R1.2 Energy cycles in reactions

Bond breaking and bond forming

A covalent bond is the electrostatic attraction between a shared pair of electrons and the nuclei of the two bonded atoms. Pulling the atoms apart means working against this attraction, so breaking a bond always absorbs energy (endothermic). When a bond forms, the atoms move into a lower-energy arrangement and the same amount of energy is released, so forming a bond always releases energy (exothermic).

The potential-energy curve for two hydrogen atoms shows why. As the atoms approach, each nucleus attracts the other atom's electron and the energy falls. At very short distances the two nuclei repel strongly and the energy rises again. The bottom of the well is the most stable arrangement: its distance is the bond length, and its depth is the energy needed to break the bond.

Whether a whole reaction is exothermic or endothermic depends on the balance between these two processes:

• If the bonds formed in the products are stronger in total than the bonds broken in the reactants, more energy is released than absorbed: the reaction is exothermic.

• If the bonds broken are stronger in total than the bonds formed, the reaction is endothermic.

Bond enthalpy

A bond enthalpy refers to the gas phase so that the value measures only the bond itself, without any energy needed to overcome intermolecular forces. The equation that represents it breaks the bond evenly (homolytic fission), so each atom keeps one electron of the shared pair and gaseous atoms are formed:

For a molecule with several identical bonds, the equation is scaled to break one mole of bonds. The ${{\mathrm{N}{-}\mathrm{H}}}$ bond enthalpy in ammonia is represented by ${{\mathchoice{\textstyle\frac{1}{3}}{\frac{1}{3}}{\frac{1}{3}}{\frac{1}{3}}\,\mathrm{NH}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathchoice{\textstyle\frac{1}{3}}{\frac{1}{3}}{\frac{1}{3}}{\frac{1}{3}}\,\mathrm{N}\mskip2mu (\mathrm{g}) {}+{} \mathrm{H}\mskip2mu (\mathrm{g})}}$.

Average bond enthalpies

The energy needed to break a particular bond depends slightly on the rest of the molecule. In water, removing the first hydrogen atom, ${{\mathrm{H}{-}\mathrm{OH}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{H}\mskip2mu (\mathrm{g}) {}+{} \mathrm{OH}\mskip2mu (\mathrm{g})}}$, needs about $497\ \text{kJ mol}^{-1}$, but breaking the remaining bond, ${{\mathrm{OH}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{O}\mskip2mu (\mathrm{g}) {}+{} \mathrm{H}\mskip2mu (\mathrm{g})}}$, needs only about $430\ \text{kJ mol}^{-1}$ (both at 298 K). The ${{\mathrm{O}{-}\mathrm{H}}}$ bonds in methanol, ethanoic acid and phenol each have slightly different values again.

The data booklet lists average bond enthalpies (a few, such as ${{\mathrm{H}{-}\mathrm{H}}}$ and ${{\mathrm{Cl}{-}\mathrm{Cl}}}$, occur in only one molecule and are exact). The average for ${{\mathrm{O}{-}\mathrm{H}}}$, $463\ \text{kJ mol}^{-1}$, is close to the mean of the two values in water, about $464\ \text{kJ mol}^{-1}$. Using an average makes calculations possible for thousands of compounds from one short table, at the cost of some accuracy.

Selected values. Data: IB Chemistry data booklet (2025), section 12.

Bond enthalpy, bond length and polarity

Bond enthalpies follow patterns in structure. Between the same two atoms, a triple bond is shorter and stronger than a double bond, which is shorter and stronger than a single bond, because more shared electrons pull the nuclei closer together: ${{\mathrm{C}{-}\mathrm{C}}}$ $346\ \text{kJ mol}^{-1}$ (154 pm), ${{\mathrm{C}{=}\mathrm{C}}}$ $614\ \text{kJ mol}^{-1}$ (134 pm), ${{\mathrm{C}{\equiv}\mathrm{C}}}$ $839\ \text{kJ mol}^{-1}$ (120 pm). A double bond is less than twice as strong as a single bond.

For single bonds between carbon and the halogens, the bond gets longer and weaker down the group. The larger halogen atom holds the shared pair further from its nucleus, so the attraction is weaker.

