Conservation of energy in a reacting system

The law of conservation of energy states that the total energy of an isolated system remains constant, even though energy can be transferred or changed from one form to another. In chemistry, this idea sits behind energy cycles: if we account for the energy absorbed and released in separate steps, we can predict the overall energy change.

A chemical reaction rearranges atoms. Some bonds in the reactants have to break; new bonds in the products then form. Breaking bonds is endothermic, since energy is needed to pull bonded atoms apart. Forming bonds is exothermic, because energy is released when attractions between nuclei and shared electrons are established.

Bond enthalpy and average values

A bond enthalpy is the enthalpy change required to break one mole of a specified covalent bond in gaseous molecules by homolytic fission under standard conditions. Its unit is $\mathrm{kJ\,mol^{-1}}$. Homolytic fission is bond breaking in which the two bonding electrons are shared equally between the atoms. A radical is a species with an unpaired electron, usually shown with a dot, such as $H\bullet$.

For example, the H–H bond enthalpy corresponds to:

The positive sign matters: it shows that energy has been absorbed.

An average bond enthalpy is a mean bond enthalpy obtained from the same type of bond measured in a range of different gaseous compounds. Data-booklet bond enthalpies are average values. A C–H bond in methane does not have exactly the same electronic environment as a C–H bond next to an oxygen atom, a double bond, or a halogen. Even in a single molecule, bonds written as the same type may not have identical surroundings.

That is why bond enthalpy calculations give estimates. They usually differ from experimental enthalpy changes because the data are averaged, and because bond enthalpy calculations treat the species as gaseous covalent molecules; they do not properly include intermolecular forces in liquids and solids.

Calculating enthalpy change from bond enthalpies

The calculation uses a simple bit of accounting: add the energy needed to break bonds, then subtract the energy released when bonds form.

Average bond enthalpy values are supplied in the data booklet.

In practice, draw full structural formulae, count every bond broken in the reactants, count every bond formed in the products, multiply by the correct bond enthalpies, then subtract formed from broken. The balanced equation matters, because the answer is for the reaction as written.

For example, in an addition reaction such as ethene reacting with hydrogen bromide, the structures make the hidden bookkeeping visible: a C=C bond and an H–Br bond are replaced by a C–C bond and a C–Br bond, with one extra C–H bond in the product. A common mark-losing mistake is to count atoms instead of bonds.

Stability, bond length and polarity

Bond enthalpy links to bond strength. In general, shorter bonds are stronger because the bonded nuclei are closer to the shared electron density, so more energy is needed to separate them. Bond polarity can matter too: if a bond has greater ionic character, electrostatic attraction may increase bond strength, although size and overlap are also important. Don’t reduce this to one rule; bond length, bond order, atom size and polarity all affect the value.

This connects neatly to carbon–halogen bonds in nucleophilic substitution. The C–I bond is weaker and longer than the C–Cl bond, so it breaks more easily; other factors being comparable, halogenoalkanes with weaker C–X bonds react faster in substitution. That is the chemical reason behind the usual reactivity trend from chloroalkanes to iodoalkanes.

Why the pathway does not matter

Hess’s law is a thermochemical law stating that the enthalpy change for a reaction is the same whichever route is taken, as long as the initial and final states are identical. It follows directly from conservation of energy.

A state function is a property whose change depends only on the initial and final states, not on the route between them. Enthalpy is one of these properties. So if reactants A form products B directly, or go by way of an intermediate C, the overall enthalpy change does not change.

Using Hess’s law in calculations

There are two standard ways to use Hess’s law. They look different on the page, but both do the same energy accounting.

In the summation of equations method, known thermochemical equations are rearranged until they add up to the target equation. Reverse an equation, and the sign of its enthalpy change reverses. Multiply an equation by a factor, and its enthalpy change is multiplied by the same factor. Any species that appears on both sides cancels, just like in algebra.

In the enthalpy cycle method, the target reaction is drawn across the top, with both sides connected to a common set of substances underneath, often combustion products or elements in their standard states. Follow one route around the cycle. If the route goes against an arrow, change the sign of that enthalpy value.

