Energetics

Class 11 Chemistry, unit 6, book pages 114–143

Every physical and chemical change involves a change of energy, but the total energy of the universe stays the same. This unit studies the heat released or absorbed in chemical reactions: how it is described with enthalpy changes (formation, combustion, neutralization, solution, bond energies, lattice energy), how it is measured in a calorimeter, and how it is calculated with Hess's law and the Born–Haber cycle. It ends with entropy and Gibbs free energy, which decide whether a reaction goes by itself, and with the calorie content of food.

Learning outcomes (30)
  • Describe that chemical reactions are accompanied by enthalpy changes and these changes can be exothermic ( is negative) or endothermic ( is positive).
  • Interpret a reaction pathway diagram, in terms of the reaction and of the activation energy.
  • Define terms such as standard conditions, enthalpy change, reaction, formation, combustion, neutralization.
  • Explain that energy transfer occurs during chemical reactions because of the breaking and making of bonds.
  • Calculate the bond energies for the enthalpy change of reaction .
  • Describe that some bond energies are exact and some bond energies are approximate.
  • Calculate enthalpy changes from approximate experimental results, including the use of the relationships and .
  • Define terms such as enthalpy change of atomization, lattice energy, , first electron affinity, EA.
  • Use terms such as enthalpy change of atomization, lattice energy, first electron affinity.
  • Explain factors affecting the electron affinities of elements.
  • Construct Born–Haber cycles for ionic solids.
  • Perform calculations involving Born–Haber cycles.
  • Explain the effect of ionic charge and ionic radius on the numerical magnitude of lattice energy.
  • Apply enthalpy change with reference to hydration and solution.
  • Construct an energy cycle involving enthalpy change of solutions, and enthalpy change of hydration.
  • Perform calculations involving energy cycles.
  • Explain the effect of ionic charge and ionic radius on the numerical magnitude of an enthalpy change of hydration.
  • Define the term entropy, , as the number of possible arrangements of the particles and their energy in a given system.
  • Explain the sign of entropy changes that occur during a change in state, temperature change and a reaction in which there is a change in the number of gaseous molecules.
  • Calculate the entropy change for a reaction, , given the standard entropies, , of the reactants and the products.
  • Explain the concept of heat as a form of energy.
  • Explain the relationship between temperature and kinetic energy of particles.
  • State that total energy is conserved in chemical reactions.
  • Explain the concept of standard conditions and standard states in measuring energy changes.
  • Explain Hess's law.
  • Apply Hess's law to calculate enthalpy changes in a reaction carried out in multiple steps.
  • Explain the relationship between bond formation energy and bond breaking energy.
  • Explain Gibbs free energy.
  • Apply the concept of Gibbs free energy to solve problems.
  • Outline how enthalpy change relates to the calorie concept of the food we eat.

6.1Energy in chemical reactions

Every process in the universe, in living cells or test tubes, in the atmosphere or in water, involves a change of energy. Some processes release energy and others need it, but the total amount of energy in the universe stays the same. Energetics is the study of the energy changes that go with all types of physical and chemical changes. This unit studies the role of energy in chemical reactions and how the heat released or absorbed in a reaction is measured and calculated.

In a chemical reaction, energy in the form of heat is either evolved or absorbed. The reason is that old bonds are broken and new bonds are formed. Bond breaking always consumes energy and bond making always releases energy.

Bond breaking and bond making decide the direction of energy transfer
CaseWhat happensNet transfer of energy
Energy released by bond formation is greater than energy consumed by bond breakingMore energy leaves than entersFrom the system to the surroundings (heat is given out)
Energy consumed by bond breaking is greater than energy released by bond formationMore energy enters than leavesFrom the surroundings to the system (heat is taken in)

6.1.1Units of thermal energy

Thermal energy is the energy in an object or system due to the movement of its molecules and atoms. The unit of heat or thermal energy in the SI system is the joule (J).

Another common unit of heat is the calorie. It is defined as the heat or thermal energy required to raise the temperature of one gram of water from to .

Units of thermal energy
UnitSymbolMeaning
JouleJSI unit of heat and thermal energy
CaloriecalHeat needed to raise of water from to ;

6.1.2Thermochemical reactionsOften examined

When a chemical change takes place, energy is exchanged between the system and its surroundings. Energy has many forms, such as heat, light and work.

Thermochemical reaction
A chemical reaction which proceeds with the evolution or absorption of heat.
Thermochemical equation
A balanced chemical equation which also shows the heat change of the reaction.
Thermochemistry
The branch of chemistry which deals with the heat or thermal energy changes in chemical reactions.

For example, the burning of carbon:

There are two types of thermochemical reactions: exothermic and endothermic reactions.

1. Exothermic reactions

A chemical reaction that proceeds with the evolution of heat is called an exothermic reaction. The chemical system transfers energy to the surroundings as the reactants are converted to products. The burning of fuels is a highly exothermic reaction; the energy released can heat a room, drive an engine or cook food. Examples:

2. Endothermic reactions

A chemical reaction that proceeds with the absorption of heat is called an endothermic reaction. Heat is transferred from the surroundings to the system. Examples:

Exothermic and endothermic reactions
Exothermic reactionEndothermic reaction
HeatEvolved (given out)Absorbed (taken in)
Direction of heat flowSystem to surroundingsSurroundings to system
Sign of NegativePositive
Energy of products compared with reactantsLowerHigher
Example, ,
Enthalpy profiles of an exothermic and an endothermic reaction. The accent arrow is the enthalpy change (products minus reactants).
Concept assessment exercise 6.1
Classify the following processes as exothermic or endothermic: (a) freezing of water, (b) combustion of methane, (c) sublimation of dry ice, (d) , (e) decomposition of limestone.
  1. Decide whether heat is given out (exothermic, ) or taken in (endothermic, ) in each process.
  2. (a) Freezing of water: liquid water becomes ice, so heat is released to the surroundings.
  3. (b) Combustion of methane: all combustion reactions release heat.
  4. (c) Sublimation of dry ice: solid carbon dioxide changes to gas, so energy has to be absorbed to separate the molecules.
  5. (d) Condensation of steam: gas becomes liquid, so heat is released.
  6. (e) Decomposition of limestone: heat must be supplied continuously to break it into lime and carbon dioxide.

6.1.3Heat of reaction

The amount of energy released or absorbed depends on the amounts of substances that react. For example, the more fuel burns, the more energy is released. So, when an energy change is reported, the amounts of the chemicals that produce it must be reported as well. For this purpose energy changes are expressed for molar quantities of reactants and products.

Heat of reaction
The amount of heat evolved or absorbed in a chemical reaction when the molar quantities of reactants and products are the same as shown in the chemical equation.
Standard enthalpy change,
The heat of reaction measured at (298 K) and one atmosphere pressure.

Reading a thermochemical equation:

This equation shows that 1 mole of solid carbon (12 g) reacts with 1 mole of oxygen gas (32 g) to give 1 mole of gas (44 g) at and 1 atmosphere pressure, and of heat is evolved.

The book asks what this second equation shows. It shows that 1 mole of hydrogen gas reacts with 1 mole of iodine gas to give 2 moles of hydrogen iodide gas, and of heat is absorbed (the sign is positive).

6.1.4Relation between enthalpy change and heat of reaction

Most chemical reactions occur at constant pressure, so the heat change in these reactions can be equated to the change in enthalpy. The heat of a reaction is therefore the change in enthalpy, \Delta H: the difference between the enthalpies of the products (final state) and the enthalpies of the reactants (initial state).

\Delta H=H_\text{products}-H_\text{reactants}
Sign of the enthalpy change
ProcessHeat flowSign of \Delta H
ExothermicReleased by the system to the surroundingsNegative
EndothermicAbsorbed by the system from the surroundingsPositive
Combustion (an exothermic process)Released by the system to the surroundingsNegative
H=E+PV

6.1.5Standard states and standard enthalpy changes

The value of \Delta H changes with the conditions, so standardised values of \Delta H are used. These values are calculated when all the substances are in their standard states (standard conditions). The superscript ^\circ in \Delta H^\circ shows that the value is for standard conditions.

6.1.6Conditions for standard heat of reaction

The standard states are as follows.

Standard states
SubstanceStandard state
A gasPressure of 1\,\mathrm{atm}
An element or a compoundIts most stable physical state at 1\,\mathrm{atm} and 25^\circ\mathrm{C} (298\,\mathrm{K})
A substance in aqueous solutionConcentration of 1\,\mathrm{M}

The book now defines eight important enthalpies. Each one is a standard enthalpy change for a particular kind of change; they are treated one by one in the sections that follow.

The eight enthalpies defined in this unit
Standard enthalpySymbolChange it describes
Reaction\Delta H^\circ_\mathrm{r}Reaction as shown by the balanced equation
Formation\Delta H^\circ_\mathrm{f}1 mol of a compound formed from its elements
Combustion\Delta H^\circ_\mathrm{c}1 mol of a substance completely burnt in excess oxygen
Atomization\Delta H^\circ_\mathrm{at}1 mol of gaseous atoms formed from the element
Neutralization\Delta H^\circ_\mathrm{n}1 mol of \ce{H+} combining with 1 mol of \ce{OH-} to form 1 mol of water
Solution\Delta H^\circ_\mathrm{sol}1 mol of a substance dissolved in so much solvent that more dilution gives no heat change
First electron affinity\Delta H^\circ_\mathrm{EA}1 mol of gaseous atoms gaining one electron each
Lattice energy\Delta H_\mathrm{l}1 mol of an ionic solid formed from its gaseous ions

6.1.6 (1)Standard enthalpy of reactionOften examined

Standard enthalpy of reaction, \Delta H^\circ_\mathrm{r}
The enthalpy change in a chemical reaction when the reactants and products are in their standard states and their molar quantities are the same as shown by the balanced chemical equation.

