Reaction-Kinetics Formulae

Introduction

Kinetics formulae describe how fast concentrations change, not where equilibrium lies. A differential rate law gives instantaneous concentration dependence; an integrated law gives concentration after time under specified conditions. First- and second-order reactions have different linear plots, rate-constant units and half-life behavior. Picking the expression from a balanced overall equation alone is usually unjustified.

Core explanation

For aA + bB → products, reaction rate can be defined r = −(1/a)d[A]/dt = −(1/b)d[B]/dt for a homogeneous system with well-defined concentrations. A measured rate law might be r = k[A]^m[B]^n. The exponents m and n are experimental orders unless the step is known to be elementary. Overall stoichiometric coefficients do not generally equal these exponents because mechanisms can have multiple steps.

For a simple first-order disappearance −d[A]/dt = k[A], integration gives [A] t = [A] 0e^(−kt), or ln([A] t/[A] 0) = −kt. Its k has time⁻¹ units. The half-life is t₁/₂ = ln2/k and is independent of starting concentration under constant conditions. A straight line of ln[A] versus time has slope −k, but a short noisy dataset can appear roughly linear in several plots.

For a simple second-order disappearance −d[A]/dt = k[A]², integration gives 1/[A] t = 1/[A] 0 + kt. The corresponding k has concentration⁻¹ time⁻¹ units, and t₁/₂ = 1/(k[A] 0). Doubling initial concentration halves the half-life under this model. The expression assumes the rate law is exactly k[A]², not a general reaction with two reactants whose concentrations change independently.

A zero-order case obeys −d[A]/dt = k, giving [A] t = [A] 0 − kt until the model ceases to apply. Its half-life is [A] 0/(2k). Saturation of catalyst sites can produce an apparent zero-order regime, but it may change as substrate concentration falls. Each integrated law assumes a constant k and reaction environment during the fitted interval.

For A + B with r = k[A][B], if B is present in great excess and remains nearly constant, r ≈ k obs[A] with k obs = k[B]. This is pseudo-first-order behavior. It does not make the molecular mechanism unimolecular. Changing the excess B concentration should change k obs approximately proportionally if that rate law remains valid.

Rate constants depend on temperature, solvent, pH, catalyst state and other conditions. Integrated expressions may fail if a reaction approaches reverse equilibrium, products inhibit catalysis, or reactants mix slowly. A log-linear plot is a diagnostic, not proof of mechanism. Confirm with initial-rate variations and a model that explains observed species.

The units of k are a strong check. If rate has units concentration/time and overall order is p, k has concentration^(1−p)/time. For first order this is s⁻¹; for second order with molarity, M⁻¹ s⁻¹. Numerical values of k cannot be compared without units and temperature.

Step-by-step reasoning

Define reaction-rate stoichiometry. Determine or assume a rate law from data, not only equation coefficients. Match its order to an integrated expression and check k units. Calculate concentration or half-life, then check whether temperature, reagent excess and reverse reaction assumptions remain reasonable.

Visual explanation

Draw three concentration-versus-time curves: zero order decreases linearly, first order decays exponentially, and second order bends with a half-life that grows as concentration falls. Beside each, show its linear diagnostic plot: [A], ln[A] or 1/[A] versus time.

Real-world analogy

An event can proceed at a fixed number per hour, a fixed fraction per hour or a rate that depends on pairs meeting. Those resemble zero-, first- and second-order patterns. Chemistry adds molecular mechanisms and changing conditions, so the analogy cannot by itself establish order.

Real-world example

In a hydrolysis experiment, water is the solvent and far more abundant than an organic substrate. If the elementary rate depends on both, water concentration may be effectively constant, making substrate disappearance look first order. A change of solvent composition can reveal the hidden water dependence.

Why?

Rate formulae allow prediction of processing time, shelf life and reaction progress. They also help identify whether a measured kinetic pattern is compatible with a proposed mechanism.

Common misconception

“Second-order overall reaction means two molecules collide in the rate-limiting step” is not guaranteed from an empirical rate law alone. Another mistake uses the first-order 0.693/k half-life for a second-order process, where initial concentration matters.

Worked example

A first-order reactant has k = 0.0200 min⁻¹ and starts at 0.100 M. After 30.0 min, [A] = 0.100e^(−0.0200×30.0) ≈ 0.0549 M. Its half-life is ln2/0.0200 ≈ 34.7 min. Because 30 min is slightly less than one half-life, a concentration slightly above 0.050 M is sensible.

Quick check

1. Which plot is linear for ideal second-order disappearance −d[A]/dt = k[A]²? Answer: A plot of 1/[A] versus time has slope k.

Exam focus

Write the differential law before choosing an integrated formula. State rate-constant units and half-life dependence. Explain why a pseudo-first-order law does not prove a one-molecule elementary step.

Advanced insight

Multiple mechanisms can produce the same macroscopic rate law. Isotope effects, intermediate detection, product inhibition and temperature dependence add constraints. A statistically good line fit is necessary for a proposed simple integrated law but is not sufficient evidence of a unique microscopic mechanism.

Summary

First-order disappearance gives exponential decay and concentration-independent half-life; simple second-order disappearance gives a linear inverse-concentration plot and concentration-dependent half-life. Rates and orders are kinetic observations under stated conditions, separate from equilibrium and overall stoichiometry.

Practice questions

1. What are the units of a first-order k if time is seconds? Answer: s⁻¹. 2. What is t₁/₂ for k = 0.10 s⁻¹ in a first-order process? Answer: ln2/0.10 ≈ 6.93 s. 3. For −d[A]/dt = k[A]², what happens to half-life if [A]₀ doubles? Answer: It halves, since t₁/₂ = 1/(k[A]₀). 4. Can overall reaction coefficients alone establish a rate law? Answer: No, unless a specified elementary step justifies that connection.

Sources

- OpenStax Chemistry 2e: Integrated Rate Laws. - OpenStax Chemistry 2e: Rate Laws.