ChemistryCBSE Class 11 · Thermodynamics

Enthalpy Changes, Hess's Law and Calorimetry Class 11

Q: Why is enthalpy useful?

Most reactions occur at constant pressure, where the heat exchanged equals ΔH.

Q: Why is Δn_g needed?

Gases do work PΔV on the surroundings, making ΔH differ from ΔU.

Q: Is a bond enthalpy always exact?

No; it is an average over compounds, so bond enthalpy estimates of ΔH are approximate.

Q: What is the standard state?

The most stable form of a substance at 1 bar and a specified temperature (usually 298 K).

Q: Why do we divide by moles in calorimetry?

To express the heat per mole of reaction as the enthalpy change.

Updated 17 October 2026 · Fact-checked

At constant pressure the heat exchanged is the enthalpy change ΔH = ΔU + Δn_gRT, where Δn_g is the change in moles of gas. Hess's law says ΔH of a reaction is the same whether it occurs in one step or several, so ΔrH can be built from ΔfH values (ΔrH = ΣΔfH products − ΣΔfH reactants), combustion data or bond enthalpies. Calorimeters give q = CΔT (bomb: ΔU; coffee-cup: ΔH).

Both routes from C + O₂ to CO₂ lose the same energy: −110.5 + (−283.0) = −393.5 kJ/mol, because H is a state function.

Illustration drawn for QuestionBuddy.

Understand Enthalpy Changes, Hess's Law and Calorimetry

Enthalpy. $H = U + P V$ is a state function. At constant pressure the heat absorbed equals $Δ H$ ; at constant volume it equals $Δ U$ . For reactions involving gases (ideal behaviour):
$ $Δ H = Δ U + Δ n_{g} R T, Δ n_{g} = n_{g} (products) - n_{g} (reactants) .$ $
For reactions of solids and liquids, $Δ H \approx Δ U$ . Exothermic means $Δ H < 0$ .

Standard enthalpies. $Δ_{f} H^{⊖}$ is the enthalpy change for forming one mole of a compound from its elements in their standard states (zero for the elements in their standard states). $Δ_{c} H^{⊖}$ is the enthalpy of combustion of one mole in excess oxygen. Others: atomisation, bond dissociation, vaporisation, fusion, neutralisation, solution, hydration, lattice.

Hess's law. Because $H$ is a state function, $Δ_{r} H$ is path-independent. Procedure: write the target equation; manipulate the given equations (reverse changes the sign of $Δ H$ ; multiplying by $n$ multiplies $Δ H$ by $n$ ); add them so that unwanted species cancel; add the $Δ H$ values.
$ $Δ_{r} H^{⊖} = \sum Δ_{f} H^{⊖} (products) - \sum Δ_{f} H^{⊖} (reactants) .$ $
From bond enthalpies (gas phase): $Δ_{r} H \approx \sum (bonds broken) - \sum (bonds formed)$ , only approximate because average values are used.

Calorimetry. Heat gained by the surroundings $q = m c Δ T$ or $C Δ T$ (with $C$ the heat capacity of the calorimeter). In a bomb calorimeter (constant volume) the heat released gives $Δ U$ , converted to $Δ H$ with $Δ n_{g} R T$ . In a coffee-cup calorimeter (constant pressure) the heat gives $Δ H$ directly. Per mole: divide the heat by the moles reacted.

Sign care. A temperature rise in the calorimeter means the reaction released heat: $Δ_{r} U$ or $Δ_{r} H$ is negative.

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

ΔH and ΔU: $Δ H = Δ U + Δ n_{g} R T$; $R = 8.314 \times 1 0^{- 3}$ kJ mol⁻¹ K⁻¹.
From formation enthalpies: $Δ_{r} H^{⊖} = \sum Δ_{f} H^{⊖} (prod) - \sum Δ_{f} H^{⊖} (react)$; Elements = 0.
Hess's law: $Δ H_{t o t a l} = Δ H_{1} + Δ H_{2} + \dots$; Sum over steps.
Bond enthalpies: $Δ_{r} H \approx \sum Δ H_{b r o k e n} - \sum Δ H_{f or m e d}$; Gas phase, approximate.
Calorimeter heat: $q = m c Δ T = C Δ T$; Heat gained by water/apparatus.
Heat per mole: $Δ_{r} H = - \frac{q}{n}$; Exothermic reaction.

How to solve Enthalpy Changes, Hess's Law and Calorimetry questions

Decide whether the data are formation, combustion or bond enthalpies; then use the matching form of Hess's law and mind the signs.

1Write balanced equations with states; note Δn_g.
2For ΔH ↔ ΔU conversions use ΔH = ΔU + Δn_gRT with R in kJ.
3For Hess cycles, reverse or multiply equations so unwanted species cancel; change ΔH accordingly.
4For formation data, use Σproducts − Σreactants.
5For calorimetry, compute q, divide by moles, and give ΔH or ΔU the sign opposite to the surroundings' gain.
6Check the units (kJ vs J, per mole).