This trend has direct consequences for reactivity. Iodoalkanes undergo nucleophilic substitution fastest because the ${{\mathrm{C}{-}\mathrm{I}}}$ bond is the easiest to break (R3.4). Chlorofluorocarbons release chlorine atoms in the upper atmosphere because ultraviolet light can break the ${{\mathrm{C}{-}\mathrm{Cl}}}$ bond but not the much stronger ${{\mathrm{C}{-}\mathrm{F}}}$ bond (R3.3).

Bond polarity also affects bond strength. The hydrogen halides become weaker from ${{\mathrm{H}{-}\mathrm{F}}}$ ($567\ \text{kJ mol}^{-1}$) to ${{\mathrm{H}{-}\mathrm{I}}}$ ($298\ \text{kJ mol}^{-1}$): the bonds become longer, and they also become less polar. A large difference in electronegativity adds an electrostatic attraction between the partial charges $\delta+$ and $\delta-$ to the attraction of the shared pair, so very polar bonds tend to be stronger.

Calculating enthalpy changes from bond enthalpies

A reaction between gases can be imagined as two steps: first every bond in the reactants is broken to give separate gaseous atoms, and then every bond in the products is formed from those atoms. The enthalpy change is the energy absorbed in the first step minus the energy released in the second:

The combustion of hydrogen shows the method. Two moles of ${{\mathrm{H}{-}\mathrm{H}}}$ bonds and one mole of ${{\mathrm{O}{=}\mathrm{O}}}$ bonds are broken; four moles of ${{\mathrm{O}{-}\mathrm{H}}}$ bonds are formed in two moles of water molecules.

Counting bonds correctly is the main skill. Every bond in every molecule is included, multiplied by the coefficient in the equation, and a bond that appears unchanged on both sides may be either counted on both sides or left out of both.

The same equation can be rearranged to find one unknown bond enthalpy when $\Delta H$ for a gas-phase reaction has been measured. This is how many of the values in data tables were obtained in the first place.

Why bond enthalpy estimates differ from measured values

The combustion of methane shows the two main limitations of the method:

The data booklet value of $\Delta H_{\text{c}}^{\ominus}$ for methane is $-891\ \text{kJ mol}^{-1}$. Two effects account for the difference:

• Average values. The ${{\mathrm{C}{-}\mathrm{H}}}$ bonds in methane and the ${{\mathrm{C}{=}\mathrm{O}}}$ bonds in carbon dioxide are not exactly the averages in the table, which are taken over many different compounds.

• States. Bond enthalpies apply to gases only. The standard enthalpy of combustion forms liquid water, and condensing two moles of steam releases about another $88\ \text{kJ}$ ($2 \times 44$). Including it brings the estimate to about $-896\ \text{kJ mol}^{-1}$.

For this reason bond enthalpies are used to estimate enthalpy changes, mainly for gas-phase reactions, while values from enthalpies of formation or combustion (HL) are more accurate.

Hess’s law

Many enthalpy changes cannot be measured directly. Carbon burns in a limited supply of oxygen to a mixture of carbon monoxide and carbon dioxide, never pure carbon monoxide, and limestone decomposes only above about 800 °C, far too hot for a calorimeter. Such values are found indirectly using Hess's law.

Hess's law follows from the law of conservation of energy. Enthalpy is a state function: its value depends only on the present state of a system, not on how it got there. If two routes between the same reactants and products had different enthalpy changes, energy could be created by going one way and returning the other, which is impossible. The change in height between the bottom and top of a hill is a useful comparison: it is the same whether a walker climbs the steep path or the long winding one.

An energy cycle (enthalpy cycle) shows the direct route and an indirect route between the same states. Following the arrows, the enthalpy change of the direct route equals the sum of the changes along the indirect route. When a step has to be followed against the direction of its arrow, its sign is changed.

Using an energy cycle

The enthalpy change for ${{\mathrm{C}\mskip2mu (\mathrm{s}) {}+{} \mathchoice{\textstyle\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{CO}\mskip2mu (\mathrm{g})}}$ cannot be measured, but the enthalpy changes of combustion of carbon and carbon monoxide can. Both burn to carbon dioxide, so they form a cycle.

