When is Delta H equal to Delta U?

When there is no change in the number of moles of gas (Delta n_g = 0), Delta H equals Delta U. Examples include reactions between only solids and liquids and gaseous reactions with unchanged gas moles.

Why is neutralization of a strong acid by a strong base always about -57.1 kJ per mole?

Both strong acids and strong bases fully ionize in dilute solution. The only chemical change is H+ combining with OH- to form water, which is identical no matter which strong acid and base are used, so the enthalpy change is constant.

Enthalpies for Different Types of Reaction

Q: Why is enthalpy more useful than internal energy for chemistry problems?

Most laboratory reactions occur at constant atmospheric pressure. At constant pressure the heat absorbed equals Delta H, so enthalpy is what a calorimeter measures directly - and it is the quantity most closely tied to reactions as we actually run them.

Q: Why is enthalpy of formation of an element in its stable state taken as zero?

It is a convention chosen so absolute enthalpies drop out of calculations. Only differences matter physically, and this convention gives self-consistent tabulated values for all compounds.

Q: Why does the enthalpy of a reaction depend on temperature?

Enthalpies rise with temperature according to heat capacities. If products and reactants have different heat capacities, the enthalpy change of the reaction also varies with temperature - quantified by Kirchhoff's equation.

ChemistryChemical ThermodynamicsFor JEE aspirants

The enthalpy change $Δ H$ of a reaction is the heat absorbed or released at constant pressure - the most useful energy quantity for chemistry because most reactions occur in open vessels at atmospheric pressure. This concept covers the definition $H = U + P V$ , the crucial relation $Δ H = Δ U + Δ n_{g} R T$ for gaseous reactions, and every important type of enthalpy tested in JEE and NEET: enthalpies of formation, combustion, neutralization, ionization, hydration, solution, precipitation, transition, and phase transitions. It also covers Kirchhoff's equation for the temperature dependence of $Δ H$ .

Key Formulas - Quick Reference

Enthalpy: $H = U + P V$
At constant pressure, $q_{p} = Δ H$ ; at constant volume, $q_{v} = Δ U$
For gaseous reactions: $Δ H = Δ U + Δ n_{g} R T$ where $Δ n_{g} =$ (moles of gaseous products) - (moles of gaseous reactants)
Enthalpy of reaction from formation enthalpies: $Δ H_{r x n}^{\circ} = \sum ν_{i} Δ H_{f}^{\circ} (products) - \sum ν_{i} Δ H_{f}^{\circ} (reactants)$
Neutralization of strong acid + strong base: $Δ H = - 57.1 kJ mol^{- 1}$ (approx $- 13.7 kcal mol^{- 1}$ )
Enthalpy of ionization: $Δ H_{i o n} = Δ H_{n e u t, weak} - Δ H_{n e u t, strong}$
Kirchhoff's equation: $\frac{Δ H _{2} - Δ H _{1}}{T _{2} - T _{1}} = Δ C_{p}$ ; similarly $\frac{Δ U _{2} - Δ U _{1}}{T _{2} - T _{1}} = Δ C_{v}$

1. Enthalpy of a system

Most chemical reactions occur at constant atmospheric pressure. Under constant pressure, the heat exchanged $q_{p}$ equals a state-function quantity called the enthalpy:

H = U + P V

(definition of enthalpy)

Since $U$ , $P$ and $V$ are state functions, $H$ is also a state function and an extensive property. Like $U$ , absolute enthalpy cannot be measured; only changes $Δ H$ can.

Relation between $Δ H$ and $Δ U$

Differentiating $H = U + P V$ :

$Δ H = Δ U + Δ (P V)$

For a chemical reaction involving gases, at constant temperature $Δ (P V) = Δ (n R T) = (Δ n_{g}) R T$ , where $Δ n_{g}$ is the change in moles of gaseous species. Hence:

Δ H = Δ U + Δ n_{g} R T

For reactions in condensed phases (only solids and liquids), $Δ n_{g} = 0$ and so $Δ H \approx Δ U$ .

