Enthalpy of Neutralisation: Calorimetry Numericals Step by Step

By Aniket Bhardwaj · 11 October 2026 · Updated 11 October 2026 · Class 11 Chemistry

When an acid and a base react, the mixture gets warm. Measuring that temperature rise lets you calculate the enthalpy of neutralisation, a standard practical and a common Class 11 numerical. The working is only three steps: find the heat absorbed by the solution, find the moles of water formed, then divide. This article shows each step with the arithmetic.

The diagram follows Example 1 from 100 g of solution and a 6.7 °C rise to 2.80 kJ, 0.050 mol of water, and ΔH = −56.0 kJ/mol. — Divide the heat absorbed by the solution by the moles of water formed, then give the reaction heat a negative sign.

Definition and key idea

The enthalpy of neutralisation is the enthalpy change when an acid and a base react to form one mole of water. For a strong acid with a strong base, the ionic reaction is the same every time:

H⁺(aq) + OH⁻(aq) → H₂O(l) ΔH ≈ −57 kJ per mole of water

The value is usually quoted as about −57 kJ/mol (some books give −57.1 or −57.3 kJ/mol; use the value in your textbook). The sign is negative because the reaction releases heat. The other ions (Na⁺, Cl⁻ and so on) are spectators and do not take part, which is why every strong acid and strong base pair gives nearly the same value.

The calorimetry method

q = m × c × ΔT
Moles of water formed = moles of limiting reagent
ΔH_{neutralisation} = − q / n (in kJ per mole)

m is the mass of the final solution in grams. For dilute solutions, take the density as 1.00 g/mL, so 100 mL of solution has a mass of 100 g.
c is the specific heat capacity. For dilute aqueous solutions use 4.18 J g⁻¹ K⁻¹, the value of water.
ΔT is the rise in temperature (final minus initial). A change of 1 °C equals a change of 1 K.
n is the number of moles of water formed. Use the limiting reagent. Moles = molarity × volume in litres.

The solution gains the heat that the reaction loses, so the reaction's heat change is the negative of q. A temperature rise therefore means a negative ΔH.

Example 1: 50 mL of 1.0 M HCl is mixed with 50 mL of 1.0 M NaOH. The temperature rises by 6.7 °C.
Step 1: Total volume = 100 mL, so m = 100 g.
Step 2: q = 100 × 4.18 × 6.7 = 2800.6 J = 2.80 kJ.
Step 3: Moles of HCl = 1.0 × 0.050 = 0.050 mol. Moles of NaOH = 0.050 mol. They are equal, so n(H₂O) = 0.050 mol.
Step 4: ΔH = −2.80 / 0.050 = −56.0 kJ/mol.
Answer: ΔH_{neutralisation} ≈ −56.0 kJ/mol.

Example 2: 100 mL of 0.50 M HCl is mixed with 100 mL of 0.50 M NaOH. The temperature rises by 3.3 °C.
Step 1: Total volume = 200 mL, so m = 200 g.
Step 2: q = 200 × 4.18 × 3.3 = 2758.8 J = 2.76 kJ.
Step 3: Moles of HCl = 0.50 × 0.100 = 0.050 mol, so n(H₂O) = 0.050 mol.
Step 4: ΔH = −2.759 / 0.050 = −55.2 kJ/mol.
Answer: ΔH ≈ −55.2 kJ/mol. The value is a little below the textbook figure because some heat escapes in a real experiment.

Example 3: What temperature rise should you expect? Use 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH, with ΔH = −57.1 kJ/mol.
Step 1: Moles of water = 0.050 mol.
Step 2: Heat released = 57.1 × 0.050 = 2.855 kJ = 2855 J.
Step 3: ΔT = q / (m c) = 2855 / (100 × 4.18) = 6.83 °C.
Answer: about 6.8 °C rise. Example 1's 6.7 °C is close to this, as it should be.

Example 4: Unequal amounts, with a limiting reagent. 40 mL of 1.0 M HCl is mixed with 60 mL of 1.0 M NaOH. The temperature rises by 5.4 °C.
Step 1: Moles of HCl = 1.0 × 0.040 = 0.040 mol. Moles of NaOH = 1.0 × 0.060 = 0.060 mol. HCl is limiting.
Step 2: n(H₂O) = 0.040 mol (not 0.060).
Step 3: q = 100 × 4.18 × 5.4 = 2257.2 J = 2.257 kJ.
Step 4: ΔH = −2.257 / 0.040 = −56.4 kJ/mol.
Answer: ΔH ≈ −56.4 kJ/mol.

Weak acids and weak bases

When a weak acid (such as ethanoic acid) or a weak base takes part, the measured heat is usually different from the strong-strong value. Part of the energy goes into ionising the weak electrolyte, which is not complete at the start. So the neutralisation enthalpy for weak acids and bases is generally numerically smaller than the strong-strong figure, though exceptions exist. For board-level questions, remember the general idea and use the values given in the question.

Common mistakes

Dividing by the wrong moles. Use the moles of water formed, which equals the limiting reagent, not the larger amount.
Using only one volume for the mass. The mass is for the total mixture (acid + base), not just one solution.
Mixing joules and kilojoules. q = m c ΔT gives joules (with c in J). Divide by 1000 before writing kJ/mol.
Forgetting the sign. The solution gets hotter, but the reaction releases heat. ΔH is negative.
Using the average temperature or the wrong ΔT. Use the maximum temperature reached minus the starting temperature of the mixture. In a careful experiment, both solutions start at the same temperature.
Ignoring heat loss and the cup. Real experiments lose some heat, so the measured value is often a little low. Foam cups with lids reduce this, but do not remove it.
Calling it "heat of reaction per mole of acid." The standard definition is per mole of water formed. For H₂SO₄ (two H⁺ per molecule), the moles of water are twice the moles of acid.

Exam relevance

Question type	Steps
Find ΔH from temperature rise	q = mcΔT, then n of water, then ΔH = −q/n
Predict temperature rise	Moles of water × ΔH gives heat, then ΔT = q/(mc)
Why the measured value differs from −57 kJ/mol	Heat loss, calorimeter absorbing heat, impure solutions, weak acid or base
Practical viva	Why a lid is used, why both solutions start at the same temperature, why stirring is needed

This calculation uses the same ideas as the specific heat capacity problems in the related article below. Check your own practical file for the exact format and the value of c your teacher expects.

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