Lesson 9: Enthalpy and Hess's Law

Reactions exchange energy with their surroundings, and chemistry keeps the accounts with enthalpy change, ΔH — the heat exchanged at constant pressure (Reactivity 1.1). An exothermic reaction releases heat (ΔH negative, the products sit lower in energy); an endothermic one absorbs it (ΔH positive). Energy-profile diagrams draw the story: reactants, products, the arrow of ΔH, and the activation barrier between. Because a laboratory cannot measure every reaction directly — some are too slow, too violent, or contaminated by side reactions — chemistry needs a bookkeeping theorem, and Hess's law provides it: the enthalpy change of a reaction is independent of the route taken, because enthalpy is a state function and energy is conserved. If you can reach the products by any path whose steps are measurable, the direct ΔH falls out of the arithmetic.

Standard quantities: formation and combustion

Two reference reactions organise the data. The standard enthalpy change of formation, ΔH_f⦵, is the enthalpy change when one mole of a compound forms from its elements in their standard states at 100 kPa and a stated temperature (usually 298 K). By definition ΔH_f⦵ of an element in its standard state is zero — there is no formation reaction. The standard enthalpy change of combustion, ΔH_c⦵, is the enthalpy released when one mole of a substance burns completely in oxygen under standard conditions. Formation data power the cycle ΔH_reaction = ΣΔH_f⦵(products) − ΣΔH_f⦵(reactants) — products minus reactants, with every value multiplied by its stoichiometric coefficient. Combustion data run the other way round (the cycle points down to common combustion products): ΔH_reaction = ΣΔH_c⦵(reactants) − ΣΔH_c⦵(products). Drawing the cycle with arrows before substituting is not pedantry; the direction of the cycle is the sign of the answer, and sign errors are this topic's biggest mark leak.

Worked example 1: Hess's law from combustion data

Calculate ΔH for the hydrogenation of ethene, C₂H₄(g) + H₂(g) → C₂H₆(g), given ΔH_c⦵ values: C₂H₄ = −1411, H₂ = −286, C₂H₆ = −1560 kJ mol⁻¹. (The direct reaction is hard to measure cleanly — mixed products, catalyst needed: exactly why Hess exists.) Route: burn the reactants and the products to the same end point (2CO₂ + 3H₂O in both cases). ΔH = ΣΔH_c⦵(reactants) − ΣΔH_c⦵(products) = [(−1411) + (−286)] − [−1560] = −1697 + 1560 = −137 kJ mol⁻¹. Sanity check the sign: hydrogenating a double bond should be exothermic (a π bond is traded for two stronger σ bonds), and −137 is a sensible magnitude — the experimental value is −137 kJ mol⁻¹. Always interrogate your sign against the chemistry; a positive hydrogenation enthalpy should send you back to the cycle before it costs you marks.

Worked example 2: the indirect Hess experiment — ΔH_f of MgO

Magnesium oxide's formation enthalpy cannot be measured by burning magnesium in a school calorimeter (the reaction is too vigorous and forms some Mg₃N₂ in air). Instead, measure two solution reactions in a polystyrene cup calorimeter with the same excess of 2 mol dm⁻³ HCl: Reaction 1: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g), ΔH₁. Magnesium ribbon (0.0500 mol scale) fizzes away; q = mcΔT on the acid solution. Reaction 2: MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l), ΔH₂, using the same acid volume and concentration. With measured values ΔH₁ = −467 kJ mol⁻¹ and ΔH₂ = −151 kJ mol⁻¹, and ΔH_f⦵(H₂O) = −286 kJ mol⁻¹ for H₂ + ½O₂ → H₂O: Hess cycle (target: Mg + ½O₂ → MgO): ΔH_f(MgO) = ΔH₁ − ΔH₂ + ΔH_f(H₂O) = (−467) − (−151) + (−286) = −602 kJ mol⁻¹ — the data-booklet value. Technique and evaluation, the parts that make this an IA-grade design: use the *same* acid batch and volume in both runs so the MgCl₂(aq) end-states genuinely match (the cycle assumes identical final states); take temperature readings every 30 s before and after mixing and extrapolate the cooling line back to the mixing time to recover the true ΔT (Lesson 18's method); use a lid and insulation; and repeat each run. The dominant error is heat loss to the cup and air during the slower MgO run, which biases ΔH₂ low in magnitude and propagates directly into the final answer — a quantifiable, discussable imperfection, not a confession of "human error".

Nature of Science — and the real world

Hess published his law in 1840, before the first law of thermodynamics was fully articulated — thermochemistry's bookkeeping anticipated the theory that later justified it, a reminder that careful measurement can run ahead of explanation. Today the same arithmetic sizes industrial reactors (an unaccounted exotherm destroyed the Bhopal plant's safety margins), formulates fuels and explosives, and lets dieticians assign energy values to food: the Calorie on a label is a Hess cycle run through a bomb calorimeter and a set of standard physiological corrections.

Common mistakes

• Reversing the combustion cycle: with combustion data it is reactants minus products; with formation data, products minus reactants. Draw the arrows; do not rely on memory.
• Forgetting stoichiometric multipliers — every ΔH_f⦵ or ΔH_c⦵ must be multiplied by its coefficient in the equation.
• Using ΔH_c⦵ for an element that does not burn (or for CO₂/H₂O, which are combustion *products* with ΔH_c = 0 in these cycles) as though it had a tabulated burning value.
• Dropping state symbols: ΔH values differ between H₂O(l) and H₂O(g) by 44 kJ mol⁻¹ per mole — enough to flip an answer's last mark.

IB Exam Experience

Question 1. What is the standard enthalpy change of formation of carbon dioxide?