Exothermic vs endothermic

Figure: Enthalpy versus progress of reaction. Exothermic (left): products sit lower than reactants; the reaction releases heat and

Δ H < 0

. Endothermic (right): products sit higher than reactants; the reaction absorbs heat and

Δ H > 0

. In both cases the pink arrow marks the activation energy - the minimum energy needed to reach the transition state at the peak.

Exothermic: $H_{products} < H_{reactants}$ , so $Δ H < 0$ . Heat is released. Examples: combustion, neutralization.
Endothermic: $H_{products} > H_{reactants}$ , so $Δ H > 0$ . Heat is absorbed. Examples: melting of ice, decomposition of CaCO $_{3}$ .

Solved Example 1

For the reaction

CH_{4} (g) + 2 O_{2} (g) ⇌ CO_{2} (g) + 2 H_{2} O (ℓ)

, express the relation between

Δ H

and

Δ U

Solution:

Gaseous moles: reactants $= 1 + 2 = 3$ , products $= 1$ (water is liquid).

$Δ n_{g} = 1 - 3 = - 2$ .

$Δ H = Δ U + Δ n_{g} R T = Δ U - 2 R T$ .

2. Standard states and standard enthalpy

Since enthalpies depend on temperature and pressure, chemists agree on reference conditions called the standard state:

Gas: ideal gas at $1 bar$ partial pressure at the specified temperature.
Liquid: pure liquid at $1 bar$ pressure at the specified temperature.
Solid: pure crystalline solid at $1 bar$ pressure at the specified temperature.
Solution species: solute at $1 molal$ (approximately $1 M$ ) at $1 bar$ .

The specified temperature is usually $298.15 K$ ( $2 5^{\circ}$ C). Enthalpies measured at standard state are denoted with a superscript circle, e.g. $Δ H^{\circ}$ .

3. Enthalpy of formation $(Δ H_{f}^{\circ})$

The standard enthalpy of formation of a compound is the enthalpy change when one mole of the compound is formed from its constituent elements in their most stable states at standard conditions.

By convention:

Δ H_{f}^{\circ}

(element in its stable state)

= 0

.
So

Δ H_{f}^{\circ} (O_{2}, g) = 0

Δ H_{f}^{\circ} (C, graphite) = 0

Δ H_{f}^{\circ} (Br_{2}, ℓ) = 0

Δ H_{f}^{\circ} (S, rhombic) = 0

, but

Δ H_{f}^{\circ} (C, diamond) \neq = 0

Examples of formation reactions:

$\frac{1}{2} H_{2} (g) + \frac{1}{2} Br_{2} (ℓ) \to HBr (g), Δ H_{f}^{\circ} (HBr, g) = - 36.4 kJ mol^{- 1}$

$S(rhombic) + O_{2} (g) \to SO_{2} (g), Δ H_{f}^{\circ} (SO_{2}, g) = - 296.9 kJ mol^{- 1}$

Enthalpy of reaction from formation enthalpies

Δ H_{r x n}^{\circ} = \sum ν_{i} Δ H_{f}^{\circ} (products) - \sum ν_{i} Δ H_{f}^{\circ} (reactants)

where $ν_{i}$ is the stoichiometric coefficient in the balanced equation.

Solved Example 2

Calculate the standard enthalpy of the reaction

ZnO (s) + CO (g) \to Zn (s) + CO_{2} (g)

. Given

Δ H_{f}^{\circ} (ZnO, s) = - 350 kJ mol^{- 1}

Δ H_{f}^{\circ} (CO_{2}, g) = - 390 kJ mol^{- 1}

Δ H_{f}^{\circ} (CO, g) = - 110 kJ mol^{- 1}

Solution:

$Δ H^{\circ} = [Δ H_{f}^{\circ} (Zn) + Δ H_{f}^{\circ} (CO_{2})] - [Δ H_{f}^{\circ} (ZnO) + Δ H_{f}^{\circ} (CO)]$

$= [0 + (- 390)] - [(- 350) + (- 110)] = - 390 - (- 460) = + 70 kJ$

4. Enthalpy of combustion $(Δ H_{C}^{\circ})$

The enthalpy of combustion is the enthalpy change when one mole of a substance undergoes complete combustion with oxygen at standard conditions, giving products in their stable states.

