IP Chemistry Upper Sec 10: Rate of Reactions

Study guide26 Nov 2025, 00:00 ZUpdated 02 Jun 2026

Factors affecting reaction rate, catalysts, and data handling for IP Sec 3-4 Chemistry (O-Level 6092, 2026).

Download PDF Join our Telegram study group

These notes align with SEAB GCE O-Level Chemistry (6092) content used in IP programmes (exams from 2026).

Status: SEAB O-Level Chemistry 6092 syllabus (exams from 2026) checked 2026-06-02 - scope unchanged; remains the reference for this note.

6092 Route Map

Use this page for Topic 10 Rate of Reactions: factor explanations, collision theory, activation energy, catalysts, and experiment data interpretation.

Need	Best Eclat page
Official 6092 paper format and topic order	O-Level Chemistry 6092 syllabus guide
Sec 3 foundation before rate graphs	Sec 3 Chemistry notes
Rate practical setups and data collection	Rate of Reaction Experiments for O-Level Chemistry
Paper 2 answer layout and timing	O-Level Chemistry Paper 2 structured question strategy

The core idea is simple: Reaction rate depends on successful collisions per second.

Use it as a working check: Higher concentration, pressure, or surface area increases collision frequency. Higher temperature also increases the fraction of particles with enough energy.

Then go one layer deeper: Example: powdered calcium carbonate reacts faster than chips because more surface is exposed, so acid particles collide with the solid more often.

What you must know

Factors: concentration/pressure, temperature, surface area, catalysts (including enzymes).

Reviewed by

Azmi·Senior Chemistry Specialist

Sources

SEAB GCE O-Level Chemistry (6092) syllabus (examinations from 2026)

Factor changed	Collision-theory phrase to use	What the graph should show
Higher concentration of a solution	More reacting particles per unit volume, so collision frequency increases.	Steeper initial gradient if the limiting reagent amount is unchanged.
Higher gas pressure	Gas particles are closer together, so collision frequency increases.	Steeper initial gradient for a gas reaction.
Smaller solid pieces or powder	Larger exposed surface area, so particles collide with the solid more often.	Faster reaction, but the same final gas volume if the same mass of solid reacts completely.
Higher temperature	Particles move faster and a larger fraction have energy at least $E_a$ .	Much steeper initial gradient; the curve reaches the plateau sooner.
Catalyst added	Alternative pathway with lower $E_a$ , so more collisions are successful.	Steeper curve; catalyst is still present after the reaction.

Question clue	What changed	Collision-theory answer move	Common trap
Temperature increases	Particles have more kinetic energy, so a larger fraction have energy $\geq E_a$ .	Say collision frequency increases and more collisions are successful.	Saying only "particles collide more often" and missing the energy fraction.
Catalyst is added	The reaction uses an alternative pathway with lower $E_a$ .	Say more collisions have enough energy for the new pathway and the catalyst is regenerated.	Saying the catalyst gives particles energy.
Concentration increases	The number of particles per unit volume increases.	Say collision frequency increases if other conditions stay fixed.	Drawing a lower activation-energy barrier for concentration.
Surface area increases	More solid particles are exposed to the other reactant.	Say collisions with the solid surface happen more often.	Saying each particle has more energy.

Graph feature	What it means	Answer move	Common trap
Area to the right of $E_a$	Particles with enough energy for successful collisions	Larger area means a larger fraction of successful collisions.	Comparing only the peak height of the curve.
Higher-temperature curve	Peak is lower and shifted right; tail beyond $E_a$ is larger	Say particles have higher kinetic energy and more collisions exceed $E_a$ .	Saying every particle now has energy above $E_a$ .
Catalyst added at same temperature	$E_a$ line shifts left for the alternative pathway	Say more particles exceed the lower activation energy.	Redrawing the whole distribution as if temperature changed.
Same temperature, no catalyst	Distribution and $E_a$ are unchanged	Do not claim the rate changes without a changed condition.	Treating a labelled diagram as proof of a new rate.

Step	What to write	Why it matters
Keep the end point fixed	Stop timing when the same cross just disappears.	The turbidity threshold must be comparable between trials.
Convert time to rate proxy	Use $\dfrac{1}{t}$ , not $t$ .	Larger $\dfrac{1}{t}$ means a faster reaction.
Compare with units	Quote values in $\pu{s-1}$ .	The unit shows that reciprocal time, not time, is being compared.

Graph clue	What it means	How to answer
Steeper initial gradient	Faster initial rate.	Compare the slope near $t = 0$ , not the final height.
Same final plateau	Same amount of product formed or same amount of reactant used.	Say the limiting reagent amount was the same, even if one reaction finished faster.
Plateau reached earlier	Reaction finished sooner.	Link this to a faster rate, not to more product.
Rate at a stated time	Gradient of the tangent at that time.	Draw or imagine a tangent and use $\text{gradient} = \dfrac{\text{change in y}}{\text{change in x}}$ .

IP Chemistry Upper Sec 10: Rate of Reactions

6092 Route Map

What you must know

Sources

Factor-choice checkpoint

Energy barrier checkpoint

Maxwell-Boltzmann sketch checkpoint

Reciprocal-time checkpoint

Rate Graph Reading Checkpoint

Fair-test design checkpoint

Detailed notes

Worked walkthroughs

Pitfalls and fixes

Practice drills

Quick applications

Exam cues

Next steps

Related Posts

Planning part	What to state	Example for marble chips and acid
Independent variable	The factor deliberately changed.	Concentration of hydrochloric acid.
Dependent variable	The result used to compare rate.	Volume of carbon dioxide collected per unit time.
Controlled variables	Other factors that affect collision frequency or energy.	Same mass and size of marble chips, same acid volume, same temperature.
Rate comparison	How the data will be compared.	Compare the initial gradient of gas-volume against time graphs.