Thermochemistry

This is Unit Four of Chemistry, Part Two: thermochemical equations and Hess's law. Part One defined enthalpy and fixed its sign. This part is about using it — and the whole of it rests on one property established there, that enthalpy is a state function depending only on where a system starts and ends. Because that is true, enthalpy changes can be added, reversed and scaled like ordinary algebra, which lets a chemist calculate the heat of a reaction nobody has ever measured. By the end you will be able to say what happens to delta H when an equation is doubled; say why physical state has to appear in a thermochemical equation; say what makes Hess's law valid; say what to do to delta H when a reaction is reversed or scaled; and say what the summation form subtracts from what.

Three entries. What a thermochemical equation reports and what happens to that number when the coefficients change, why the physical states are part of the information rather than decoration, and how a real measurement gets scaled to the equation as written.

Double it.

Chem 0682 Scaling Enthalpy Change with a Thermochemical Equation's Coefficients

A thermochemical equation reports delta H beside a balanced equation, and multiplying every coefficient by a factor multiplies delta H by that factor too. If forming one mole of water releases 286 kilojoules, forming two moles by doubling every coefficient releases 572. Reading a doubled reaction's unchanged lab temperature as proof the heat did not scale confuses delta H with a measured value.

Liquid, or vapour.

Chem 0685 Physical State in a Thermochemical Equation

A reaction's enthalpy change depends on the physical states of every reactant and product, so a thermochemical equation has to show them. Forming liquid water from hydrogen and oxygen releases 286 kilojoules per mole, but forming water vapour from the same elements releases only 242. Omitting a state label leaves the equation's delta H ambiguous between two different quantities.

Which one runs out.

Chem 0688 Determining Molar Enthalpy Change from a Limiting Reactant

Determining a molar enthalpy change from measured data means identifying which reactant limits the reaction, then scaling the measured heat up to the amount the balanced equation would produce. Dropping 0.240 grams of magnesium into 25 millilitres of 0.500-molar acid gives 0.00987 moles of metal against 0.0125 of acid — and since the equation consumes two acid units per magnesium atom, the acid runs out first. Dividing the measured 2.89 kilojoules by the 0.00625 moles of reaction gives about minus 462. Scaling to the reactant in excess produces a value the data does not support.

Three entries. The fixed reference condition that lets values measured in different places be compared at all, the combustion quantity defined against it, and the calculation that converts a mass of fuel into a quantity of heat.

One shared footing.

Chem 0691 Standard State and ΔH°

A standard state is a fixed reference condition — materials at one bar, solutions at one molar — and delta H with a degree symbol marks a change measured under it. A table of standard values lets chemists compare reactions on one shared footing regardless of when each was measured. Delta H with its standard-state superscript and an ordinary delta H under unstated conditions are not interchangeable labels.

One mole, completely.

Chem 0694 Standard Enthalpy of Combustion

A standard enthalpy of combustion is the enthalpy change when one mole of a substance burns completely under standard-state conditions. Ethanol's is a large negative value, reflecting the substantial heat released when one mole burns completely. A tabulated value describes one mole burning completely, not whatever partial amount a real burn happens to consume.

Mass to moles first.

Chem 0697 Calculating Heat Released by Combustion of a Given Mass

Calculating the heat released by burning a given mass means converting that mass to moles using the molar mass, then multiplying by the standard enthalpy of combustion — which is defined per mole rather than per gram. A litre of isooctane at 0.692 grams per millilitre masses about 692 grams, which is roughly 6.06 moles; multiplying by about minus 5460 kilojoules per mole gives around minus 3.3 times ten to the fourth kilojoules. Multiplying the mass in grams directly by a per-mole value skips the conversion and is wrong by the molar mass.

Two entries. The reference quantity that the whole summation method depends on, defined against elements in their standard states, and the procedure for solving one out of a measured reaction.

From the elements.

Chem 0700 Standard Enthalpy of Formation

A standard enthalpy of formation is the heat change forming one mole of a compound from its elements in their standard states. Carbon dioxide's is the change when graphite and oxygen gas combine to form one mole of it. Assigning an element a nonzero formation enthalpy in some form other than its stable standard state breaks the reference the whole system rests on.

Exactly one mole.

Chem 0703 Writing and Solving Formation-Reaction Equations

Writing a formation-reaction equation means balancing elements in their standard states so the equation forms exactly one mole of the target compound, which lets a known overall enthalpy change be scaled down to give an unknown formation value. Ozone forms through three O2 going to two O3 with a known overall delta H — and because formation enthalpy is defined for exactly one mole of product, that value must be divided by two. Reporting the full reaction's delta H miscounts a two-mole value as a one-mole one.

