ChemistryISC Class 11 · Chemical Thermodynamics

Enthalpy Changes: ΔH and ΔU, Hess's Law, Bond Enthalpy and Calorimetry

Q: Why do we use enthalpy rather than internal energy for most reactions?

Reactions in open vessels occur at constant pressure; the heat exchanged is then exactly ΔH.

Q: Is bond enthalpy the same as bond energy?

They are used almost interchangeably at this level; bond enthalpy is the enthalpy needed to break one mole of a bond in the gas phase, averaged over compounds.

Q: Why is the enthalpy of neutralisation of weak acids less than 57 kJ mol⁻¹?

Part of the energy is used to ionise the weak acid or base completely.

Updated 17 October 2026 · Fact-checked

Enthalpy H = U + pV, so ΔH = ΔU + Δn_g RT. By Hess's law the enthalpy change of a reaction is the same by any route; hence ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants) or, for gases, ΔrH ≈ Σ bond enthalpies(broken) − Σ bond enthalpies(formed).

Understand Enthalpy Changes, Hess's Law and Bond Enthalpy

State functions and sign convention. U, H, S and G are state functions: changes depend only on the initial and final states. Heat absorbed by the system is positive (endothermic, ΔH > 0); heat released is negative (exothermic, ΔH < 0).

ΔH and ΔU. At constant pressure q_p = ΔH and at constant volume q_v = ΔU. They are related by ΔH = ΔU + Δn_g RT, where Δn_g = moles of gaseous products − moles of gaseous reactants; R = 8.314 J mol⁻¹ K⁻¹. If Δn_g = 0, ΔH ≈ ΔU.

Standard enthalpies. The standard enthalpy of formation ΔfH° is the enthalpy change when one mole of a compound is formed from its elements in their standard states (zero for elements in their standard state, for example graphite and O₂(g)). The standard enthalpy of combustion ΔcH° is the enthalpy change when one mole of a substance burns completely in oxygen. Neutralisation of a strong acid and a strong base releases about −57 kJ per mole of water formed, because the net change is H⁺ + OH⁻ → H₂O.

Hess's law. The enthalpy change of a reaction is the same whether it takes place in one step or several steps. Add the equations (reverse them or multiply them as needed, changing the sign or the factor of ΔH accordingly). From formation data: ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants).

Bond enthalpy. For gas-phase reactions: ΔrH ≈ Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed). Because bond enthalpies are averages, this is only approximate.

Calorimetry. q = m c ΔT for a solution calorimeter, or q = C_cal ΔT for a bomb calorimeter (at constant volume, giving ΔU).

Enthalpy Changes, Hess's Law and Bond Enthalpy: formulas and key facts

ΔH and ΔU: $Δ H = Δ U + Δ n_{g} R T$; Δn_g counts gases only.
Hess's law (formation data): $Δ r H ° = ΣΔ f H ° (p r o d u c t s) - ΣΔ f H ° (r e a c t an t s)$; Elements in standard state: ΔfH° = 0.
Bond enthalpies: ΔrH ≈ Σ ΔH(bonds broken) − Σ ΔH(bonds formed); Gas-phase reactions; approximate.
Calorimetry: $q = m c Δ T ∣ q = C_{c} a l Δ T$; q is the heat gained by the calorimeter and contents.

Enthalpy Changes, Hess's Law and Bond Enthalpy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Heat from burning natural gas

Suppose a kitchen burns methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Take ΔfH° (kJ mol⁻¹): CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8.

Find the enthalpy of combustion of methane and the heat from 8 g.

Try it first, then show the solutionSolution

ΔcH° = [−393.5 + 2(−285.8)] − [−74.8 + 0] = −965.1 + 74.8.
= −890.3 kJ mol⁻¹.
8 g is 0.5 mol (CH₄ = 16 g mol⁻¹): heat = 0.5 × 890.3 = 445 kJ.

Answer: −890.3 kJ mol⁻¹; 445 kJ from 8 g.

Memory hook: Multiply ΔfH° by the stoichiometric coefficients.

Tricks to remember Enthalpy Changes, Hess's Law and Bond Enthalpy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Products minus reactants
ΔrH = Σ products − Σ reactants for formation data; the reverse for bond enthalpies.
Use it when: Avoiding sign errors.
Flip means sign flip
Reversing a reaction changes the sign of ΔH; multiplying scales it.
Use it when: Hess's law manipulation.