Combustion is always exothermic, so $Δ H_{C}^{\circ}$ is always negative.

Examples:

$CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (ℓ), Δ H_{C}^{\circ} = - 890.36 kJ mol^{- 1}$

$C_{6} H_{6} (ℓ) + \frac{15}{2} O_{2} (g) \to 6 CO_{2} (g) + 3 H_{2} O (ℓ), Δ H_{C}^{\circ} = - 3268 kJ mol^{- 1}$

Uses. Enthalpies of combustion are experimentally accurate and widely tabulated. They are used to (i) calculate calorific values of fuels, (ii) calculate enthalpies of formation of hydrocarbons (which cannot be measured directly).

5. Enthalpy of neutralization

The enthalpy of neutralization is the enthalpy change when one gram-equivalent of an acid is neutralized by one gram-equivalent of a base in dilute solution to form one mole of water:

H^{+} (a q) + OH^{-} (a q) \to H_{2} O (ℓ), Δ H = - 57.1 kJ mol^{- 1} \approx - 13.7 kcal mol^{- 1}

Why is it constant for strong acid + strong base?

Both strong acids and strong bases dissociate completely in dilute solution. The only chemical change is the combination of $H^{+}$ and $OH^{-}$ to form water - which is the same regardless of the identity of the strong acid or base. Hence the enthalpy change is a constant.

Weak acids and bases: enthalpy of ionization

For a weak acid (or base), some heat is absorbed during ionization of the weak acid, so the net heat released on neutralization is less than $57.1 kJ mol^{- 1}$ . The difference gives the enthalpy of ionization of the weak acid or base.

Two-step picture: Ionization:

HCN \to H^{+} + CN^{-}

Δ H_{1} = Δ H_{i o n}

(positive)
Neutralization:

H^{+} + OH^{-} \to H_{2} O

Δ H_{2} = - 57.1 kJ mol^{- 1}

Overall:

HCN + OH^{-} \to H_{2} O + CN^{-}

Δ H = Δ H_{1} + Δ H_{2}

Example: $HCN + OH^{-} \to H_{2} O + CN^{-}$ , $Δ H = - 12 kJ mol^{- 1}$ .

Hence $Δ H_{i o n} (HCN) = - 12 - (- 57.1) = + 45.1 kJ mol^{- 1}$ .

Solved Example 3

Enthalpy of neutralization of HCl by NaOH is

- 57.1 kJ mol^{- 1}

and by NH

_{4}

OH is

- 51.1 kJ mol^{- 1}

. Calculate the enthalpy of dissociation of NH

_{4}

OH.

Solution:

Split the neutralization of NH $_{4}$ OH into ionization + neutralization:

(i) $NH_{4} OH (a q) \to NH_{4}^{+} (a q) + OH^{-} (a q), Δ H_{1} = ?$

(ii) $H^{+} (a q) + OH^{-} (a q) \to H_{2} O (ℓ), Δ H_{2} = - 57.1 kJ mol^{- 1}$

Total: $H^{+} + NH_{4} OH \to NH_{4}^{+} + H_{2} O, Δ H = - 51.1 kJ mol^{- 1}$

$Δ H = Δ H_{1} + Δ H_{2} \Rightarrow Δ H_{1} = - 51.1 - (- 57.1) = + 6.0 kJ mol^{- 1}$

6. Enthalpy of hydration

Hydration enthalpy is defined in two closely related ways:

(a) Hydration of anhydrous or partially hydrated salt

The enthalpy change when one mole of an anhydrous salt combines with the requisite moles of water to form a stable hydrated salt.

