DSE Chemistry | Enthalpy Change and Hess Law: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary Chemistry

Students add signs at random and end with the wrong sign on delta H. This lesson holds two rules: q = mc delta T always measures heat released, so delta H must be negative, and Hess law means aligning the direction and coefficients of known equations before adding them.

Objectives — Students can: (1) find q = 2730 J for 100 g of water rising 6.5 degrees; (2) convert 2730 J from 0.0500 mol into delta H of minus 54.6 kJ per mole; (3) use minus 394 for carbon to carbon dioxide and minus 283 for carbon monoxide to carbon dioxide to get minus 111 for carbon to carbon monoxide; (4) use enthalpies of formation to get minus 891 kJ per mole for methane combustion.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Specific heat capacity	How much energy raises 1 g of water by 1 degree?	4.2 J
Exothermic and endothermic	Does a temperature rise mean heat is released?	Yes, so delta H is negative
Enthalpy of formation	What is the enthalpy of formation of an element in its standard state?	Zero
Mole conversion	0.0500 mol releases 2730 J; what is that per mole?	54.6 kJ, so minus 54.6

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening question	Ask where the heat goes when something burns	Say the water absorbs it
8-22 min	Calorimetry	100 g of water from 22.0 to 28.5 degrees	Compute q and delta H per mole
22-34 min	Sign convention	Why delta H must be negative here	Decide the sign before calculating
34-46 min	Hess by equations	Two combustion values to get carbon to carbon monoxide	Reverse the second equation then add
46-54 min	Formation method	Methane at minus 891	Sum the formation values
54-60 min	Independent practice	Three questions in eight minutes	Check and label errors

03 Teaching script (read it out)

Scene 1: Calorimetry gives q; the sign on delta H is your job

q = mc delta T, so 100 g rising 6.5 degrees is 100 times 4.2 times 6.5, which is 2730 J. Water warming means the reaction released heat, so delta H is negative. Dividing 2730 J by 0.0500 mol gives 54 600 J per mole, or 54.6 kJ per mole, written as minus 54.6. Make students write the sign sentence before touching numbers: water warms, heat released, delta H below zero. Judging the sign first beats patching it afterwards.

Ask the student for the sign of delta H when the water cools from 30 to 24 degrees, and why.

Scene 2: Hess law treats equations as blocks that must be aligned first

The target is carbon plus half an oxygen giving carbon monoxide. Known: carbon to carbon dioxide is minus 394, and carbon monoxide to carbon dioxide is minus 283. Reverse the second to carbon dioxide giving carbon monoxide plus half an oxygen, so its value becomes plus 283. Adding cancels carbon dioxide and leaves the target, with minus 394 plus 283 equal to minus 111. Every change to an equation changes its value: reverse flips the sign, multiply scales it. Write those two rules in the margin.

Ask the student why carbon dioxide cancels once the second equation is reversed.

Scene 3: Enthalpy is a state function, so the path is free to choose

Enthalpy change depends only on the start and end states, which is the whole reason Hess law works. Using formation values for methane combustion: products give minus 394 plus two times minus 286, or minus 966; the reactant gives minus 75; subtracting gives minus 891 kJ per mole. Students must own the phrase products minus reactants and use the fact that a combustion must be negative to catch a reversed order.

Ask the student why the formation method needs no energy diagram.

04 Question types and worked steps

Type	Example	Steps	Answer
Calorimetry	100 g of water up 6.5 degrees, 0.0500 mol	q = mc delta T, divide by moles, fix the sign	Minus 54.6 kJ per mole
Two-equation Hess	Carbon to carbon dioxide minus 394, monoxide minus 283	Reverse the second, cancel, flip its sign	Minus 111 kJ per mole
Formation method	Methane with values minus 75, minus 394, minus 286	Sum products, sum reactants, subtract	Minus 891 kJ per mole
Neutralisation or solution	0.100 mol releases 4200 J	q = mc delta T, divide by moles, negative	Minus 42.0 kJ per mole

Sanity check (teach this): Order check: 100 g of water rising 6.5 degrees needs about 2700 J, and dividing by 0.05 mol gives roughly fifty-something kJ, which is the right order for a combustion value. Carbon to carbon monoxide at minus 111 should lie between minus 394 and minus 283 because it is partial combustion; minus 677 means the second equation was added without reversing. Methane at minus 891 matches the textbook near minus 890, and plus 891 means products and reactants were swapped.

05 Independent practice (3 questions)

Question	Answer	What to watch
Delta H when 200 g rises 5.0 degrees for 0.100 mol	Minus 42.0 kJ per mole	Divide by moles before converting to kJ
Methane combustion from formation values minus 394 and minus 286	Minus 891 kJ per mole	Products minus reactants
Delta H when 0.25 mol releases 13.65 kJ	Minus 54.6 kJ per mole	Scale by four and keep the sign

06 Common mistakes and what to say

Mistake	Student writes	Say this
Dropping the sign on delta H	\”The water warms, so delta H is plus 54.6\”	\”The water gained heat, so the reaction released it; releasing means negative. A plus sign says the reaction absorbed its own heat.\”
Reversing an equation without flipping its value	\”I reversed the second equation and kept minus 283\”	\”Reversing means travelling the other way, so the energy changes sign too; without that the sum is guaranteed wrong.\”
Products and reactants swapped	\”Reactants minus products gives plus 891\”	\”It is products minus reactants, and a combustion must be negative; your plus sign already disproves the answer.\”
Wrong mass in q = mc delta T	\”I used 0.0500 mol as the mass\”	\”m is the mass of the water being heated, not the amount of reactant; the units are not even the same.\”

07 Homework (paper, section and question numbers)

Chemistry chapter index — Revise mole calculations, then questions 1 and 2
Mole and stoichiometry — Question 1 needs accurate moles before dividing energy
Rates of reaction — Question 3 contrasts rate with enthalpy, which are different ideas

08 Exit check (last 5 minutes)

Question	Answer	Decision
q for 100 g of water up 6.5 degrees	2730 J	Error: mass left out
Delta H for 0.0500 mol	Minus 54.6 kJ per mole	Error: missing negative
Delta H for carbon to carbon monoxide	Minus 111 kJ per mole	Error: equations added directly
Methane combustion delta H	Minus 891 kJ per mole	Error: order reversed

09 Tutor FAQ

Why use delta H rather than q?

q belongs to one experiment and scales with amounts, while delta H is the enthalpy change per mole and can be compared across reactions.

What if the calorimeter loses heat?

Measured q falls short; calibrate with a substance of known combustion enthalpy or accept the question assumption of no loss, and state it.

How does Hess law relate to energy diagrams?

The diagram plots start, intermediate and end heights; going around equals going direct, which is Hess law drawn as a picture.

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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.