Enthalpy Changes, Hess's Law and Bond Enthalpy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Elements in standard state · 1 mark · MCQBoard level

The standard enthalpy of formation of O₂(g) and of graphite at 298 K is:
(a) positive
(b) negative
(c) zero
(d) equal to their bond enthalpy

Try it first, then show the solutionSolution

By definition ΔfH° of an element in its standard state is zero.

Answer: (c) zero

Example 2: Relation between ΔH and ΔU · 3 marks · Short answerBoard level

For 2H₂(g) + O₂(g) → 2H₂O(l) at 298 K, find ΔH − ΔU. If ΔH = −571.6 kJ, find ΔU. (R = 8.314 J mol⁻¹ K⁻¹)

Try it first, then show the solutionSolution

Δn_g = 0 − 3 = −3. (1 mark)
ΔH − ΔU = Δn_g RT = −3 × 8.314 × 298 = −7433 J = −7.43 kJ. (1 mark)
ΔU = ΔH + 7.43 = −571.6 + 7.43 = −564.2 kJ. (1 mark)

Answer: ΔH − ΔU = −7.43 kJ; ΔU = −564.2 kJ.

Example 3: Hess's law · 3 marks · Short answerHarder board level

Given C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ, find ΔfH° of CO(g).

Try it first, then show the solutionSolution

Required: C + ½O₂ → CO. (0.5 mark)
Equation 1 plus the reverse of equation 2: C + O₂ + CO₂ → CO₂ + CO + ½O₂. (1 mark)
ΔH = −393.5 − (−283.0) = −110.5 kJ mol⁻¹. (1.5 marks)

Answer: ΔfH°(CO) = −110.5 kJ mol⁻¹.

Example 4: Bond enthalpy calculation · 3 marks · Short answerHarder board level

Use bond enthalpies (kJ mol⁻¹): C–H 414, Cl–Cl 243, C–Cl 339 and H–Cl 431 to find ΔH for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g).

Try it first, then show the solutionSolution

Bonds broken: one C–H and one Cl–Cl = 414 + 243 = 657 kJ. (1 mark)
Bonds formed: one C–Cl and one H–Cl = 339 + 431 = 770 kJ. (1 mark)
ΔH = 657 − 770 = −113 kJ mol⁻¹ (exothermic). (1 mark)

Answer: −113 kJ mol⁻¹

Example 5: Bomb calorimeter · 5 marks · Long answerHOTS / competency

Burning 0.16 g of methane in a bomb calorimeter with heat capacity 8.0 kJ K⁻¹ raises the temperature by 1.11 K at 298 K. Find the molar ΔU and ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). (C = 12, H = 1, R = 8.314 J mol⁻¹ K⁻¹)

Try it first, then show the solutionSolution

q = C ΔT = 8.0 × 1.11 = 8.88 kJ released. (1 mark)
n = 0.16/16 = 0.01 mol. (1 mark)
ΔU = −8.88/0.01 = −888 kJ mol⁻¹. (1 mark)
Δn_g = 1 − 3 = −2, so Δn_g RT = −2 × 8.314 × 298/1000 = −4.96 kJ. (1 mark)
ΔH = ΔU + Δn_g RT = −888 − 4.96 = −893 kJ mol⁻¹. (1 mark)

Answer: ΔU = −888 kJ mol⁻¹; ΔH = −893 kJ mol⁻¹.

Where students lose marks in Enthalpy Changes, Hess's Law and Bond Enthalpy

The slips examiners see most often, and what to write instead.

Mistake: Sign error in ΔH
Giving a positive value for combustion.
Do this: Heat is released, so ΔH is negative.
Usually costs: 1 to 2 marks.
Mistake: Counting liquids in Δn_g
Including H₂O(l) in the gas count.
Do this: Δn_g counts only gaseous species.
Usually costs: 2 marks.
Mistake: Forgetting the coefficient
Using ΔfH° once for 2 mol of a product.
Do this: Multiply each ΔfH° by its coefficient.
Usually costs: 1 to 2 marks.