Examples:

\begin{aligned}\text{(i)}\quad&\ce{2H2(g) + O2(g) -> 2H2O(l)} &&\Delta H^\circ_\mathrm{r}=-571.6\,\mathrm{kJ}\\ \text{(ii)}\quad&\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} &&\Delta H^\circ_\mathrm{r}=-890.4\,\mathrm{kJ}\\ \text{(iii)}\quad&\ce{2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s)} &&\Delta H^\circ_\mathrm{r}=-850\,\mathrm{kJ}\end{aligned}

6.1.6 (2)Standard enthalpy of formationOften examined

Standard enthalpy of formation, \Delta H^\circ_\mathrm{f}
The enthalpy change that accompanies the formation of one mole of a compound from its elements, with all substances in their standard states.

To write a thermochemical equation for the enthalpy of formation:

  1. Write the elements as reactants and 1 mole of the compound as the product.
  2. Show the standard states of all the substances.
  3. Finally balance the atoms (fractions such as \tfrac12 are allowed so that the product stays at one mole).
Example 6.1
Write the thermochemical equations for the formation of \ce{H2O}, \ce{SO3} and \ce{H2O2}. Use the data given in Table 6.1 (the book says table 11.1, an old number).
  1. Water: the elements are hydrogen gas and oxygen gas. One mole of liquid water needs one hydrogen molecule and half an oxygen molecule. \Delta H^\circ_\mathrm{f} is read from Table 6.1.
    \ce{H2(g) + 1/2 O2(g) -> H2O(l)}\qquad \Delta H^\circ_\mathrm{f}=-285.8\,\mathrm{kJ\,mol^{-1}}
  2. Sulphur trioxide: the elements are solid sulphur and oxygen gas. One \ce{S} atom and three \ce{O} atoms give one \ce{SO3}, so \tfrac32 mol of \ce{O2} is needed.
    \ce{S(s) + 3/2 O2(g) -> SO3(g)}\qquad \Delta H^\circ_\mathrm{f}=-395.7\,\mathrm{kJ\,mol^{-1}}
  3. Hydrogen peroxide: one mole of \ce{H2O2} contains two H atoms and two O atoms, so one mole of \ce{H2} and one mole of \ce{O2} are needed.
    \ce{H2(g) + O2(g) -> H2O2(l)}\qquad \Delta H^\circ_\mathrm{f}=-187.8\,\mathrm{kJ\,mol^{-1}}

Standard enthalpies of formation of some substances are given in Table 6.1.

Table 6.1 (first part): standard enthalpies of formation, as printed in the book
Substance\Delta H^\circ_\mathrm{f} (\mathrm{kJ\,mol^{-1}})
\ce{H2O(g)}-245.1
\ce{H2O(l)}-285.8
\ce{H2O2(l)}-187.8
\ce{NH3(g)}-46.1
\ce{NH3(aq)}-80.3
\ce{N2H4(l)}50.6
\ce{HF(g)}-271.1
\ce{HCl(g)}-92.3
\ce{HCl(aq)}167.2
\ce{HBr(g)}-36.4
\ce{HI(g)}26.5
\ce{H2S(g)}-20.6
\ce{CO(g)}-110.5
\ce{CO2(g)}-393.5
\ce{NO(g)}90.3
\ce{NO2(g)}33.2
\ce{N2O4(g)}9.2
\ce{SO2(g)}-296.8
\ce{SO3(g)}-395.7
\ce{MgO(s)}-601.7
\ce{CaO(s)}-635.1
Table 6.1 (second part)
Substance\Delta H^\circ_\mathrm{f} (\mathrm{kJ\,mol^{-1}})
\ce{Fe2O3(s)}-824.2
\ce{SiO2(s)}-910.9
\ce{CaCO3(s)}-1207
\ce{BaCO3(s)}-1219
\ce{NaOH(s)}-425.6
\ce{KOH(s)}-424.8
\ce{HNO3(l)} (printed \ce{HONO2})174.1
\ce{HNO3(aq)}-207.4
\ce{H2SO4(l)} (printed \ce{(HO)2SO2})-814.0
\ce{H2SO4(aq)}-909.3
\ce{NH4Cl(s)}-314.4
\ce{NaCl(s)}-411.2
\ce{NaBr(s)}-361.1
\ce{NaI(s)}-287.8
\ce{KCl(s)}-436.8
\ce{KBr(s)}-393.8
\ce{KI(s)}327.9
\ce{AgCl(s)}-127.1
\ce{CaCl2(s)}-795.8
\ce{AlCl3(s)}-704.2
Table 6.1 (third part)
Substance\Delta H^\circ_\mathrm{f} (\mathrm{kJ\,mol^{-1}})
\ce{Na+(aq)}-240.1
\ce{K+(aq)}-252.4
\ce{Mg^2+(aq)}-466.9
\ce{Ca^2+(aq)}-542.8
\ce{Al^3+(aq)}-524.7
\ce{C}(s, diamond)1.9
\ce{CH4(g)}-74.8
\ce{C2H2(g)}226.7
\ce{C2H4(g)}52.3
\ce{C2H6(g)}-84.7
\ce{C3H8(g)}-103.9
\ce{C4H10(g)}-126.2
\ce{C6H6(l)}49.0
\ce{C6H5CH3(l)}12.1
\ce{C8H18(l)}-249.9
\ce{CH3OH(l)}-238.7
\ce{C2H5OH(l)}-277.7
\ce{CH3CHO(l)}-192.3
\ce{CH3COCH3(l)}-248.1
\ce{CH3COOH(l)}-484.5

6.1.6 (3)Standard enthalpy of combustionOften examined

Standard enthalpy of combustion, \Delta H^\circ_\mathrm{c}
The enthalpy change when one mole of a substance is completely burnt in excess of oxygen under standard conditions.

Standard enthalpies of combustion of some substances are given in Table 6.2.

Table 6.2: standard enthalpies of combustion
Substance\Delta H^\circ_\mathrm{c} (\mathrm{kJ\,mol^{-1}})Substance\Delta H^\circ_\mathrm{c} (\mathrm{kJ\,mol^{-1}})Substance\Delta H^\circ_\mathrm{c} (\mathrm{kJ\,mol^{-1}})
\ce{H2(g)}-285.8\ce{CH4(g)}-890.4\ce{C6H6(l)}-3268
\ce{C}(graphite)-393.5\ce{C2H6(g)}-1560\ce{C2H5OH(l)}-1367
\ce{C}(diamond)-395.4\ce{C8H18(l)}-5512\ce{CH3CHO(l)}-1167
\ce{S}(rhombic)-296.9\ce{C2H4(g)}-1411\ce{CH3COOH(l)}-875
\ce{S}(monoclinic)-297.2\ce{C2H2(g)}-1300\ce{C6H12O6(s)}-2802

To write a thermochemical equation for the enthalpy of combustion:

  1. Write 1 mole of the element or compound and oxygen as reactants.
  2. Write the oxides of the given element, or of the elements present in the compound, as products.
  3. Show the standard states of all the substances.
  4. Finally balance the atoms.
Example 6.2
Write the thermochemical equations for the combustion of \ce{C}, \ce{CH4} and \ce{H2}. Use the data given in Table 6.2 (the book says table 11.2).
  1. Carbon burns to carbon dioxide. One mole of carbon is burnt.
    \ce{C(s) + O2(g) -> CO2(g)}\qquad \Delta H^\circ_\mathrm{c}=-393.5\,\mathrm{kJ\,mol^{-1}}
  2. Methane burns to carbon dioxide and water. The four H atoms need two water molecules, so four O atoms for water plus two for \ce{CO2} means 2\,\ce{O2} in all.
    \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}\qquad \Delta H^\circ_\mathrm{c}=-890.4\,\mathrm{kJ\,mol^{-1}}
  3. Hydrogen burns to water. One mole of \ce{H2} needs half a mole of \ce{O2}.
    \ce{H2(g) + 1/2 O2(g) -> H2O(l)}\qquad \Delta H^\circ_\mathrm{c}=-285.8\,\mathrm{kJ\,mol^{-1}}

6.1.6 (4)Standard enthalpy of atomization

Standard enthalpy of atomization, \Delta H^\circ_\mathrm{at}
The enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions.

Examples:

\begin{aligned}\text{(i)}\quad&\ce{1/2 H2(g) -> H(g)} &&\Delta H^\circ_\mathrm{at}=+218\,\mathrm{kJ\,mol^{-1}}\\ \text{(ii)}\quad&\ce{1/2 Cl2(g) -> Cl(g)} &&\Delta H^\circ_\mathrm{at}=+121\,\mathrm{kJ\,mol^{-1}}\end{aligned}

6.1.6 (5)Standard enthalpy of neutralizationOften examined

Standard enthalpy of neutralization, \Delta H^\circ_\mathrm{n}
The amount of heat evolved when one mole of \ce{H+} ions from an acid combine with one mole of \ce{OH-} ions from a base to form one mole of water under standard conditions.
\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)}\qquad \Delta H^\circ_\mathrm{n}=-57.4\,\mathrm{kJ\,mol^{-1}}

Strong acids and strong bases ionise completely in their aqueous solutions. So when solutions of \ce{NaOH} and \ce{HCl} are mixed, the only change that occurs is the formation of water; the \ce{Na+} and \ce{Cl-} ions remain in solution. The heat of neutralization is therefore due to the formation of water from \ce{H+} and \ce{OH-} ions.