Going from ${{\mathrm{C}\mskip2mu (\mathrm{s}) {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ to ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ directly ($\Delta H_1$) must give the same enthalpy change as going first to carbon monoxide ($\Delta H$) and then burning it ($\Delta H_2$):

The ${{\mathchoice{\textstyle\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ written beside carbon monoxide does not take part in the target reaction; it is there so that the same atoms appear in every box of the cycle.

Combining equations

The same result can be reached by treating the thermochemical equations like algebra. The given equations are rearranged so that, when they are added together, everything cancels except the target equation. Two rules apply:

• Reversing an equation changes the sign of its $\Delta H$.

• Multiplying an equation by a number multiplies its $\Delta H$ by the same number.

Questions often give the data as letters rather than numbers, so that the answer is an expression such as $x + 2y - z$. The method is the same: decide for each given equation whether to reverse it and what to multiply it by.

The same approach works for a compound that cannot be made directly from its elements in a calorimeter. Carbon disulfide, a volatile liquid solvent, burns readily, and so do carbon and sulfur, so three combustion equations are enough.

Measuring an enthalpy change indirectly

Cement and lime are made by heating limestone, calcium carbonate, in a kiln. The decomposition ${{\mathrm{CaCO}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{s})\xrightarrow{}\mathrm{CaO}\mskip2mu (\mathrm{s}) {}+{} \mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ is strongly endothermic, which is one reason cement manufacture uses so much fuel. Its enthalpy change can be determined in a school laboratory with Hess's law, because calcium carbonate and calcium oxide both react quickly with dilute hydrochloric acid at room temperature to give the same products.

1. Add a weighed sample of calcium carbonate (about 2.5 g) to a measured volume of excess $2\ \text{mol dm}^{-3}$ hydrochloric acid in a polystyrene cup and record the temperature change; calculate $\Delta H_1$ per mole of ${{\mathrm{CaCO}{\vphantom{X}}_{\smash[t]{3}}}}$.

2. Repeat with a weighed sample of calcium oxide (about 1.4 g) to find $\Delta H_2$ per mole of ${{\mathrm{CaO}}}$.

3. Apply Hess's law: $\Delta H = \Delta H_1 - \Delta H_2$.

The same method gives the enthalpy change for hydrating anhydrous copper(II) sulfate to the blue pentahydrate, ${{\mathrm{CuSO}{\vphantom{X}}_{\smash[t]{4}}\mskip2mu (\mathrm{s}) {}+{} 5\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})\xrightarrow{}\mathrm{CuSO}{\vphantom{X}}_{\smash[t]{4}}\,{\cdot}\,5\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{s})}}$, from the enthalpy changes of dissolving each solid in water.

Standard enthalpy changes of formation and combustion (HL)

Tables of data are only useful if every value refers to the same kind of process under the same conditions. Two standard enthalpy changes are tabulated in the data booklet: enthalpies of formation (section 13) and of combustion (section 14).

Standard enthalpy change of formation

The equation for a standard enthalpy change of formation must form exactly one mole of the substance, so fractional coefficients are often needed, and each element appears in its standard state at 298 K and 100 kPa:

Most compounds have negative values: they are more stable than the elements they are made from. A few, such as ethyne ($+228\ \text{kJ mol}^{-1}$) and benzene ($+49\ \text{kJ mol}^{-1}$), have positive values.

Allotropes of an element have different energies, so only the most stable form at 298 K is the standard state. For carbon this is graphite; diamond has $\Delta H_{\text{f}}^{\ominus} = +1.9\ \text{kJ mol}^{-1}$, which shows that diamond is very slightly less stable than graphite. The values for the two forms of oxygen differ much more: ozone has $\Delta H_{\text{f}}^{\ominus} = +143\ \text{kJ mol}^{-1}$.

