JEE Mains Chemistry · Chemical Thermodynamics

Enthalpies of Phase Change, Solution and Neutralisation

Enthalpy changes that make no new compound: melting, boiling and sublimation, dissolving and diluting, hydrating a salt, and neutralising H⁺ with OH⁻.

All Chemical Thermodynamics notes JEE Mains Chemistry strategy

Why this matters

Sixteen PYQs, eleven of them numerical, none yet from 2026. Nine deal with phase changes, heating paths, solution, dilution and hydration; seven with the heat of neutralisation and the temperature rise it causes, where the limiting reagent decides the answer.

Concept 1 of 2: Enthalpies of phase change, solution, dilution and hydration

At its transition temperature a substance takes in heat without getting hotter. The energy loosens the attractions between molecules (potential energy) instead of speeding them up (kinetic energy). Enthalpy is a state function, so any path can be split into warming steps and phase steps and added up.

Definition

$\Delta_{\mathrm{sub}}H = \Delta_{\mathrm{fus}}H + \Delta_{\mathrm{vap}}H$ , all at the same temperature.
Heating or cooling path: add $nC_p\Delta T$ for each phase and $\pm\Delta H$ for each change of phase. Freezing and condensing take the negative sign.
Enthalpy is extensive: heat = moles × molar enthalpy.
The heat of solution depends on how much solvent is used. Heat of dilution = $\Delta H$ (more dilute) − $\Delta H$ (more concentrated).
Hydration of an anhydrous salt: $\Delta_{\mathrm{hyd}}H = \Delta_{\mathrm{sol}}H(\text{anhydrous}) - \Delta_{\mathrm{sol}}H(\text{hydrate})$ .
Endothermic: breaking bonds, melting, vaporising, subliming, dissolving $\mathrm{NH_4Cl}$ . Exothermic: freezing, condensing, burning, dissolving a gas such as HCl.

Hess cycles for phase changes and hydration

\Delta_{\mathrm{sub}}H = \Delta_{\mathrm{fus}}H + \Delta_{\mathrm{vap}}H,\qquad \Delta_{\mathrm{hyd}}H = \Delta_{\mathrm{sol}}H_{\mathrm{anhydrous}} - \Delta_{\mathrm{sol}}H_{\mathrm{hydrate}}

Worked example

Find the enthalpy change to turn 1 mol of ice at 0 °C into steam at 100 °C.

\Delta_{\mathrm{fus}}H = 6.0

kJ mol⁻¹,

C_p

(liquid water) = 75 J K⁻¹ mol⁻¹,

\Delta_{\mathrm{vap}}H = 40.7

kJ mol⁻¹.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q51Moderate

Example 1 · Chemical Thermodynamics · Enthalpies of Phase Change, Solution and Neutralisation

The heat of solution of anhydrous

CuSO_{4}

and

CuSO_{4}\cdot 5H_{2}O

are

- 70\text{ }kJ{\text{ }mol}^{- 1}

and

+ 12\text{ }kJ{\text{ }mol}^{- 1}

respectively. The heat of hydration of

CuSO_{4}

CuSO_{4}\cdot 5H_{2}O

-x\text{ }kJ

. The value of

x

is ________

The sign of a heat of dilution

Heat of dilution is the more dilute value minus the more concentrated one. Going from −60 to −65 kJ mol⁻¹ gives −5 kJ mol⁻¹, not +5.

Kilojoules beside joules in a heating path

\Delta_{\mathrm{fus}}H

is quoted in kJ mol⁻¹ but

C_p

in J K⁻¹ mol⁻¹. Convert one of them before adding, or the latent heat comes out a thousand times too small.

Drill 8 more on enthalpies of phase change, solution, dilution and hydration

Concept 2 of 2: Enthalpy of neutralisation and the temperature rise

A strong acid and a strong base are fully ionised, so the only reaction on mixing is H⁺ + OH⁻ → H₂O. That gives the same heat, about 57 kJ, for every mole of water formed. Count the water, not the acid: whichever of H⁺ or OH⁻ runs out first sets it.

Definition

$\mathrm{H^+(aq) + OH^-(aq) \to H_2O(l)}$ , $\Delta H = -57.1$ kJ mol⁻¹ for a strong acid with a strong base.
Moles of water = the smaller of mol $\mathrm{H^+}$ and mol $\mathrm{OH^-}$ . $\mathrm{H_2SO_4}$ gives two $\mathrm{H^+}$ per mole.
Heat $q = n_{\mathrm{water}} \times 57.1$ kJ; temperature rise $\Delta T = \frac{q}{mc}$ , with $m$ the TOTAL mass of the mixture (1 g per mL).
A weak acid or weak base releases less, because part of the heat is spent ionising it. Ionisation enthalpy = 57.1 − | $\Delta H$ (weak)| kJ mol⁻¹.

Temperature rise on neutralisation

\Delta T = \frac{n_{\mathrm{H_2O}}\,|\Delta_{\mathrm{neut}}H|}{m\,c}

Worked example

150 mL of 0.4 M HCl is mixed with 100 mL of 0.25 M NaOH. Find the temperature rise. (

\Delta_{\mathrm{neut}}H = -57.3

kJ mol⁻¹,

c = 4.2

J g⁻¹ K⁻¹, density 1 g mL⁻¹)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 20 · Q58Moderate

Example 2 · Chemical Thermodynamics · Enthalpies of Phase Change, Solution and Neutralisation

200\text{ }mL

0.2MHCl

is mixed with

300\text{ }mL

0.1MNaOH

. The molar heat of neutralization of this reaction is

- 57.1\text{ }kJ

. The increase in temperature in

\ ^{\circ}C

of the system on mixing is

x \times10^{- 2}

. The value of

x

\_\_\_\_

. (Nearest integer) [Given: Specific heat of water

= 4.18\text{ }J{\text{ }g}^{- 1}{\text{ }K}^{- 1}

Density of water

\left. \ = 1.00\text{ }g{\text{ }cm}^{- 3} \right\rbrack

(Assume no volume change on mixing)

Only one solution's volume

The heat warms the whole mixture. 100 mL of acid plus 100 mL of base is 200 g of solution, not 100 g.

More acid is not more heat

Heat follows the limiting reagent. 50 mL of acid with 20 mL of base neutralises less than 30 mL with 30 mL, and it warms a larger volume, so its temperature rise is smaller.

Drill 6 more on enthalpy of neutralisation and the temperature rise

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Enthalpies of phase change, solution, dilution and hydration
Hess cycles for phase changes and hydration
$\Delta_{\mathrm{sub}}H = \Delta_{\mathrm{fus}}H + \Delta_{\mathrm{vap}}H,\qquad \Delta_{\mathrm{hyd}}H = \Delta_{\mathrm{sol}}H_{\mathrm{anhydrous}} - \Delta_{\mathrm{sol}}H_{\mathrm{hydrate}}$
Enthalpy of neutralisation and the temperature rise
Temperature rise on neutralisation
$\Delta T = \frac{n_{\mathrm{H_2O}}\,|\Delta_{\mathrm{neut}}H|}{m\,c}$

Watch out for (4)

The sign of a heat of dilution
→ Enthalpies of phase change, solution, dilution and hydration
Kilojoules beside joules in a heating path
→ Enthalpies of phase change, solution, dilution and hydration
Only one solution's volume
→ Enthalpy of neutralisation and the temperature rise
More acid is not more heat
→ Enthalpy of neutralisation and the temperature rise

Test yourself on Chemical Thermodynamics

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