From the full equation to the ionic equation
  1. Write the full ionic equation, with all strong electrolytes as ions.
    \ce{Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) -> Na+(aq) + Cl-(aq) + H2O(l)}
  2. Cancel the spectator ions \ce{Na+} and \ce{Cl-} from both sides. What is left is the net reaction.
    \ce{H+(aq) + OH-(aq) -> H2O(l)}\qquad \Delta H^\circ_\mathrm{n}=-57.4\,\mathrm{kJ\,mol^{-1}}

6.1.6 (6)Standard enthalpy of solution

Standard enthalpy of solution, \Delta H^\circ_\mathrm{sol}
The enthalpy change when one mole of a substance is dissolved in so much solvent that further dilution results in no detectable heat change, under standard conditions.

Examples:

\begin{aligned}&\ce{NH4Cl(s) ->[H2O] NH4+(aq) + Cl-(aq)} &&\Delta H^\circ_\mathrm{sol}=+15.1\,\mathrm{kJ\,mol^{-1}}\\ &\ce{HCl(g) ->[H2O] H+(aq) + Cl-(aq)} &&\Delta H^\circ_\mathrm{sol}=-75\,\mathrm{kJ\,mol^{-1}}\end{aligned}

The first dissolving is endothermic (the solution cools) and the second is exothermic (the solution warms).

Concept assessment exercise 6.2
(1) Write the thermochemical equation from the given information: (a) the standard enthalpy of formation of benzene (l) is +49.0\,\mathrm{kJ\,mol^{-1}}; (b) the standard enthalpy of formation of ethanol (l) is -277.7\,\mathrm{kJ\,mol^{-1}}; (c) the standard enthalpy of combustion of acetic acid (l) is -876\,\mathrm{kJ\,mol^{-1}}; (d) when ethanol is burned in air, 1367\,\mathrm{kJ\,mol^{-1}} of energy is released at 1\,\mathrm{atm} and 25^\circ\mathrm{C}. (2) Which of the following equations describes a reaction for which \Delta H^\circ equals the enthalpy of formation of a compound? (a) \ce{4Al(s) + 2O2(g) -> 2Al2O3(s)}; (b) \ce{4Al(s) + 3/2 O2(g) -> Al2O3(s)}; (c) \ce{CO(g) + 1/2 O2(g) -> CO2(g)}.
Question 1
  1. (a) Formation: elements as reactants, one mole of \ce{C6H6(l)} as product. Carbon is graphite. Six C atoms and six H atoms need 6\,\ce{C} and 3\,\ce{H2}.
    \ce{6C(s) + 3H2(g) -> C6H6(l)}\qquad \Delta H^\circ_\mathrm{f}=+49.0\,\mathrm{kJ\,mol^{-1}}
  2. (b) Ethanol \ce{C2H5OH} contains 2 C, 6 H and 1 O, so 2\,\ce{C}, 3\,\ce{H2} and \tfrac12\,\ce{O2} are needed.
    \ce{2C(s) + 3H2(g) + 1/2 O2(g) -> C2H5OH(l)}\qquad \Delta H^\circ_\mathrm{f}=-277.7\,\mathrm{kJ\,mol^{-1}}
  3. (c) Combustion: one mole of acetic acid burns in oxygen to give \ce{CO2} and \ce{H2O}. \ce{CH3COOH} has 2 C, 4 H and 2 O. The products need 4 O for 2\,\ce{CO2} and 2 O for 2\,\ce{H2O}, 6 O in all. The acid supplies 2 of them, so 4 O (that is 2\,\ce{O2}) come from oxygen gas.
    \begin{aligned}&\ce{CH3COOH(l) + 2O2(g) -> 2CO2(g) + 2H2O(l)}\\ &\Delta H^\circ_\mathrm{c}=-876\,\mathrm{kJ\,mol^{-1}}\end{aligned}
  4. (d) Combustion of one mole of ethanol uses 3\,\ce{O2}. Energy is released, so \Delta H is negative.
    \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)}\qquad \Delta H^\circ_\mathrm{c}=-1367\,\mathrm{kJ\,mol^{-1}}
Question 2

A formation reaction forms exactly one mole of one compound from its elements in their standard states. Check each equation against this.

  1. (a) forms 2 mol of \ce{Al2O3}, so its \Delta H^\circ is twice the enthalpy of formation. Not correct.
  2. (b) starts from the elements \ce{Al} and \ce{O2} and forms 1 mol of \ce{Al2O3}, so it is the formation equation. As printed it is not balanced (4\,\ce{Al} cannot give one \ce{Al2O3}); the correct formation equation is \ce{2Al(s) + 3/2 O2(g) -> Al2O3(s)}. It is the intended answer.
  3. (c) starts from a compound (\ce{CO}), not from the elements. It is the combustion of \ce{CO}. Not correct.
\text{Question 2: equation (b)}

6.1.6 (7)Standard enthalpy of first electron affinity

Standard enthalpy of first electron affinity, \Delta H^\circ_\mathrm{EA}
The energy released when one mole of gaseous atoms gains an electron each to form one mole of gaseous ions with a single negative charge, under standard conditions.
\ce{Cl(g) + e- -> Cl-(g)}\qquad \Delta H^\circ_\mathrm{EA}=-349\,\mathrm{kJ\,mol^{-1}}

6.1.6 (8)Lattice energy

Lattice energy
The energy released when one mole of an ionic compound (solid) is formed from its constituent gaseous ions.
\ce{Na+(g) + Cl-(g) -> NaCl(s)}\qquad \Delta H=-787\,\mathrm{kJ\,mol^{-1}}

6.2Bond energyOften examined

When a chemical reaction occurs, old bonds are broken and new bonds are formed. Breaking bonds always requires energy, and forming a bond always releases energy.

Bond dissociation energy
The energy required when one mole of a particular bond breaks apart to form neutral atoms.
Bond energy
The amount of energy released when one mole of that bond forms from neutral atoms.

The difference between the energy needed to break the bonds and the energy released on forming the new bonds decides whether the reaction absorbs or releases energy. For any chemical reaction, the enthalpy change is the sum of the bond dissociation energies of the reactants minus the sum of the bond energies of the products.

\Delta H=\sum H_\text{(bonds broken)}-\sum H_\text{(bonds formed)}
Example 6.3: bond energy of H–Cl
Calculate the bond energy of the \ce{H-Cl} bond from known bond energy data. Bond dissociation energy of \ce{H2}=436\,\mathrm{kJ\,mol^{-1}}; bond dissociation energy of \ce{Cl2}=243\,\mathrm{kJ\,mol^{-1}}; bond energy of \ce{HCl}=\,?
  1. Write the reaction. One mole of \ce{H2} and one mole of \ce{Cl2} break their bonds and give two moles of \ce{H-Cl} bonds.
    \ce{H2 + Cl2 -> 2HCl}
  2. The book takes the energy absorbed in breaking the bonds equal to the energy released in forming the new bonds.
    E(\ce{H-H})+E(\ce{Cl-Cl})=2\,E(\ce{H-Cl})
  3. Put in the data.
    436+243=2\,E(\ce{H-Cl})
  4. Add and divide by two.
    679=2\,E(\ce{H-Cl})\;\Rightarrow\;E(\ce{H-Cl})=\dfrac{679}{2}=339.5\,\mathrm{kJ\,mol^{-1}}

The book gives 339\,\mathrm{kJ\,mol^{-1}}. The experimentally determined bond energy of \ce{HCl} is 431.5\,\mathrm{kJ\,mol^{-1}}. The difference arises because the theoretical values are averages and do not take into account situations where the bond strength differs from the average, for example the unequal sharing of the electron pair due to the difference in electronegativity values.

E(\ce{H-Cl})\approx 339\,\mathrm{kJ\,mol^{-1}}\text{ (experimental }431.5\text{)}
Enthalpy change from bond energies

The same formula gives the enthalpy change of the reaction between hydrogen and oxygen to produce water.

Worked example: formation of water from bond energies
Find \Delta H for \ce{2H2(g) + O2(g) -> 2H2O(l)} from bond energies. Bond dissociation energy of \ce{H2}=436\,\mathrm{kJ} per mole; of \ce{O2}=493.6\,\mathrm{kJ} per mole; bond energy of the \ce{O-H} bond =460\,\mathrm{kJ} per mole.
  1. Count the bonds broken. Two moles of \ce{H-H} bonds and one mole of \ce{O=O} bonds are broken.
    \text{energy absorbed}=2(436)+493.6=872+493.6=1365.6\,\mathrm{kJ}
  2. Count the bonds formed. Two molecules of \ce{H2O} contain four \ce{O-H} bonds.
    \text{energy released}=4(460)=1840\,\mathrm{kJ}
  3. Apply \Delta H=\sum H_\text{broken}-\sum H_\text{formed}.
    \Delta H^\circ=2(436)+493.6-4(460)=872+493.6-1840=-474.4\,\mathrm{kJ}
  4. The sign is negative, so the reaction between hydrogen and oxygen to form water is exothermic.
Energy changes for \ce{2H2 + O2 -> 2H2O}: breaking bonds costs 1365.6\,\mathrm{kJ}, forming bonds returns 1840\,\mathrm{kJ}. The accent arrow is the net \Delta H.
\Delta H^\circ=-474.4\,\mathrm{kJ}
Calculation of bond energy for the enthalpy of reaction

You can calculate the bond energy per mole of methane by breaking all the \ce{C-H} bonds:

