Calorimetry and Enthalpy
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Direct answer
Heat measured in a sealed bomb calorimeter at constant volume equals the internal energy change ΔU; heat measured in an open vessel at constant (atmospheric) pressure equals the enthalpy change ΔH, and the two connect through ΔH = ΔU + ΔngRT, where Δng is the change in moles of gaseous species only. Enthalpy is a state function, so it depends purely on initial and final states, which is what makes standard tabulations possible. NEET's canonical worked case is the combustion of benzene in a bomb calorimeter releasing −3263.9 kJ/mol, which converts to ΔH = −3267.6 kJ/mol after applying Δng = −1.5 at 298 K. Standard enthalpies — of formation (zero for elements in their reference state), combustion (always exothermic), and neutralisation of a strong acid with a strong base (about −57.3 kJ per mole, commonly quoted from NCERT) — are the tabulated values questions draw on.
What you must remember
- Two heat identities: qp = ΔH (constant pressure, coffee-cup calorimeter) and qv = ΔU (constant volume, bomb calorimeter).
- Bridge equation: ΔH = ΔU + ΔngRT; count only gaseous moles — for C(graphite) + O2(g) → CO2(g), Δng = 1 − 1 = 0, so ΔH equals ΔU exactly.
- Enthalpy of formation: ΔfH° = 0 for an element in its most stable form at standard state (O2 gas, C graphite, Br2 liquid).
- Neutralisation constant: strong acid + strong base in dilute solution gives about −57.3 kJ per mole of water formed, because it is always H+ + OH− → H2O.
- Combustion is exothermic by definition, so ΔcH° is always negative — a quick eliminator in matching questions.
- Extensive property arithmetic: if a reaction is doubled, ΔH doubles; if reversed, sign flips — Hess's law machinery in one line.
- Units discipline: ΔH in kJ/mol, R = 8.314 J K−1 mol−1, so convert the RT term to kJ before adding.
Converting a bomb-calorimeter reading
Work the NCERT benzene example the way the paper expects. Combustion: C6H6(l) + 15/2 O2(g) → 6CO2(g) + 3H2O(l). The bomb fixes volume, so the measured −3263.9 kJ/mol is ΔU. Gaseous moles: products 6, reactants 7.5, so Δng = −1.5. Then ΔH = ΔU + ΔngRT = −3263.9 + (−1.5 × 8.314 × 10−3 × 298) kJ = −3263.9 − 3.7 = −3267.6 kJ/mol. The enthalpy is more negative because the reaction contracts gas volume, so the system does less expansion work at constant pressure and must release extra heat.
The reverse logic saves marks: if Δng is positive, ΔH is less negative than ΔU; if zero (C + O2 → CO2), the two are identical and any option showing a difference is wrong. Note that benzene and water being liquids is precisely why they do not enter Δng — the commonest slip is counting all stoichiometric coefficients instead of gaseous ones only.
Where candidates drop easy marks
The chapter's questions are procedural, and the slips are predictable. The Δng count includes only gases — condensed phases never enter. The unit trap: R in joules against ΔH in kilojoules; the RT correction is tiny (3.7 kJ above) but exact answers hinge on it. Enthalpy of formation questions hide the reference state: ΔfH° of O2 is zero but of O3 is not; of Br2(l) zero but Br2(g) not; diamond is not carbon's reference state, so ΔfH°(diamond) is about 1.9 kJ/mol, a classic option. Neutralisation questions with a weak acid have smaller magnitude than −57.3 kJ because part of the heat drives ionisation — a conceptual favourite. And exothermic always means ΔH negative, ΔU negative, and system heating surroundings — assertion-reason items test that chain with "combustion in a bomb releases heat at constant volume" welded to "ΔH equals ΔU here", true only when Δng = 0.
Frequently asked questions
Why is the heat from a bomb calorimeter equal to ΔU and not ΔH?
The sealed bomb holds volume constant, and at constant volume all heat exchanged changes internal energy with no pressure-volume work term.
When do ΔH and ΔU become equal?
When Δng = 0 — no change in gaseous moles, as in the combustion of graphite to carbon dioxide — the RT term vanishes and ΔH = ΔU.
Why is the standard enthalpy of formation of an element zero?
The element's reference state (O2 gas, C graphite, Br2 liquid) is defined as the formation product from itself, needing no change and hence zero enthalpy.
Why is neutralisation enthalpy lower when a weak acid is used?
Part of the liberated energy is consumed ionising the weak acid before its proton can meet hydroxide, so less than the ~57.3 kJ emerges as heat.
If a reaction's ΔH is −100 kJ, what is ΔH when the equation is halved and reversed?
Halving gives −50 kJ, and reversing flips the sign, so the new reaction carries +50 kJ — state-function arithmetic NEET tests directly.