Chemical Thermodynamics · Lesson 5 of 10

Formation Enthalpies and Hess’s Law

“Calculate otherwise inaccessible reaction heats by combining consistent reference data and reaction paths.”

Learning Objectives

• Write a standard formation reaction for one mole of a compound. • Identify elemental reference states and interpret zero formation enthalpy. • Calculate reaction enthalpy from tabulated formation values. • Scale and reverse thermochemical equations consistently. • Use Hess’s law to construct an indirect enthalpy route.

Building a shared reference

We can measure an enthalpy difference without knowing an absolute enthalpy. To organise thousands of such differences, chemists choose a common reference: the elements in their reference states at the temperature of interest. Formation enthalpy then measures the change from those elements to a particular compound. The amount formed must be exactly one mole.

Definition

Standard molar enthalpy of formation

The enthalpy change when one mole of a substance forms from its constituent elements in their reference states, with all species in their specified standard states at the stated temperature. It is written $Δ_{f} H^{\circ}$ , usually in kJ mol−1.

At 298 K and 1 bar, reference examples are hydrogen gas as $H_{2}$ , oxygen gas as $O_{2}$ , graphite for carbon, and rhombic sulphur. By convention, each element in its reference state has $Δ_{f} H^{\circ} = 0$ . This sets a reference zero; it does not mean the element has no internal energy or zero entropy. Diamond and gaseous carbon atoms are not the carbon reference state, so their formation values need not be zero.

Example — Recognising a formation reaction

Problem
Which equation defines standard formation enthalpy of hydrogen bromide gas: $H_{2} (g) + B r_{2} (l) \to 2HBr (g)$ or half of that equation?

1.Hydrogen and liquid bromine are appropriate elemental reference states at 298 K.
2.The full equation forms two moles of HBr, so its reaction enthalpy is twice the molar formation enthalpy.
3.The defining equation is $\frac{1}{2} H_{2} (g) + \frac{1}{2} B r_{2} (l) \to HBr (g)$ . Fractional stoichiometric coefficients are allowed.
4.Using $Δ_{f} H^{\circ} (HBr, g) = - 36.40 kJ mo l^{- 1}$ , the full two-mole reaction has $Δ_{r} H^{\circ} = - 72.80 kJ mo l^{- 1}$ per reaction as written.

Making calcium carbonate from calcium oxide and carbon dioxide is not a formation reaction from reference elements: its starting materials are already compounds. Likewise, the physical state matters. Forming liquid water and forming water vapour have different enthalpy changes even though the molecular formula is the same.

Reading and using a formation table

The table below groups reference data by type so that their meaning is easier to see. Values are the chapter’s rounded standard formation data near 298 K. A positive entry means formation from reference elements requires enthalpy; a negative one means it releases enthalpy. These signs alone do not establish a reaction rate or complete thermodynamic stability.