Standard enthalpy change of combustion

The equation for a standard enthalpy change of combustion burns one mole of the fuel, so the oxygen may need a fractional coefficient. Complete combustion of a compound of carbon, hydrogen and oxygen gives ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ and ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$; water is a liquid under standard conditions:

Combustion values are always negative. Some substances appear in both tables: the combustion of hydrogen, ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} \mathchoice{\textstyle\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$, is also the formation of liquid water, so $\Delta H_{\text{c}}^{\ominus}({{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}) = \Delta H_{\text{f}}^{\ominus}({{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}) = -286\ \text{kJ mol}^{-1}$. In the same way, $\Delta H_{\text{c}}^{\ominus}$ of graphite equals $\Delta H_{\text{f}}^{\ominus}$ of carbon dioxide, $-394\ \text{kJ mol}^{-1}$.

Data: IB Chemistry data booklet (2025), sections 13 and 14.

Writing the equation that a standard enthalpy change refers to is a skill of its own. The checks are the same every time: exactly one mole of the named substance, every species in its standard state at 298 K, and the equation balanced for atoms.

Hess’s law with formation and combustion data (HL)

Using enthalpies of formation

Every substance in a reaction can, in principle, be made from the same elements. A cycle through the elements therefore links reactants and products. The arrows from the elements point up to the reactants and the products, because each is a formation.

Going round the cycle, $\sum\Delta H_{\text{f}}^{\ominus}(\text{reactants}) + \Delta H^{\ominus} = \sum\Delta H_{\text{f}}^{\ominus}(\text{products})$, which rearranges to (data booklet):

Each value is multiplied by the coefficient of that substance in the equation, and elements in their standard states contribute zero.

Formation data also give the enthalpy changes of reactions that are hard to study directly. The thermal decomposition of calcium carbonate, with $\Delta H_{\text{f}}^{\ominus}$ values of $-1207$ for ${{\mathrm{CaCO}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{s})}}$, $-635$ for ${{\mathrm{CaO}\mskip2mu (\mathrm{s})}}$ and $-394\ \text{kJ mol}^{-1}$ for ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$, gives $\Delta H^{\ominus} = (-635 - 394) - (-1207) = +178\ \text{kJ mol}^{-1}$, the value the two calorimetry experiments with hydrochloric acid aim to reproduce.

Using enthalpies of combustion

For organic compounds, enthalpies of combustion are easier to measure than enthalpies of formation. When both reactants and products burn to the same combustion products, the arrows point down to them.

Going round this cycle, $\sum\Delta H_{\text{c}}^{\ominus}(\text{reactants}) = \Delta H^{\ominus} + \sum\Delta H_{\text{c}}^{\ominus}(\text{products})$, so (data booklet):

A combustion cycle can also be used backwards to find an enthalpy of formation. For ethanol, ${{2\,\mathrm{C}\mskip2mu (\mathrm{s}) {}+{} 3\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} \mathchoice{\textstyle\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}\mathrm{OH}\mskip2mu (\mathrm{l})}}$: $\Delta H_{\text{f}}^{\ominus} = [2(-394) + 3(-286)] - (-1367) = -279\ \text{kJ mol}^{-1}$, in agreement with the tabulated $-278\ \text{kJ mol}^{-1}$.

When data of both kinds are available, the two methods should agree to within the rounding of the tabulated values, which makes a useful check on an answer. The trimerization of ethyne to benzene, a reaction studied in the nineteenth century as a route to aromatic compounds, shows both cycles side by side.

Born–Haber cycles (HL)

The lattice enthalpy of an ionic compound measures the strength of its ionic bonding (S2.1), but it cannot be measured directly: a crystal cannot simply be broken into gaseous ions in a calorimeter. A Born–Haber cycle finds it indirectly. The formation of the solid from its elements is compared with a route in which the elements are first turned into gaseous atoms, then into gaseous ions, which then come together to form the lattice.

The enthalpy changes in the cycle

Several of these terms need care:

• Atomization of a diatomic element. The enthalpy of atomization makes one mole of atoms, so for chlorine it is half the ${{\mathrm{Cl}{-}\mathrm{Cl}}}$ bond enthalpy: $\frac{1}{2}(242) = +121\ \text{kJ mol}^{-1}$. For a metal it is the enthalpy of sublimation. Bromine at 298 K is a liquid, so its atomization also includes vaporization.

