DSE Chemistry | Rates of Reaction Tutor Prep Guide

DSE Chemistry | Rates of Reaction Tutor Prep Guide | Collision Theory and Catalysts

2026-11-01

Tutor lesson-prep guides · Secondary Chemistry

Students can recite that higher concentration reacts faster but cannot read the slope of a rate curve. This lesson makes the curve the centre: rate is the slope, and the slope falls because the reactants are being used up.

Objectives — Students can: (1) calculate the average rate over a stated interval from volume against time data; (2) explain concentration, surface area, temperature and catalysts with one collision-theory sentence each; (3) state that a catalyst changes the time taken, never the final amount.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Molar volume	Volume of 1 mol of gas at room conditions	24 dm3
Gradient	What does the slope of a curve mean	rate of change
Collisions	What two things a collision needs	enough energy and the right orientation

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Recall mol and gas volume	24 dm3 = 24000 cm3	convert 24 cm3
8-20 min	Average rate	change in volume over change in time	student gets 0.80 cm3/s
20-32 min	Why the rate falls	concentration of acid drops	student points at the slope
32-44 min	Four factors	collision theory one at a time	student explains one
44-54 min	Catalysts	lower activation energy	student draws the energy profile
54-60 min	Exit check	three questions collected and marked	done independently

03 Teaching script (read it out)

Scene 1: Rate is just the slope written as a number

Marble chips with hydrochloric acid collect 24 cm3 in 30 s, 40 cm3 at 60 s and 48 cm3 at 90 s. First 30 s: (24) / 30 = 0.80 cm3/s. Next 30 s: (40 – 24) / 30 = 0.53 cm3/s. The rate drops because the acid is being used up.

Board focus: Rate is just the slope written as a number

Scene 2: Collision theory: four factors, one sentence

Higher concentration, larger surface area, higher temperature and a catalyst all mean the same thing: more successful collisions every second. Make students say it four times in one pattern instead of memorising four paragraphs.

Board focus: Collision theory: four factors, one sentence

Scene 3: A catalyst changes the road, not the destination

The 24 cm3 collected is 0.001 mol of CO2 at 24000 cm3 per mol. With a catalyst more gas arrives in the same time, but the final volume is identical. Draw two curves, one steep and one shallow, ending at the same height.

Board focus: A catalyst changes the road, not the destination

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	average rate	change in volume over time	0.80 cm3/s
Calculation	volume to moles	V divided by 24000	0.001 mol
Explanation	why heating is faster	energy distribution shifts right	more particles pass Ea
Explanation	catalyst	alternative route, lower Ea	no change to enthalpy

Sanity check (teach this): Use one interval at a time; cumulative volume divided by total time is not the rate of that interval. Bring all gas volumes to the same temperature and pressure. A catalyst never moves the position of equilibrium, only the speed of reaching it.

05 Independent practice (3 questions)

Question	Answer	What to watch
24 cm3 converted to moles at room conditions	0.001 mol	divide by 24000
Average rate over the first 30 s	0.80 cm3/s	24 / 30
Average rate from 30 s to 60 s	0.53 cm3/s	(40 – 24) / 30
Why the curve goes flat	reactants used up	slope is zero

06 Common mistakes and what to say

Mistake	Student writes	Say this
Cumulative volume divided by total time	40 / 60 = 0.67 cm3/s quoted as the first rate	“0.67 is the average over the whole run. The first 30 s is 0.80.”
Skipping the unit conversion	treating 24 cm3 as 24 mol	“24 cm3 is 0.001 mol. That is a factor of twenty thousand.”
Thinking a catalyst increases yield	saying more product forms	“A catalyst buys speed, not amount. The final volume is the same.”
Confusing heat with concentration	saying there are more particles	“Heating does not add particles, it gives each one more energy.”

07 Homework (paper, section and question numbers)

DSE Chemistry practice paper (rates and energy) — Q5-Q6
DSE Chemistry practice paper (industrial processes) — Q10-Q11

08 Exit check (last 5 minutes)

Question	Answer	Decision
Average rate from 30 s to 60 s	0.53 cm3/s	correct means ready for equilibrium
Moles in 24 cm3 of gas	0.001 mol	wrong means redrill 24000
What a catalyst changes	the rate, not the yield	wrong means redraw the energy profile

09 Tutor FAQ

Why not use instantaneous rate?

DSE asks for average rate over a stated interval; instantaneous rate needs a tangent and only appears when the data are continuous.

How fine must the powder be?

Questions compare powder with lumps; the mark is for contact area, not for a particle size number.

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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.