Chemistry · Lesson
Measuring reaction rates and interpreting rate data
Turn a concentration change into an interval-average rate. Distinguish the disappearance of one species from the rate normalized to a written reaction.
Back to the Chemistry mapBy the end, you can
- Read a concentration–time change with its sign and units.
- Distinguish average and instantaneous rates.
- Apply stoichiometric factors under a stated constant-volume model.
Start with a quantity that can be measured
A reaction rate describes how quickly a chemical change proceeds. A useful experiment follows a quantity connected to the amounts of reactants or products: for example a calibrated light-absorption signal, gas amount or electrical conductivity. The instrument signal needs an appropriate model or calibration before it becomes a concentration.
Record the conditions and time scale. Temperature, mixing, pressure, volume and the sampled material can affect interpretation. An original concentration table can teach a calculation, but its invented values should not be presented as measured evidence.
A finite interval gives an average
If reactant A falls from 0.100 mol L⁻¹ to 0.0800 mol L⁻¹ during 10.0 s, its interval-average concentration slope is Δ[A]/Δt = −0.00200 mol L⁻¹ s⁻¹. Its positive average consumption rate is −Δ[A]/Δt = 0.00200 mol L⁻¹ s⁻¹. A product’s appearance rate is positive when its concentration rises.
The instantaneous rate comes from the local slope at a particular time. Two endpoint samples supply an average over their interval; they do not establish that the slope stayed constant or give its exact value at the midpoint. Shorter intervals and an appropriate fitted model can help estimate local rates.
State the equation before comparing species rates
For the model net reaction 2 A → B at constant volume, define r = −(1/2)d[A]/dt = d[B]/dt when this reaction alone accounts for the species changes and no accumulating intermediate disrupts that balance. A disappears twice as quickly in concentration units as B appears, although both describe the same normalized reaction rate.
Our hypothetical endpoints are [A] = 0.100 → 0.0800 mol L⁻¹ and [B] = 0.000 → 0.0100 mol L⁻¹ over 10.0 s. The average r is 0.00100 mol L⁻¹ s⁻¹. The factor 1/2 follows the written equation; it is not an arbitrary correction to the instrument.
| Quantity | Initial concentration / mol L⁻¹ | Final concentration / mol L⁻¹ | Average positive rate / mol L⁻¹ s⁻¹ |
|---|---|---|---|
| A consumed | 0.100 | 0.0800 | 0.00200 |
| B formed | 0.000 | 0.0100 | 0.00100 |
| Written-reaction rate r | Use the same two endpoints | Normalize A consumption by 2 | 0.00100 |
Original hypothetical constant-volume 2 A → B example over 10.0 s. Only the written reaction contributes; no intermediate accumulates. Endpoints determine interval averages, not instantaneous rates.
Check what the concentration change actually represents
Changing volume can change concentration without the corresponding chemical consumption. Side reactions or accumulating intermediates can also prevent one overall equation from matching every instantaneous species slope. A rate based on moles, a variable-volume balance or a more complete reaction network may then be needed.
Keep the question precise: which species, which interval, which units and which reaction basis? A rate calculation summarizes the stated observations and model. By itself it does not reveal the sequence of molecular steps.
Worked examples
Normalize an average consumption rate
For the stated constant-volume 2 A → B model, A decreases from 0.100 to 0.0800 mol L⁻¹ in 10.0 s. Find A’s average consumption rate and the normalized average reaction rate.
- Consumption concentration change = 0.100 − 0.0800 = 0.0200 mol L⁻¹.
- Divide by 10.0 s to obtain 0.00200 mol L⁻¹ s⁻¹.
- Divide that consumption rate by the coefficient 2.
A consumption: 0.00200 mol L⁻¹ s⁻¹. Normalized r: 0.00100 mol L⁻¹ s⁻¹, equal to B’s average appearance rate under the model.
Try it yourself
If [A] decreases with time, what sign has its concentration slope?
Show the answer and reasoning
Negative.
A positive consumption rate uses the negative of that declining slope.
In the stated 2 A → B model, A is consumed at 0.00200 mol L⁻¹ s⁻¹. What is normalized r?
Show the answer and reasoning
0.00100 mol L⁻¹ s⁻¹.
Divide A’s consumption rate by its coefficient 2.
Do two concentration samples establish an exact instantaneous rate at the midpoint?
Show the answer and reasoning
No.
Their difference gives an interval average; local behavior needs further evidence or a model.
Can a changing volume alter concentration even without chemical consumption?
Show the answer and reasoning
Yes.
A concentration rate must account for volume changes when they matter.
What to carry forward
- Connect the measured signal to a defined chemical quantity.
- Keep signed slopes distinct from positive consumption rates.
- State the written equation and volume assumption before normalizing.
- A rate observation does not by itself determine a mechanism.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.