The heat change of a reaction in solution is found from the temperature rise of the solution. You calculate the heat Q with Q = mcθ, divide by the moles that reacted, and then give the sign from the thermometer evidence.
This lesson is part of SPM Chemistry thermochemistry. It relies on the ideas in distinguishing exothermic and endothermic changes.
What is the layout of the working?
Write each quantity on its own labelled line. This makes each slip visible.
- Mass m of the solution in g.
- Temperature change θ in °C.
- Heat Q = m × c × θ in J, with c = 4.2 J g⁻¹ °C⁻¹ unless the question gives another value.
- Moles of the reactant that is limiting, or of the product named in the definition.
- ΔH = −Q ÷ moles for a temperature rise, converted to kJ/mol.
Worked example: neutralisation
An original experiment mixes 50 cm³ of 1.0 mol/dm³ hydrochloric acid with 50 cm³ of 1.0 mol/dm³ sodium hydroxide solution. The temperature rises by 6.5 °C.
Mass: 50 + 50 = 100 cm³, so m = 100 g.
Heat: Q = 100 × 4.2 × 6.5 = 2 730 J.
Moles of water formed: 50 ÷ 1 000 × 1.0 = 0.050 mol.
ΔH: the temperature rose, so ΔH is negative. ΔH = −2 730 ÷ 0.050 = −54 600 J/mol = −54.6 kJ/mol.
The mistakes that cost marks
Two slips account for most lost marks.
| Quantity | Wrong | Right |
|---|---|---|
| Mass | 50 g (only the acid) | 100 g (whole mixture) |
| Moles | 0.10 mol (acid plus alkali) | 0.050 mol (water formed) |
| Unit | −54 600 kJ/mol | −54.6 kJ/mol |
With the wrong mass of 50 g, Q would be 1 365 J, and the final answer would be half the correct value. A labelled line for each quantity makes such a slip easy to see.
Why does a real result differ from a book value?
Some heat escapes to the air and the cup, so the measured temperature rise is smaller than it should be. This makes the calculated size of ΔH smaller than the true value. The sources of error are covered in explaining sign conventions and sources of error.
Check yourself
100 cm³ of 0.50 mol/dm³ copper(II) sulfate solution reacts with excess zinc powder and the temperature rises by 10.5 °C. Use c = 4.2 J g⁻¹ °C⁻¹ and treat the mass as 100 g. Calculate ΔH per mole of copper(II) sulfate.
Answer
Q = 100 × 4.2 × 10.5 = 4 410 J.
Moles of CuSO₄ = 100 ÷ 1 000 × 0.50 = 0.050 mol.
Temperature rose, so ΔH is negative: ΔH = −4 410 ÷ 0.050 = −88 200 J/mol = −88.2 kJ/mol.
The numbers are invented for practice, so do not compare this value with a data book.
What to study next
Continue with explaining sign conventions and sources of error, then try the thermochemistry practice set.
The graph evidence and fair-comparison lab helps with temperature-time graphs. If you want a teacher to check your working line by line, see online one-to-one Chemistry tuition.