Saying a reaction is “faster” is not an answer. This page turns your readings into a number with units — including the tangent method, which is the one students most often get wrong.
📚 What you need to know
Rate = the change in amount of substrate or product ÷ the time it took.
If you can only time an endpoint, use rate = 1 ÷ time, with units of s−1.
A graph of product against time starts as a straight line, then curves and plateaus as substrate runs out.
The gradient of the graph at any point is the rate at that moment.
To find the rate on a curve you must draw a tangent and find its gradient.
The initial rate is the gradient of the tangent drawn through the origin — the fairest value to compare between conditions.
Steeper line, faster reaction. Always give units.
Two ways to write a rate
Which one you use depends on what you were able to measure.
When you can measure an amount
rate = change in amount of substrate or product ÷ time
The amount is usually a volume or a mass, so typical units are cm3 s−1, cm3 min−1 or g s−1. If concentration is used, mol dm−3 s−1.
When you can only time the end point
rate = 1 ÷ time taken (units: s−1)
This second one is handy for experiments like the amylase and iodine test, where all you know is how long the starch took to disappear. It is not measuring a quantity, but it still lets you compare conditions fairly.
Watch the direction of the logic. A high rate means the reaction happened in less time. So a shorter time gives a bigger value of 1 ÷ time. Students often write this the wrong way round.
Why the graph bends
Plot the volume of product against time and you always get the same shape: a straight climb at the start, then a bend, then a flat plateau.
At the start there is plenty of substrate, so every active site is kept busy and product appears at a steady rate. As the reaction proceeds, substrate is used up, so collisions with active sites become less frequent, the rate falls and the line flattens. When the plateau is reached, the substrate has gone.
The triangle must sit on the straight red tangent. If you take your readings from the blue curve you will get a rate that is too low.
🧩 Drawing a tangent and getting a rate
Put a ruler on the origin. For an initial rate the tangent must pass through (0, 0).
Line it up with the early curve. Rotate the ruler until the edge lies along the first part of the curve without cutting through it.
Draw the line long. Extend it well across the graph — a longer line makes the triangle easier to read accurately.
Build a big triangle on the tangent. Choose points that sit on clear gridlines.
Read off a and b.a is the change up the y-axis, b is the change along the x-axis.
Divide and add units. Rate = a ÷ b, with the y-axis unit over the x-axis unit.
The biggest single mark-loser here is reading the triangle off the curve instead of the tangent. Draw the tangent in a different colour if you can — then there is no chance of mixing them up under exam pressure.
Comparing two conditions
Once you can measure a rate, comparing conditions is easy: the steeper the line, the faster the reaction. Notice that both curves below end up at the same plateau. They have the same amount of substrate, so they make the same amount of product in the end — one just gets there sooner.
A warmer reaction is not a bigger reaction. It reaches the same finish line, just sooner.
Worked examples
WORKED EXAMPLE
A tangent drawn through the origin passes through the point (25 s, 60 cm3). Calculate the initial rate of reaction.
Step 1: identify a and b from the tangenta = 60 cm³ (change in volume), b = 25 s (change in time)Step 2: dividerate = 60 ÷ 25 = 2.4Step 3: units come from the axes
y-axis is cm³, x-axis is s, so the unit is cm³ per second.
Initial rate = 2.4 cm³ s−¹no units, no mark — even when the number is perfect
WORKED EXAMPLE
Oxygen volume was recorded as: 0 s → 0 cm3, 30 s → 42 cm3, 90 s → 78 cm3, 120 s → 84 cm3. Compare the mean rate over the first 30 s with the mean rate over the last 30 s, and explain the difference. [4]
Step 1: first interval(42 − 0) ÷ 30 = 1.4 cm³ s−¹Step 2: last interval(84 − 78) ÷ 30 = 0.2 cm³ s−¹Step 3: compare1.4 ÷ 0.2 = 7, so the start is seven times fasterStep 4: explain
Substrate has been used up, so fewer collisions with active sites and fewer complexes form.
1.4 cm³ s−¹ falling to 0.2 cm³ s−¹ as substrate runs out“the enzyme got tired” is not a thing — the enzyme is fine, the substrate has gone
WORKED EXAMPLE
In an amylase experiment, iodine stopped turning blue-black after 40 s at 20°C and after 25 s at 30°C. Calculate both rates and say which condition was faster.
Step 1: use rate = 1 ÷ timeat 20°C: 1 ÷ 40 = 0.025 s−¹at 30°C: 1 ÷ 25 = 0.040 s−¹Step 2: compare the numbers0.040 ÷ 0.025 = 1.6 times faster30°C is faster, at 0.040 s−¹shorter time, bigger rate — check your answer makes sense before you write it
💡 Exam tip
Units, every time. Take them straight off the axes: y-axis unit divided by x-axis unit.
For an initial rate the tangent goes through the origin. For a rate at some later time, the tangent touches the curve at that time instead.
Make the triangle as large as the graph allows. Small triangles magnify your reading errors.
If the question gives a table rather than a graph, just subtract the values: change in amount ÷ change in time.
Describe a graph in two parts: what happens at first, and what happens later. Quote figures from the axes.
Round sensibly and keep the significant figures consistent with the data you were given.
⚠ Common mix-up
Reading the triangle off the curve instead of the tangent. This is the classic error and it always gives too small a rate.
Drawing the tangent as a chord. A tangent touches the curve; it should not cut through it.
Thinking the plateau means the enzyme has denatured. The plateau means the substrate has run out.
Getting 1 ÷ time upside down. A longer time gives a smaller rate.
Comparing the final volumes instead of the gradients. Rate is about steepness, not height.
Leaving off units, or mixing them — cm3 min−1 is not the same number as cm3 s−1.
Up next: Activation Energy (Skills) — why reactions need a push to get started, and how an enzyme makes that push smaller without changing the energy released.
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