• Electron affinities. The first electron affinity is usually exothermic, because the incoming electron is attracted by the nucleus. The second electron affinity is always endothermic: the electron is added to an ion that is already negative, and the repulsion must be overcome. For oxygen, $EA_1 = -141$ and $EA_2 = +753\ \text{kJ mol}^{-1}$

• Lattice enthalpy. The data booklet defines lattice enthalpy for the breaking up of the lattice into gaseous ions, so its values are positive (endothermic). The reverse process, forming the lattice from gaseous ions, releases the same amount of energy.

• Ionization energies. The second ionization energy is always larger than the first, because the electron is removed from a positive ion. Only first ionization energies are in the data booklet; second ionization energies are given in questions.

The Born–Haber cycle for sodium chloride

The Born–Haber cycle is usually drawn as an energy-level diagram. Endothermic steps go up and exothermic steps go down, and the elements in their standard states are the zero level. For sodium chloride, the steps from the elements to the gaseous ions are:

1. atomize sodium: ${{\mathrm{Na}\mskip2mu (\mathrm{s})\xrightarrow{}\mathrm{Na}\mskip2mu (\mathrm{g})}}$, $+107\ \text{kJ mol}^{-1}$

2. atomize chlorine: ${{\mathchoice{\textstyle\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}{\frac{1}{2}}\,\mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{Cl}\mskip2mu (\mathrm{g})}}$

3. ionize sodium: ${{\mathrm{Na}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{Na}{\vphantom{X}}^{+}\mskip2mu (\mathrm{g}) {}+{} \mathrm{e}{\vphantom{X}}^{-}}}$

4. add an electron to chlorine: ${{\mathrm{Cl}\mskip2mu (\mathrm{g}) {}+{} \mathrm{e}{\vphantom{X}}^{-}\xrightarrow{}\mathrm{Cl}{\vphantom{X}}^{-}\mskip2mu (\mathrm{g})}}$

Hess's law says that the direct route from the elements to solid sodium chloride ($\Delta H_{\text{f}}^{\ominus} = -411\ \text{kJ mol}^{-1}$) has the same enthalpy change as the indirect route through the gaseous ions. Forming the lattice from the ions is the reverse of the lattice enthalpy, so:

The data booklet lists $790\ \text{kJ mol}^{-1}$; values from different sources differ by a few $\text{kJ mol}^{-1}$ because each step is measured separately. The cycle shows that forming gaseous ions from sodium and chlorine is endothermic in total ($+375\ \text{kJ mol}^{-1}$). Sodium chloride is stable only because so much energy is released when the ions come together in the lattice.

Atomizing an element that is not a gas at 298 K takes more than one step. Bromine is a liquid, so forming one mole of bromine atoms means first vaporizing half a mole of ${{\mathrm{Br}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{l})}}$ and then breaking half a mole of ${{\mathrm{Br}{-}\mathrm{Br}}}$ bonds. The two steps add up to the enthalpy of atomization used in the cycle for potassium bromide.

Compounds with divalent ions

For compounds containing ions with a charge of 2, extra steps appear, and the amounts in the formula must be counted carefully:

• ${{\mathrm{MgO}}}$: two ionization energies of magnesium ($IE_1 + IE_2$) and two electron affinities of oxygen ($EA_1 + EA_2$).

• ${{\mathrm{MgCl}{\vphantom{X}}_{\smash[t]{2}}}}$: $IE_1 + IE_2$

• ${{\mathrm{Na}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$: two moles of sodium, so $2 \times \Delta H_{\text{at}}({{\mathrm{Na}}})$

The magnesium oxide cycle also explains why magnesium forms ${{\mathrm{Mg}{\vphantom{X}}^{2+}}}$ rather than ${{\mathrm{Mg}{\vphantom{X}}^{+}}}$ in its compounds, even though removing the second electron costs a further $1451\ \text{kJ mol}^{-1}$. Doubling the charge on both ions increases the attraction in the lattice so much that the extra lattice enthalpy more than repays the cost of the second ionization energy and the second electron affinity.

In magnesium chloride the formula contains two chloride ions, so every chlorine step in the cycle is needed twice. Writing the equation for each step with its amounts avoids the most common slip, which is to include only one electron affinity.

