AS Chemistry: Hess Law and Enthalpy Cycles — Revision Notes

Condensed for the final weeks. For the full explanation, use the Hess’s Law and Enthalpy Cycles study guide.

Sign convention

EXOTHERMIC   delta-H NEGATIVE   heat released to surroundings   products lower
ENDOTHERMIC  delta-H POSITIVE   heat absorbed from surroundings  products higher

Bond breaking is endothermic. Bond making is exothermic. Every enthalpy explanation reduces to the balance between the two.

Definitions

Term	Definition
Standard enthalpy of formation ΔH_f	Enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions
Standard enthalpy of combustion ΔH_c	Enthalpy change when one mole of a substance is completely burned in oxygen
Standard enthalpy of neutralisation	Enthalpy change when an acid and alkali react to form one mole of water
Bond enthalpy	Energy required to break one mole of a specified bond in the gaseous state

Standard conditions: 298 K and 100 kPa, all substances in standard states.

Two marks live in the small print: “one mole” appears in all four, and ΔH_f of any element in its standard state is zero by definition.

Hess’s law

The total enthalpy change is independent of the route taken, provided the initial and final conditions are the same.

Two cycle patterns, and choosing the right one is most of the work:

Given formation data — arrows point up from elements:

delta-H_r = sum(delta-H_f products) - sum(delta-H_f reactants)

Given combustion data — arrows point down to combustion products:

delta-H_r = sum(delta-H_c reactants) - sum(delta-H_c products)

They are opposite subtractions. Remember which by drawing the arrows: formation arrows point into the reaction line from below, combustion arrows point away downwards.

Multiply by the stoichiometric coefficients, and remember ΔH_f(O₂) = ΔH_f(C, graphite) = 0.

Worked example. Find ΔH_f⦵ for methane, CH₄(g), given ΔH_c(C, graphite) = −394 kJ/mol, ΔH_c(H₂) = −286 kJ/mol, ΔH_c(CH₄) = −890 kJ/mol.

formation reaction: C(graphite) + 2H2(g) -> CH4(g)

route via combustion:
[ΔHc(C) + 2 x ΔHc(H2)] - ΔHc(CH4)
= [(-394) + 2(-286)] - (-890)
= -966 - (-890)
= -76 kJ/mol

ΔH_f⦵(CH₄) = −76 kJ/mol. Combustion data given → elements and target compound go at the top of the cycle, combustion products at the bottom, since every substance can reach the same combustion products by a measurable route.

Bond enthalpy calculations

delta-H = sum(bonds broken) - sum(bonds made)

Bond enthalpies are mean values averaged across many compounds, so results are approximate. They also apply only to gaseous species, which is why a bond-enthalpy answer differs from the experimental value when liquids are involved — the enthalpy of vaporisation is unaccounted for.

Worked example. Calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g), given bond enthalpies H–H = +436, Cl–Cl = +243, H–Cl = +432 (all kJ/mol).

bonds broken: 436 + 243 = 679 kJ/mol
bonds made:   2 x 432 = 864 kJ/mol
delta-H = 679 - 864 = -185 kJ/mol

Bond enthalpy values are always quoted positive (the energy needed to break that bond); the sign of the final answer comes from the subtraction, not from the individual values.

Calorimetry

q = m c delta-T          c(water) = 4.18 J g^-1 K^-1
delta-H = -q / n         (n = moles of the LIMITING reactant)

Four points that decide the marks:

• m is the mass of water (or solution) heated, not the mass of fuel burned.
• 1 cm³ of solution ≈ 1 g.
• q comes out in joules; ΔH is usually quoted in kJ mol⁻¹ — divide by 1000.
• The minus sign converts “heat gained by the water” into “enthalpy change of the reaction”. Omitting it reverses the sign of the answer.

Why experimental values are less exothermic than data-book values: heat loss to the surroundings, incomplete combustion, evaporation of the fuel, heat absorbed by the apparatus, and non-standard conditions. A good answer names two or three and says how each would be reduced — insulation, a lid, a shorter distance between flame and beaker.

Exam traps

• Omitting “one mole” or “standard states” from a definition.
• Using the mass of fuel rather than the mass of water in q = mcΔT.
• Dropping the minus sign in ΔH = −q/n.
• Subtracting the wrong way round for the type of data given.
• Forgetting to divide by the moles of the limiting reactant.
• Treating bond enthalpy answers as exact.
• Not converting J to kJ.

Self-test

1. Define standard enthalpy of formation, and give its value for O₂.
2. Give the two Hess’s law expressions and say when each applies.
3. In q = mcΔT for a burning-fuel experiment, what is m?
4. Why are bond enthalpy calculations only approximate?
5. Give three reasons an experimental enthalpy of combustion is less exothermic than the data-book value.

Answers: 1. The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions; zero for O₂, as for any element in its standard state. 2. With formation data, ΣΔH_f(products) − ΣΔH_f(reactants); with combustion data, ΣΔH_c(reactants) − ΣΔH_c(products). 3. The mass of water (or solution) being heated, not the mass of fuel. 4. Bond enthalpies are mean values averaged over many different compounds, and apply only to gaseous species. 5. Heat lost to the surroundings, incomplete combustion, evaporation of the fuel, heat absorbed by the apparatus, non-standard conditions — any three.