Definition

Heat of neutralization (ΔHₙ) is the heat released when 1 equivalent (or 1 mole of H⁺) reacts with 1 equivalent (or 1 mole of OH⁻) to form 1 mole of water in a neutralization reaction.

General reaction:$$\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$$

• It is exothermic.
• ΔH is negative.

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Standard Heat of Neutralization

For a strong acid + strong base:$$\boxed{\Delta H = -57.1 \text{ kJ mol}^{-1}}$$

or approximately$$\boxed{-13.7 \text{ kcal mol}^{-1}}$$

Examples

$$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$$$$\mathrm{HNO_3 + KOH \rightarrow KNO_3 + H_2O}$$$$\mathrm{HBr + LiOH \rightarrow LiBr + H_2O}$$

All have$$\boxed{\Delta H \approx -57.1\ \text{kJ mol}^{-1}}$$

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Why is it always −57.1 kJ mol⁻¹?

Strong acids and strong bases completely ionize.

Example:$$\mathrm{HCl \rightarrow H^+ + Cl^-}$$$$\mathrm{NaOH \rightarrow Na^+ + OH^-}$$

Actual reaction:$$\mathrm{H^+ + Cl^- + Na^+ + OH^-}$$

↓$$\mathrm{Na^+ + Cl^- + H_2O}$$

Net ionic equation:$$\boxed{\mathrm{H^+ + OH^- \rightarrow H_2O}}$$

The spectator ions (Na⁺, Cl⁻) do not affect the heat released.

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Strong Acid + Weak Base

Example:$$\mathrm{HCl + NH_4OH \rightarrow NH_4Cl + H_2O}$$

Observed heat:

≈ −51 to −55 kJ mol⁻¹

Why smaller?

Weak base is not completely ionized.

Energy is required for ionization:$$\mathrm{NH_4OH \rightarrow NH_4^+ + OH^-}$$

Some heat is used in ionization.

Therefore,$$\boxed{\text{Heat released} < 57.1\ \text{kJ}}$$

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Weak Acid + Strong BaseExample:$$\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O}$$

Heat released:

≈ −55 kJ mol⁻¹

Reason:

Acetic acid is weak.$$\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}$$

Energy is required for ionization before H⁺ reacts.

Hence,$$\boxed{\Delta H < -57.1\ \text{kJ}}$$

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Weak Acid + Weak Base

Example:$$\mathrm{CH_3COOH + NH_4OH}$$

Heat released is the smallest among the four combinations because both reactants require ionization.

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Order of Heat Released

$$\boxed{ \text{Strong acid + Strong base} > \text{Strong acid + Weak base} \approx \text{Weak acid + Strong base} > \text{Weak acid + Weak base} }$$Strong acid + Strong base>Strong acid + Weak base≈Weak acid + Strong base>Weak acid + Weak base​

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Effect of Concentration

Heat of neutralization depends on the number of moles neutralized, not simply on concentration.

Example:

100 mL of 1 M HCl + 100 mL of 1 M NaOH

Moles reacted:$$0.1\ \text{mol}$$0.1 mol

Heat evolved:$$0.1 \times 57.1 = 5.71\ \text{kJ}$$

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Formula for Numerical Problems

Method 1

$$\boxed{q = n \times 57.1\ \text{kJ}}$$

where $n$n = moles of water formed (or moles of H⁺ neutralized).

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Method 2 (Calorimetry)

$$\boxed{q = mc\Delta T}$$

where

• $m$ = mass of solution
• $c$ = specific heat (≈ 4.18 J g⁻¹ K⁻¹)
• $\Delta T$ = temperature rise

Then,$$\Delta H = \frac{-q}{\text{moles reacted}}$$

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Polybasic Acids

H₂SO₄ + NaOH

$$\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}$$

Water formed = 2 mol

Heat released$$2 \times 57.1 = 114.2\ \text{kJ}$$

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H₃PO₄ + 3NaOH

$$\mathrm{H_3PO_4 + 3NaOH \rightarrow Na_3PO_4 + 3H_2O}$$H3​PO4​+3NaOH→Na3​PO4​+3H2​O

Maximum heat:$$3 \times 57.1 = 171.3\ \text{kJ}$$

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Heat Released per Mole of Acid

Acid	Water formed	Heat released
HCl	1	−57.1 kJ
HNO₃	1	−57.1 kJ
H₂SO₄	2	−114.2 kJ
H₃PO₄ (complete)	3	−171.3 kJ

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Important JEE/NEET Points

• Heat of neutralization is always negative (exothermic).
• It is defined per mole of water formed (or per mole of H⁺ neutralized).
• Strong acid + strong base → −57.1 kJ mol⁻¹.
• Weak acids/bases show a lower magnitude because part of the energy is used for ionization.
• Spectator ions do not affect the heat of neutralization.
• Use the limiting reagent to determine the moles of water formed.

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Previous JEE/NEET-Type Questions

Q1

100 mL of 1 M HCl is mixed with 100 mL of 1 M NaOH. Heat evolved is:

Solution:

• Moles HCl = 0.1 mol
• Moles NaOH = 0.1 mol
• Water formed = 0.1 mol

Heat:$$0.1 \times 57.1 = \boxed{5.71\ \text{kJ}}$$

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Q2

One mole of H₂SO₄ is completely neutralized by NaOH. Heat evolved is:

Water formed = 2 mol$$2 \times 57.1 = \boxed{114.2\ \text{kJ}}$$

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Q3

Which reaction has the greatest heat of neutralization?

A. HCl + NaOH
B. CH₃COOH + NaOH
C. HCl + NH₄OH
D. CH₃COOH + NH₄OH

Answer: A. HCl + NaOH (strong acid + strong base)

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JEE/NEET Revision Box

• Definition: Heat released when 1 mol of H₂O is formed by neutralization.
• Strong acid + strong base: −57.1 kJ mol⁻¹
• Weak acid/base: Lower magnitude due to ionization.
• Formula: $q = n \times 57.1\ \text{kJ}$q=n×57.1 kJ
• H₂SO₄ + 2NaOH: −114.2 kJ
• H₃PO₄ + 3NaOH: −171.3 kJ (complete neutralization)
• Always calculate using the moles of water formed (or H⁺ neutralized) and the limiting reagent.