The sign of ΔH comes from the direction of the temperature change. A temperature rise means exothermic and a negative ΔH. A temperature fall means endothermic and a positive ΔH.
This lesson follows reading energy-level diagrams in the thermochemistry chapter. It connects the diagram to real measurements.
Worked example: dissolving a salt
A student dissolves 0.05 mol of a salt in 100 cm³ of water. The temperature falls by 3.0 °C. Take the specific heat capacity of water as 4.2 J g⁻¹ °C⁻¹ and the density of water as 1 g cm⁻³.
Particles. As the salt dissolves, the particles absorb energy from the water, so the water cools.
Heat. Q = mcθ = 100 × 4.2 × 3.0 = 1 260 J.
Moles. The amount dissolved is 0.05 mol.
ΔH. ΔH = Q ÷ n = 1 260 ÷ 0.05 = 25 200 J mol⁻¹ = +25.2 kJ mol⁻¹.
The sign is positive because the temperature fell, so the change is endothermic. The figure θ = 3.0 was used as a positive size, not as −3.0.
Worked example: a neutralisation
50 cm³ of 1.0 mol dm⁻³ hydrochloric acid is mixed with 50 cm³ of 1.0 mol dm⁻³ sodium hydroxide. The temperature rises by 6.5 °C.
The ionic equation is H⁺ + OH⁻ → H₂O. Each solution contains 0.05 mol, so 0.05 mol of water forms.
Q = 100 × 4.2 × 6.5 = 2 730 J. Then ΔH = −2 730 ÷ 0.05 = −54 600 J mol⁻¹ = −54.6 kJ mol⁻¹. The sign is negative because the temperature rose.
Why might the experimental value differ?
Some of the heat released warms the cup and the air, not only the solution. The measured temperature rise is therefore smaller than it would be in perfect insulation.
A smaller rise gives a smaller Q, so the size of ΔH is smaller than the theoretical value. This is why a neutralisation result is usually less negative than the accepted figure.
To reduce the loss:
- Use a polystyrene cup, which is a poor conductor of heat.
- Add a lid to cut heat loss by convection.
- Stir gently and record the highest steady temperature.
The mistake that costs marks
A common slip is to write ΔH = −25.2 kJ mol⁻¹ for a dissolving process because “the temperature went down, so negative”. The fall in temperature means energy came in from the water, so ΔH is positive.
| Observation | Heat flow | Sign of ΔH |
|---|---|---|
| Temperature rises | out of the system | negative |
| Temperature falls | into the system | positive |
Check yourself
50 cm³ of 0.50 mol dm⁻³ copper(II) sulfate solution reacts with excess zinc. The temperature rises by 25.0 °C. Find ΔH per mole of copper(II) sulfate.
Answer
Moles of copper(II) sulfate = 0.050 × 0.50 = 0.025 mol.
Q = 50 × 4.2 × 25.0 = 5 250 J.
ΔH = −5 250 ÷ 0.025 = −210 000 J mol⁻¹ = −210 kJ mol⁻¹.
The sign is negative because the temperature rose.
What to study next
Test the whole chapter with the thermochemistry practice set. The mole and stoichiometry steps tool can help you check the moles. Log slips in the mistake log.
If sign and error questions keep losing marks, see online one-to-one Chemistry tuition.