What lattice enthalpies show

Lattice enthalpies from Born–Haber cycles follow the pattern expected from the electrostatic attraction between ions: they are larger for higher ionic charges and smaller ions (shorter distance between the ion centres).

Data: IB Chemistry data booklet (2025), section 16.

Applying the ideas in exam questions

Paper 1A: a bond-enthalpy expression

Paper 1A: combining equations

Paper 2: making methanol

Paper 1B: hydrating copper(II) sulfate

Paper 1A: an enthalpy of formation equation (HL)

Paper 2: calcium oxide (HL)

Self-check

• To do:

  I can explain why bond breaking is endothermic and bond forming is exothermic.

• To do:

  I can define bond enthalpy and write the equation that represents it, with gaseous atoms as products.

• To do:

  I can explain why bond enthalpies are average values and why ΔH calculated from them differs from measured values.

• To do:

  I can calculate ΔH from average bond enthalpies as bonds broken minus bonds formed, counting every bond, and work backwards to an unknown bond enthalpy.

• To do:

  I can relate bond enthalpy to bond length, bond multiplicity and polarity.

• To do:

  I can state Hess’s law and explain it as a consequence of the conservation of energy.

• To do:

  I can use an energy cycle or combine thermochemical equations to calculate an unknown enthalpy change.

• To do:

  I can describe how Hess’s law is used to find an enthalpy change that cannot be measured directly, such as the decomposition of calcium carbonate.

• To do:

  I can define standard enthalpy changes of formation and combustion and write the equations that represent them, with one mole of the named substance and correct state symbols (HL).

• To do:

  I can explain why ΔH⦵f of an element in its standard state is zero, and why allotropes have different values (HL).

• To do:

  I can calculate ΔH⦵ from ΔH⦵f data (products minus reactants) and from ΔH⦵c data (reactants minus products) (HL).

• To do:

  I can define and write equations for atomization, ionization energies, electron affinities and lattice enthalpy (HL).

• To do:

  I can interpret a Born–Haber cycle and determine an unknown value, such as a lattice enthalpy or an electron affinity, for compounds of univalent and divalent ions, including the factors of 2 and the atomization of liquid bromine (HL).

• To do:

  I can explain how ionic charge and ionic radius affect lattice enthalpy (HL).

C uses the whole Cl–Cl bond enthalpy instead of half. D adds the electron affinity with the wrong sign. A uses the enthalpy of formation with the wrong sign.

The answer is B.

And 179 more in the data booklet.

This is $-241\ \text{kJ}$ per mole of water. The data booklet value for the formation of steam is $-242\ \text{kJ mol}^{-1}$, so here the estimate is very close: ${{\mathrm{H}{-}\mathrm{H}}}$ and ${{\mathrm{O}{=}\mathrm{O}}}$ are exact values and the ${{\mathrm{O}{-}\mathrm{H}}}$ average is close to the real bonds in water.

Each ${{\mathrm{NH}{\vphantom{X}}_{\smash[t]{3}}}}$ has three ${{\mathrm{N}{-}\mathrm{H}}}$ bonds and two molecules form, so six ${{\mathrm{N}{-}\mathrm{H}}}$ bonds are made.

The measured value is $-92\ \text{kJ}$ for this equation. Using the ${{\mathrm{N}{-}\mathrm{N}}}$ single bond ($158\ \text{kJ mol}^{-1}$) instead of ${{\mathrm{N}{\equiv}\mathrm{N}}}$ would give a wildly wrong answer, $-880\ \text{kJ}$.

Solve for $x$.

This agrees with the data booklet value of $431\ \text{kJ mol}^{-1}$. Here the result is exact rather than an average, because all three molecules are diatomic and each contains only the one bond.

Match the target. ${{\mathrm{C}\mskip2mu (\mathrm{s})}}$ is a reactant in the target and in (1): keep (1). Two ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ are needed: multiply (2) by 2. ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4}}\mskip2mu (\mathrm{g})}}$ is a product in the target but a reactant in (3): reverse (3).

