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MHT-CET Chemistry · Chemical Thermodynamics and Energetics

Thermochemistry, Hess's Law and Bond Enthalpy

The enthalpy of a reaction is products minus reactants in enthalpies of formation (elements count zero), is independent of the route (Hess's law), and can be estimated from bond enthalpies as bonds broken minus bonds formed.

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Why this matters

21 PYQs, 1 HARD. Two thirds are ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants) for a combustion, or the per-mole scaling of a formation enthalpy (the equation makes 2 mol, the answer wants 1). The rest: bond enthalpies, Hess's law as a statement, and the enthalpy of solution as lattice plus hydration.

Concept 1 of 3

Enthalpy of Formation and the Reaction Enthalpy From It

Intuition

ΔfH° is the enthalpy change when ONE mole of a compound forms from its elements in their standard states; for an element it is zero by definition. Because enthalpy is a state function, any reaction's ΔH is the formation enthalpies of what it makes minus those of what it uses.

Definition

  • ΔrH∘=∑ΔfH∘(products)−∑ΔfH∘(reactants)\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}); ΔfH∘(O2,H2,N2,C)=0\Delta_f H^\circ(\text{O}_2, \text{H}_2, \text{N}_2, \text{C}) = 0.
  • Methane combustion: [−394+2(−286)]−[−75]=−891[-394 + 2(-286)] - [-75] = -891 kJ (with CO₂ at −390: −887). Ethene: [2(−390)+2(−286)]−(−52)=−1300[2(-390) + 2(-286)] - (-52) = -1300. Ethyne: [2(−393)−286]−227=−1299[2(-393) - 286] - 227 = -1299. Ethanol: [2(−390)+3(−285)]+280=−1355[2(-390) + 3(-285)] + 280 = -1355.
  • Per mole: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3, −92 kJ → ΔfH(NH3)=−46\Delta_f H(\text{NH}_3) = -46. H2+Cl2→2HCl\text{H}_2 + \text{Cl}_2 \to 2\text{HCl}, −194 → −97. Decomposition of water per mole: +573.2/2=+286.6+573.2/2 = +286.6 kJ.
  • Scaling by amount: 12 g C → 1 mol CH₄, ΔH = −75 kJ. 149.6 kJ released at −74.8 kJ mol⁻¹ → 2 mol = 32 g CH₄. 3 g ethane releasing 8.84 kJ → −88.4 kJ mol⁻¹.
  • Exothermic: neutralisation (KOH + HNO₃), combustion. Endothermic: melting, dissolving NaCl, N2+2O2→2NO2\text{N}_2 + 2\text{O}_2 \to 2\text{NO}_2.

Reaction enthalpy from formation enthalpies

ΔrH∘=∑νp ΔfHproducts∘−∑νr ΔfHreactants∘\Delta_r H^\circ = \sum \nu_p\,\Delta_f H^\circ_{\text{products}} - \sum \nu_r\,\Delta_f H^\circ_{\text{reactants}}

Worked example

Find ΔrH° for C3H8(g)+5O2→3CO2+4H2O(l)\text{C}_3\text{H}_8(g) + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}(l) given ΔfH°: propane −104, CO₂ −394, water −286 kJ mol⁻¹.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Chemical Thermodynamics and EnergeticsMODERATE
Calculate the standard enthalpy change of following reaction
CH4( g)+2O2( g)→CO2( g)+2H2O(l)CH_{4(\text{ }g)}+ 2O_{2(\text{ }g)}\rightarrow CO_{2(\text{ }g)}+ 2H_{2}O_{(\mathcal{l})}
If ΔfH∘(CH4)=−75 kJ mol−1\Delta_{f}H^{\circ}\left( CH_{4} \right)= - 75\text{ }kJ{\text{ }mol}^{- 1}
ΔfH∘(CO2)=−394 kJ mol−1ΔfH∘(H2O)=−286 kJ mol−1{\Delta_{f}H^{\circ}\left( CO_{2} \right)= - 394\text{ }kJ{\text{ }mol}^{- 1} }{\Delta_{f}H^{\circ}\left( H_{2}O \right)= - 286\text{ }kJ{\text{ }mol}^{- 1} }

[Q96 · 22 April Shift I · 2025]

Reporting the equation's ΔH as the formation enthalpy

ΔfH\Delta_f H is per MOLE of compound. An equation that makes 2 mol of NH₃ at −92 kJ gives −46 kJ mol⁻¹; −92 is the first option every time.
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Concept 2 of 3

Hess's Law and the Enthalpy of Solution

Intuition

Enthalpy is a state function, so the heat of a reaction does not depend on the number of steps — add the equations and add their ΔH values. Dissolving an ionic solid is two steps: pull the lattice apart (lattice enthalpy, endothermic) and hydrate the ions (exothermic); the sum is the enthalpy of solution.

Definition

  • Hess's law: ΔH\Delta H of a reaction is the same whether it occurs in one step or several. It DOES depend on physical states, temperature, and constant-P versus constant-V; it does NOT depend on the path.
  • Adding equations: C+12O2→CO\text{C} + \tfrac{1}{2}\text{O}_2 \to \text{CO} (−x) plus CO+12O2→CO2\text{CO} + \tfrac{1}{2}\text{O}_2 \to \text{CO}_2 (−y) gives C+O2→CO2\text{C} + \text{O}_2 \to \text{CO}_2, Q=−(x+y)Q = -(x + y).
  • ΔsolnH=ΔLH+ΔhydH\Delta_{\text{soln}}H = \Delta_L H + \Delta_{\text{hyd}}H. KCl: 700+(−680)=+20700 + (-680) = +20 kJ mol⁻¹ (endothermic — the solution cools).
  • Reversing an equation reverses the sign; multiplying it multiplies ΔH.