Enthalpy Changes, Hess's Law and Calorimetry 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.

A hot-pack

Suppose a hand-warmer releases 12.5 kJ of heat while 0.050 mol of a salt dissolves in a closed pack at constant pressure.

Find ΔH per mole.

Try it first, then show the solutionSolution

q released = 12.5 kJ for 0.050 mol.
ΔH = −12.5/0.050 = −250 kJ/mol.

Answer: −250 kJ/mol.

Memory hook: Exothermic means negative ΔH.

Cooking gas cylinder

Suppose the combustion of 1 mol of propane (44 g) releases 2220 kJ.

Estimate the heat from 4.4 kg of propane.

Try it first, then show the solutionSolution

n = 4400/44 = 100 mol.
q = 100 × 2220 = 2.22 × 10⁵ kJ.

Answer: 2.22 × 10⁵ kJ.

Memory hook: Scale by the number of moles.

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

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

Gas moles only
Δn_g counts only gaseous species.
Use it when: ΔH–ΔU conversion.
Reverse flips the sign
Reverse an equation, change the sign of ΔH.
Use it when: Hess cycles.
Products minus reactants
Σ ΔfH(products) − Σ ΔfH(reactants).
Use it when: Formation data.
Bonds broken minus formed
Opposite order from the formation rule.
Use it when: Bond enthalpy estimates.

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

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

Example 1: ΔH to ΔU · 2 marks · Very short answerBoard level

For 2NO₂(g) → N₂O₄(g), ΔH = −57 kJ at 298 K (suppose). Find ΔU (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

Δn_g = 1 − 2 = −1. (1)
ΔU = ΔH − Δn_gRT = −57 + (8.314 × 10⁻³ × 298) = −57 + 2.48 = −54.5 kJ. (1)

Answer: −54.5 kJ.

Example 2: Enthalpy of formation of CO · 3 marks · Short answerBoard level

Given C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ/mol and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ/mol, find the enthalpy of formation of CO(g).

Try it first, then show the solutionSolution

Target: C + ½O₂ → CO. Add the first equation to the reverse of the second: CO₂ → CO + ½O₂ with ΔH = +283.0 kJ. (1)
C + O₂ → CO₂ (−393.5); CO₂ → CO + ½O₂ (+283.0). (1)
ΔfH(CO) = −393.5 + 283.0 = −110.5 kJ/mol. (1)

Answer: −110.5 kJ/mol.

Example 3: Energy from ethanol · 3 marks · Short answerHarder board level

Using ΔfH values (kJ/mol) C₂H₅OH(l) −277.7, CO₂(g) −393.5 and H₂O(l) −285.8, find the enthalpy of combustion of ethanol. Hence find the heat from 11.5 g and the rise in temperature of 5.0 kg of water that absorbs it (c = 4.18 J g⁻¹ K⁻¹).

Try it first, then show the solutionSolution

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O; ΔcH = 2(−393.5) + 3(−285.8) − (−277.7) = −1366.7 kJ/mol. (1)
n = 11.5/46 = 0.25 mol; q = 0.25 × 1366.7 = 341.7 kJ. (1)
ΔT = 341700/(5000 × 4.18) = 16.4 K. (1)

Answer: −1366.7 kJ/mol; 341.7 kJ; 16.4 K.

Example 4: Bomb calorimeter · 4 marks · Long answerHarder board level

Suppose 0.46 g of liquid ethanol (46 g/mol) is burnt completely in a bomb calorimeter of heat capacity 10.0 kJ/K, and the temperature rises by 1.37 K. C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). Find ΔU and ΔH per mole at 298 K.

Try it first, then show the solutionSolution

q = CΔT = 10.0 × 1.37 = 13.7 kJ released by 0.010 mol. (1)
ΔU = −13.7/0.010 = −1370 kJ/mol. (1)
Δn_g = 2 − 3 = −1. (1)
ΔH = ΔU + Δn_gRT = −1370 − 2.48 = −1372.5 kJ/mol. (1)

Answer: ΔU = −1370 kJ/mol; ΔH ≈ −1372.5 kJ/mol.

Example 5: Hydrogenation of ethyne · 4 marks · Long answerHOTS / competency

Given (kJ/mol) combustion enthalpies of C₂H₂(g) −1300, C(graphite) −393.5 and H₂(g) −285.8, and ΔfH of C₂H₄(g) = +52.4 and of C₂H₆(g) = −84.0. (a) Find ΔfH of ethyne. (b) Find ΔH for C₂H₂ + H₂ → C₂H₄ and for C₂H₄ + H₂ → C₂H₆. (c) Find the heat released in converting 5.2 g of ethyne completely to ethane (C₂H₂ = 26 g/mol).

