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HKDSE Chemistry | Energetics and Enthalpy Change Prep Guide (for tutors) | q = mcT, Signs and Calorimetry Errors
Students can quote q = mcT but start guessing the sign of the enthalpy change, and in experiments they treat the temperature reading as the temperature rise. This lesson pins down three things: who absorbs the heat, that enthalpy is per mole, and where the errors come from.
Objectives — Students can: (1) compute heat with q = mcT and convert to kJ; (2) divide by moles to obtain the enthalpy change with the correct sign; (3) explain how exothermic and endothermic reactions differ in temperature; (4) name three major sources of error in calorimetry.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Amount of substance | How mass relates to moles | n = m / M |
| Specific heat capacity | The value for water | 4.18 J per g per K |
| Temperature change | What a positive value means | a rise in temperature |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-12 min | Exothermic and endothermic | Read the thermometer | a rise means heat released |
| 12-28 min | q = mcT | m is the solution mass | 50 x 4.18 x 12.4 = 2,592 J |
| 28-44 min | From q to enthalpy | Convert to kJ, then divide by moles | enthalpy change -51.8 kJ/mol |
| 44-54 min | Calorimetry and errors | heat loss, heat capacity assumption, incomplete reaction | insulate and repeat readings |
| 54-60 min | Practice and exit check | three questions unaided | a sign error means redo act two |
03 Teaching script (read it out)
Scene 1: Act one: identify what absorbs the heat
In neutralisation the heat released is absorbed by the solution, so m is the total mass of solution and c is the specific heat capacity of the solution, usually taken as 4.18. Warming 50 g of water by 12.4 C absorbs 50 x 4.18 x 12.4 = 2,592 J, which is 2.59 kJ. Substituting 1 mol instead of 0.0500 mol makes the answer twenty times too large, the most common order-of-magnitude error.
Ask the student: in an exothermic reaction, what happens to the solution temperature? (It rises because heat flows from the reaction to the solution.) Student task: calculate the heat absorbed when 100 g of solution warms by 6.5 C (2.72 kJ).
Ask the student: in an exothermic reaction, what happens to the solution temperature? (It rises because heat flows from the reaction to the solution.) Student task: calculate the heat absorbed when 100 g of solution warms by 6.5 C (2.72 kJ).
Scene 2: Act two: enthalpy change is always per mole
Enthalpy change equals minus q divided by moles. Here q is 2.59 kJ and n is 0.0500 mol, giving -51.8 kJ/mol. Exothermic reactions carry a negative sign, endothermic a positive one, and writing “releases 51.8 kJ” without a sign still loses marks. If the equation coefficients change, the enthalpy change scales with them: forming 2 mol of water releases twice as much as forming 1 mol.
Ask the student: does a negative enthalpy change mean the products are higher or lower in energy? (Lower, so the surplus leaves as heat.) Student task: does the enthalpy change double if only 0.0250 mol reacts? (No, it is a per-mole quantity.)
Ask the student: does a negative enthalpy change mean the products are higher or lower in energy? (Lower, so the surplus leaves as heat.) Student task: does the enthalpy change double if only 0.0250 mol reacts? (No, it is a per-mole quantity.)
Scene 3: Act three: errors must be explained mechanistically
A calorimeter is never perfectly insulated, so heat escapes, the measured temperature rise is too small and the magnitude of the enthalpy change is underestimated. Warming 150 g of solution by 4.7 C absorbs 2.95 kJ; losing a tenth of the heat means measuring only nine tenths of it. Other errors: counting the solute mass twice, assuming a density of exactly 1 g per cm3, incomplete reaction, and delayed thermometer readings.
Ask the student: why use one thermometer for both initial and final readings? (To avoid the systematic offset between two instruments.) Student task: list three ways to improve insulation and say which error each one reduces.
Ask the student: why use one thermometer for both initial and final readings? (To avoid the systematic offset between two instruments.) Student task: list three ways to improve insulation and say which error each one reduces.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Heat calculation | given mass, specific heat and temperature rise | q = mcT, keep joules | 2,592 J |
| Heat to enthalpy | given heat and moles | convert to kJ then divide, add a minus sign for exothermic | -51.8 kJ/mol |
| Classifying reactions | given temperature data | a rise means exothermic | a rise gives a negative value |
| Error analysis | given an experimental set-up | look for heat loss and assumptions | insulate, repeat readings |
Sanity check (teach this): 2,592 J for 50 g of water rising 12.4 C is the right order of magnitude, since 1 g needs 4.18 J per degree. -51.8 kJ/mol sits close to the standard strong acid-strong base value of about minus 57 kJ/mol; an answer of minus 570 or minus 5.7 means the mole division is wrong. Water has a much larger specific heat capacity than metals, which is why a water calorimeter is more stable than a metal one. Kelvin and Celsius give the same numerical temperature difference, but Fahrenheit does not.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Heat absorbed by 50 g of water rising 12.4 C | 2.59 kJ | q = mcT |
| Enthalpy change when 0.0500 mol releases that heat | -51.8 kJ/mol | exothermic takes a minus sign |
| Heat absorbed by 150 g of solution rising 4.7 C | 2.95 kJ | unit conversion |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Using the solute mass for m | substituting 2 g of sodium hydroxide only | “The solution absorbs the heat, so m is the total mass of solution, not the solute.” |
| Dropping the negative sign | answering plus 51.8 kJ/mol | “Exothermic must be negative; a plus sign claims heat is absorbed.” |
| Mixing joules and kilojoules | dividing joules by moles directly | “Convert to kJ before dividing by moles, otherwise the answer is out by a factor of 1000.” |
| Using the final temperature as the change | writing a change of 32.4 C | “The change is final minus initial, not the final reading itself.” |
07 Homework (paper, section and question numbers)
- Secondary Chemistry chapter index — energetics set, questions 1 and 3
- Prep guide master index — other chapters in this subject
- HKDSE Chemistry practice papers — enthalpy questions 1 to 4
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| In q = mcT, what does m stand for | the total mass of solution absorbing the heat | three correct means move to combustion |
| Enthalpy change when 0.0500 mol releases 2.59 kJ | -51.8 kJ/mol | one wrong means reteach the sign |
| Main source of error in calorimetry | heat lost to the surroundings | two or more wrong means return to act one |
09 Tutor FAQ
Why is the enthalpy of neutralisation nearly constant?
Strong acids and bases are fully ionised in water, so the reaction is always H+ plus OH- forming water, giving about minus 57 kJ/mol.
Why is the value smaller for a weak acid?
A weak acid is only partly ionised, so energy is first absorbed to ionise it, offsetting part of the heat released.
Can the temperature change alone give the enthalpy change?
No; the heat must also be divided by the number of moles, because temperature change reflects heat only, not amount of substance.
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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.