Substance and state	Formation enthalpy / kJ mol−1	Substance and state	Formation enthalpy / kJ mol−1
$A l_{2} O_{3} (s)$	−1675.7	$HI (g)$	+26.48
$BaC O_{3} (s)$	−1216.3	$KCl (s)$	−436.75
$B r_{2} (l)$	0	$KBr (s)$	−393.8
$B r_{2} (g)$	+30.91	$MgO (s)$	−601.70
Carbon, diamond	+1.89	$Mg (OH)_{2} (s)$	−924.54
Carbon, graphite	0	$NaF (s)$	−573.65
$CaC O_{3} (s)$	−1206.92	$NaCl (s)$	−411.15
$CaO (s)$	−635.09	$NaBr (s)$	−361.06
$C H_{4} (g)$	−74.81	$NaI (s)$	−287.78
$C_{2} H_{4} (g)$	+52.26	$N H_{3} (g)$	−46.11
$C H_{3} OH (l)$	−238.86	$NO (g)$	+90.25
$C_{2} H_{5} OH (l)$	−277.69	$N O_{2} (g)$	+33.18
$C_{6} H_{6} (l)$	+49.0	$PC l_{3} (l)$	−319.70
$CO (g)$	−110.53	$PC l_{5} (s)$	−443.5
$C O_{2} (g)$	−393.51	$Si O_{2} (s, quartz)$	−910.94
$C_{2} H_{6} (g)$	−84.68	$SnC l_{2} (s)$	−325.1
$C l_{2} (g)$	0	$SnC l_{4} (l)$	−511.3
$C_{3} H_{8} (g)$	−103.85	$S O_{2} (g)$	−296.83
n-Butane gas	−126.15	$S O_{3} (g)$	−395.72
Red $HgS (s)$	−58.2	$Si H_{4} (g)$	+34
$H_{2} (g)$	0	$SiC l_{4} (g)$	−657.0
$H_{2} O (g)$	−241.82	$C (g)$	+716.68
$H_{2} O (l)$	−285.83	$H (g)$	+217.97
$HF (g)$	−271.1	$Cl (g)$	+121.68
$HCl (g)$	−92.31	$F e_{2} O_{3} (s)$	−824.2
$HBr (g)$	−36.40	—	—

Imagine forming every reactant from reference elements, then converting those reactants to products. An alternative route forms the products directly from the same reference elements. Because both routes have the same endpoints, subtracting the reactant formation enthalpies from the product formation enthalpies gives the reaction enthalpy.

Reaction enthalpy from formation dataLaTeX

Δ_{r} H^{\circ} = i \sum a_{i} Δ_{f} H_{i}^{\circ} (products) - j \sum b_{j} Δ_{f} H_{j}^{\circ} (reactants)

Multiply each molar formation value by its coefficient in the balanced equation, add each side, then subtract the reactant total from the product total.

Example — Decomposing calcium carbonate

Problem
Calculate $Δ_{r} H^{\circ}$ for $CaC O_{3} (s) \to CaO (s) + C O_{2} (g)$ using −1206.9, −635.1, and −393.5 kJ/mol respectively.

1.The coefficients are all one. The product sum is $- 635.1 - 393.5 = - 1028.6 kJ mo l^{- 1}$ .
2.The reactant sum is $- 1206.9 kJ mo l^{- 1}$ .
3.Subtract: $Δ_{r} H^{\circ} = - 1028.6 - (- 1206.9) = + 178.3 kJ mo l^{- 1}$ .
4.The positive sign means the decomposition needs heat under the stated constant-pressure conditions.

Thermochemical equations carry amounts

A thermochemical equation combines a balanced reaction with its enthalpy change. Its coefficients specify mole ratios, and its quoted change refers to the reaction amount defined by those coefficients. Multiplying the whole equation by a factor multiplies its enthalpy change by the same factor. Reversing the equation changes the sign because the original initial and final states exchange roles.

Example — Scaling iron oxide reduction

Problem
Use formation data to evaluate $F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2Fe (s) + 3 H_{2} O (l)$ . What happens if every coefficient is halved?

1.The elemental reference values for iron and hydrogen are zero. The product sum is $3 (- 285.83) = - 857.49 kJ mo l^{- 1}$ .
2.The reactant sum is $- 824.2 kJ mo l^{- 1}$ . Thus the reaction change is $- 857.49 - (- 824.2) = - 33.29 kJ mo l^{- 1}$ .
3.The written equation consumes one mole of iron oxide. Halving all coefficients defines a half-sized reaction and gives $- 16.645 kJ mo l^{- 1}$ per new reaction as written.
4.The energy scales with reaction amount. Reversing either equation would reverse its enthalpy sign.

The same rule applies to ammonia: if $N_{2} + 3 H_{2} \to 2N H_{3}$ releases 91.8 kJ per written reaction, its reverse absorbs 91.8 kJ, and formation of one mole releases 45.9 kJ using that dataset. Always keep a reported numerical value attached to its balanced equation rather than memorising it beside a chemical name.

