3.1.4Inquiry Process
Enzyme Reaction Rates- Skills — IB Biology HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 104 of 160
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5What Rate Means
A rate is always a change in amount divided by the time taken.
Rate of reaction
- Rate of reaction The change in amount of product, or substrate, divided by the time taken, for example cm³ of oxygen per second.
Units
- Units Rate is written with units such as cm³ s⁻¹ or g min⁻¹, and the unit must always be included in the answer.
Initial rate
- Initial rate The rate at the very start, found from the steepest part of the graph, when substrate is plentiful and little product has formed.
Tangent method
- Tangent method A tangent drawn to the curve at time zero has a gradient equal to the initial rate, found as change in y divided by change in x.
Part 2 of 5Reading a Product Curve
1
At the start the curve is steep because many substrate molecules and free active sites are available.
2
As substrate is used up, collisions between substrate and active sites become less frequent and the gradient decreases.
3
Product may also affect the rate, for example if it changes the pH of the mixture.
4
Eventually all of the substrate has been converted and the curve becomes horizontal, so the rate is zero.
5
Comparing the gradients at different times, or for different conditions, shows which condition gives the faster reaction.
Part 3 of 5Comparing Conditions
Comparing Conditions
Steeper curve
- Steeper curve A steeper line at the start means a higher initial rate under that condition.
Same final level
- Same final level If the same amount of substrate was used, all curves reach the same final amount of product even though they get there at different speeds.
Different final level
- Different final level A lower plateau shows that less substrate was available, not just that the enzyme was slower.
Rate against concentration
- Rate against concentration Plotting initial rate against substrate concentration shows the rate rising and then reaching a maximum.
| Time interval (s) | Volume of O₂ (cm³) | Change (cm³) | Rate (cm³ s⁻¹) |
|---|---|---|---|
| 0 to 30 | 8.0 | 8.0 | 0.27 |
| 90 to 120 | 20.0 (from 18.0) | 2.0 | 0.067 |
Part 4 of 5Worked Example
Question
- Oxygen from catalase breaking down hydrogen peroxide was collected.
- After 30 s the volume was 8.0 cm³, after 90 s it was 18.0 cm³, and after 120 s it was 20.0 cm³.
- Calculate the mean rate for the first 30 s, the rate between 90 s and 120 s, and the mean rate for the whole 120 s.
- First 30 s: 8.0 / 30 = 0.27 cm³ s⁻¹ (to two significant figures).
- From 90 s to 120 s: (20.0 - 18.0) / (120 - 90) = 2.0 / 30 = 0.067 cm³ s⁻¹.
- Whole experiment20.0 / 120 = 0.17 cm³ s⁻¹.
- The rate falls with time because the substrate is being used up.
Answer 0.27 cm³ s⁻¹ at the start, 0.067 cm³ s⁻¹ at the end, and 0.17 cm³ s⁻¹ as the overall mean.
Part 5 of 5Exam Checklist & Tips
3.1.4 – Key Points for the Exam
- Initial rate The rate of reaction at the start, when substrate concentration is highest.
- Tangent A straight line that touches a curve at one point and has the same gradient there.
- Gradient Change in y divided by change in x on a graph.
- Plateau The flat part of a graph where there is no further change.
Common Mistakes
- Leaving out units from a rate, or writing them as cm³ instead of cm³ s⁻¹.
- Taking the mean rate over the whole experiment when the question asks for the initial rate.
- Drawing a tangent that cuts the curve instead of touching it at one point.
- Explaining a fall in rate as the enzyme wearing out rather than the substrate being used up.
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