A rate of reaction tells you how fast a quantity changes, so from a graph you calculate change in quantity ÷ change in time. This appears in rates of reaction whenever a question supplies a gas-volume or mass-loss graph and asks for a rate.
All the data on this page is invented for practice and does not come from a real experiment.
What does the gradient of the graph mean?
On a graph of product volume against time, the gradient is the rate. A steep line means a fast reaction and a flat line means almost no change per second.
Two kinds of rate are asked for. The average rate covers an interval. The rate at a moment uses a tangent, which is a straight line that touches the curve at one point without crossing it there.
How do you work it out, step by step?
- Read the axis units first: for example volume in cm³ and time in s.
- Choose the interval or the point the question names.
- For an average rate, subtract the two readings: change in volume ÷ change in time.
- For a rate at a moment, draw a tangent, pick two points far apart on that line and divide the change in y by the change in x.
- Write the unit as quantity per time, for example cm³/s.
Worked example
Invented data for carbon dioxide collected from a reaction:
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|---|---|
| Volume (cm³) | 0 | 24 | 40 | 50 | 56 | 59 | 60 | 60 |
Question A: Find the average rate from 0 to 10 s and from 20 to 30 s.
0 to 10 s: change in volume = 24 − 0 = 24 cm³, change in time = 10 s, so rate = 24 ÷ 10 = 2.4 cm³/s.
20 to 30 s: change in volume = 50 − 40 = 10 cm³, change in time = 10 s, so rate = 10 ÷ 10 = 1.0 cm³/s.
The rate fell because the reactants were being used up.
Question B: A tangent drawn to the curve at 20 s passes through (10 s, 28 cm³) and (30 s, 52 cm³). Find the rate at 20 s.
Gradient = (52 − 28) ÷ (30 − 10) = 24 ÷ 20 = 1.2 cm³/s.
Question C: What happens after 60 s? The volume stays at 60 cm³, so the rate is 0 cm³/s. The reaction has finished, which means a reactant has run out, and the particles have not disappeared.
What mistake should you watch for?
Mistaken working: “Rate at 20 s = 40 ÷ 20 = 2.0 cm³/s.”
This divides the volume at one time by that time. It gives the average rate from the start up to 20 s, not the rate at 20 s. The curve is steeper early on, so the answer is too high.
The correction is to use a tangent for a rate at a moment, or two readings for an interval, and to show both values you subtracted. Also check that the two points on a tangent are well apart, because close points magnify reading errors.
Check yourself
1. Using the table above, find the average rate from 30 s to 50 s.
Show answer
Change in volume = 59 − 50 = 9 cm³. Change in time = 20 s. Rate = 9 ÷ 20 = 0.45 cm³/s.
2. In another invented experiment a flask loses mass as gas escapes. Its mass falls from 50.0 g to 49.2 g in 40 s. Find the average rate.
Show answer
Change in mass = 0.8 g. Rate = 0.8 ÷ 40 = 0.020 g/s.
3. Why is the average rate for 0 to 10 s larger than for 50 to 60 s in the table?
Show answer
At the start the reactants are at their highest concentration, so successful collisions are most frequent. By 50 to 60 s most reactant has been used up, so the change per second is small.
Where does this lead next?
Next, use what the graph shows to explain the cause, starting with concentration effects using collision ideas. The rate and energy graph interpreter lets you practise reading gradients on labelled data, and the mole and equation-ratio tutor helps when you need the expected final volume.
Some students can divide correctly but hesitate over which two points to use. Our teachers look at that choice in online one-to-one Chemistry tuition.