Example: $CuSO_{4} (s) + 5 H_{2} O (ℓ) \to CuSO_{4} \cdot 5 H_{2} O (s), Δ H^{\circ} = - 78 kJ mol^{- 1}$

(b) Hydration of gaseous ions

The enthalpy change when one mole of a gaseous ion dissolves in a large amount of water to form the aqueous ion:

$Cl^{-} (g) + aq \to Cl^{-} (a q), Δ H = Δ H_{f}^{\circ} (Cl^{-}, a q)$

By convention, $Δ H_{f}^{\circ} (H^{+}, a q) = 0$ , which lets us tabulate hydration enthalpies of all other ions.

7. Enthalpy of solution

The integral enthalpy of solution is the enthalpy change when one mole of solute dissolves in a definite amount of solvent to give a solution of specified concentration.

HCl (g) + 10 H_{2} O (ℓ) \to HCl (10 H_{2} O), Δ H_{1} = - 69.5 kJ mol^{- 1}

HCl (g) + 200 H_{2} O (ℓ) \to HCl (200 H_{2} O), Δ H = - 74.2 kJ mol^{- 1}

HCl (g) + aq \to HCl (a q), Δ H = - 75.0 kJ mol^{- 1}

(infinite dilution)

When the amount of solvent is not specified, it is taken to be very large (infinite dilution).

8. Enthalpy of transition

The enthalpy change when one mole of a substance is transformed from one allotropic form to another at standard conditions:

$C(graphite) \to C(diamond), Δ H_{t r s}^{\circ} = + 1.90 kJ mol^{- 1}$

Solved Example 4

Calculate the enthalpy of transition of diamond to graphite from the following data:

C(diamond) + O_{2} (g) \to CO_{2} (g), Δ H = - 395.41 kJ mol^{- 1}

C(graphite) + O_{2} (g) \to CO_{2} (g), Δ H = - 393.51 kJ mol^{- 1}

Solution:

Subtracting the second from the first:

$C(diamond) - C(graphite) \to 0, Δ H = - 395.41 - (- 393.51) = - 1.90 kJ mol^{- 1}$

Reversing: $C(graphite) \to C(diamond), Δ H_{t r s}^{\circ} = + 1.90 kJ mol^{- 1}$

9. Phase transition enthalpies

Enthalpy changes accompanying phase transitions at constant $T$ and $P$ :

Transition	Symbol	Direction	Sign of $Δ H$
Fusion (melting)	$Δ H_{f u s}$	solid $\to$ liquid	positive (endothermic)
Vaporization	$Δ H_{v a p}$	liquid $\to$ gas	positive (endothermic)
Sublimation	$Δ H_{s u b}$	solid $\to$ gas	positive (endothermic)
Freezing (solidification)	$- Δ H_{f u s}$	liquid $\to$ solid	negative (exothermic)
Condensation	$- Δ H_{v a p}$	gas $\to$ liquid	negative (exothermic)

These are connected by Hess's law: $Δ H_{s u b} = Δ H_{f u s} + Δ H_{v a p}$ .

10. Enthalpy of atomization $(Δ H_{a}^{\circ})$

The enthalpy of atomization is the enthalpy change when one mole of a substance is completely dissociated into isolated gaseous atoms.

For a diatomic gas like $H_{2}$ : $H_{2} (g) \to 2 H (g), Δ H_{a}^{\circ} = 436 kJ mol^{- 1}$ (same as bond dissociation enthalpy in this case).

For a solid: $Na (s) \to Na (g), Δ H_{a}^{\circ} = 108 kJ mol^{- 1}$ (same as enthalpy of sublimation for a monoatomic solid).

