DSE Chemistry: Reaction Rates — Tutor Lesson Plan (rate calculation, collision theory, catalysts)

2026-10-28

Tutor lesson-prep guides · Secondary chemistry

Students know heating speeds a reaction up but cannot say why. This lesson forces the phrase effective collision frequency, then explains why a catalyst cannot change the yield.

Objectives — the student can compute an average rate with units, explain four factors through collision theory, and state what a catalyst does and does not change.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Concentration units	What does mol/dm3 mean?	Moles of solute per dm3
Time units	How many minutes in 30 s?	0.5 minutes
Energy profile	Which gap is activation energy?	Reactants to the transition state

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Defining rate	Change in concentration over time	State the units
8-24 min	Rate calculation	2.0 to 1.4 mol/dm3 in 30 s	Get 0.020 mol/dm3/s
24-40 min	Collision theory	Effective collisions and activation energy	Draw the profile
40-52 min	Four factors	Concentration, temperature, surface area, catalyst	Explain each
52-56 min	Measurement	Gas volume, mass loss, colour	Choose the method
56-60 min	Exit check	Three questions marked	Finish alone

03 Teaching script (read it out)

Scene 1: A rate always carries units

Average rate is the change in concentration divided by time. From 2.0 to 1.4 mol/dm3 in 30 s: 0.6 / 30 = 0.020 mol/dm3/s.
Wrong units lose the mark, so write the whole string.

Student explains why the rate falls as the reaction proceeds.

Scene 2: Collision theory: three conditions at once

Particles must collide, collide with at least the activation energy, and collide in the right orientation to react.
Heating raises mean kinetic energy and shifts the energy distribution right, so far more particles clear the barrier.

Note: more collisions is not enough; say effective collisions.

Scene 3: A catalyst lowers the barrier, not the yield

A catalyst offers a route with lower activation energy and lowers it for the reverse reaction by the same amount, so the equilibrium position and yield are unchanged and only the time to reach it falls.
A rough rule is a doubling of rate for every 10 K, so 20 K gives about four times.

Student answers whether a catalyst raises the yield (no).

04 Question types and worked steps

Type	Example	Steps	Answer
Rate calculation	concentration and time	Change over time, with units	0.020 mol/dm3/s
Gas method	40 cm3 down to 16 cm3 in 20 s	Volume change over time	1.2 cm3/s
Explanation	powder faster than lumps	Larger surface area, more contact	Higher effective collision frequency
Catalyst	against temperature	Lower barrier against higher mean energy	No shift against possible shift

Sanity check (teach this): A rate is positive. A negative answer means the initial and final concentrations were written the wrong way round.

05 Independent practice (3 questions)

Question	Answer	What to watch
2.0 to 1.4 mol/dm3 in 30 s	0.020 mol/dm3/s	Units complete
Gas volume 40 to 16 cm3 in 20 s	1.2 cm3/s	The decrease was used
Rate increase for 20 K rise	about 4 times	Doubling rule applied

06 Common mistakes and what to say

Mistake	Student writes	Say this
No units on a rate	just 0.020	A rate is concentration over time, write the units.
Product concentration used	wrong direction	Reactants fall, so take the magnitude.
Catalyst said to raise yield	yield increases	It only speeds the approach to equilibrium.
Only more collisions mentioned	missing energy condition	Effective collisions need energy and orientation.

07 Homework (paper, section and question numbers)

DSE Chemistry practice paper (average rate) — Section A (concentration and rate)
Secondary chemistry chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Rate for 2.0 to 1.4 mol/dm3 in 30 s	0.020 mol/dm3/s	All three right: move to equilibrium
Catalyst effect on equilibrium position	none	One wrong: re-teach collision theory
Rate rise for 20 K	about 4 times	Two wrong: repeat scene 2

09 Tutor FAQ

Rate or equilibrium?

Rate is how fast, equilibrium is the final ratio; a catalyst touches only the first.

Must I draw the energy profile?

For explanation questions it saves words, especially with and without a catalyst.

Which measurement method?

Gas produced means volume, colour change means colorimetry or timing.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.