IB Biology HLEnzymes & MetabolismPaper 1 & 2~11 min read
Enzyme Reaction Rates (Skills)
Describing a curve is one thing. Getting a number off it is another, and it is the part students lose marks on. This page is about turning a graph into a rate, with the right units attached.
📚 What you need to know
Rate can be found from the substrate disappearing or the product accumulating in a given time.
Rate = change in amount ÷ time. Units follow whatever you measured, for example cm3 s−1.
If you only have a time, rate = 1 ÷ time taken, with units of s−1.
A high rate means the reaction finished in less time.
A product-against-time graph starts as a straight line and then plateaus, because substrate is being used up as the reaction proceeds.
The gradient at any point is the rate at that moment. The steeper the line, the faster the reaction.
To find the initial rate, draw a tangent through the origin along the early part of the curve, then calculate its gradient.
Gradient = change in y ÷ change in x, read off as far along the tangent as is convenient.
Two ways to write a rate
If you can measure how much substrate disappeared or how much product appeared, use the amount directly.
Rate from an amount
rate of reaction = change in amount of reactant or product ÷ time
Sometimes you cannot measure an amount at all. In the iodine test for starch, for example, all you really record is how long the reaction took. In that case:
Rate from a time
rate of reaction = 1 ÷ time taken (s), giving units of s−1
This second version catches people out because it feels backwards. It is not: a reaction finishing in a short time is a fast reaction, and dividing by a small number gives a big answer. A reaction taking 20 s has a rate of 0.05 s−1; one taking 50 s has a rate of 0.02 s−1. The first is faster, and the number says so.
If the question gives you volumes or masses, use amount ÷ time. If all you have is a stopwatch reading, use 1 ÷ time. Look at what you were actually given before you pick a formula.
Why the graph bends
Plot the volume of product against time and you get the same shape almost every time: a straight climb that gradually flattens out.
At the start, substrate is plentiful and active sites are constantly occupied, so product appears at a steady rate — a directly proportional straight line. As the reaction proceeds the substrate concentration falls, collisions with active sites become less frequent, and the reaction slows. Eventually the substrate runs out completely and the line goes flat.
Read the flat part correctly. A plateau on a product-against-time graph means the substrate has run out. A plateau on a rate-against-substrate-concentration graph means the active sites are saturated. Same shape, completely different reason — check which axes you are looking at.
Finding the initial rate
The initial rate is the rate at the very start, when nothing has been used up yet. It is the fairest number to compare between experiments, which is why examiners ask for it.
🧩 Drawing the tangent
Put a ruler on the origin.
Line it up so it follows the curve during the early part of the reaction.
Draw the line and extend it as far as is convenient to read from — further along means smaller reading errors.
Pick a point on the tangent and drop lines to both axes to make a triangle.
Read off the change in y (call it a) and the change in x (call it b).
Initial rate = a ÷ b. Write the units.
The tangent and the curve part company almost immediately — that separation is the reaction slowing down as substrate disappears.
Draw the triangle big. If you read a rise of 8 over a run of 2, one millimetre of ruler error wrecks your answer. Read 80 over 20 instead and the same slip barely matters.
Comparing two reactions
Once you can read a gradient, comparing conditions becomes easy: put both curves on the same axes and look at which one is steeper at the start.
A common exam trap: students say the cooler reaction “made less product”. It made the same amount — it just took longer.
Worked examples
WE 1
Calculate an initial rate from a tangent
A tangent drawn through the origin of a product-against-time graph passes through the point (20 s, 80 cm3). Calculate the initial rate of reaction. (2 marks)
Step 1: read the triangle
Change in volume, a = 80 cm3. Change in time, b = 20 s.
Step 2: divideinitial rate = 80 ÷ 20 = 4.0Step 3: units
Volume per second, so the units are cm3 s⁻¹.
initial rate = 4.0 cm³ s⁻¹the tangent starts at the origin, so you do not subtract anything — the point itself gives both changes
WE 2
Calculate a rate from a time
Using the iodine test, starch was fully broken down in 40 s at 20°C and in 25 s at 30°C. Calculate both rates and comment. (3 marks)
Step 1: at 20°Crate = 1 ÷ 40 = 0.025 s⁻¹Step 2: at 30°Crate = 1 ÷ 25 = 0.040 s⁻¹Step 3: comment
The rate is higher at 30°C because the molecules have more kinetic energy, so successful collisions with the active site are more frequent.
0.025 s⁻¹ and 0.040 s⁻¹ — warmer is fastergive the same number of decimal places for both so the comparison is fair
WE 3
Find a rate part-way through
On a product-against-time graph, the volume of oxygen rises from 20 cm3 at 10 s to 60 cm3 at 40 s. Calculate the mean rate over this interval and explain why it is lower than the initial rate. (3 marks)
Step 1: the two changes
Change in volume = 60 − 20 = 40 cm3. Change in time = 40 − 10 = 30 s.
Step 2: dividerate = 40 ÷ 30 = 1.33 cm³ s⁻¹Step 3: why it is lower
Substrate has been used up, so its concentration has fallen. There are fewer collisions with active sites, so the reaction has slowed.
1.33 cm³ s⁻¹ — slower because substrate is running outsubtract both values here; only a tangent through the origin lets you skip that step
💡 Exam tips
Write the units every time. A correct number with no units usually drops a mark.
For an initial rate, the tangent must pass through the origin.
Use a big triangle and state the coordinates you read from, so an examiner can follow you.
If you are given a time only, reach for 1 ÷ time and use s−1.
Say steeper = faster when comparing curves — it is the phrase mark schemes use.
Watch the axes before you explain a plateau. Product-vs-time means substrate ran out; rate-vs-substrate means saturation.
⚠ Common mistakes
Thinking a longer time means a faster reaction. More time means slower.
Measuring the gradient of the curve instead of the tangent. You cannot read a gradient off a curved line directly.
Drawing a tiny triangle. Small readings magnify every error.
Forgetting to subtract when the interval does not start at zero.
Saying a slower reaction produces less product. With the same substrate it produces the same amount, just later.
Mixing up the axes. Time goes on the x-axis; amount of product or substrate goes on the y-axis.
Up next: Activation Energy (Skills). You can now measure how fast a reaction goes. The next page explains what an enzyme is actually doing to the energy of the reaction to make that speed possible.
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