Three entries. Why enthalpy changes can be added across steps at all — which follows directly from the state-function property — and then the two mechanical operations that build a target equation out of reactions somebody else already measured.

Sum of the steps.

Chem 0706 Hess's Law

Hess's law states that if a process is written as the sum of stepwise processes, the overall enthalpy change equals the sum of the steps' enthalpy changes — because enthalpy is a state function depending only on starting and ending conditions. Carbon burning directly to carbon dioxide releases the same total heat as carbon first forming carbon monoxide and that burning further, since both paths start and end at the same states. Adding known reactions without cancelling shared species gives a total matching no actual reaction.

Cancel the middle.

Chem 0709 Stepwise Hess's Law Addition

A stepwise calculation adds two or more known reactions, cancelling any species appearing on both sides of the combined total so the sum matches the target exactly, then sums the corresponding delta H values. Iron and chlorine forming FeCl3 can be built from a reaction forming FeCl2 plus one converting FeCl2 onward — adding them cancels the intermediate entirely and gives about minus 399.5 kilojoules per mole. An uncancelled intermediate signals the reactions were not combined correctly.

Reverse, and flip.

Chem 0712 Manipulating Reactions to Reach a Target Equation

Manipulating known reactions means reversing them, scaling them, or both, whichever combination sums exactly to the target — applying the matching sign change or multiplication factor to each reaction's own delta H as that reaction is adjusted. Building chlorine trifluoride's formation reaction takes halving one known reaction, reversing and halving a second, and reversing a third outright. Reversing an equation without flipping the sign of its delta H smuggles a wrong-signed value into the total.

Three entries to close the unit. A shortcut that skips the algebra entirely by going through formation enthalpies, that shortcut applied to a reaction with several substances on each side, and then the capstone — which measures a value in a calorimeter and checks it against the calculation.

Products minus reactants.

Chem 0715 The Summation Form of Hess's Law

A reaction's standard enthalpy change equals the summed formation enthalpies of its products minus those of its reactants, each weighted by its coefficient. The form treats a reaction as decomposing every reactant to its elements and reforming them into every product, in one equation. Summing formation enthalpies without weighting each by its own coefficient ignores how much of each substance actually reacts.

Weight each one.

Chem 0718 Applying the Summation Form to a Multi-Reactant Reaction

Applying the summation form to a reaction with several reactants and products means multiplying each substance's formation enthalpy by its coefficient, summing those weighted values separately across products and across reactants, and subtracting the reactant sum from the product sum. For nitrogen dioxide and water forming nitric acid and nitric oxide, that gives about minus 138.4 kilojoules. Adding a reactant's contribution into the product sum instead of subtracting gives roughly twice the magnitude, wrongly signed.

Measure, then check.

Chem 0729 Capstone: From Calorimetry Data to a Hess's-Law Enthalpy Change

A chemist measures a neutralisation in a coffee-cup calorimeter: fifty millilitres each of one-molar acid and base at twenty-two degrees warm the combined solution to 28.9. The solution's heat gain is about 2.9 kilojoules, so the reaction released 2.9 — negative delta H, exothermic, matching the observed rise. Dividing by the 0.0500 moles reacted gives near minus 58 per mole. That measured value is then checked against the summation form using tabulated formation enthalpies, and the two agree closely. Expecting them to match exactly rather than within the calorimeter's known imprecision misreads what a reference calculation can promise.

Five things to carry out of this part, and out of the unit. Doubling a thermochemical equation doubles its delta H, because delta H is reported for the equation exactly as written — and the lab temperature staying the same does not contradict that, since delta H is a per-equation quantity rather than a reading off a thermometer. Physical state has to appear in a thermochemical equation because the enthalpy change depends on it: forming liquid water releases 286 kilojoules per mole while forming water vapour releases 242, so an equation without state labels is ambiguous between two different numbers. Hess's law is valid because enthalpy is a state function — it depends only on where a system starts and ends, not on the route, which is exactly what permits adding steps together. When you reverse a reaction you flip the sign of its delta H, and when you scale a reaction you scale its delta H by the same factor; every structural change to an equation has to be carried through to its enthalpy value. And the summation form subtracts the summed formation enthalpies of the reactants from those of the products, each weighted by its coefficient in the balanced equation. Unit Five takes up electronic structure and the periodic properties that follow from it.