Determining Reaction Orders from Trials

Introduction

The balanced equation tells you what reacts, but not how the rate depends on each concentration. That dependence, expressed in the rate equation, must be found by experiment. The most common approach is to run several trials with different starting concentrations and compare their initial rates. The skill lies in choosing the right pairs of trials so that only one variable changes at a time, just as a controlled experiment isolates a single factor.

Core explanation

The rate equation. For a reaction involving A and B, rate = k[A]^m[B]^n. The orders m and n are usually 0, 1 or 2, but can be fractional; they are not generally equal to the coefficients in the balanced equation.

Isolating one reactant. Pick two trials in which [B] is identical and only [A] differs. Dividing the rate equations cancels k and [B]^n:

rate₂/rate₁ = ([A]₂/[A]₁)^m

So if [A] doubles and the rate doubles, m = 1; if the rate quadruples, m = 2; if the rate is unchanged, m = 0.

Using logarithms. When the ratios are not simple, take logs:

m = log(rate₂/rate₁) ÷ log([A]₂/[A]₁)

For example, a concentration ratio of 1.5 with a rate ratio of 2.25 gives m = log 2.25/log 1.5 = 2.

Example data set for 2NO + 2H₂ → N₂ + 2H₂O at a fixed temperature:

Trial [NO] / mol dm⁻³ [H₂] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹ --- --- --- --- 1 0.0100 0.0100 1.2 × 10⁻⁵ 2 0.0200 0.0100 4.8 × 10⁻⁵ 3 0.0100 0.0300 3.6 × 10⁻⁵

Trials 1 and 2: [NO] doubles, rate × 4, so second order in NO. Trials 1 and 3: [H₂] triples, rate × 3, so first order in H₂. Rate = k[NO]²[H₂], overall third order — note that the order in H₂ is not its coefficient of 2.

The rate constant. Substitute any trial: k = 1.2 × 10⁻⁵ ÷ (0.0100² × 0.0100) = 12 dm⁶ mol⁻² s⁻¹. Units follow from rearranging: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹. Calculating k from every trial and checking that the values agree is an excellent test of the orders.

When no clean pair exists. If both concentrations change between two trials, first determine one order from another pair, then divide out its known effect before finding the other.

Formulae

rate₂/rate₁ = ([A]₂/[A]₁)^m; m = log(rate ratio)/log(concentration ratio); units of k = (mol dm⁻³)^(1 − overall order) s⁻¹.

Step-by-step reasoning

1. Find two trials where only [A] changes; calculate the concentration and rate ratios. 2. Deduce m by inspection or with logarithms. 3. Repeat for each other reactant. 4. Write the rate equation and calculate k from one trial, with units. 5. Check k with another trial, or predict a rate for a new composition.

Visual explanation

Draw a grid with [A] on one axis and [B] on the other. Each trial is a point. A useful pair of trials lies on a horizontal or vertical line: moving along it changes only one concentration. Diagonal moves mix two effects and need more careful unpicking.

Real-world analogy

A baker wants to know whether more yeast or more sugar makes dough rise faster. Changing both at once reveals nothing about either. Changing only the yeast in one pair of batches, and only the sugar in another, isolates each effect — exactly the logic of initial-rate trials.

Real-world example

The acid-catalysed iodination of propanone is a classic: the rate depends on [propanone] and [H⁺] but is independent of [I₂]. The zero order in iodine shows that iodine is not involved in the slow step, providing direct evidence about the reaction mechanism.

Why?

Why use initial rates rather than rates later in the reaction? At t = 0 the concentrations are known exactly from how the mixture was made, and products have not yet accumulated to cause a reverse reaction or interfere. Later in the run both effects add uncertainty.

Common misconception

"The orders are the coefficients in the balanced equation." They are found only from experiment. In the NO/H₂ reaction, H₂ has coefficient 2 but order 1, because the rate reflects the mechanism's slow step rather than the overall stoichiometry.

Worked example

Question: For the iodination of propanone, trials give: (1) [propanone] 0.40, [H⁺] 0.20, [I₂] 0.0020, rate 4.0 × 10⁻⁶; (2) [propanone] 0.80, other values as trial 1, rate 8.0 × 10⁻⁶; (3) [I₂] 0.0040, other values as trial 1, rate 4.0 × 10⁻⁶; (4) [propanone] 0.80, [H⁺] 0.40, [I₂] 0.0020, rate 1.6 × 10⁻⁵ (all mol dm⁻³ and mol dm⁻³ s⁻¹). Find the rate equation and k.

Reasoning: Trials 1 and 2: propanone doubles, rate doubles, so first order. Trials 1 and 3: I₂ doubles, rate unchanged, so zero order. Trials 2 and 4: only H⁺ doubles, rate doubles, so first order. k = 4.0 × 10⁻⁶ ÷ (0.40 × 0.20) = 5.0 × 10⁻⁵ dm³ mol⁻¹ s⁻¹.

Answer: Rate = k[CH₃COCH₃][H⁺], with k = 5.0 × 10⁻⁵ dm³ mol⁻¹ s⁻¹.

Quick check

1. When [A] is increased by a factor of 3 with all else constant, the initial rate rises by a factor of 9. What is the order in A? Answer: Since 3^m = 9, the order is m = 2, second order in A.

Exam focus

State which trials you compare and why ("[B] is constant between trials 1 and 2"). Give k with units derived from the overall order. When asked to predict a rate, use the full rate equation with k rather than chaining ratios, to avoid compounding errors.

Advanced insight

When one reactant is in large excess its concentration barely changes, so the rate equation collapses to pseudo-first-order form, rate = k′[A] with k′ = k[B]ⁿ. Measuring k′ at several excess concentrations of B and plotting log k′ against log[B] gives n as the gradient — a graphical version of the log-ratio method.

Summary

Orders are found by comparing initial rates in trials where only one concentration changes. The rate ratio equals the concentration ratio raised to the order, and logarithms handle awkward numbers. Orders need not match coefficients. The rate constant is then calculated from any trial, with units set by the overall order, and checked against the others.

Practice questions

1. Using the NO/H₂ table, predict the initial rate when [NO] = 0.0300 and [H₂] = 0.0200 mol dm⁻³. Answer: Rate = 12 × (0.0300)² × 0.0200 = 2.16 × 10⁻⁴ mol dm⁻³ s⁻¹. 2. Increasing [B] by a factor of 2.0 raises the rate by a factor of 2.83. Find the order in B. Answer: Order = log 2.83/log 2.0 ≈ 1.5. 3. Give the units of k for an overall second-order reaction with concentrations in mol dm⁻³ and time in seconds. Answer: dm³ mol⁻¹ s⁻¹. 4. Doubling [C] leaves the initial rate unchanged. What does this suggest about C and the mechanism? Answer: The reaction is zero order in C, suggesting that C is not involved in, or before, the rate-determining step.