11. Enthalpy of precipitation

The enthalpy change when one mole of a precipitate forms in solution:

$BaCl_{2} (a q) + Na_{2} SO_{4} (a q) \to BaSO_{4} (s) + 2 NaCl (a q), Δ H^{\circ} = - 24.27 kJ mol^{- 1}$

12. Enthalpy of ionization (for weak acids/bases)

Already introduced under neutralization - the enthalpy change when one mole of a weak electrolyte ionizes completely in dilute solution.

Δ H_{i o n} = Δ H_{n e u t} (weak) - Δ H_{n e u t} (strong-strong) = Δ H_{n e u t} (weak) + 57.1 kJ mol^{- 1}

The larger the enthalpy of ionization, the weaker the acid or base.

13. Enthalpy of formation of ions

The enthalpy change when one mole of gaseous ions is obtained from elements in their standard state:

$\frac{1}{2} Cl_{2} (g) + aq \to Cl^{-} (a q), Δ H = Δ H_{f}^{\circ} (Cl^{-}, a q)$

By convention $Δ H_{f}^{\circ} (H^{+}, a q) = 0$ , which lets us tabulate the enthalpies of formation of all other ions.

Solved Example 5

The enthalpy of formation of

H_{2} O (ℓ)

- 285 kJ mol^{- 1}

and enthalpy of neutralization of a strong acid and a strong base is

- 55 kJ mol^{- 1}

. What is the enthalpy of formation of

OH^{-}

ions?

Solution:

Neutralization: $H^{+} (a q) + OH^{-} (a q) \to H_{2} O (ℓ), Δ H_{n e u t} = - 55 kJ mol^{- 1}$

$Δ H_{n e u t} = Δ H_{f}^{\circ} (H_{2} O, ℓ) - Δ H_{f}^{\circ} (H^{+}, a q) - Δ H_{f}^{\circ} (OH^{-}, a q)$

Using convention $Δ H_{f}^{\circ} (H^{+}, a q) = 0$ :

$- 55 = - 285 - 0 - Δ H_{f}^{\circ} (OH^{-}, a q)$

$Δ H_{f}^{\circ} (OH^{-}, a q) = - 285 - (- 55) = - 230 kJ mol^{- 1}$

14. Kirchhoff's equation: variation of $Δ H$ with temperature

Enthalpies of individual substances depend on temperature (higher $T$ gives higher $H$ ), so the enthalpy change of a reaction also varies with temperature. Consider $P + Q \to R + S$ :

At $T_{1}$ : $Δ H_{1}^{\circ} = H_{m}^{\circ} (R, T_{1}) + H_{m}^{\circ} (S, T_{1}) - H_{m}^{\circ} (P, T_{1}) - H_{m}^{\circ} (Q, T_{1})$

At $T_{2}$ : $Δ H_{2}^{\circ} = H_{m}^{\circ} (R, T_{2}) + H_{m}^{\circ} (S, T_{2}) - H_{m}^{\circ} (P, T_{2}) - H_{m}^{\circ} (Q, T_{2})$

The temperature dependence of enthalpy for each species: $H_{m}^{\circ} (X, T_{2}) - H_{m}^{\circ} (X, T_{1}) = C_{P, X} (T_{2} - T_{1})$ .

Kirchhoff's equation:

\frac{Δ H _{2} - Δ H _{1}}{T _{2} - T _{1}} = Δ C_{p}, where Δ C_{p} = \sum ν_{i} C_{p, i} (products) - \sum ν_{i} C_{p, i} (reactants)

At constant volume:

\frac{Δ U _{2} - Δ U _{1}}{T _{2} - T _{1}} = Δ C_{v}

If $Δ C_{p}$ is temperature-dependent, use the integral form: $Δ H_{2} = Δ H_{1} + \int_{T_{1}}^{T_{2}} Δ C_{p} d T$ .