Add. ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$ and ${{2\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ cancel, leaving the target equation:

The accepted value is $-74\ \text{kJ mol}^{-1}$; the difference of $1\ \text{kJ}$ comes from rounding in the data.

Route through the combustion products. Burn the reactants (down the left), then go back up to ethane against its combustion arrow (reversed, so the sign changes):

Hydrogenation is exothermic, as expected: a ${{\mathrm{C}{=}\mathrm{C}}}$ and an ${{\mathrm{H}{-}\mathrm{H}}}$ bond are replaced by a ${{\mathrm{C}{-}\mathrm{C}}}$ and two ${{\mathrm{C}{-}\mathrm{H}}}$ bonds.

Check that only the target is left. $3{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ appears on each side, and ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$ and $2{{\mathrm{SO}{\vphantom{X}}_{\smash[t]{2}}}}$ cancel.

The positive value means carbon disulfide is less stable than the carbon and sulfur it could be made from. The measured value is $+89\ \text{kJ mol}^{-1}$.

Reaction 2. $Q = 50.0 \times 4.18 \times 22.1 = 4619\ \text{J}$; $n = 1.40/56.08 = 0.02496\ \text{mol}$; $\Delta H_2 = -4.619/0.02496 = -185\ \text{kJ mol}^{-1}$.

The accepted value is $+178\ \text{kJ mol}^{-1}$. The small temperature rise in reaction 1 has a large percentage uncertainty, and heat loss is greatest in reaction 2, where the temperature rise is largest; making $\Delta H_2$ less negative makes the final answer too small.

(b) One mole of propan-1-ol burns completely. Three carbon atoms give $3{{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$; eight hydrogen atoms give $4{{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$, which is liquid under standard conditions. The products contain $6 + 4 = 10$ oxygen atoms, one of which comes from the alcohol, so $9$ O atoms, $\frac{9}{2}{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$, are needed.

(c) Nitrogen and hydrogen are diatomic gases, so fractions appear: 2 N (${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}$), 4 H ($2{{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}$) and 3 O ($\frac{3}{2}{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$).

In every case the coefficient of the substance named is 1, and the other coefficients are fixed by balancing; they may be fractions.

(b) Formation data (products minus reactants; $\Delta H_{\text{f}}^{\ominus}({{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}) = 0$):

The two methods agree. Adding hydrogen to the three double bonds of a hypothetical cyclohexatriene would be expected to release about three times the value for one double bond; the much smaller value for benzene is evidence for its extra stability (S2.2, HL).

(b) Combustion cycle. Both burn to ${{6\,\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} 3\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$. The arrows point down to the combustion products, so the answer is reactants minus products:

The two routes agree exactly here. The large negative value reflects the very positive $\Delta H_{\text{f}}^{\ominus}$ of ethyne: its ${{\mathrm{C}{\equiv}\mathrm{C}}}$ bonds make it far less stable than benzene.

Substitute and solve.

The data booklet value is $-349\ \text{kJ mol}^{-1}$. The small difference comes from rounding and from the different sources of the five values; a Born–Haber cycle is only as accurate as its least accurate step.

Energy to form the gaseous ions ${{\mathrm{K}{\vphantom{X}}^{+}\mskip2mu (\mathrm{g}) {}+{} \mathrm{Br}{\vphantom{X}}^{-}\mskip2mu (\mathrm{g})}}$ from the elements:

Apply Hess's law. $\Delta H_{\text{f}}^{\ominus} = 295 - \Delta H_{\text{lattice}}^{\ominus}$:

The data booklet value is $691\ \text{kJ mol}^{-1}$. Using the ${{\mathrm{Br}{-}\mathrm{Br}}}$ bond enthalpy alone and forgetting the vaporization would give $+674\ \text{kJ mol}^{-1}$.

Energy to form the gaseous ions from the elements:

Apply Hess's law. $\Delta H_{\text{f}}^{\ominus} = 3197 - \Delta H_{\text{lattice}}^{\ominus}$, so:

The data booklet value is $3791\ \text{kJ mol}^{-1}$. It is almost five times the lattice enthalpy of sodium chloride, because the ions have twice the charge and are smaller.