\ce{CH4(g) -> C(s) + 4H(g)}\qquad \Delta H^\circ=+1662\,\mathrm{kJ}
Worked example: average C–H bond energy in methane
In the reaction above, four moles of \ce{C-H} bonds are broken and 1662\,\mathrm{kJ} of energy is absorbed. Find the average bond energy of the \ce{C-H} bond.
  1. The bond dissociation energy per mole of \ce{CH4} is +1662\,\mathrm{kJ}, and this is for breaking 4 mol of \ce{C-H} bonds.
  2. Divide by the number of bonds broken.
    E(\ce{C-H})=\dfrac{+1662}{4}=+415.5\,\mathrm{kJ\,mol^{-1}}
E(\ce{C-H})_\text{average}=+415.5\,\mathrm{kJ\,mol^{-1}}
Worked example: enthalpy change of ammonia formation from bond energies
Find \Delta H for \ce{N2(g) + 3H2(g) -> 2NH3(g)} from these bond energies in \mathrm{kJ\,mol^{-1}}: \ce{N#N} 945, \ce{H-H} 436, \ce{N-H} 391. State whether the reaction is exothermic.
  1. Count the bonds broken in the reactants: 1 \ce{N#N} and 3 \ce{H-H}.
    \sum H_{\text{broken}}=945+3\times436=945+1308=2253\,\mathrm{kJ}
  2. Count the bonds formed in the products. Each \ce{NH3} has 3 \ce{N-H} bonds and there are 2 molecules, so 6 bonds.
    \sum H_{\text{formed}}=6\times391=2346\,\mathrm{kJ}
  3. Apply \Delta H=\sum H_{\text{broken}}-\sum H_{\text{formed}}.
    \Delta H=2253-2346=-93\,\mathrm{kJ}
  4. More energy is released in forming bonds than is spent in breaking them, so the reaction is exothermic. The result is close to 2\times(-46.1)=-92.2\,\mathrm{kJ} from Table 6.1 because bond energies are averages.
Energy path of the reaction: 2253 kJ is taken in to break bonds and 2346 kJ is given out when bonds form, so the net change is a release of 93 kJ (drawn to scale: 1 px = 10 kJ).
\Delta H=-93\,\mathrm{kJ}\ \text{(exothermic)}

6.2.1Variability in bond energies

Some bond energies are relatively well defined and precise. Others are approximate and can vary depending on the specific molecules involved and the environment in which the bond occurs.

Exact and approximate bond energies
Exact (precise) bond energiesApproximate bond energies
Where foundSimple diatomic molecules such as \ce{H2}, \ce{O2}, \ce{N2} and \ce{Cl2}Complex molecules, or molecules in different environments
ReasonThe molecules are made of identical atoms, so the bond dissociation energy is the same for every bond in the moleculeBond strength is affected by neighbouring atoms, molecular geometry and the inductive effect of functional groups
ExampleThe \ce{H-H} bond in \ce{H2} is about 436\,\mathrm{kJ\,mol^{-1}}\ce{C-C} in ethane, the \ce{C-H} bond in different molecules, and the \ce{O-H} bond in methanol compared with water

The \ce{C-C} bond energy in ethane (\ce{C2H6}) can be considered approximate because it depends on the general structure of the molecule. Similarly, the energy of a \ce{C-H} bond can vary with the chemical environment and the molecule in question. Some bonds have different strengths in different compounds: the \ce{O-H} bond energy in methanol (\ce{CH3OH}) is different from that in water because of differences in molecular structure and chemical environment.

6.3Heat and temperature

Heat as a form of energy

When water boils in a cooking pot, the lid jumps in the air after some time. The mechanical energy that makes the lid jump comes from the heat energy of the burner flame. This shows that heat is an energy that can be converted into other forms.

The law of conservation of energy states that energy cannot be produced in any physical or chemical process; it can only be changed from one form to another. Different forms of energy include light, mechanical, chemical, thermal, nuclear and magnetic energy.

Heat is all the energy produced by the movement of the molecules of a substance. Heat is a form of energy. When a body absorbs heat it becomes hotter, and when it releases heat it becomes colder. The effect of heat on a body is indicated by its temperature, which refers to the intensity of the heat of a body. Heat is measured in units of energy: the SI unit is the joule (J); it is also measured in calories.

Temperature is a measure of the average kinetic energy of the molecules of a substance. It describes the hotness or coldness of a substance or object. The SI unit of temperature is the kelvin (K); temperature is also measured on the Fahrenheit (^\circ\mathrm{F}) and Celsius (^\circ\mathrm{C}) scales. The amount of heat absorbed by a body is directly proportional to the temperature change.

Heat and temperature
HeatTemperature
What it isA form of energy: the energy of the movement of the particlesA measure of the average kinetic energy of the particles
Depends onParticle velocity, size and number of particlesAverage kinetic energy only
SI unitjoule (J)kelvin (K)

Heat capacity and specific heat capacity

The amount of heat absorbed (q) by a substance is proportional to its temperature change. The constant of proportionality is the heat capacity.

q\propto\Delta T\;\Rightarrow\; q=\text{heat capacity}\times\Delta T
\text{Heat capacity}=\dfrac{\text{heat}}{\Delta T}
(1)
Heat capacity
The amount of heat required to raise the temperature of a given amount of a substance by 1 kelvin. Unit: joule per kelvin (\mathrm{J\,K^{-1}}).
Specific heat capacity, c
The amount of heat required to raise the temperature of one gram of a substance by 1 kelvin. Unit: joule per gram per kelvin (\mathrm{J\,g^{-1}\,K^{-1}}).
c=\dfrac{\text{heat}}{\text{mass}\times\Delta T}
(2)
q=m\times c\times\Delta T

Knowing the specific heat capacity of a substance, its mass and the temperature change when it is heated or cooled, the heat absorbed or released can be found.

Example: molar heat capacity of copper
The specific heat capacity of copper is 0.387\,\mathrm{J\,g^{-1}\,K^{-1}}. Find its molar heat capacity (molar mass of \ce{Cu}=63.54\,\mathrm{g\,mol^{-1}}).
  1. The molar heat capacity is the heat needed to raise the temperature of one mole by 1 K. Multiply the heat per gram by the mass of one mole.
    \text{molar heat capacity}=0.387\,\mathrm{J\,g^{-1}\,K^{-1}}\times63.54\,\mathrm{g\,mol^{-1}}
  2. Multiply.
    =24.59\,\mathrm{J\,mol^{-1}\,K^{-1}}
24.59\,\mathrm{J\,mol^{-1}\,K^{-1}}

6.3.1Measurement of heat of reactionOften examined

Calorimeter
A device that measures heat flow.

Calorimeters measure the heat released from a system either at constant pressure or at constant volume, so there are two types of calorimeter.

Constant pressure calorimeter

In a constant pressure calorimeter the pressure of the system is fixed. We need a thermally insulated container with a thermometer and a stirrer. For most purposes a coffee cup calorimeter is used.

Figure 6.1: Coffee cup calorimeter. The accent mark is the thermometer, which gives \Delta T.
q=m\times c\times\Delta T
Symbols used with the calorimeter
SymbolMeaning
mMass of the reactants (the reaction mixture)
cSpecific heat capacity of the reaction mixture
\Delta TChange in temperature

The heat gained by the solution is the heat released by the reaction, and the reaction happens at constant pressure, so the heat of neutralization (or other reaction) per mole is

\Delta H=-\dfrac{q}{n}=-\dfrac{m\,c\,\Delta T}{n}

where n is the number of moles of the reactant that is used up (or of the product formed). The minus sign is there because a temperature rise means the reaction released heat, so \Delta H is negative.

Example 6.4: heat of neutralization
When 50\,\mathrm{cm^3} of 1.0\,\mathrm{M} \ce{NaOH} neutralizes 50\,\mathrm{cm^3} of 1.0\,\mathrm{M} \ce{HCl} at 25^\circ\mathrm{C}, the temperature rises from 25^\circ\mathrm{C} to 31.9^\circ\mathrm{C}. Calculate the heat of neutralization. (Specific heat of water =4.2\,\mathrm{J\,g^{-1}\,K^{-1}}. Density of water =1\,\mathrm{g\,cm^{-3}}.)
Given
  • V(\ce{NaOH}) = 50\,\mathrm{cm^3}
  • V(\ce{HCl}) = 50\,\mathrm{cm^3}
  • M = 1.0\,\mathrm{M}\text{ for both}
  • c = 4.2\,\mathrm{J\,g^{-1}\,K^{-1}}
  • \rho = 1\,\mathrm{g\,cm^{-3}}
  • T_i,\;T_f = 25^\circ\mathrm{C},\;31.9^\circ\mathrm{C}
To find
  • heat of neutralization \Delta H^\circ_\mathrm{n} per mole of water
  1. Total volume of the reaction mixture.
    V=50+50=100\,\mathrm{cm^3}
  2. Mass of the mixture, using the density of water.
    m=100\,\mathrm{cm^3}\times1\,\mathrm{g\,cm^{-3}}=100\,\mathrm{g}
  3. Rise in temperature (a rise of 6.9^\circ\mathrm{C} is a rise of 6.9\,\mathrm{K}).
    \Delta T=31.9-25=6.9\,\mathrm{K}
  4. Heat evolved, q=mc\Delta T.
    q=100\,\mathrm{g}\times4.2\,\mathrm{J\,g^{-1}\,K^{-1}}\times6.9\,\mathrm{K}=2.9\times10^{3}\,\mathrm{J}
  5. Moles of \ce{NaOH} and \ce{HCl}, from n=\dfrac{M\times V(\mathrm{cm^3})}{1000}.
    n(\ce{NaOH})=\dfrac{1\times50}{1000}=0.05\,\mathrm{mol},\qquad n(\ce{HCl})=0.05\,\mathrm{mol}
  6. From the equation, 1 mol of acid and 1 mol of base give 1 mol of water, so 0.05 mol of water forms.
    \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)}
  7. 0.05 mol of water releases 2.9\times10^3\,\mathrm{J}, so 1 mol releases
    \dfrac{2.9\times10^{3}}{0.05}=5.8\times10^{4}\,\mathrm{J\,mol^{-1}}=58.0\,\mathrm{kJ\,mol^{-1}}
  8. Heat is evolved at constant pressure, so q_p=\Delta H and the sign is negative.
    q_p=\Delta H^\circ_\mathrm{n}=-58.0\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_\mathrm{n}=-58.0\,\mathrm{kJ\,mol^{-1}}
Concept assessment exercise 6.3
Experimental data show that when 87.5\,\mathrm{cm^3} of a 0.10\,\mathrm{M} \ce{HCl} solution is mixed with 25.0\,\mathrm{cm^3} of 0.35\,\mathrm{M} \ce{NaOH} in a calorimeter, complete neutralization occurs. The temperature of the calorimeter changes from 25^\circ\mathrm{C} to 26.07^\circ\mathrm{C}. Use these data to determine the heat of neutralization for the reaction.
Given
  • V(\ce{HCl}),\;M = 87.5\,\mathrm{cm^3},\;0.10\,\mathrm{M}
  • V(\ce{NaOH}),\;M = 25.0\,\mathrm{cm^3},\;0.35\,\mathrm{M}
  • T_i,\;T_f = 25^\circ\mathrm{C},\;26.07^\circ\mathrm{C}
  • c,\;\rho = 4.2\,\mathrm{J\,g^{-1}\,K^{-1}},\;1\,\mathrm{g\,cm^{-3}}\text{ (as in Example 6.4)}
To find
  • heat of neutralization \Delta H_\mathrm{n} per mole
  1. Moles of each reactant. They are equal, which is why neutralization is complete.
    \begin{aligned}n(\ce{HCl})&=\dfrac{0.10\times87.5}{1000}=8.75\times10^{-3}\,\mathrm{mol}\\ n(\ce{NaOH})&=\dfrac{0.35\times25.0}{1000}=8.75\times10^{-3}\,\mathrm{mol}\end{aligned}
  2. Total volume and mass of the mixture (density 1\,\mathrm{g\,cm^{-3}}).
    V=87.5+25.0=112.5\,\mathrm{cm^3}\;\Rightarrow\; m=112.5\,\mathrm{g}
  3. Temperature rise.
    \Delta T=26.07-25=1.07\,\mathrm{K}
  4. Heat evolved.
    q=mc\Delta T=112.5\times4.2\times1.07=505.6\,\mathrm{J}
  5. Heat per mole of water formed (8.75\times10^{-3} mol). The reaction releases heat, so the sign is negative.
    \Delta H_\mathrm{n}=-\dfrac{505.6\,\mathrm{J}}{8.75\times10^{-3}\,\mathrm{mol}}=-5.778\times10^{4}\,\mathrm{J\,mol^{-1}}=-57.78\,\mathrm{kJ\,mol^{-1}}
\Delta H_\mathrm{n}=-57.78\,\mathrm{kJ\,mol^{-1}}

