A fair comparison changes one factor and keeps everything else the same. In a table of several runs, your first job is to find the pair that differ in only one way.
This lesson closes rates of reaction. It combines graph rates, concentration, surface area and catalysts. All numbers here are fictional.
How do you test a dataset for fairness?
Ask three questions in order.
- What is the independent variable? This is the one factor the question says is being investigated.
- What is being measured? This is the dependent variable, such as time to collect a fixed volume of gas.
- Which other factors are controlled? Mass of solid, particle size, volume and concentration of acid, and temperature are common controls.
Then pick the two runs that differ in the independent variable only.
Worked example
Invented data. Four runs react calcium carbonate with dilute hydrochloric acid.
Each collects 40 cm³ of gas. The time is recorded.
| Run | Acid (mol/dm³) | CaCO₃ form (1.0 g) | Temperature (°C) | Time (s) |
|---|---|---|---|---|
| 1 | 1.0 | lumps | 30 | 40 |
| 2 | 0.5 | lumps | 30 | 85 |
| 3 | 0.5 | powder | 30 | 35 |
| 4 | 0.5 | lumps | 40 | 50 |
Question: Which runs give a fair test of concentration, and what do they show?
Step 1. Concentration is the independent variable. Runs 1 and 2 differ in concentration only, since both use lumps at 30 °C. Runs 1 and 3 differ in two factors, so they do not make a fair test.
Step 2, compare the rates. Rate can be judged as 1 ÷ time. Run 1: 1 ÷ 40 = 0.025 s⁻¹. Run 2: 1 ÷ 85 = 0.0118 s⁻¹, about 0.012 s⁻¹.
Step 3, conclude. The higher concentration gave the higher rate. Run 1 took 40 s and run 2 took 85 s.
Step 4, explain. More particles per unit volume means more frequent collisions, so more successful collisions each second.
Other fair pairs: Runs 2 and 3 test particle size (powder faster). Runs 2 and 4 test temperature (higher temperature faster). At a higher temperature particles have more energy, so collisions are more frequent and a larger proportion have enough energy.
How do repeats help?
Suppose run 1 was repeated three times and gave 40 s, 42 s and 38 s. The mean is (40 + 42 + 38) ÷ 3 = 120 ÷ 3 = 40 s.
If one repeat had given 70 s, it would be far from the others and be treated as anomalous. You would say why you excluded it and calculate the mean from the rest.
What mistake should you watch for?
Mistaken answer: “Runs 1 and 3 show that concentration matters because run 3 was faster.”
Run 3 has lower acid concentration than run 1 and yet is faster, which might even seem to contradict the concentration effect. The real difference is that run 3 used powder. Two factors changed, so the pair cannot be used.
The correction is to cross out every pair that differs in more than the variable under test before you read any times.
Check yourself
1. Which runs would you compare to test temperature, and what is the conclusion?
Show answer
Runs 2 and 4. Only temperature differs (30 °C and 40 °C). Run 4 took 50 s against 85 s, so the higher temperature gave a faster rate.
2. A student repeats a fourth run and gets 52 s, 49 s, 75 s and 51 s. Give a sensible mean.
Show answer
75 s is anomalous and is excluded with a reason. Mean = (52 + 49 + 51) ÷ 3 = 152 ÷ 3 = 50.7 s (3 s.f.).
3. Why can run 1 and run 3 not be used to test concentration?
Show answer
Both the concentration and the form of the solid differ, so you cannot tell which caused the change in time.
Where does this lead next?
Put everything together in the rates of reaction mixed practice. The rate and energy graph interpreter is useful for checking how a dataset looks as a graph, and the mole and equation-ratio tutor for amounts.
If planning a fair comparison on paper still feels slippery, a teacher can go through it step by step in online one-to-one Chemistry tuition.