Rate from Concentration-Time Data

Introduction

In a reaction such as 2N₂O₅ → 4NO₂ + O₂, nitrogen dioxide appears four times as fast as oxygen, and dinitrogen pentoxide disappears twice as fast as oxygen appears. Which of these is "the rate"? Chemists solve the ambiguity by dividing each observed slope by its stoichiometric coefficient, giving a single rate of reaction for the whole equation. This page shows how to extract slopes from data and how to normalise them so that different measurements agree.

Core explanation

Observed slopes. An experiment usually follows one species — by colour, pressure, conductivity or titration — and produces a concentration–time table or graph. The average rate of change over an interval is Δ[X]/Δt. The instantaneous rate is the gradient of the tangent to the curve at a particular time. For reactants the slope is negative; for products it is positive.

Normalising by coefficients. For a general reaction aA + bB → cC + dD, the rate of reaction is defined as

rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = (1/c) d[C]/dt = (1/d) d[D]/dt

The negative signs make the rate positive for reactants; dividing by coefficients makes all four expressions equal. The rate of reaction is thus a property of the balanced equation, and its value depends on how that equation is written.

Example. In 2N₂O₅ → 4NO₂ + O₂, if [N₂O₅] falls by 3.0 × 10⁻³ mol dm⁻³ in 100 s, then −d[N₂O₅]/dt = 3.0 × 10⁻⁵ mol dm⁻³ s⁻¹. The rate of reaction is half of this, 1.5 × 10⁻⁵ mol dm⁻³ s⁻¹. NO₂ forms at 4 × 1.5 × 10⁻⁵ = 6.0 × 10⁻⁵ mol dm⁻³ s⁻¹, and O₂ forms at 1.5 × 10⁻⁵ mol dm⁻³ s⁻¹.

Average versus instantaneous. Because rates usually fall as reactants are used up, the average rate over an interval lies between the instantaneous rates at its ends. A good estimate of the instantaneous rate at time t is the central difference: the change between a point just before and a point just after t, divided by that time interval. Consider:

t / s 0 50 100 200 --- --- --- --- --- [A] / mol dm⁻³ 0.100 0.0819 0.0670 0.0449

The average rate of loss of A from 0 to 50 s is (0.100 − 0.0819)/50 = 3.62 × 10⁻⁴ mol dm⁻³ s⁻¹, and from 50 to 100 s it is 2.98 × 10⁻⁴. The central-difference estimate at 50 s is (0.100 − 0.0670)/100 = 3.30 × 10⁻⁴ mol dm⁻³ s⁻¹, close to the tangent value.

Initial rate. The tangent at t = 0 gives the initial rate, when concentrations are known exactly. Initial rates are the basis for finding reaction orders.

Formulae

rate = −(1/a) Δ[A]/Δt = (1/c) Δ[C]/Δt; units mol dm⁻³ s⁻¹.

Step-by-step reasoning

1. Write the balanced equation and note each coefficient. 2. Calculate the observed slope Δ[X]/Δt for the measured species, with its sign. 3. Divide by the coefficient of X (and change sign for a reactant) to get the rate of reaction. 4. Multiply the rate of reaction by any other coefficient to find that species' rate of change. 5. State units: mol dm⁻³ s⁻¹.

Visual explanation

On a single graph, draw falling curves for reactants and rising curves for products. The curves have different steepness at the same moment. Scaling each vertical axis by its coefficient makes all the curves collapse onto one shape, whose gradient is the rate of reaction.

Real-world analogy

A sandwich shop uses two slices of bread and one slice of cheese per sandwich. Bread disappears twice as fast as cheese, yet both describe the same thing: sandwiches made per minute. Dividing by the "coefficient" converts ingredient usage into the single rate of sandwich production.

Real-world example

In the bromate–bromide reaction, BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O, the formation of bromine can be followed by its colour. The rate at which bromide is consumed is not measured directly but is calculated as five thirds of the rate of bromine formation.

Why?

Why define the rate of reaction with coefficients rather than choose one species? A single, species-independent rate makes it possible to write one rate equation and one rate constant for the reaction, whichever species is measured in the laboratory.

Common misconception

"The rate of the reaction is the rate at which the measured species changes." That is true only when the measured species has a coefficient of 1. Otherwise the observed slope must be divided by the coefficient.

Worked example

Question: For BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O, bromine forms at 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Find the rate of reaction and the rates of consumption of Br⁻ and H⁺.

Reasoning: Rate of reaction = (1/3) × 3.0 × 10⁻⁴ = 1.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Br⁻ is consumed at 5 × 1.0 × 10⁻⁴ = 5.0 × 10⁻⁴ mol dm⁻³ s⁻¹, and H⁺ at 6 × 1.0 × 10⁻⁴ = 6.0 × 10⁻⁴ mol dm⁻³ s⁻¹.

Answer: Rate = 1.0 × 10⁻⁴; Br⁻ 5.0 × 10⁻⁴; H⁺ 6.0 × 10⁻⁴ mol dm⁻³ s⁻¹.

Quick check

1. In N₂ + 3H₂ → 2NH₃, ammonia forms at 0.040 mol dm⁻³ s⁻¹. How fast is hydrogen consumed? Answer: Rate of reaction = 0.040/2 = 0.020 mol dm⁻³ s⁻¹, so H₂ is consumed at 3 × 0.020 = 0.060 mol dm⁻³ s⁻¹.

Exam focus

Examiners test the coefficient conversion almost every time rates are given. Write the defining expression with fractions and signs, then substitute. When reading slopes from graphs, draw a clear tangent, quote the two points used and include units in the gradient.

Advanced insight

When volume changes during a reaction, as in an open gas-phase flow system, the rate of change of concentration is not purely chemical because dilution also alters concentration. The rigorous definition then uses the extent of reaction: rate = (1/V) dξ/dt, which reduces to the concentration form only at constant volume.

Summary

Observed slopes from concentration–time data give rates of change for individual species. Dividing each by its stoichiometric coefficient, with a sign change for reactants, gives one rate of reaction for the balanced equation. Average rates cover an interval, instantaneous rates come from tangents or central differences, and the initial rate is the tangent at time zero.

Practice questions

1. For 2H₂O₂ → 2H₂O + O₂, oxygen forms at 2.5 × 10⁻⁵ mol dm⁻³ s⁻¹. How fast is hydrogen peroxide consumed? Answer: Rate of reaction = 2.5 × 10⁻⁵, so H₂O₂ is consumed at 2 × 2.5 × 10⁻⁵ = 5.0 × 10⁻⁵ mol dm⁻³ s⁻¹. 2. Using the table on this page, calculate the average rate of loss of A between 100 s and 200 s. Answer: (0.0670 − 0.0449)/100 = 2.21 × 10⁻⁴ mol dm⁻³ s⁻¹. 3. In 4NH₃ + 5O₂ → 4NO + 6H₂O, oxygen is consumed at 0.10 mol dm⁻³ s⁻¹. Find the rate of formation of water. Answer: Rate of reaction = 0.10/5 = 0.020; water forms at 6 × 0.020 = 0.12 mol dm⁻³ s⁻¹. 4. Explain why the average rate from 0 to 50 s in the table exceeds the instantaneous rate at 50 s. Answer: The rate falls as A is used up, so the average over the interval includes faster early rates and is larger than the rate at the end of the interval.