It’s worth writing state symbols whenever the data depend on them. For example, forming $H_2O(l)$ and forming $H_2O(g)$ are not the same enthalpy change. Hess’s law only works cleanly when the substances and their states match exactly.

Experimental Hess cycles

Hess’s law is especially useful when a reaction is hard to measure directly. A thermal decomposition, for instance, may be too hot, too slow or too messy for simple calorimetry. Instead, two related reactions can be measured in a calorimeter, then their enthalpy changes combined in a cycle to find the missing value.

When calorimetry data are put into a Hess calculation, the usual assumptions still matter. Heat loss to the surroundings, heat absorbed by the cup or thermometer, incomplete reaction, evaporation and uncertain extrapolation of temperature all affect the measured enthalpy changes. The cycle is exact in principle; the experimental numbers placed into it may not be.

Standard conditions and standard states

A standard state is the physical state and form in which a pure substance is most stable under standard conditions. In this syllabus, standard conditions are normally 298 K and 100 kPa. Don’t treat the standard state as a tiny detail: carbon as graphite and carbon as diamond are different starting substances.

The symbol $\Delta H_c^\ominus$ means standard enthalpy change of combustion. Here, $\Delta H_c^\ominus$ is the molar enthalpy change when one mole of a substance burns completely in oxygen under standard conditions ($\mathrm{kJ\ mol^{-1}}$). Combustion enthalpies are usually negative because complete combustion releases energy.

A standard enthalpy change of combustion is an enthalpy change when one mole of a substance in its standard state undergoes complete combustion in oxygen under standard conditions. For hydrocarbons and many oxygen-containing organic compounds, complete combustion forms $CO_2(g)$ and $H_2O(l)$, unless the question specifies otherwise.

For example, the standard combustion equation for butane is:

Fractional coefficients are fine in these definitions, because the equation has to refer to exactly one mole of the substance being combusted.

Standard enthalpy of formation

The symbol $\Delta H_f^\ominus$ means standard enthalpy change of formation. In other words, $\Delta H_f^\ominus$ is the molar enthalpy change when one mole of a compound forms from its elements in their standard states under standard conditions ($\mathrm{kJ\ mol^{-1}}$).

A standard enthalpy change of formation is an enthalpy change when one mole of a substance is formed from its constituent elements in their standard states under standard conditions. For butane, the formation equation is:

The standard enthalpy of formation of an element in its standard state is zero. That does not mean no energy is stored in the element. It just means the enthalpy scale needs a reference point, and standard-state elements are chosen as that reference.

Key differences between standard enthalpy changes of combustion and formation.

An allotrope is a structural form of an element in the same physical state but with different bonding or arrangement of atoms. Diamond and graphite are allotropes of carbon, so you should expect different formation enthalpies for them. Graphite is the standard state of carbon, so $\Delta H_f^\ominus$ for graphite is $0\ \mathrm{kJ\ mol^{-1}}$; diamond is not the reference form, so its value is not zero.

Writing equations from definitions

For combustion, begin with one mole of the substance plus oxygen, then balance the equation to give the complete-combustion products. For formation, begin with elements in standard states and end with one mole of the compound. Always include state symbols, since the data-booklet values depend on them.

Combustion and formation data are supplied in the data booklet. The skill is not memorising the numbers. It is turning the definition into the correct balanced equation, then using the data consistently.

Formation data route

With formation data, Hess’s law compares the enthalpy needed to form the products from elements with the enthalpy needed to form the reactants from elements.

Include the coefficients from the balanced equation in the sums. If the equation has 3 mol of $CO_2$, multiply the $CO_2$ formation value by 3. Elements in their standard states contribute zero.

A good sign check: formation data use products minus reactants. You’re comparing the position of the products on the enthalpy scale with the position of the reactants on that same scale.

Combustion data route

With combustion data, the shared endpoint is complete combustion products. For many organic reactions, all carbon ends up as $CO_2$ and all hydrogen ends up as $H_2O$. Hess’s law gives:

As before, multiply each data-booklet value by its coefficient in the balanced equation. The order has switched to reactants minus products, the opposite of the formation expression. That’s where careless errors often creep in.

