DSE Chemistry | Chemical Equilibrium Lesson-Prep Guide (Tutor Edition)

DSE Chemistry | Chemical Equilibrium Lesson-Prep Guide (Tutor Edition) | Kc, Le Chatelier and Haber

2026-11-01

Tutor lesson-prep guides · Secondary Chemistry

Students hear equilibrium and assume equal concentrations. This lesson turns equilibrium into numbers with one ICE table, then explains why industrial conditions are always a compromise.

Objectives — Students can (1) use an ICE table to move from initial to equilibrium concentrations; (2) write a Kc expression with gases and aqueous species only; (3) use Le Chatelier for concentration, pressure and temperature changes and explain the Haber trade-offs.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Molarity	Concentration when 1 mol dissolves in 2 dm3	0.5 mol dm-3
Quadratics	Positive root of x squared equals 0.04	x = 0.2
Reversible reactions	What does the reversible arrow mean	Forward and backward both occur

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Definition	Equal rates, constant concentrations, not equal values	State three features
8-22 min	ICE table	Initial, change, equilibrium rows	Fill in 0.20 mol dm-3
22-36 min	Kc	Products over reactants, coefficients as powers	Kc = 64
36-50 min	Le Chatelier	Concentration, pressure, temperature, catalyst	Predict four shifts
50-56 min	Haber process	Yield against rate	Give three reasons
56-60 min	Exit check	Three questions marked on the spot	Work independently

03 Teaching script (read it out)

Scene 1: Equilibrium means equal rates, not equal amounts

At equilibrium the forward and reverse rates are equal and every concentration stays constant, but the values are usually different. The reaction has not stopped; the net change is zero.
A catalyst only shortens the time taken to reach equilibrium. It changes neither the position nor Kc.

Ask students: does adding a catalyst change Kc? (No, only the time.)

Scene 2: The ICE table: three rows and the answer falls out

H2 + I2 giving 2HI, starting at 1.00 mol dm-3 each, with [HI] = 1.60 mol dm-3 at equilibrium.
HI rose by 1.60, so by the 1:1:2 ratio H2 and I2 each fell by 0.80 to 0.20 mol dm-3.
Kc = [HI] squared divided by [H2][I2] = 1.60 squared over 0.20 times 0.20 = 64. Conversion is 0.80 over 1.00, that is 80 per cent.

Ask students: starting with 1.00 of HI and none of the others, how much HI remains at equilibrium? (Work back from the same Kc.)

Scene 3: Le Chatelier: four situations, one table

Add a reactant shifts right; remove a product shifts right; raise pressure shifts towards fewer gas moles; raise temperature shifts towards the endothermic side.
For N2 + 3H2 giving 2NH3 there are 4 gas moles on the left and 2 on the right, so pressure helps, and Kc = [NH3] squared over [N2][H2] cubed = 1.00 squared over 0.50 times 1.50 cubed = 0.593.
Haber: 450 C balances rate against yield, 200 atm pushes the yield up, and iron is the catalyst that shortens the time to equilibrium.

Reminder: raising temperature on an exothermic reaction lowers the yield but raises the rate. Answer the two questions separately.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Kc from equilibrium concentrations	Expression first, coefficients as powers	64
Calculation	ICE table to equilibrium values	Initial, change, equilibrium	0.20 mol dm-3
Decision	Direction of shift	Four Le Chatelier situations	Right or left
Explanation	Haber trade-offs	Yield, rate and cost	450 C, 200 atm, iron

Sanity check (teach this): Kc includes gases and aqueous species only; pure solids and pure liquids are left out. An equilibrium concentration must be below its starting value unless the reaction ran backwards, so a value above the start usually means the ICE table signs are reversed. Kc has no units in DSE and its size only shows the product ratio.

05 Independent practice (3 questions)

Question	Answer	What to watch
Kc when H2 and I2 start at 1.00 and [HI] = 1.60	64	Halve the change first
Kc for [N2] = 0.50, [H2] = 1.50, [NH3] = 1.00	0.593	Coefficients become powers
Effect of raising pressure on N2 + 3H2 giving 2NH3	Shifts right	Count gas moles

06 Common mistakes and what to say

Mistake	Student writes	Say this
Equilibrium treated as equal concentrations	Answered that [H2] equals [HI] at equilibrium	Equal rates, not equal concentrations.
Missing powers in Kc	Kc written as [NH3] over [N2][H2]	The coefficients 2 and 3 become squared and cubed.
Solid included in Kc	CaCO3 written into the expression	A pure solid has fixed concentration and is omitted.
Heating assumed to raise yield	Said heating raises the yield of an exothermic reaction	Heating shifts to the endothermic side, so an exothermic yield falls.

07 Homework (paper, section and question numbers)

DSE Chemistry practice paper (equilibrium and rate) — Section B Kc and explanation questions
Secondary Chemistry chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Kc when H2 and I2 start at 1.00 and [HI] = 1.60	64	Three correct means move on to acid-base equilibrium
Effect of raising pressure on ammonia synthesis	Shifts right	One wrong means re-teach Le Chatelier
Effect of a catalyst on Kc	No change	Two or more wrong means restart act one

09 Tutor FAQ

Does Kc have units?

DSE treats it as unitless; give the number only.

Why not run Haber at a lower temperature?

The yield would be higher but far too slow to be economic.

Has the reaction really stopped at equilibrium?

No. Both directions continue at equal rates with zero net change.

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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.