6.4Hess's lawOften examined

Enthalpy is a state function, so the enthalpy change of a chemical reaction in going from some initial state to some final state does not depend on the path followed by the reaction. In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in many steps. This principle is known as Hess's law.

Hess's law
G. H. Hess stated this law in 1840. It states that the enthalpy change in a chemical reaction is the same whether the reaction takes place in a single step or in several steps.

Mathematically, the sum of the enthalpy changes round a complete cycle is zero:

\sum\Delta H(\text{cycle})=0

Suppose a reactant A changes into the product B in one step, with enthalpy change \Delta H. Now suppose the change takes place in three steps: A to C, C to D, and finally D to B.

Figure 6.3: The direct route A to B (accent) and the three-step route A to C to D to B give the same total enthalpy change.

If \Delta H_1, \Delta H_2 and \Delta H_3 are the enthalpy changes of the steps, then according to Hess's law:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3
Example 6.6: burning carbon in one or two steps
The combustion of \ce{C} to \ce{CO2} evolves 393.5\,\mathrm{kJ}: \ce{C(s) + O2(g) -> CO2(g)}, \Delta H^\circ=-393.5\,\mathrm{kJ}. The reaction may also take place in two steps: \ce{C(s) + 1/2 O2(g) -> CO(g)}, \Delta H^\circ_1=-110.52\,\mathrm{kJ} and \ce{CO(g) + 1/2 O2(g) -> CO2(g)}, \Delta H^\circ_2=-282.98\,\mathrm{kJ}. Show that the net enthalpy change is the same.
  1. Write the two steps and add them. \ce{CO} appears on the right of step 1 and on the left of step 2, so it cancels. The \tfrac12\,\ce{O2}+\tfrac12\,\ce{O2} adds to 1\,\ce{O2}.
    \begin{aligned}&\ce{C(s) + 1/2 O2(g) -> CO(g)} &&\Delta H^\circ_1=-110.52\,\mathrm{kJ}\\ &\ce{CO(g) + 1/2 O2(g) -> CO2(g)} &&\Delta H^\circ_2=-282.98\,\mathrm{kJ}\\ \hline &\ce{C(s) + O2(g) -> CO2(g)} &&\Delta H^\circ_1+\Delta H^\circ_2\end{aligned}
  2. Add the enthalpy changes.
    \Delta H^\circ_1+\Delta H^\circ_2=-110.52-282.98=-393.50\,\mathrm{kJ}
  3. This equals the enthalpy change of the one-step reaction, -393.5\,\mathrm{kJ}. So the enthalpy change in a chemical reaction is independent of the path followed.
Figure 6.4: Energy cycle for the reaction between \ce{C(s)} and \ce{O2(g)} to produce \ce{CO2(g)}. The accent arrow is the one-step change.
\Delta H^\circ_1+\Delta H^\circ_2=-393.5\,\mathrm{kJ}
Why Hess's law is useful

Many compounds cannot be prepared directly from their elements, and some of them cannot be decomposed into their constituent elements, for example \ce{CCl4}. Some elements do not burn completely because a protective covering forms on their surface, such as \ce{Al} and \ce{B}. So the enthalpies of formation of \ce{CCl4}, \ce{Al2O3} and \ce{B2O3} cannot be determined directly in a calorimeter. Hess's law is particularly useful for finding the enthalpies of formation of such compounds.

Example 6.7: enthalpy of formation of methane
The enthalpy of formation of methane cannot be measured directly. By Hess's law it can be found indirectly from the enthalpies of combustion of \ce{CH4}, \ce{H2} and \ce{C}: (i) \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}, \Delta H^\circ=-890.4\,\mathrm{kJ\,mol^{-1}}; (ii) \ce{H2(g) + 1/2 O2(g) -> H2O(l)}, \Delta H^\circ=-285.8\,\mathrm{kJ\,mol^{-1}}; (iii) \ce{C(s) + O2(g) -> CO2(g)}, \Delta H^\circ=-393.5\,\mathrm{kJ\,mol^{-1}}; (iv) \ce{C(s) + 2H2(g) -> CH4(g)}, \Delta H^\circ=\,?

Combine equations (i), (ii) and (iii) to give equation (iv), and add the corresponding \Delta H^\circ values. Look at the reactants and products of the required equation: reactants \ce{C(s)} and 2\,\ce{H2(g)}, product \ce{CH4(g)}.

  1. Equation (iii) has \ce{C(s)} as a reactant, as the required equation needs. Use it as it is.
    \ce{C(s) + O2(g) -> CO2(g)}\qquad \Delta H^\circ=-393.5\,\mathrm{kJ}
  2. Equation (ii) has \ce{H2(g)} as a reactant, but the required equation needs 2\,\ce{H2(g)}. Multiply (ii) by 2, and double its \Delta H^\circ too.
    \ce{2H2(g) + O2(g) -> 2H2O(l)}\qquad \Delta H^\circ=2(-285.8)=-571.6\,\mathrm{kJ}
  3. Equation (i) has \ce{CH4(g)} as a reactant, but it is needed as a product. Reverse (i) and change the sign of \Delta H^\circ.
    \ce{CO2(g) + 2H2O(l) -> CH4(g) + 2O2(g)}\qquad \Delta H^\circ=+890.4\,\mathrm{kJ}
  4. Add the three equations and cancel the species that occur on both sides (\ce{CO2}, 2\,\ce{H2O} and 2\,\ce{O2} cancel).
    \ce{C(s) + 2H2(g) -> CH4(g)}
  5. Add the enthalpy changes.
    \Delta H^\circ=+890.4-571.6-393.5=-74.7\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_\mathrm{f}(\ce{CH4})\approx-74.7\,\mathrm{kJ\,mol^{-1}}
Worked example: nitrogen dioxide from its elements through nitrogen monoxide
Use Hess's law to find \Delta H^\circ for \ce{N2(g) + 2O2(g) -> 2NO2(g)} from (i) \ce{N2(g) + O2(g) -> 2NO(g)}, \Delta H^\circ=+180.5\,\mathrm{kJ} and (ii) \ce{2NO(g) + O2(g) -> 2NO2(g)}, \Delta H^\circ=-114.1\,\mathrm{kJ}. Then compare with Table 6.1.
  1. Write the target and look at the given equations. \ce{N2} is on the left in (i) and \ce{NO2} is on the right in (ii), both as in the target, so neither equation is reversed or multiplied.
    \ce{N2 + 2O2 -> 2NO2}
  2. Add the equations. The 2\ce{NO} appears on the right of (i) and on the left of (ii), so it cancels. One \ce{O2} from (i) and one from (ii) add to 2\ce{O2}.
    \ce{N2 + O2 + 2NO + O2 -> 2NO + 2NO2}
  3. Add the enthalpy changes.
    \Delta H^\circ=+180.5+(-114.1)=+66.4\,\mathrm{kJ}
  4. Check with Table 6.1. The equation makes 2 mol, and \Delta H^\circ_\mathrm{f}(\ce{NO2})=+33.2\,\mathrm{kJ\,mol^{-1}}.
    2\times33.2=+66.4\,\mathrm{kJ}\ \checkmark
Two routes to 2NO₂ give the same total change. The direct arrow (accent) equals the sum of the two steps: +180.5 − 114.1 = +66.4 kJ.
\Delta H^\circ=+66.4\,\mathrm{kJ}

6.4.1Enthalpies of reactions from enthalpies of formationOften examined

It is often convenient to calculate \Delta H^\circ for a reaction from the standard enthalpies of formation, \Delta H^\circ_\mathrm{f}, of the reactants and products. For a given reaction:

\Delta H^\circ=\sum\text{coeff}\,\Delta H^\circ_\mathrm{f}\,(\text{products})-\sum\text{coeff}\,\Delta H^\circ_\mathrm{f}\,(\text{reactants})

where coeff is the coefficient of the substance in the balanced equation.

