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A bomb calorimeter gives you ΔU. A question asks for ΔH. The bridge between them is one short equation, ΔH = ΔU + ΔngRT, and most errors with it come from counting the wrong moles or mixing joules with kilojoules. This post derives the equation, works four examples (two straight from NCERT), and lists the slips to watch for.
Where does the equation come from?
The NCERT Class 11 chapter Thermodynamics builds it in section 5.2.2. Enthalpy is defined as H = U + pV, so at constant pressure ΔH = ΔU + pΔV. Solids and liquids barely change volume, so pΔV matters only for gases. For ideal gases at constant p and T, pΔV = ΔngRT, which gives NCERT's equation 5.10:
ΔH = ΔU + ΔngRT
Here Δng is "the number of moles of gaseous products minus the number of moles of gaseous reactants". The JEE (Main) 2026 syllabus lists internal energy, enthalpy and the first law in its thermodynamics unit, so this relation is tested in board exams and JEE Main.
Why the two values differ: a bomb calorimeter is sealed, so it measures heat at constant volume, which is ΔU. A reaction in an open flask can push the air back, so the heat measured there is ΔH.
Which value of R, and in what units?
NIST gives the molar gas constant as 8.314 462 618 J mol⁻¹ K⁻¹. Use 8.314 J mol⁻¹ K⁻¹ and T in kelvin. Your answer for ΔngRT then comes out in joules per mole, so divide by 1000 before adding it to a ΔU given in kJ.
Worked example 1: boiling water (NCERT Problem 5.5)
The molar enthalpy of vaporisation of water at 1 bar and 100 °C is 41 kJ mol⁻¹. Find ΔU.
H₂O(l) → H₂O(g), so Δng = 1 − 0 = 1.
ΔngRT = 1 × 8.3 × 373 = 3096 J = 3.10 kJ mol⁻¹. (NCERT rounds R to 8.3 here.)
ΔU = ΔH − ΔngRT = 41.00 − 3.10 = 37.9 kJ mol⁻¹.
About 3 kJ of each mole's heat goes into pushing back the atmosphere as the vapour forms. The rest raises the internal energy.
Worked example 2: making ammonia at 298 K
N₂(g) + 3H₂(g) → 2NH₃(g).
Δng = 2 − (1 + 3) = −2.
ΔH − ΔU = −2 × 8.314 × 298 = −4955 J, about −4.96 kJ mol⁻¹.
You do not need either value to find the gap. Four moles of gas become two, the surroundings push the system in, and ΔH is about 4.96 kJ more negative than ΔU.
Worked example 3: burning graphite (NCERT Problem 5.6)
C(graphite) + O₂(g) → CO₂(g). One mole of gas is used and one is made, so Δng = 1 − 1 = 0. Graphite is a solid and is not counted. NCERT finds ΔU = −2.48 × 10² kJ mol⁻¹ from the calorimeter's temperature rise, and since Δng = 0, ΔH = ΔU.
Worked example 4: the phase-label trap (NCERT exercise 5.8)
Cyanamide reacts with oxygen in a bomb calorimeter, and ΔU = −742.7 kJ mol⁻¹ at 298 K. The products are N₂(g) + CO₂(g) + H₂O(l), which is 2 mol of gas. Water is a liquid and does not count.
The exercise text describes cyanamide as a solid, NH₂CN(s), but the printed equation labels it (g). The answer depends on which you use:
Solid cyanamide: gaseous reactant is 3/2 mol O₂ only, so Δng = 2 − 1.5 = +0.5. ΔngRT = 0.5 × 8.314 × 298 = 1239 J = 1.24 kJ. ΔH = −742.7 + 1.24 = −741.5 kJ mol⁻¹.
Gaseous cyanamide: Δng = 2 − 2.5 = −0.5, so ΔH = −742.7 − 1.24 = −743.9 kJ mol⁻¹.
The lesson is the habit, not the answer key. Read the state symbol of every species before you count, and say in your answer which phase you assumed.
Common mistakes and how to avoid them
Counting liquids and solids. Only (g) species go into Δng.
Reversing the subtraction. It is products minus reactants. Making fewer gas moles gives a negative Δng.
Adding joules to kilojoules. Convert ΔngRT to kJ first.
Using Celsius. 25 °C is 298 K.
Using the wrong amount of reaction. Δng comes from the balanced equation as written. If the question asks per mole of a reactant whose coefficient is 2, halve everything.
A practice routine
For ten reactions from your textbook, write only Δng next to each, with no other working. Check the state symbols every time.
Compute ΔngRT at 298 K for each and confirm it is a small multiple of 2.48 kJ.
Solve NCERT Problems 5.5 and 5.6 and exercise 5.8 without looking at the solutions.
Mix these with the sign questions from our guide to thermodynamics sign conventions. The reason ΔH is path-independent is explained in state and path functions.
If you set up the equation correctly but the numbers still go wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the line where your units slipped.
Start with the ammonia, graphite and water examples above: write Δng for each from memory, then check.