Solved Example 6

The standard heat of formation of gaseous NH

_{3}

- 11.02 kcal mol^{- 1}

298 K

. Given the constant-pressure molar heat capacities of

N_{2}

H_{2}

and

NH_{3}

are respectively

6.96, 6.89, 8.38 cal mol^{- 1} K^{- 1}

, find

Δ H^{\circ}

773 K

for

\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to NH_{3} (g)

Solution:

$Δ H_{1} = - 11.02 kcal mol^{- 1}$ at $T_{1} = 298 K$ , want $Δ H_{2}$ at $T_{2} = 773 K$ .

$Δ C_{p} = C_{p} (NH_{3}) - \frac{1}{2} C_{p} (N_{2}) - \frac{3}{2} C_{p} (H_{2})$

$= 8.38 - \frac{1}{2} (6.96) - \frac{3}{2} (6.89) = 8.38 - 3.48 - 10.335 = - 5.435 cal mol^{- 1} K^{- 1}$

Kirchhoff: $Δ H_{2} = Δ H_{1} + Δ C_{p} \cdot (T_{2} - T_{1})$

$= - 11020 + (- 5.435) (773 - 298) = - 11020 - 2581.6 = - 13602 cal mol^{- 1} \approx - 13.6 kcal mol^{- 1}$

15. Common Mistakes to Avoid

Watch out

Applying $Δ H = Δ U + Δ n_{g} R T$ to reactions with no gaseous species. $Δ n_{g}$ counts only gaseous moles - ignore solids and liquids.
Forgetting the stoichiometric coefficients. $Δ H_{r x n}^{\circ}$ uses $ν_{i} Δ H_{f}^{\circ}$ - so if the equation has $2 H_{2} O (ℓ)$ , use $2 \times Δ H_{f}^{\circ} (H_{2} O, ℓ)$ .
Confusing enthalpy of formation with enthalpy of combustion. Formation forms the compound from elements. Combustion burns the compound in oxygen.
Assuming enthalpy of neutralization is always $- 57.1 kJ mol^{- 1}$ . Only strong acid + strong base gives this constant value. Weak acid or weak base neutralizations release less heat.
Forgetting that $Δ H_{f}^{\circ}$ of an element in its stable form is zero. But $Δ H_{f}^{\circ}$ of an allotrope that is not the standard form is not zero (e.g. diamond, monoclinic sulphur).
Ignoring the phase symbols in enthalpy problems. $Δ H_{f}^{\circ} (H_{2} O, ℓ) \neq = Δ H_{f}^{\circ} (H_{2} O, g)$ . Always check the phase.

Frequently Asked Questions

Q1. Why is enthalpy more useful than internal energy for chemistry problems?

Most laboratory reactions occur in open vessels at constant atmospheric pressure. At constant pressure, the heat absorbed equals $Δ H$ , not $Δ U$ . So $Δ H$ is what a coffee-cup calorimeter directly measures, and it is the thermodynamic quantity most closely tied to reactions as we actually run them.

Q2. When is $Δ H$ equal to $Δ U$ ?

When there is no change in the number of moles of gas ( $Δ n_{g} = 0$ ), then $Δ H = Δ U$ . This includes reactions between only solids and liquids, and gaseous reactions where the moles of gas are unchanged (e.g. $H_{2} + Cl_{2} \to 2 HCl$ ).

Q3. Why is the enthalpy of formation of an element in its stable state taken as zero?

It is a convention chosen so that absolute enthalpies (which cannot be measured) drop out of calculations. The differences of formation enthalpies are what matter physically, and this convention gives self-consistent, tabulated values for all compounds.

Q4. Why is neutralization of a strong acid by a strong base always $- 57.1 kJ mol^{- 1}$ ?

Both strong acids and strong bases are fully ionized in dilute solution. The only chemical change is $H^{+} + OH^{-} \to H_{2} O$ , which is identical regardless of which strong acid and strong base were used. Hence the enthalpy change is the same, approximately $- 57.1 kJ mol^{- 1}$ .