Example 6.8: burning of gasoline
Calculate \Delta H^\circ for the following reaction, which takes place when gasoline burns in internal combustion engines. The values of \Delta H^\circ_\mathrm{f} are -269\,\mathrm{kJ}, 0\,\mathrm{kJ}, -393.5\,\mathrm{kJ} and -285\,\mathrm{kJ} for \ce{C8H18(l)}, \ce{O2(g)}, \ce{CO2(g)} and \ce{H2O(l)} respectively. \ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(l)}
  1. Write the formula with the coefficients from the equation.
    \Delta H^\circ_\text{reaction}=\left[16\,\Delta H^\circ_\mathrm{f}(\ce{CO2})+18\,\Delta H^\circ_\mathrm{f}(\ce{H2O})\right]-\left[2\,\Delta H^\circ_\mathrm{f}(\ce{C8H18})+25\,\Delta H^\circ_\mathrm{f}(\ce{O2})\right]
  2. Substitute the values. The working uses -285.8 for water. \ce{O2} is an element, so its term is zero.
    =16(-393.5)+18(-285.8)-2(-269)-25(0)
  3. Multiply out each term.
    =-6296-5144.4+538+0
  4. Add.
    =-11440.4+538=-10902.4\,\mathrm{kJ}\approx-1.09\times10^{4}\,\mathrm{kJ}
\Delta H^\circ\approx-1.09\times10^{4}\,\mathrm{kJ}
Example 6.9: oxidation of glucose in living tissues
Calculate \Delta H^\circ_\text{reaction} for the following reaction, which takes place in the tissues of living organisms. The values of \Delta H^\circ_\mathrm{f} are -1258.18\,\mathrm{kJ}, 0\,\mathrm{kJ}, -393.5\,\mathrm{kJ} and -285\,\mathrm{kJ} for \ce{C6H12O6(s)}, \ce{O2(g)}, \ce{CO2(g)} and \ce{H2O(l)} respectively. \ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(l)}
  1. Write the formula for the equation.
    \Delta H^\circ_\text{reaction}=\left[6\,\Delta H^\circ_\mathrm{f}(\ce{CO2})+6\,\Delta H^\circ_\mathrm{f}(\ce{H2O})\right]-\left[\Delta H^\circ_\mathrm{f}(\ce{C6H12O6})+6\,\Delta H^\circ_\mathrm{f}(\ce{O2})\right]
  2. Substitute the values (the working uses -285.8 for water; \ce{O2} is 0).
    =\left[6(-393.5)+6(-285.8)\right]-\left[-1258.18+6(0)\right]
  3. Multiply out.
    =\left[-2361.0-1714.8\right]-\left[-1258.18\right]=-4075.8+1258.18
  4. Add.
    \Delta H^\circ_\text{reaction}=-2817.62\,\mathrm{kJ}

This means that one mole (180 g) of glucose provides 2817.62\,\mathrm{kJ} of energy. The energy given by one gram of glucose is therefore:

\dfrac{2817.62\,\mathrm{kJ\,mol^{-1}}}{180\,\mathrm{g\,mol^{-1}}}=15.65\,\mathrm{kJ\,g^{-1}}
\Delta H^\circ=-2817.62\,\mathrm{kJ};\;15.65\,\mathrm{kJ\,g^{-1}}
Worked example: heat needed to decompose limestone
Calculate \Delta H^\circ for \ce{CaCO3(s) -> CaO(s) + CO2(g)} from Table 6.1 (\ce{CaCO3} -1207, \ce{CaO} -635.1, \ce{CO2} -393.5, all in \mathrm{kJ\,mol^{-1}}). Then find the heat absorbed when 50\,\mathrm{g} of \ce{CaCO3} decomposes (M=100\,\mathrm{g\,mol^{-1}}).
  1. Use \Delta H^\circ=\sum\Delta H_\mathrm{f}^\circ(\text{products})-\sum\Delta H_\mathrm{f}^\circ(\text{reactants}). All coefficients are 1.
    \Delta H^\circ=\left[(-635.1)+(-393.5)\right]-\left[-1207\right]
  2. Evaluate the brackets.
    =-1028.6+1207
  3. Subtract.
    \Delta H^\circ=+178.4\,\mathrm{kJ}
  4. The sign is positive, so the decomposition is endothermic: heat must be supplied, which is why limestone is heated strongly.
  5. Heat for 50\,\mathrm{g}: moles first, then scale the equation.
    n=\dfrac{50}{100}=0.50\,\mathrm{mol},\qquad q=0.50\times178.4=89.2\,\mathrm{kJ\ absorbed}
\Delta H^\circ=+178.4\,\mathrm{kJ};\; q=89.2\,\mathrm{kJ}

6.4.2Calculation of enthalpy of combustion

The enthalpy of combustion, \Delta H^\circ_\mathrm{c}, is the heat evolved when 1 mole of a substance burns completely in oxygen at standard conditions. It can be calculated by the same general formula as in section 6.4.1, with the combustion equation:

\Delta H^\circ_\mathrm{c}=\sum\text{coeff}_\mathrm{p}\,\Delta H^\circ_\mathrm{f}\,(\text{products})-\sum\text{coeff}_\mathrm{r}\,\Delta H^\circ_\mathrm{f}\,(\text{reactants})
Example 6.10: combustion of ethyne
Calculate \Delta H^\circ_\mathrm{c} for ethyne (\ce{C2H2}). The standard enthalpies of formation of \ce{C2H2(g)}, \ce{CO2(g)} and \ce{H2O(l)} are +226.73, -393.5 and -285.8\,\mathrm{kJ\,mol^{-1}} respectively.
  1. Write the combustion equation for one mole of ethyne. The two C atoms give 2\,\ce{CO2}, the two H atoms give one \ce{H2O}, and the O atoms needed are 4+1=5, that is \tfrac52\,\ce{O2}.
    \ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(l)}
  2. Apply the formula; \ce{O2} is an element in its standard state, so its term is zero.
    \Delta H^\circ_\mathrm{c}=\left[2(-393.5)+1(-285.8)\right]-\left[(+226.73)+\tfrac52(0)\right]
  3. Evaluate the brackets.
    =\left[-787.0-285.8\right]-\left[+226.73\right]=-1072.8-226.73
  4. Add.
    \Delta H^\circ_\mathrm{c}=-1299.53\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_\mathrm{c}(\ce{C2H2})=-1299.53\,\mathrm{kJ\,mol^{-1}}

6.5Born–Haber cycleOften examined

The Born–Haber cycle is a special application of Hess's law to binary ionic compounds (\ce{M+X-}). It helps us to calculate the lattice energies of binary ionic compounds.

Lattice energy
The change in energy that takes place when separated gaseous ions are packed together to form one mole of an ionic solid.
\ce{M+(g) + X-(g) -> MX(s)}

Lattice energy cannot be determined directly. However, it can be determined indirectly by means of the Born–Haber cycle. Consider sodium chloride, whose standard enthalpy of formation is \Delta H^\circ_\mathrm{f}=-411\,\mathrm{kJ\,mol^{-1}}. The formation reaction can be thought of as taking place in several steps, one of which is the formation of the lattice. This complete sequence of reactions is called a cycle (Figure 6.5).

The five steps for \ce{NaCl}
The five steps of the Born–Haber cycle for \ce{NaCl}
StepWhat happensEquation\Delta H^\circ
ISublimation of solid sodium: \Delta H_\mathrm{s}\ce{Na(s) -> Na(g)}+108\,\mathrm{kJ\,mol^{-1}}
IIIonization of \ce{Na(g)} to \ce{Na+(g)}: the first ionization energy, \Delta H_\mathrm{IE}\ce{Na(g) -> Na+(g) + e-}+496\,\mathrm{kJ\,mol^{-1}}
IIIDissociation of \ce{Cl2}: atomization of chlorine, \Delta H_\mathrm{at} (half a mole of \ce{Cl2} gives one mole of \ce{Cl} atoms)\ce{1/2 Cl2(g) -> Cl(g)}+121\,\mathrm{kJ\,mol^{-1}}
IVFormation of \ce{Cl-(g)}: the electron affinity of chlorine, \Delta H_\mathrm{EA}\ce{Cl(g) + e- -> Cl-(g)}-349\,\mathrm{kJ\,mol^{-1}}
VFormation of solid \ce{NaCl} from the gaseous ions: the lattice energy \Delta H_\mathrm{l}, which is to be calculated\ce{Na+(g) + Cl-(g) -> NaCl(s)}\Delta H_\mathrm{l}=\,?

The first three steps are all endothermic (energy is taken in), so they are drawn upward in the figure.