Exam tips

Write the balanced equation with states before using ΔfH° data.
Show the arithmetic of each sum separately.
Give units: kJ mol⁻¹ (per mole of reaction as written).
Use the correct R (8.314 J mol⁻¹ K⁻¹) when converting ΔH and ΔU.

Practice questions on Enthalpy Changes, Hess's Law and Bond Enthalpy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Hess's law of constant heat summation is a consequence of:
(a) the second law
(b) the first law, since H is a state function
(c) Le Chatelier's principle
(d) the law of mass action
Hint
The result is path independent.
Answer
(b) the first law, since H is a state function
Q2 · 2 marks · Short answerBoard level
Using ΔfH°(NH₃) = −46 kJ mol⁻¹, find ΔH for N₂(g) + 3H₂(g) → 2NH₃(g).
Hint
Multiply by the coefficient.
Answer
ΔH = 2 × (−46) = −92 kJ.
Q3 · 3 marks · Short answerHarder board level
100 mL of 1.0 M HCl is mixed with 100 mL of 1.0 M NaOH in a calorimeter and the temperature rises by 6.7 K. Taking 4.18 J g⁻¹ K⁻¹ and a total mass of 200 g, find the enthalpy of neutralisation per mole of water.
Hint
q = mcΔT; moles of water = 0.1.
Answer
q = 200 × 4.18 × 6.7 = 5601 J. n = 0.1 mol. ΔH = −56.0 kJ mol⁻¹.
Q4 · 3 marks · Short answerHarder board level
Given the enthalpies of combustion of C₂H₄(g), H₂(g) and C₂H₆(g) as −1411, −285.8 and −1560 kJ mol⁻¹, find ΔH for C₂H₄ + H₂ → C₂H₆.
Hint
ΔH = Σ ΔcH(reactants) − Σ ΔcH(products).
Answer
ΔH = (−1411 − 285.8) − (−1560) = −1696.8 + 1560 = −136.8 kJ.
Q5 · 3 marks · Short answerHarder board level
The enthalpy of vaporisation of water at 298 K is 44.0 kJ mol⁻¹. Find ΔU for the vaporisation of 1 mol. (R = 8.314 J mol⁻¹ K⁻¹)
Hint
Δn_g = +1.
Answer
ΔU = ΔH − Δn_g RT = 44.0 − 1 × 8.314 × 298/1000 = 44.0 − 2.48 = 41.5 kJ.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a school compares fuels using ΔcH (kJ mol⁻¹): H₂ −285.8, CH₄ −890.3 and C₂H₅OH −1367 (H = 1, C = 12, O = 16).
(i) Find the heat per gram for each fuel. (1 mark)
(ii) Which gives the most heat per gram? (1 mark)
(iii) What mass of methane is needed for 1000 kJ? (1 mark)
(iv) Give one reason why the ranking per gram may differ from the ranking per mole. (1 mark)
Hint
Divide by the molar mass.
Answer
(i) H₂: 285.8/2 = 142.9 kJ g⁻¹; CH₄: 890.3/16 = 55.6 kJ g⁻¹; C₂H₅OH: 1367/46 = 29.7 kJ g⁻¹. (ii) Hydrogen. (iii) 1000/55.6 = 18.0 g. (iv) The molar masses differ widely; a light fuel such as H₂ gives much more heat per gram than a heavy one even if ΔcH per mole is smaller.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write ΔH in terms of ΔU.Show answer
ΔH = ΔU + Δn_g RT.
2.State Hess's law.Show answer
The enthalpy change of a reaction is the same whether it occurs in one or more steps.
3.ΔfH° of an element in its standard state?Show answer
Zero.
4.Enthalpy of neutralisation (strong acid and base) per mole of water?Show answer
About −57 kJ mol⁻¹.

Enthalpy Changes, Hess's Law and Bond Enthalpy: frequently asked questions

Why do we use enthalpy rather than internal energy for most reactions?

Reactions in open vessels occur at constant pressure; the heat exchanged is then exactly ΔH.

Is bond enthalpy the same as bond energy?

They are used almost interchangeably at this level; bond enthalpy is the enthalpy needed to break one mole of a bond in the gas phase, averaged over compounds.

Why is the enthalpy of neutralisation of weak acids less than 57 kJ mol⁻¹?

Part of the energy is used to ionise the weak acid or base completely.