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Chemical Energetics: Thermochemistry

Topic 3 of 3

Indirect routes

Use supplied Hess and bond-energy routes.

A-Level 8873, revised syllabus (2026-2027)

Follow the arrows in the supplied cycle

The total enthalpy change depends on the initial and final states, not the route.

Hess' Law allows two routes between the same initial and final states to be equated. Read a supplied cycle by reversing signs for reversed arrows and multiplying energy values when an equation is scaled. The H1 calculation scope uses supplied simple cycles, formation/combustion/neutralisation data and bond energies; constructing a Born-Haber cycle is not an H1 requirement.

A supplied methane formation cycle

Elements and methane at the top both burn to the same products below. The left route releases 394 plus twice 286 kJ; the right combustion releases 890 kJ. The top arrow is the unknown methane formation enthalpy.

Given cycle: carbon is graphite; hydrogen, oxygen, methane and carbon dioxide are gases; water is liquid. Both downward routes consume 2O2. Values are kJ per reaction as drawn.

Worked example

A formation enthalpy that is difficult to measure directly

Use the supplied cycle and combustion values C = -394, H2 = -286 and CH4 = -890 kJ mol-1.

  1. Direct left route: -394 + 2(-286) = -966 kJ.
  2. Top then right route: Delta Hf(CH4) + (-890).
  3. Equate them: Delta Hf - 890 = -966.
Answer

ΔHf(CH4) = -76 kJ mol-1. The multiplier two belongs to the two moles of H2 burned.

The equivalent algebraic rule is ΔH(reaction) = sum of product formation enthalpies - sum of reactant formation enthalpies, each multiplied by its coefficient. When using combustion data instead, both sides lead down to common combustion products, so the subtraction direction differs. Following the actual arrows prevents memorised-sign errors.

Neutralisation data are quoted per mole of water formed. If a supplied cycle step neutralises two moles of water, use twice the listed neutralisation enthalpy. A reverse neutralisation arrow has the opposite sign. Check species, amount and physical states before treating two routes as having the same endpoint.

A supplied cycle using neutralisation data

The direct neutralisation route releases 53.6 kJ. An alternative route first separates HA into H+ and A-, then neutralises H+ with OH- to release 57.1 kJ. Equal endpoints allow the unknown first step to be calculated.

Supplied illustrative data, in kJ for one mole of HA. All solutes are aqueous and water is liquid at the same stated temperature. Positions show routes, not enthalpy levels; OH- is carried unchanged through the first step.

Worked example

Use neutralisation data without constructing a new cycle

In the supplied cycle, find the enthalpy for HA(aq) → H+(aq) + A-(aq).

  1. The direct route is HA + OH- to A- + H2O, with Delta H = -53.6 kJ for one mole of HA.
  2. Along the alternative route, HA first forms H+ and A-. The OH- is unchanged at this stage, so this arrow is the requested dissociation process.
  3. The second arrow is H+ + OH- to H2O, releasing 57.1 kJ. A- is unchanged in that step.
  4. Equate routes: -53.6 = Delta H(dissociation) + (-57.1). Therefore Delta H(dissociation) = -53.6 + 57.1 = +3.5 kJ mol-1.
Answer

Dissociation is endothermic by +3.5 kJ mol-1 for this supplied example. Some of the neutralisation energy is used in dissociation, so the net weak-acid neutralisation is less exothermic. The conclusion follows from these data; do not assume every weak acid has the same dissociation enthalpy.

Worked example

Average bond energies provide another route

Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g), given H-H = 436, Cl-Cl = 243 and H-Cl = 431 kJ mol-1.

  1. Break one H-H and one Cl-Cl bond: energy in = 436 + 243 = 679 kJ.
  2. Form two H-Cl bonds: energy out = 2 × 431 = 862 kJ.
  3. Delta H = energy for bonds broken - energy released by bonds formed.
Answer

ΔH = 679 - 862 = -183 kJ per mole of reaction as written, or -91.5 kJ per mole of HCl formed. Average bond-energy estimates may differ from measured values because the molecular environment matters.