Born–Haber cycle for NaCl: the lattice energy
Use the five steps above and \Delta H^\circ_\mathrm{f}(\ce{NaCl})=-411\,\mathrm{kJ\,mol^{-1}} to find the lattice energy of \ce{NaCl}.
  1. Add the five equations. Everything cancels except the elements on the left and \ce{NaCl} on the right, so the sum is the formation of \ce{NaCl}.
    \begin{aligned}&\ce{Na(s) -> Na(g)}\\ &\ce{Na(g) -> Na+(g) + e-}\\ &\ce{1/2 Cl2(g) -> Cl(g)}\\ &\ce{Cl(g) + e- -> Cl-(g)}\\ &\ce{Na+(g) + Cl-(g) -> NaCl(s)}\\ \hline &\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}\end{aligned}
  2. By Hess's law the sum of the five energy changes is the overall energy change.
    \Delta H^\circ_\mathrm{f}=\Delta H^\circ_\mathrm{s}+\Delta H^\circ_\mathrm{IE}+\Delta H^\circ_\mathrm{at}+\Delta H^\circ_\mathrm{EA}+\Delta H_\mathrm{l}
  3. Put in the numbers.
    -411=+108+496+121+(-349)+\Delta H_\mathrm{l}
  4. Rearrange for \Delta H_\mathrm{l}. The four known steps add to +376.
    \Delta H_\mathrm{l}=-411-(108+496+121-349)=-411-376
  5. Evaluate.
    \Delta H_\mathrm{l}=-787\,\mathrm{kJ\,mol^{-1}}

So the lattice energy of \ce{NaCl} is 787\,\mathrm{kJ\,mol^{-1}} (released when the lattice forms).

\Delta H_\mathrm{l}(\ce{NaCl})=-787\,\mathrm{kJ\,mol^{-1}}
Figure 6.5: Born–Haber cycle for the formation of \ce{NaCl(s)}. The accent arrow is the lattice energy, found by difference.
Worked example: Born–Haber cycle and lattice energy of potassium bromide
Draw the Born–Haber cycle for \ce{KBr(s)} and find its lattice energy. Data in \mathrm{kJ\,mol^{-1}}: sublimation of \ce{K} +89; first ionization energy of \ce{K} +419; \ce{1/2 Br2(l) -> Br(g)} +112; electron affinity of \ce{Br} -325; \Delta H^\circ_\mathrm{f}(\ce{KBr})=-393.8.
  1. List the steps in order, starting from the elements: sublimation of K, ionization of K(g), atomization of bromine, electron affinity of Br(g), and last the formation of the lattice from the gaseous ions.
  2. By Hess's law the enthalpy of formation is the sum of all five steps.
    \Delta H_\mathrm{f}=\Delta H_\mathrm{s}+\Delta H_\mathrm{IE}+\Delta H_\mathrm{at}+\Delta H_\mathrm{EA}+\Delta H_\mathrm{l}
  3. Add the four known steps.
    89+419+112+(-325)=+295\,\mathrm{kJ}
  4. Rearrange for the lattice energy.
    \Delta H_\mathrm{l}=\Delta H_\mathrm{f}-295=-393.8-295=-688.8\,\mathrm{kJ\,mol^{-1}}
  5. Draw the cycle: the first three steps and the ionization are endothermic, so they go up; the electron affinity goes down; the lattice energy goes down to \ce{KBr(s)}, which lies at the level of \Delta H_\mathrm{f} below the elements.
Born–Haber cycle for KBr (to scale, 1 px = 3.3 kJ). The accent arrow is the lattice energy, found by difference.
\Delta H_\mathrm{l}=-688.8\,\mathrm{kJ\,mol^{-1}}
Concept assessment exercise 6.5: Born–Haber cycle for MgO
Draw a complete Born–Haber cycle for the formation of \ce{MgO(s)}. Calculate the lattice energy of \ce{MgO} from the following data: standard enthalpy of formation of \ce{MgO}=-602\,\mathrm{kJ\,mol^{-1}}; standard enthalpy of sublimation of \ce{Mg}=150\,\mathrm{kJ\,mol^{-1}}; ionization energy of \ce{Mg(g)} to \ce{Mg^2+(g)}=2180\,\mathrm{kJ\,mol^{-1}}; standard enthalpy of atomization of \ce{O2}=247\,\mathrm{kJ\,mol^{-1}}; electron affinity of \ce{O(g)} to \ce{O^-(g)}=-141\,\mathrm{kJ\,mol^{-1}}; electron affinity of \ce{O^-(g)} to \ce{O^2-(g)}=+878\,\mathrm{kJ\,mol^{-1}}.
  1. Write the steps of the cycle. Magnesium forms a 2+ ion and oxygen a 2- ion, so there are two ionization stages (given together as 2180) and two electron-gain stages.
    \begin{aligned}&\ce{Mg(s) -> Mg(g)} &&\Delta H_\mathrm{s}=+150\\ &\ce{Mg(g) -> Mg^2+(g) + 2e-} &&\Delta H_\mathrm{IE}=+2180\\ &\ce{1/2 O2(g) -> O(g)} &&\Delta H_\mathrm{at}=+247\\ &\ce{O(g) + e- -> O-(g)} &&\Delta H_\mathrm{EA1}=-141\\ &\ce{O-(g) + e- -> O^2-(g)} &&\Delta H_\mathrm{EA2}=+878\\ &\ce{Mg^2+(g) + O^2-(g) -> MgO(s)} &&\Delta H_\mathrm{l}=\,?\end{aligned}
  2. By Hess's law the sum of the steps is the formation of \ce{MgO}.
    \Delta H^\circ_\mathrm{f}=\Delta H_\mathrm{s}+\Delta H_\mathrm{IE}+\Delta H_\mathrm{at}+\Delta H_\mathrm{EA1}+\Delta H_\mathrm{EA2}+\Delta H_\mathrm{l}
  3. Substitute.
    -602=150+2180+247+(-141)+878+\Delta H_\mathrm{l}
  4. Add the known steps: 150+2180+247-141+878=3314.
    \Delta H_\mathrm{l}=-602-3314
  5. Evaluate.
    \Delta H_\mathrm{l}=-3916\,\mathrm{kJ\,mol^{-1}}
Born–Haber cycle for \ce{MgO(s)} (levels are not drawn to scale). The accent arrow is the lattice energy.
\Delta H_\mathrm{l}(\ce{MgO})=-3916\,\mathrm{kJ\,mol^{-1}}

6.5.1Factors affecting lattice energyOften examined

The size of the lattice energy is influenced by the charge density of an ion. Charge density is the ratio of the ionic charge to the ionic radius of the ion.

\text{charge density}=\dfrac{\text{ionic charge}}{\text{ionic radius}}

Two factors have a significant effect on lattice energy.

1. Ionic charge

As the magnitude of the ionic charge increases, the lattice energy also increases. Ions with higher charges have stronger electrostatic attractions, and they release more energy when they come together to form an ionic lattice. For example, \ce{MgO} has a higher lattice energy than \ce{NaCl}, because \ce{Mg^2+} and \ce{O^2-} ions have a larger charge than \ce{Na+} and \ce{Cl-} ions.

2. Ionic radius

As the ionic radius increases, the lattice energy decreases. Larger ions keep the opposite ions at a greater distance, giving a weaker electrostatic attraction between the ions. Larger ions therefore need less energy to separate, which means a lower lattice energy. For example, \ce{LiF} and \ce{KF} have the same ionic charge but different ionic radii. The \ce{Li+} ion is smaller than the \ce{K+} ion, so \ce{LiF} has the stronger electrostatic attraction between its ions. So \ce{LiF} has a higher lattice energy than \ce{KF}.

Factors affecting lattice energy
FactorChangeEffect on lattice energyExample
Ionic chargeLarger chargeLarger (more energy released)\ce{MgO}>\ce{NaCl}
Ionic radiusLarger ionSmaller\ce{LiF}>\ce{KF}
Concept assessment exercise 6.6
Which of the following compounds is expected to have the higher lattice energy? Why? (a) \ce{LiF} or \ce{NaF}; (b) \ce{NaCl} or \ce{MgCl2}.
  1. (a) Both have ions of charge 1+ and 1-, so charge does not decide. The fluoride ion is common. \ce{Li+} is smaller than \ce{Na+} (it is above sodium in group 1), so the ions in \ce{LiF} are closer and attract more strongly. \ce{LiF} has the higher lattice energy.
    \ce{LiF}>\ce{NaF}
  2. (b) The chloride ion is common. \ce{Mg^2+} has a charge of 2+ against 1+ for \ce{Na+}, and it is also smaller than \ce{Na+}. Higher charge and smaller radius both increase the electrostatic attraction, so \ce{MgCl2} has the higher lattice energy.
    \ce{MgCl2}>\ce{NaCl}
\text{(a) }\ce{LiF};\quad\text{(b) }\ce{MgCl2}

6.5.2Role of hydration in the dissolving process

The energy change when a solution forms depends on three types of interaction: between solute and solute particles, between solvent and solvent particles, and between solute and solvent particles.

The three interactions when a solid dissolves
InteractionWhat happensEnergy
Solute–soluteThe interactions between solute particles are broken (the solute leaves its lattice)Absorbed: endothermic
Solvent–solventThe solvent molecules move apart to make room for the solute particlesAbsorbed: endothermic
Solute–solventSolvent molecules surround the solute particles on all sides (hydration when the solvent is water)Released: exothermic

The strength of these two types of effect decides whether the dissolving is endothermic or exothermic. When a solute dissolves in water, the attractive forces between the solute particles must be overcome to break them out of the solid lattice. A strong force must bind these separate solute particles to the water molecules in solution. So the heat of solution measures the net flow of energy when a substance dissolves.

The energy needed to break the solute particles apart is the lattice energy (\Delta H_\text{lattice}) of the solute. The energy released when solute particles bind to water molecules is the enthalpy of hydration (\Delta H_\text{hyd}); for other solvents this energy is called the heat of dissolution. So the enthalpy of solution is the difference between the lattice energy and the enthalpy of hydration:

\Delta H_\text{solution}=\Delta H_\text{lattice}-\Delta H_\text{hyd}
Endothermic and exothermic solution

If \Delta H_\text{solution} is very large and positive, the solute is unlikely to be soluble. If \Delta H_\text{solution} is moderately positive, the solute dissolves.