• (A) The enthalpy change when 1 mol of carbon burns in oxygen
• (B) The enthalpy change when 1 mol of CO₂ forms from its elements in their standard states under standard conditions
• (C) The enthalpy change when 1 mol of CO₂ decomposes into its elements
• (D) The heat released by any reaction producing CO₂

Question 2. Using ΔH_f⦵ values (kJ mol⁻¹): CaCO₃ = −1207, CaO = −635, CO₂ = −394, calculate ΔH for CaCO₃(s) → CaO(s) + CO₂(g).

• (A) +178 kJ mol⁻¹
• (B) −178 kJ mol⁻¹
• (C) −2236 kJ mol⁻¹
• (D) +1029 kJ mol⁻¹

Question 3. A reaction has ΔH = −75 kJ mol⁻¹. Which statement is correct?

• (A) The products have more energy than the reactants and heat is absorbed
• (B) The products have less energy than the reactants and heat is released to the surroundings
• (C) The reaction cannot occur without heating
• (D) The activation energy is −75 kJ mol⁻¹

Answer key. 1. (B) — the definition requires formation from elements, standard states and standard conditions; (A) describes ΔH_c⦵ of carbon, which happens to be numerically equal here but is a different quantity. 2. (A) — [(−635) + (−394)] − (−1207) = −1029 + 1207 = +178 kJ mol⁻¹ (endothermic — limestone decomposition needs a kiln). 3. (B) — negative ΔH means energy leaves the system; the surroundings warm.

Structured question (Paper 2 style). The standard enthalpy changes of combustion are: CH₄ = −890, C₂H₆ = −1560 and H₂ = −286 kJ mol⁻¹. (a) Define the term standard enthalpy change of combustion. [2] (b) The dehydrogenation of ethane, C₂H₆(g) → C₂H₄(g) + H₂(g), is difficult to measure directly. Given additionally that ΔH_c⦵(C₂H₄) = −1411 kJ mol⁻¹, construct a Hess cycle and calculate ΔH for this reaction. [3] (c) State why this enthalpy change is difficult to measure directly by calorimetry. [1] Marking notes. (a) The enthalpy change when one mole of a substance [1] is burned completely in oxygen under standard conditions (100 kPa, 298 K) [1]. (b) ΔH = ΣΔH_c⦵(reactants) − ΣΔH_c⦵(products) = (−1560) − [(−1411) + (−286)] = −1560 + 1697 = +137 kJ mol⁻¹; award [1] for the correct cycle direction, [1] for substitution, [1] for the value with sign (the reverse of Worked example 1 — an endothermic dehydrogenation, consistent with needing a hot catalyst industrially). (c) The reaction is slow/requires a catalyst and high temperature, and side reactions (cracking, combustion) contaminate the heat measurement — accept any one sound reason [1].

IA idea: Determine the enthalpy of formation of a metal oxide indirectly by Hess's law, comparing solution calorimetry of the metal and its oxide in acid, and evaluate how cooling corrections change the result.

Two more enthalpy changes you must be able to define — and a second cycle tool

The standard enthalpy change of neutralisation ΔH_neut⦵ is the enthalpy change when one mole of water forms from the neutralisation of an acid by an alkali under standard conditions. For any strong acid with any strong alkali it is the same number, ≈ −57 kJ mol⁻¹, because the reaction is always the same ionic event: H⁺(aq) + OH⁻(aq) → H₂O(l) — the spectator ions cancel, and the identical value across HCl/NaOH, HNO₃/KOH and friends is beautiful evidence for the ionic model. With a weak acid it is less exothermic (some energy is consumed completing the acid's ionisation — Lesson 14's fingerprint again). The enthalpy change of solution prices dissolving: NaOH dissolves exothermically (lattice breaking costs less than hydration pays out), NH₄NO₃ endothermically (the basis of instant cold packs — and of Lesson 22's entropy story, since the process is spontaneous anyway).

A second worked cycle cements the method in the other common dress — a formation-data problem where the target is a displacement. Calculate ΔH for Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) using ΔH_f⦵(CuSO₄, aq) = −844 and ΔH_f⦵(ZnSO₄, aq) = −1063 kJ mol⁻¹ (elements zero). ΔH = ΣΔH_f(products) − ΣΔH_f(reactants) = (−1063 + 0) − (0 + (−844)) = −219 kJ mol⁻¹. The strongly negative value predicts what the bench shows — zinc displaces copper vigorously and the solution warms (this is also a voltaic cell in a beaker, Lesson 15) — and the same arithmetic prices every displacement in the reactivity series. Remember the division of labour: formation data, products minus reactants; combustion data, reactants minus products; and whenever a question supplies one set, it has silently told you which cycle to draw.

Enthalpy changes of solution: two giants nearly cancelling

Dissolving an ionic solid can be dissected by a Hess cycle of its own: first separate the lattice into gaseous ions (cost: the lattice enthalpy, large and positive), then hydrate those ions (gain: the hydration enthalpies, large and negative). For sodium chloride the two terms are each near 780 kJ mol⁻¹ and of opposite sign, so ΔH_solution is the small difference of two giants — a few kJ mol⁻¹, slightly endothermic — and its *sign* is decided by details of ion size and charge, not by any general rule. Magnesium sulfate's terms imbalance the other way (the small, doubly charged Mg²⁺ hydrates fiercely) and dissolution is exothermic enough to warm the beaker. Two consequences follow. Small percentage errors in either giant wreck the small difference — solution enthalpies are measured by calorimetry, not trusted to cycles, a live example of Lesson 10's theme that method follows measurability. And processes with ΔH_solution near zero or positive can still dissolve spontaneously, which enthalpy alone cannot explain — the loose thread that Lesson 22 ties off with entropy. Hold the question until then: *why does ammonium nitrate dissolve in cold water at all?*