Hess’s law connects alternative routes

Some reaction enthalpies are difficult to measure directly because several products form at once. Carbon monoxide formation is an example: burning carbon does not reliably stop at pure carbon monoxide. Hess’s law lets us combine measurable reactions to reach the desired initial and final states without needing the direct route to happen experimentally.

Definition

Hess’s law

The enthalpy change between specified initial and final states is independent of the route. If a reaction is represented by a sum of steps, its enthalpy change is the sum of the signed enthalpy changes of those steps at the same temperature.

Adding reaction stepsLaTeX

Δ_{r} H = Δ H_{1} + Δ H_{2} + \dots + Δ H_{n}

Balance and scale the equations first. Species that cancel must have the same physical state and conditions.

Hess’s law as an enthalpy route— Both routes connect the same reactants and products. Adding the step changes must reproduce the direct change.

Example — Finding carbon monoxide formation indirectly

Problem
Use $C (graphite) + O_{2} \to C O_{2}$ with −393.5 kJ/mol and $CO + \frac{1}{2} O_{2} \to C O_{2}$ with −283.0 kJ/mol to find $C (graphite) + \frac{1}{2} O_{2} \to CO$ .

1.Reverse the second equation so CO appears as a product: $C O_{2} \to CO + \frac{1}{2} O_{2}$ , with enthalpy $+ 283.0 kJ mo l^{- 1}$ .
2.Add it to graphite combustion. Carbon dioxide cancels, and half the oxygen remains on the reactant side.
3.The net equation is the required formation reaction of one mole of CO.
4.Add the enthalpies: $- 393.5 + 283.0 = - 110.5 kJ mo l^{- 1}$ .

Common mistake

Only elements in their reference states have zero standard formation enthalpy by convention. Also, Hess’s law describes endpoint energy differences; the chosen intermediate reactions need not be the actual chemical mechanism.

Quick check

Which is a defining formation reaction for one mole of liquid water?

Quick check

What does zero formation enthalpy of graphite mean?

Quick check

A thermochemical equation is reversed and multiplied by two. What happens to its enthalpy change?

Quick check

Which formula uses standard formation enthalpies correctly?

Quick check

What makes Hess’s law valid?

Quick check

A reaction releases 80 kJ when two moles of product form. What change corresponds to the same reaction scaled to one mole of product?

Practice Problems

Write standard formation equations for one mole each of methane gas, ethanol liquid, and ammonia gas from their reference elements.
Explain why carbon atoms in the gas phase and diamond do not share graphite’s zero formation value.
Calculate $Δ_{r} H^{\circ}$ for $N_{2} O_{4} (g) + 3CO (g) \to N_{2} O (g) + 3C O_{2} (g)$ using formation enthalpies 9.7, −110, 81, and −393 kJ/mol respectively.
Given $N_{2} + 3 H_{2} \to 2N H_{3}$ with $Δ H = - 92.4 kJ mo l^{- 1}$ , find the standard molar formation enthalpy of ammonia.
Construct a Hess cycle connecting graphite, carbon monoxide, and carbon dioxide and label the direction of each enthalpy change.
Explain why a reaction forming calcium carbonate from calcium oxide and carbon dioxide is not the defining formation reaction from elements.

Key Takeaways

Formation enthalpy refers to one mole made from reference-state elements. Zero elemental formation values are a convention, not zero absolute energy. Formation-data sums and Hess’s law both rely on path-independent enthalpy changes. Scaling, reversing, and combining equations require matching operations on their enthalpy values.

Previous · Lesson 4

Reaction Enthalpy, Standard States, and Phase Changes

Next · Lesson 6

Combustion, Atomisation, and Bond Enthalpy

Building a shared reference

Reading and using a formation table

Thermochemical equations carry amounts

Hess’s law connects alternative routes

Quiz

Practice Problems

Key Takeaways