HKDSE Chemistry | Chemical equilibrium and Le Chatelier teaching guide

HKDSE Chemistry | Chemical equilibrium and Le Chatelier teaching guide | Kc = 64, 80% conversion, 15% Haber yield

2026-11-03

Tutor lesson-prep guides · Secondary chemistry

Students can recite that rates become equal at equilibrium, then plug the initial moles straight into Kc. One set of numbers, 0.50 and 0.50 going to 0.80, takes the class from the definition to Kc = 64 and then back to why industry accepts a 15% yield.

Objectives — Students can: (1) explain that equilibrium is dynamic, with equal forward and reverse rates rather than a stopped reaction; (2) build an ICE table and calculate Kc = 64 from the equilibrium concentrations; (3) predict the shift under concentration, pressure and temperature changes, and explain why the Haber process runs at only about 15% conversion per pass.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Reversible reactions	What does the symbol mean?	Both directions proceed at once
Molar concentration	How is concentration calculated?	Moles divided by volume in dm3
Exothermic reactions	What does a negative enthalpy change mean?	Energy is released

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	What equilibrium is	Dynamic, equal rates, constant concentrations	Stress that it never stops
8-20 min	ICE table	Initial, change, equilibrium rows	Change row follows coefficients
20-32 min	Kc calculation	Substitute equilibrium values to get 64	Never use initial values
32-42 min	Le Chatelier	Concentration, pressure, temperature	No shift when gas moles are equal
42-52 min	Haber trade-off	Low temperature and high pressure help yield, not rate	Catalysts do not change Kc
52-60 min	Exit check	Three questions in class	Every student states 80%

03 Teaching script (read it out)

Scene 1: Act 1: equilibrium is dynamic, not stopped

Ask the class what equilibrium means. Most will say the reaction has stopped. Correct that first: both directions still run, but at equal rates, so the concentrations stop changing.
Use a door analogy: ten people enter and ten leave every minute, so the number inside is constant while people keep moving.
Then give the exam wording: forward and reverse rates are equal but not zero, the system is closed, temperature is fixed, and macroscopic properties such as colour, pressure and concentration stay steady.

Blackboard line: equal rates, constant concentrations.

Scene 2: Act 2: ICE tables and Kc, one line at a time

Write H2(g) + I2(g) ⇌ 2HI(g) with 0.50 mol dm-3 of each starting material and 0.80 mol dm-3 of HI at equilibrium.
The change row follows the coefficients: HI goes from 0 to 0.80, so 0.80 is formed and each reactant loses 0.40, leaving 0.10 of each.
Kc = [HI]2 / ([H2][I2]) = 0.64 / 0.01 = 64.
Conversion = 0.40 / 0.50 x 100% = 80%. Note that Kc happens to be unitless here; in general it carries units.

Make them compute Kc = 64 from 0.80 themselves.

Scene 3: Act 3: Le Chatelier and the industrial trade-off

Le Chatelier in one line: a system shifts to oppose the disturbance.
Adding a reactant or removing a product shifts right. Increasing pressure shifts towards the side with fewer gas moles; here both sides have 2, so there is no shift, which is the most common error. Raising temperature shifts towards the endothermic direction; this forward reaction is exothermic, so Kc falls and the mixture shifts left.
Haber process: N2 + 3H2 ⇌ 2NH3, enthalpy change -92 kJ mol-1. Low temperature favours yield but is far too slow, so plants use about 450 C, 200 atm and an iron catalyst, achieving roughly 15% per pass with unreacted gas recycled. A catalyst speeds up reaching equilibrium only; it changes neither Kc nor the final yield.

Remember: equal gas moles means pressure does not shift it.

04 Question types and worked steps

Type	Example	Steps	Answer
ICE table	Find Kc from initial and equilibrium data	Three rows, convert by coefficients, use equilibrium values	0.64 / 0.01 = 64
Direction of shift	Effect of increasing pressure	Count gas moles on both sides first	Two against two, so no shift
Temperature effect	How Kc changes on heating	Decide whether the forward reaction is exothermic	Exothermic, so Kc decreases
Industrial question	Why not run the Haber process colder	Trade yield against rate	Too slow, so 450 C is used

Sanity check (teach this): Kc uses equilibrium values only; substituting 0.50 gives 2.56, not 64.
Count gas moles before predicting a pressure effect: when the change is zero there is no shift, only a faster approach.
A catalyst changes neither Kc nor the yield.
Conversion is amount reacted over amount started: 0.40 / 0.50 = 80%.

05 Independent practice (3 questions)

Question	Answer	What to watch
Value of Kc	64	0.64 divided by 0.01
Conversion of H2	80%	0.40 divided by 0.50
Moles of HI at equilibrium in 1 dm3	0.80 mol	Concentration times volume
Doubling pressure: relation between Q and Kc	Q equals Kc, no shift	Numerator and denominator both times four

06 Common mistakes and what to say

Mistake	Student writes	Say this
Using initial concentrations in Kc	Answering 2.56	“Kc only reads the equilibrium row; the initial row is history.”
Assuming pressure always shifts right	Answering that pressure shifts to products	“Count the gas molecules first: two against two means no shift.”
Treating a catalyst as raising yield	Answering that the catalyst shifts equilibrium	“A catalyst gets you there faster, it does not move the destination.”
Calculating conversion from product	Answering 160%	“Conversion measures reactant used up, not product formed.”

07 Homework (paper, section and question numbers)

Secondary chemistry chapter index — Questions 1-4, ICE tables and Kc
All secondary prep guides — Questions 5-6, predicting shifts
DSE chemistry practice — Question 7, Haber trade-off in 60 words

08 Exit check (last 5 minutes)

Question	Answer	Decision
Value of Kc	64	Wrong means redo the ICE table
Effect of pressure when gas moles are equal	No shift, only faster	Wrong means repeat act three
Haber conversion per pass	About 15%	Cannot answer means revisit the trade-off

09 Tutor FAQ

Does the reaction really stop at equilibrium?

No. Both directions continue at equal rates, which is why concentrations stay constant.

Does Kc change with concentration?

No. Kc depends on temperature only; a concentration change moves Q away from Kc and the system shifts back.

Why does a catalyst not change the yield?

It speeds up both directions equally, so equilibrium arrives sooner at the same position.

Find a tutor — free Tutor Registration →

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.