Rate Laws from Experimental Data

Introduction

A balanced equation tells you what reacts and in what ratio, but it says nothing reliable about how the rate depends on concentration. That dependence must be measured. The most direct approach is the method of initial rates: run the reaction several times with different starting concentrations and compare how fast it begins. This page shows how to extract orders and a rate constant from such data, and how to recognise when an experiment simply cannot tell you what you want to know.

Core explanation

The empirical rate law. For a reaction A + B → products, the rate is usually written as

rate = k[A]^m[B]^n

where m and n are the orders with respect to A and B, and m + n is the overall order. The orders are experimental quantities. They are often small integers (0, 1, 2) but may be fractions or even negative, and they need not match the stoichiometric coefficients. A complex mechanism can produce a rate law that looks nothing like the overall equation.

Why initial rates? At the very start of a run the concentrations are known exactly — they are the values you prepared. Products have not yet built up, so reverse reactions and product inhibition are negligible. Measuring the gradient of concentration against time at t = 0 therefore gives a rate that corresponds to a known set of concentrations.

Comparing experiments. The key design principle is to change one concentration at a time. If [A] is doubled while [B] is held constant, then

rate₂ ÷ rate₁ = (2)^m

so a doubling of rate means m = 1, a fourfold increase means m = 2, and no change means m = 0. For non-integer ratios, take logarithms: m = log(rate₂/rate₁) ÷ log([A]₂/[A]₁).

Finding k. Once all the orders are known, substitute any single experiment into the rate law and solve for k. In good data, each experiment gives nearly the same k; the scatter tells you about experimental precision. The units of k depend on overall order: s⁻¹ for first order, mol⁻¹ dm³ s⁻¹ for second order, mol⁻² dm⁶ s⁻¹ for third order.

Parameter identifiability. A rate law has parameters (k, m, n). Data identify those parameters only if the experiments actually vary the relevant concentrations independently. If every experiment changes [A] and [B] together in the same ratio, you can find m + n but never m and n separately — the two effects are confounded. Likewise, if [B] is always in huge excess, the data identify only a pseudo-rate constant k′ = k[B]^n, not k and n individually. Recognising what a design can and cannot reveal is as important as the arithmetic.

Method of isolation. A deliberate use of this effect is to flood the system with excess B so that its concentration barely changes. The reaction then behaves as pseudo-order m in A, which is easy to analyse. Repeating the flooded experiment at several fixed [B] values then recovers n from how k′ varies.

Step-by-step reasoning

1. Tabulate each experiment's initial concentrations and initial rate. 2. Find two runs in which only [A] changes; compute the rate ratio and hence m. 3. Find two runs in which only [B] changes; compute n in the same way. 4. If no such pair exists, use logarithms and simultaneous equations, or admit the order cannot be separated. 5. Write the rate law, then calculate k from each run and average. 6. Check the units of k match the overall order.

Visual explanation

Plot concentration of A against time for several runs. Each curve starts at a different height; draw the tangent at t = 0 on each. The steeper the tangent, the larger the initial rate. A log–log plot of initial rate against [A]₀ (with [B]₀ fixed) is a straight line whose gradient equals the order m.

Real-world analogy

Imagine testing which ingredient makes a cake rise. If you always add more flour and more baking powder together, you can never tell which caused the extra height. Changing one ingredient at a time is exactly the logic of initial-rate experiments.

Real-world example

Pharmaceutical process chemists screen reaction conditions using automated reactors that sample the first few percent of conversion. From these initial-rate measurements they build a rate law, which is then used to predict how the reaction will behave when scaled up from grams to hundreds of kilograms.

Why?

Why can orders differ from stoichiometric coefficients? The overall equation is a sum of elementary steps. The rate depends only on the slowest steps and on any equilibria before them, so species that appear in the equation may be absent from the rate law, and intermediates or catalysts may appear instead.

Common misconception

"The order of each reactant equals its coefficient in the balanced equation." This is only guaranteed for a single elementary step. For example, 2NO + O₂ → 2NO₂ is second order in NO, but the reaction of H₂ with Br₂ has a fractional order in Br₂. Orders must come from experiment.

Worked example

Question: For A + B → P at 298 K: run 1, [A] = 0.10, [B] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹; run 2, [A] = 0.20, [B] = 0.10, rate = 8.0 × 10⁻⁴; run 3, [A] = 0.10, [B] = 0.30, rate = 6.0 × 10⁻⁴. Find the rate law and k.

Reasoning: Runs 1 and 2: [A] doubles, rate × 4, so m = 2. Runs 1 and 3: [B] × 3, rate × 3, so n = 1. Rate = k[A]²[B]. From run 1: k = 2.0 × 10⁻⁴ ÷ (0.10² × 0.10) = 0.20.

Answer: Rate = k[A]²[B], k = 0.20 mol⁻² dm⁶ s⁻¹.

Quick check

1. If doubling [A] alone leaves the initial rate unchanged, what is the order with respect to A? Answer: Zero order, because the rate does not depend on the concentration of A at all.

Exam focus

Show the ratio working explicitly for each order and state which runs you compared. Always give units for k derived from the overall order. When concentrations change by awkward factors, use the logarithm method rather than guessing.

Advanced insight

Modern kinetic analysis often fits all runs simultaneously by non-linear regression rather than comparing pairs. A correlation matrix of the fitted parameters then reveals poor identifiability: if two parameters are correlated close to ±1, the data cannot separate them, and new experiments must be designed that vary the concentrations in a different direction.

Summary

Rate laws are found by experiment, not from the balanced equation. The initial-rate method changes one concentration at a time and compares rates to find each order, then uses any run to calculate k with units set by the overall order. A data set can only identify parameters whose effects it varies independently; flooding gives pseudo-rate constants that must be unpicked with further runs.

Practice questions

1. Tripling [B] while [A] is constant increases the initial rate ninefold. What is the order in B? Answer: Second order, because 3ⁿ = 9 gives n = 2. 2. A reaction is first order in A and first order in B. Give the units of k. Answer: The overall order is 2, so k has units of mol⁻¹ dm³ s⁻¹. 3. Every run in a study doubles [A] and [B] together. Explain what can and cannot be concluded. Answer: Only the overall order m + n can be found; the separate orders are confounded because the concentrations never vary independently. 4. Why are initial rates preferred to rates measured halfway through a reaction? Answer: At the start the concentrations are known exactly and products, reverse reactions and inhibition are negligible, so the rate corresponds to a well-defined composition.