Try it first, then show the solutionSolution

(a) 2C + H₂ → C₂H₂: ΔfH = 2(−393.5) + (−285.8) − (−1300) = +227.2 kJ/mol. (1)
(b) C₂H₂ → C₂H₄: 52.4 − 227.2 = −174.8 kJ. (1)
C₂H₄ → C₂H₆: −84.0 − 52.4 = −136.4 kJ. (1)
(c) Total −311.2 kJ/mol; n = 0.20 mol, so heat released = 62.2 kJ. (1)

Answer: (a) +227.2 kJ/mol; (b) −174.8 and −136.4 kJ; (c) 62.2 kJ.

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

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

Mistake: Using Σ reactants − Σ products.
Reversing the formation rule.
Do this: Σ ΔfH(products) − Σ ΔfH(reactants).
Usually costs: The sign.
Mistake: Counting all moles, not only gases, in Δn_g.
Including solids and liquids.
Do this: Only gaseous species count.
Usually costs: The answer mark.
Mistake: Forgetting to change the sign when reversing an equation.
Hess's law slip.
Do this: Reverse means the opposite sign.
Usually costs: The final answer.
Mistake: Giving ΔH in J with R in kJ.
Mixed units.
Do this: Use R = 8.314 × 10⁻³ kJ mol⁻¹ K⁻¹ with kJ.
Usually costs: The numerical.

Exam tips

Write the balanced equation with states before each calculation.
Show the Hess cycle with arrows for the manipulations.
Give ΔH per mole with its sign.
State that ΔfH of elements in their standard states is zero.
Mention that bond enthalpy results are approximate.

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

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
For the reaction H₂(g) + Cl₂(g) → 2HCl(g), ΔH = −184.6 kJ. The standard enthalpy of formation of HCl(g) is
(a) −184.6 kJ/mol (b) −92.3 kJ/mol (c) +92.3 kJ/mol (d) −369.2 kJ/mol
Hint
Formation is per mole of HCl.
Answer
(b) −92.3 kJ/mol.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For reactions involving only solids and liquids, ΔH is nearly equal to ΔU.
Reason (R): The change in volume, and hence PΔV, is negligible for solids and liquids.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
ΔH = ΔU + PΔV.
Answer
(a).
Q3 · 2 marks · Very short answerHarder board level
Hydrogen burns to give liquid water with ΔcH = −286 kJ/mol; methane has ΔcH = −890 kJ/mol. Compare the heat released per gram of each fuel (H₂ = 2, CH₄ = 16 g/mol).
Hint
Divide by molar mass.
Answer
H₂: 286/2 = 143 kJ/g; CH₄: 890/16 = 55.6 kJ/g. Hydrogen releases about 2.6 times more energy per gram.
Q4 · 3 marks · Short answerHarder board level
The enthalpies of combustion of graphite and diamond are −393.5 and −395.4 kJ/mol. Find ΔH for C(graphite) → C(diamond) and say which allotrope is more stable.
Hint
Reverse the diamond equation.
Answer
C(gr) + O₂ → CO₂ (−393.5); CO₂ → C(diamond) + O₂ (+395.4); ΔH = +1.9 kJ/mol. Graphite is more stable (lower enthalpy).
Q5 · 4 marks · Long answerHOTS / competency
Estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) from bond enthalpies (C–H 414, O=O 498, C=O in CO₂ 799, O–H 463 kJ/mol). Compare with the value from ΔfH (CH₄ −74.8, CO₂ −393.5, H₂O(g) −241.8 kJ/mol) and find the heat from 8.0 g of methane.
Hint
Broken minus formed.
Answer
Broken: 4(414) + 2(498) = 2652 kJ; formed: 2(799) + 4(463) = 3450 kJ; ΔH ≈ −798 kJ. From ΔfH: −393.5 + 2(−241.8) + 74.8 = −802.3 kJ. They agree closely; bond enthalpy values are averages. 8.0 g = 0.50 mol releases about 401 kJ.

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.ΔH and ΔU?Show answer
ΔH = ΔU + Δn_gRT.
2.ΔfH of an element?Show answer
Zero.
3.Hess's law?Show answer
ΔH is independent of the path.
4.Bomb calorimeter gives?Show answer
ΔU.
5.Units of ΔH?Show answer
kJ mol⁻¹.
6.Bond enthalpy estimate?Show answer
Broken minus formed.

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

Why is enthalpy useful?

Most reactions occur at constant pressure, where the heat exchanged equals ΔH.

Why is Δn_g needed?

Gases do work PΔV on the surroundings, making ΔH differ from ΔU.

Is a bond enthalpy always exact?

No; it is an average over compounds, so bond enthalpy estimates of ΔH are approximate.

What is the standard state?

The most stable form of a substance at 1 bar and a specified temperature (usually 298 K).

Why do we divide by moles in calorimetry?

To express the heat per mole of reaction as the enthalpy change.