GATE Thermochemistry — Hess's Law Applications
Enthalpy, being a state function, does not care how a reaction happened — only where it started and where it ended. Hess's law is that idea turned into an exam tool: any enthalpy change can be found by adding up a convenient path of known steps, even one that is impossible to carry out in a lab directly. This guide covers the four ways GATE Chemistry uses that idea — from formation values, from a reaction cycle, from bond enthalpies, and its temperature and constant-volume corrections — each with a complete worked example.
Hess's law, stated
If a target reaction can be written as the algebraic sum of other reactions,
ΔHtarget = Σ (ΔH of each step), with each step's coefficient and sign matching how it was combined
Elements in their standard states have ΔHf = 0 by definition — O₂(g), N₂(g), C(graphite), H₂(g) and similar all contribute nothing to the sum.
Worked example 1 — ΔH from formation values. Find the standard enthalpy of
combustion of propane, C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), given ΔHf[C₃H₈(g)] =
−103.8 kJ/mol, ΔHf[CO₂(g)] = −393.5 kJ/mol, ΔHf[H₂O(l)] = −285.8 kJ/mol.
Products: 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ
Reactants: (−103.8) + 5(0) = −103.8 kJ
ΔHrxn = −2323.7 − (−103.8) = −2219.9 kJ/mol
Worked example 2 — a Hess's law cycle for an unmeasurable reaction. Find
ΔH for C(graphite) + ½O₂(g) → CO(g), which cannot be measured directly because carbon burned
in limited oxygen always gives some CO₂ as well. Given:
(i) C(graphite) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ/mol
(ii) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ/mol
Subtracting (ii) from (i) cancels CO₂ on both sides and leaves exactly the target reaction:
[C + O₂ → CO₂] − [CO + ½O₂ → CO₂] gives C + ½O₂ → CO
ΔHtarget = ΔH₁ − ΔH₂ = −393.5 − (−283.0) = −110.5 kJ/mol
Worked example 3 — bond enthalpy calculation. Estimate ΔH for
H₂(g) + Cl₂(g) → 2HCl(g), given bond enthalpies H−H = 436 kJ/mol, Cl−Cl = 242 kJ/mol,
H−Cl = 431 kJ/mol.
ΔH = Σ(bonds broken) − Σ(bonds formed) = (436 + 242) − (2 × 431) = 678 − 862 =
−184 kJ/mol
Bond enthalpies are average values across many compounds, so this method is a good estimate
for simple gas-phase reactions but is noticeably less reliable for resonance-stabilised
systems (a bond-enthalpy estimate for benzene hydrogenation, for instance, overshoots the
real value because it has no way to account for the extra stabilisation) — treat it as an
estimate, not as accurate as a formation-value calculation.
Kirchhoff's equation — ΔH at a different temperature
Worked example 4. A reaction has ΔH = −50.0 kJ/mol at 298 K, and
ΔCp = 20 J mol⁻¹ K⁻¹ = 0.020 kJ mol⁻¹ K⁻¹. Find ΔH at 350 K.
ΔH(350) = −50.0 + 0.020 × (350 − 298) = −50.0 + 0.020 × 52 = −50.0 + 1.04 =
−48.96 kJ/mol
ΔH vs ΔU — what a bomb calorimeter actually measures
A bomb calorimeter runs at constant volume, so it directly measures ΔU, not ΔH. The two are related through the change in moles of gas:
Worked example 5 — when Δn = 0. Combustion of glucose,
C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), releases ΔU = −2801 kJ/mol at 298 K in a bomb
calorimeter. Find ΔH.
Δn(g) = 6 (CO₂) − 6 (O₂) = 0 — moles of gas are unchanged, since glucose and water
are not gaseous.
ΔH = ΔU + (0)RT = −2801 kJ/mol — identical to ΔU whenever Δn = 0.
Worked example 6 — when Δn ≠ 0. Combustion of one mole of liquid octane,
C₈H₁₈(l) + 12.5 O₂(g) → 8CO₂(g) + 9H₂O(l), gives ΔU = −5460 kJ/mol at 298 K. Find ΔH.
Δn(g) = 8 − 12.5 = −4.5
ΔnRT = (−4.5)(8.314 × 10⁻³ kJ mol⁻¹ K⁻¹)(298 K) = (−4.5)(2.4776) = −11.15 kJ
ΔH = ΔU + ΔnRT = −5460 + (−11.15) = −5471.2 kJ/mol — a small but real
correction, and the sign of the correction depends entirely on whether gas moles increase or
decrease.
Common mistakes that cost marks
- Forgetting to flip the sign of ΔH when a step is reversed. If a reaction in a Hess's law cycle needs to run backwards to build the target, its ΔH must be negated, not reused as given.
- Forgetting to scale ΔH when a step is multiplied. Doubling a reaction's stoichiometry doubles its ΔH — enthalpy is extensive, unlike an intensive quantity such as cell potential.
- Treating bond enthalpies as exact. They are averaged over many compounds; use ΔHf-based calculations when accurate values are available and bond enthalpies are given only as a fallback.
- Assigning a nonzero ΔHf to an element in its standard state. O₂(g), N₂(g), H₂(g), C(graphite) and similar are all zero by definition — a nonzero value for one of these is always a sign of a data-reading error, not a real answer.
- Getting the sign of the Δn(g)RT correction backwards. Δn is products minus reactants; reversing that flips a positive correction into a negative one and changes the final answer's direction.
Where this appears in GATE Chemistry
| You are asked for | Use | Watch for |
|---|---|---|
| ΔH of a reaction | ΔHf(products) − ΔHf(reactants) | Elements in standard state contribute zero |
| ΔH not directly measurable | Combine given reactions (Hess's law cycle) | Sign flips on reversal; coefficients scale ΔH |
| Rough ΔH from structure alone | Bond enthalpy: bonds broken − bonds formed | Estimate only, unreliable for resonance-stabilised molecules |
| ΔH at a different temperature | Kirchhoff's equation with ΔCp | ΔCp can itself be temperature-dependent in harder problems |
| Converting a calorimeter's ΔU to ΔH | ΔH = ΔU + Δn(g)RT | Only gas-phase species count toward Δn |
Cross-check Gibbs energy alongside enthalpy. The ABC Chemistry Calculator Suite's Gibbs free energy calculator is a natural next step once ΔH and ΔS are known.
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