Hess's law; enthalpy of solution

ΔHoverall=∑ΔHsteps,ΔsolnH=ΔLH+ΔhydH\Delta H_{\text{overall}} = \sum \Delta H_{\text{steps}},\qquad \Delta_{\text{soln}}H = \Delta_L H + \Delta_{\text{hyd}}H

Worked example

Given S+O2→SO2\text{S} + \text{O}_2 \to \text{SO}_2 (−297 kJ) and S+32O2→SO3\text{S} + \tfrac{3}{2}\text{O}_2 \to \text{SO}_3 (−396 kJ), find ΔH for SO2+12O2→SO3\text{SO}_2 + \tfrac{1}{2}\text{O}_2 \to \text{SO}_3.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Chemical Thermodynamics and EnergeticsEASY
Identify the factor from following on which heat of reaction does not depend.

[Q80 · 9th May Shift 2 · 2023]

Subtracting hydration from lattice enthalpy

Hydration enthalpy is already NEGATIVE; the enthalpy of solution is the SUM. 700 − (−680) = 1380 is the planted option.
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Concept 3 of 3

Bond Enthalpy: Bonds Broken Minus Bonds Formed

Intuition

Breaking bonds costs energy, forming them releases it. Add the enthalpies of every bond in the reactants (broken) and subtract those of every bond in the products (formed). Bond dissociation energy is positive; the bond FORMATION energy is its negative. Per bond means divide by the count of that bond in the molecule.

Definition

  • ΔrH=∑ΔH(bonds broken)−∑ΔH(bonds formed)\Delta_r H = \sum \Delta H(\text{bonds broken}) - \sum \Delta H(\text{bonds formed}).
  • C2H4+H2→C2H6\text{C}_2\text{H}_4 + \text{H}_2 \to \text{C}_2\text{H}_6: broken 4 C–H + C=C + H–H = 4(414)+615+435=27064(414) + 615 + 435 = 2706; formed 6 C–H + C–C = 2484+347=28312484 + 347 = 2831; ΔH=−125\Delta H = -125 kJ.
  • ΔfH(NH3)\Delta_f H(\text{NH}_3) per mole: 12(941)+32(436)−3(389)=−42.5\tfrac{1}{2}(941) + \tfrac{3}{2}(436) - 3(389) = -42.5 kJ mol⁻¹ — use HALF an N₂, not a whole one.
  • Per bond: C(g)+4H(g)→CH4\text{C}(g) + 4\text{H}(g) \to \text{CH}_4, −1665 kJ → C–H bond enthalpy =1665/4≈416= 1665/4 \approx 416 kJ mol⁻¹.
  • Bond formation energy of H–H = −433: dissociation of 0.5 mol H₂ needs +216.5+216.5 kJ.

Reaction enthalpy from bond enthalpies

ΔrH=∑ΔHbroken−∑ΔHformed\Delta_r H = \sum \Delta H_{\text{broken}} - \sum \Delta H_{\text{formed}}

Worked example

Estimate ΔH for H2+Cl2→2HCl\text{H}_2 + \text{Cl}_2 \to 2\text{HCl} with bond enthalpies H–H 436, Cl–Cl 243, H–Cl 431 kJ mol⁻¹.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Chemical Thermodynamics and EnergeticsMODERATE
Calculate ΔrH∘\Delta_r H^\circ for H2C=CH2(g)+H2(g)→H3C-CH3(g)\text{H}_2\text{C=CH}_2(g) + \text{H}_2(g) \to \text{H}_3\text{C-CH}_3(g) using bond enthalpies (ΔH(C-H)=414 kJ/mol\Delta H(\text{C-H}) = 414\,\text{kJ/mol}, ΔH(C=C)=615 kJ/mol\Delta H(\text{C=C}) = 615\,\text{kJ/mol}, ΔH(H-H)=435 kJ/mol\Delta H(\text{H-H}) = 435\,\text{kJ/mol}, ΔH(C-C)=347 kJ/mol\Delta H(\text{C-C}) = 347\,\text{kJ/mol}).

[Q83 · 2nd May Shift 2 · 2023]

Using a whole N₂ for the formation enthalpy of NH₃

Formation enthalpy is per mole of product, so the equation is 12N2+32H2→NH3\tfrac{1}{2}\text{N}_2 + \tfrac{3}{2}\text{H}_2 \to \text{NH}_3. The whole-equation answer, −85 kJ, is offered beside the right −42.5.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Enthalpy of Formation and the Reaction Enthalpy From It

    Reaction enthalpy from formation enthalpies

    ΔrH∘=∑νp ΔfHproducts∘−∑νr ΔfHreactants∘\Delta_r H^\circ = \sum \nu_p\,\Delta_f H^\circ_{\text{products}} - \sum \nu_r\,\Delta_f H^\circ_{\text{reactants}}
  • Hess's Law and the Enthalpy of Solution

    Hess's law; enthalpy of solution

    ΔHoverall=∑ΔHsteps,ΔsolnH=ΔLH+ΔhydH\Delta H_{\text{overall}} = \sum \Delta H_{\text{steps}},\qquad \Delta_{\text{soln}}H = \Delta_L H + \Delta_{\text{hyd}}H
  • Bond Enthalpy: Bonds Broken Minus Bonds Formed

    Reaction enthalpy from bond enthalpies

    ΔrH=∑ΔHbroken−∑ΔHformed\Delta_r H = \sum \Delta H_{\text{broken}} - \sum \Delta H_{\text{formed}}

Watch out for (3)

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