Bomb Calorimeter: Converting Delta U to Delta H

By Aniket Bhardwaj · 21 October 2026 · Updated 21 October 2026 · Physical Chemistry

A bomb calorimeter burns a sample in a sealed steel vessel, so the volume cannot change. The heat it measures is therefore the change in internal energy, ΔU, and not the change in enthalpy, ΔH. Tables of combustion data are given as ΔH, so the conversion between the two is a standard step in IIT-JAM, GATE and CSIR-NET thermodynamics. This page derives the relation, shows how to calibrate the calorimeter, and works four numericals with the arithmetic done in full.

A flow diagram shows bomb-calorimeter heat yielding ΔU, then the gas-mole correction converting it to ΔH; naphthalene gives Δn_g = −2 and a −4.96 kJ mol⁻¹ correction. — At constant volume, calorimeter heat gives ΔU; correct for gaseous mole change to find ΔH.

Why the heat at constant volume is ΔU

From the first law, ΔU = q + w. In the sign convention where w is work done on the system, w = −P_extΔV. In a rigid bomb ΔV = 0, so w = 0 and q_V = ΔU. (In books that write ΔU = q − w, the working is the same; only the sign of w differs. State which convention you use.)

Enthalpy is defined as H = U + PV. For a reaction at constant temperature involving ideal gases, PV = n_gRT. Solids and liquids have a negligible PV change, so only the gases count.

ΔH = ΔU + Δn_gRT
Δn_g = (moles of gaseous products) − (moles of gaseous reactants)

R = 8.314 J K⁻¹ mol⁻¹, so R T at 298.15 K is 2.479 kJ mol⁻¹. Keep R and the energy unit consistent: if ΔU is in kJ, use R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹.

Calibration and the heat capacity of the calorimeter

The temperature rise ΔT is measured, and the heat capacity of the whole assembly (water, steel bomb, stirrer and thermometer), C_cal, converts it to heat:

q_released = C_cal × ΔT
ΔU (per mole) = − C_cal ΔT ÷ n

The minus sign appears because the heat gained by the calorimeter is lost by the reaction. C_cal is found by burning a standard sample whose heat of combustion is known, and then used for unknown samples.

Worked example 1: finding C_cal

Question. A 0.8000 g sample of a standard substance whose heat of combustion is given as 26.4 kJ per gram is burned in the bomb. The temperature rises by 2.64 K. Find the heat capacity of the calorimeter.

Heat released. q = 0.8000 × 26.4 = 21.12 kJ.

Heat capacity. C_cal = q ÷ ΔT = 21.12 ÷ 2.64 = 8.00 kJ K⁻¹.

Worked example 2: naphthalene, Δn_g < 0

Question. Burning 0.6400 g of naphthalene, C₁₀H₈(s), in the same calorimeter (C_cal = 8.00 kJ K⁻¹) raises the temperature by 3.21 K. Find ΔU and ΔH of combustion per mole at 298.15 K.

Reaction. C₁₀H₈(s) + 12 O₂(g) → 10 CO₂(g) + 4 H₂O(l).

Molar mass. 10(12.011) + 8(1.008) = 128.174 g mol⁻¹. Moles: 0.6400 ÷ 128.174 = 4.993 × 10⁻³ mol.

Heat released. q = 8.00 × 3.21 = 25.68 kJ.

ΔU. −25.68 ÷ 4.993 × 10⁻³ = −5143 kJ mol⁻¹.

Δn_g. Gaseous products = 10, gaseous reactants = 12, so Δn_g = −2 (water is liquid and naphthalene is solid, so neither counts).

ΔH. ΔH = ΔU + Δn_gRT = −5143 + (−2)(8.314 × 10⁻³)(298.15) = −5143 − 4.96 = −5148 kJ mol⁻¹.

The correction is under 0.1 per cent of the total. It is small but it is exactly what a numerical asks for.

Worked example 3: glucose, Δn_g = 0

Question. A 1.500 g sample of glucose, C₆H₁₂O₆(s), raises the temperature in the same calorimeter by 2.91 K. Find ΔU and ΔH of combustion.

Reaction. C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(l).

Moles. M = 6(12.011) + 12(1.008) + 6(15.999) = 180.156 g mol⁻¹; n = 1.500 ÷ 180.156 = 8.326 × 10⁻³ mol.

Heat. q = 8.00 × 2.91 = 23.28 kJ, so ΔU = −23.28 ÷ 8.326 × 10⁻³ = −2796 kJ mol⁻¹.

Δn_g. 6 mol CO₂ out, 6 mol O₂ in, so Δn_g = 0 and ΔH = ΔU = −2796 kJ mol⁻¹.

Worked example 4: ethanol, a liquid fuel

Question. Burning 0.9200 g of liquid ethanol in the same calorimeter raises the temperature by 3.41 K. Find ΔU and ΔH.

Reaction. C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l).

Moles. M = 2(12.011) + 6(1.008) + 15.999 = 46.069 g mol⁻¹; n = 0.9200 ÷ 46.069 = 0.01997 mol.

ΔU. q = 8.00 × 3.41 = 27.28 kJ; ΔU = −27.28 ÷ 0.01997 = −1366 kJ mol⁻¹.

Δn_g. 2 − 3 = −1.

ΔH. −1366 + (−1)(8.314 × 10⁻³)(298.15) = −1366 − 2.48 = −1368.5 kJ mol⁻¹.

Going the other way: ΔU from ΔH

Question. For N₂(g) + 3 H₂(g) → 2 NH₃(g), ΔH = −92.2 kJ at 298.15 K (value given in the problem). Find ΔU.

Δn_g. 2 − 4 = −2.

ΔU. ΔU = ΔH − Δn_gRT = −92.2 − (−2)(8.314 × 10⁻³)(298.15) = −92.2 + 4.96 = −87.2 kJ.

When the gas moles fall, the surroundings compress the system and do work on it, so less heat is given out at constant volume than at constant pressure. That is why ΔU is less negative than ΔH here.

Common mistakes

Counting liquids and solids in Δn_g. Only gas moles count. Liquid water in the products is not included.
Ignoring the sign of the heat. The calorimeter gains heat, the reaction loses it, so ΔU is negative for combustion.
Mixing J and kJ. Use R = 8.314 J K⁻¹ mol⁻¹ with joules and 8.314 × 10⁻³ with kilojoules.
Using Δn instead of Δn_g. Total moles in the equation are not needed; only the gaseous ones.
Per gram against per mole. ΔU and ΔH are asked per mole of the substance burned; divide by moles, not grams, unless asked.

Exam relevance

Situation	Relation
Δn_g = 0	ΔH = ΔU
Δn_g < 0	ΔH is more negative than ΔU
Δn_g > 0	ΔH is less negative (higher) than ΔU
Calorimeter calibration	C_cal = q_standard ÷ ΔT

Check the current official syllabus for your exam to see whether calorimetry is tested as a numerical or a concept, and which sign convention for work it uses.

No single tool in the suite performs bomb-calorimeter conversions. The scientific calculator can still check the arithmetic of each step.

Open the Chemistry Calculator Suite →

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