Choosing and checking the method

Use formation data when the data table gives $\Delta H_f^\ominus$ values for the species in the reaction. Use combustion data when all the species can be linked to common combustion products. Both methods use Hess’s law; they just choose different common reference substances for the cycle.

The data booklet gives the equations for using formation and combustion data, but it won’t rescue poor chemistry. You still need a balanced equation, correct state symbols, and correct multiplication by stoichiometric coefficients.

Values calculated from tabulated standard enthalpies may still be approximate if the data are rounded or if the actual reaction conditions do not match the standard states. They are usually more reliable than average bond enthalpy estimates for reactions involving liquids and solids, because standard formation and combustion values refer to real substances in specified states rather than averaged gas-phase bonds.

What a Born–Haber cycle does

A Born–Haber cycle is a Hess cycle that links the formation of an ionic solid from its elements to a step-by-step route through gaseous atoms and gaseous ions. We use it to calculate an unknown energy change in ionic bonding, usually lattice enthalpy or enthalpy of formation.

You need to interpret Born–Haber cycles and determine values from them for ionic compounds with univalent ions, such as NaCl, and divalent ions, such as MgO. You won’t be asked to build a full cycle from scratch, but you do need to know what every step in a given cycle represents.

The steps in the cycle

An ionization energy is a standard enthalpy change for removing one mole of electrons from one mole of gaseous atoms or gaseous positive ions. The symbol $IE_1$ means first ionization energy: the enthalpy change for removing the first electron from one mole of gaseous atoms (kJ mol$^{-1}$). The symbol $IE_2$ means second ionization energy: the enthalpy change for removing one mole of electrons from one mole of gaseous $1+$ ions (kJ mol$^{-1}$).

Ionization is endothermic because the electron is pulled away from its attraction to a positive nucleus.

An enthalpy of atomization is a standard enthalpy change when one mole of gaseous atoms forms from an element in its standard state. The symbol $\Delta H_{at}^{\ominus}$ means standard enthalpy of atomization, where $\Delta H_{at}^{\ominus}$ is measured in kJ mol$^{-1}$. For a solid metal, this is usually sublimation, for example:

A sublimation enthalpy is a standard enthalpy change when one mole of gaseous atoms or molecules forms directly from the solid element. With a diatomic non-metal, atomization uses half a mole of molecules, so it links to bond enthalpy:

An electron affinity is a standard enthalpy change when one mole of electrons is added to one mole of gaseous atoms or gaseous negative ions. The symbol $EA$ means electron affinity, where $EA$ is measured in kJ mol$^{-1}$. The first electron affinity of many non-metals is exothermic:

For divalent anions, the second electron affinity is usually endothermic because the electron is added to an ion that is already negative.

A lattice enthalpy is a standard enthalpy change when one mole of an ionic solid is separated into its gaseous ions. The symbol $\Delta H_{lattice}^{\ominus}$ means standard lattice enthalpy, where $\Delta H_{lattice}^{\ominus}$ is measured in kJ mol$^{-1}$.

In this course, lattice enthalpy means the endothermic separation of the ionic lattice, so its value is positive. Some sources use the opposite process, formation of the lattice from gaseous ions, and give it as a negative value. Always check the definition being used.

Reading values from a Born–Haber cycle

To find an unknown, take an alternative route around the cycle with the same start and finish. Add the enthalpy changes along that route. If you go against the direction of an arrow, change the sign of that enthalpy change. If the formula contains two ions of one type, such as two chloride ions in $MgCl_2$, include the factor of 2 where the cycle shows two atoms, two electron affinities or two gaseous ions.

For a univalent compound such as KBr, the route includes atomization of K, ionization of K, atomization of Br, electron affinity of Br, and the reverse of the formation enthalpy if you are calculating lattice enthalpy. For a divalent compound such as MgO, the cycle includes both $IE_1$ and $IE_2$ for magnesium, plus the first and second electron affinity steps for oxygen.

Ionic charge and ionic radius mainly control the strength of lattice enthalpy. Higher ionic charges produce stronger electrostatic attractions and larger lattice enthalpies. Smaller ions let oppositely charged ions get closer, which also increases attraction. That is why compounds containing $Mg^{2+}$ and $O^{2-}$ have much stronger lattice enthalpies than compounds containing singly charged, larger ions.