Q5. Why is neutralization of a weak acid by a strong base less exothermic?

A weak acid is not fully ionized in solution; some energy is consumed to ionize it. The heat released on neutralization is therefore the standard $- 57.1 kJ mol^{- 1}$ minus the energy spent on ionization. The stronger the weak acid, the smaller its ionization enthalpy, and the closer the neutralization heat is to $- 57.1 kJ mol^{- 1}$ .

Q6. What is the enthalpy of atomization and how is it different from bond dissociation energy?

Enthalpy of atomization is the total energy needed to break all bonds and separate one mole of substance into gaseous atoms. For a diatomic gas like $H_{2}$ , this equals the bond dissociation energy of the H-H bond. But for polyatomic molecules, atomization means breaking every bond, giving a sum of average bond enthalpies.

Q7. Why does the enthalpy of a reaction depend on temperature?

Enthalpies of individual substances rise with temperature according to their heat capacities $C_{p}$ . If products and reactants have different heat capacities ( $Δ C_{p} \neq = 0$ ), the enthalpy change of the reaction also depends on temperature. Kirchhoff's equation quantifies this: $Δ H_{2} = Δ H_{1} + Δ C_{p} (T_{2} - T_{1})$ .

Q8. What is the difference between integral and differential enthalpy of solution?

Integral enthalpy of solution is the heat change when one mole of solute dissolves in a specified amount of solvent to give a solution of stated concentration. Differential (or partial molar) enthalpy of solution is the heat change per mole of solute added to a very large volume of an existing solution, so that its concentration essentially does not change. In JEE/NEET, "enthalpy of solution" almost always means the integral version at high dilution.

Previous year questions on Enthalpies for Different Types of Reaction

5 questions from past papers, each with a step-by-step solution.

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1. Enthalpy of a system

Relation between ΔH and ΔU

Exothermic vs endothermic

2. Standard states and standard enthalpy

3. Enthalpy of formation (ΔHf∘​)

Enthalpy of reaction from formation enthalpies

4. Enthalpy of combustion (ΔHC∘​)

5. Enthalpy of neutralization

Why is it constant for strong acid + strong base?

Weak acids and bases: enthalpy of ionization

6. Enthalpy of hydration

(a) Hydration of anhydrous or partially hydrated salt

(b) Hydration of gaseous ions

7. Enthalpy of solution

8. Enthalpy of transition

9. Phase transition enthalpies

10. Enthalpy of atomization (ΔHa∘​)

11. Enthalpy of precipitation

12. Enthalpy of ionization (for weak acids/bases)

13. Enthalpy of formation of ions

14. Kirchhoff's equation: variation of ΔH with temperature

15. Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is enthalpy more useful than internal energy for chemistry problems?

Q2. When is ΔH equal to ΔU?

Q3. Why is the enthalpy of formation of an element in its stable state taken as zero?

Q4. Why is neutralization of a strong acid by a strong base always −57.1 kJ mol−1?

Q5. Why is neutralization of a weak acid by a strong base less exothermic?

Q6. What is the enthalpy of atomization and how is it different from bond dissociation energy?

Q7. Why does the enthalpy of a reaction depend on temperature?

Q8. What is the difference between integral and differential enthalpy of solution?

Previous year questions on Enthalpies for Different Types of Reaction

Ready to master Chemical Thermodynamics?

Relation between $Δ H$ and $Δ U$

3. Enthalpy of formation $(Δ H_{f}^{\circ})$

4. Enthalpy of combustion $(Δ H_{C}^{\circ})$

10. Enthalpy of atomization $(Δ H_{a}^{\circ})$

14. Kirchhoff's equation: variation of $Δ H$ with temperature

Q2. When is $Δ H$ equal to $Δ U$ ?

Q4. Why is neutralization of a strong acid by a strong base always $- 57.1 kJ mol^{- 1}$ ?