Enthalpy of neutralisation

A-Level (AQA) · unreviewed

Equal volumes of acid and alkali of the same concentration mixed; temperature rise measured.

Worked example

Equal volumes (40.0 cm³ each) of HCl and NaOH, both 1.61 mol dm⁻³, were mixed in a polystyrene cup and the temperature increased by 11.1 K. Taking the density of the mixture as 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹, find ΔneutH.

1. Multiply by a coefficient: 2 × V(each solution) → Total volume of the mixture
= 80.00 cm³
2. Multiply Total volume of the mixture × density → Mass of the mixture
= 80.00 g
3. Heat from specific heat capacity: solve for Heat released
$q = mc\Delta T$
= 3712 J
4. Concentration from moles and volume: solve for Moles of water formed
$n = cV$
= 0.06440 mol
5. Divide Heat released ÷ Moles of water formed
= 57.64 kJ mol⁻¹
6. Multiply by a coefficient: -1 × qn → ΔneutH
= −57.64 kJ mol⁻¹

Answer: ΔneutH = −57.6 kJ mol⁻¹

Method

The mass heated is the whole mixture: 2V × density. q = mcΔT. The moles of water formed equal the moles of HCl (= moles of NaOH), n = cV. ΔneutH = −q/n, negative because heat is released.

Common mistakes

Both solutions are heated: the mass in q = mcΔT is that of the whole mixture (both volumes).
The temperature rose, so the reaction released heat: ΔH is negative.

Formulas used

Heat from specific heat capacity
$q = mc\Delta T$
Concentration from moles and volume
$c = \dfrac{n}{V}$

Practise this with fresh numbers

Enthalpy of neutralisation

Worked example

Method

Common mistakes

Formulas used

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