.40 s⁻¹ and k r = 0.10 s⁻¹, find K and the relaxation time. Answer: K = 4.0 and τ = 1/(0.50 s⁻¹) = 2.0 s. 3. Explain how a catalyst affects k f, k r and K. Answer: A catalyst increases k f and k r by the same factor, so their ratio K is unchanged while equilibrium is reached faster. 4. A reaction with K = 1.0 has k obs = 0.50 s⁻¹. Find k f. Answer: k f = k obs K/(1 + K) = 0.50 × 1.0/2.0 = 0.25 s⁻¹.