When ammonium nitrate dissolves in water, 25.7\,\mathrm{kJ\,mol^{-1}} of energy is absorbed and the flask becomes cold. This is because the crystal binds the \ce{NH4+} and \ce{NO3-} ions more tightly than the solution does.

\begin{aligned}&\ce{NH4NO3(s) + xH2O(l) -> NH4+(aq) + NO3-(aq)}\\ &\Delta H_\text{solution}=+25.7\,\mathrm{kJ\,mol^{-1}}\end{aligned}

On the other hand, if the solvent binds the solute particles more tightly than the crystal does, energy is released as the solute dissolves and the solution process is exothermic. When \ce{NaOH} dissolves in water, 44.5\,\mathrm{kJ\,mol^{-1}} of energy is released and the flask becomes hot.

\begin{aligned}&\ce{NaOH(s) + xH2O(l) -> Na+(aq) + OH-(aq)}\\ &\Delta H_\text{soln}=-44.5\,\mathrm{kJ\,mol^{-1}}\end{aligned}

So the solution process can be endothermic or exothermic depending on the difference between the lattice energy of the solute and the heat of hydration of its particles. Other examples are:

\begin{aligned}&\ce{CaCl2(s) + xH2O(l) -> Ca^2+(aq) + 2Cl-(aq)} &&\Delta H_\text{soln}=-83.0\,\mathrm{kJ\,mol^{-1}}\\ &\ce{MgSO4(s) + xH2O(l) -> Mg^2+(aq) + SO4^2-(aq)} &&\Delta H_\text{soln}=-91.2\,\mathrm{kJ\,mol^{-1}}\end{aligned}
Table 6.3: enthalpies of solution of some ionic solids
SubstanceEnthalpy of solution (\mathrm{kJ\,mol^{-1}})
\ce{LiCl}-37.0
\ce{NaCl}+2.98
\ce{KCl}+17.2
\ce{KI}+20.3
\ce{NH4NO3}+25.7
\ce{AlCl3}-321.0

6.5.3Energy cycle involving enthalpy of solution, lattice energy and enthalpy of hydrationOften examined

A solid dissolving can be drawn as a cycle. The solid \ce{MX(s)} can go to the aqueous ions \ce{M+(aq) + X-(aq)} directly (the enthalpy of solution), or in two steps: first the lattice is broken into gaseous ions (the reverse of the lattice formation), and then the gaseous ions are hydrated.

Figure 6.6: Energy cycle involving enthalpy of solution, lattice energy and enthalpy of hydration. The accent arrow is the enthalpy of solution.

According to Hess's law:

\Delta H_\text{solution}=\Delta H_\text{reverse lattice}+\Delta H_\text{hydration}
Concept assessment exercise 6.7
Construct an energy cycle and an energy level diagram to calculate \Delta H_\text{solution} for \ce{NaCl}, and calculate its enthalpy of solution from the following data: \Delta H_\text{lattice}=-786\,\mathrm{kJ\,mol^{-1}}; \Delta H_\text{hydration}=-783\,\mathrm{kJ\,mol^{-1}}.

Energy cycle.

Energy cycle for \ce{NaCl}. The accent arrow is the enthalpy of solution, the unknown.
  1. The lattice energy of \ce{NaCl} is -786 (energy released when the lattice forms). The reverse process, breaking the lattice into gaseous ions, has the opposite sign.
    \Delta H_\text{reverse lattice}=+786\,\mathrm{kJ\,mol^{-1}}
  2. Apply Hess's law round the cycle.
    \Delta H_\text{solution}=\Delta H_\text{reverse lattice}+\Delta H_\text{hydration}
  3. Substitute.
    \Delta H_\text{solution}=+786+(-783)=+3\,\mathrm{kJ\,mol^{-1}}

Energy level diagram. Starting from \ce{NaCl(s)}, the energy rises by 786\,\mathrm{kJ} to the separated gaseous ions, then falls by 783\,\mathrm{kJ} to the aqueous ions, which end up 3\,\mathrm{kJ} above the solid.

Energy level diagram for the solution of \ce{NaCl} (the +3\,\mathrm{kJ} step is exaggerated). The accent arrow is the enthalpy of solution.
\Delta H_\text{solution}=+3\,\mathrm{kJ\,mol^{-1}}

6.5.4Factors affecting enthalpy of hydration

The size of the enthalpy of hydration is influenced by the charge density of an ion, which is the ratio of the ionic charge to the ionic radius. So two factors affect the enthalpy of hydration.

1. Ionic charge

The charge of an ion affects the electrostatic interaction between the ion and the water molecules. Ions with a higher charge attract water molecules more strongly (ion–dipole interactions), so the enthalpy of hydration is larger: as the ionic charge increases, the hydration process becomes more exothermic. For example, the hydration enthalpy of \ce{MgO} is more exothermic than that of \ce{NaCl}, because \ce{Mg^2+} and \ce{O^2-} have a greater charge than \ce{Na+} and \ce{Cl-}.

2. Ionic radius

Smaller ions have a higher charge density and attract water molecules more strongly, so more energy is released when they hydrate and the enthalpy of hydration is more exothermic. Larger ions spread their charge out and attract water molecules less strongly, so less energy is released. For example, the hydration enthalpy of \ce{MgSO4} is greater than that of \ce{CaSO4}. The sulphate ion is common to both, so the difference must be due to \ce{Mg^2+} and \ce{Ca^2+}. Both are in group 2 and magnesium is above calcium, so \ce{Mg^2+} is smaller than \ce{Ca^2+}. Water molecules are attracted more strongly to \ce{Mg^2+}, so the hydration of \ce{MgSO4} is more exothermic than that of \ce{CaSO4}.

Factors affecting enthalpy of hydration
FactorChangeEffect on the enthalpy of hydrationExample
Ionic chargeLarger chargeMore exothermic\ce{MgO} more than \ce{NaCl}
Ionic radiusSmaller ionMore exothermic\ce{MgSO4} more than \ce{CaSO4}
Concept assessment exercise 6.8
Which has the higher enthalpy of hydration? Justify your answer. (a) \ce{LiF} or \ce{NaF}; (b) \ce{KNO3} or \ce{NaNO3}.
  1. (a) Both compounds have 1+ and 1- ions and the fluoride ion is common. \ce{Li+} is smaller than \ce{Na+}, so it has the higher charge density and attracts water molecules more strongly. \ce{LiF} has the higher (more exothermic) enthalpy of hydration.
    \ce{LiF}>\ce{NaF}
  2. (b) The nitrate ion is common and both cations are 1+. \ce{Na+} is smaller than \ce{K+} (sodium is above potassium in group 1), so its charge density is higher and it attracts water more strongly. \ce{NaNO3} has the higher enthalpy of hydration.
    \ce{NaNO3}>\ce{KNO3}
\text{(a) }\ce{LiF};\quad\text{(b) }\ce{NaNO3}

6.5.5Factors affecting electron affinity

Three factors affect electron affinity: nuclear charge, atomic size and shielding effect. The book gives the three names here and refers to section 10.5.1 for the details.

How the three factors change the electron affinity
FactorEffect on the electron affinity
Nuclear chargeA larger nuclear charge pulls the incoming electron in more strongly, so more energy is released (a more negative electron affinity).
Atomic sizeIn a smaller atom the incoming electron gets closer to the nucleus, so it is attracted more strongly and more energy is released.
Shielding effectInner electrons shield (screen) the incoming electron from the nuclear charge. More shielding means weaker attraction and less energy released.

6.6EntropyOften examined

Which state of matter has more disorder, random order and minimum particle energy, solid or gas? The degree of disorder and randomness of a system is called its entropy. It is marked with the symbol S. Entropy is also described as the number of possible arrangements of the particles and their energy in a given system.

As a system moves from a more ordered state to a less ordered state, the entropy of the system tends to increase. Melting ice is a good example. In ice the molecules have a fixed and ordered arrangement and minimum energy. As the ice melts, the energy of its molecules increases, they gain more freedom of movement and become more disordered, so the entropy of the system increases. On the other hand, when water vapour condenses into a liquid, the molecules become more ordered and less energetic, so the entropy of the system decreases.

The entropy change of the system is denoted by \Delta S.

  • If the entropy of the system increases, this is a positive change: \Delta S>0.
  • If the entropy of the system decreases, this is a negative change: \Delta S<0.
  • In general, when the temperature of a system increases, its entropy also increases, \Delta S>0.
  • As the temperature of the system decreases, \Delta S<0.
Sign of the entropy change
ChangeMore or less disorderSign of \Delta S
Solid to liquid (melting)MorePositive
Liquid to gasMorePositive
Gas to liquid (condensation)LessNegative
Temperature increasesMorePositive
Temperature decreasesLessNegative
Number of gaseous molecules increases in a reactionMorePositive
Number of gaseous molecules decreases in a reactionLessNegative

6.6.1Calculation of the entropy change

The entropy change for a chemical reaction is calculated from the standard entropies, S^\circ, of the products and reactants:

\Delta S^\circ=\sum S^\circ_\text{products}-\sum S^\circ_\text{reactants}
Entropy change of a reaction
Calculate \Delta S^\circ for the reaction \ce{A + 2B -> C}, given S^\circ(\ce{A})=40, S^\circ(\ce{B})=50 and S^\circ(\ce{C})=100 (the book gives the unit as \mathrm{kJ\,mol^{-1}}).
  1. Add the entropies of the products, each multiplied by its coefficient, and subtract the same sum for the reactants. \ce{B} has coefficient 2.
    \Delta S^\circ=\left[1(100)\right]-\left[1(40)+2(50)\right]
  2. Evaluate the brackets.
    =100-(40+100)=100-140
  3. Subtract.
    \Delta S^\circ=-40

The result is negative: two molecules of B and one of A combine into a single product, so the system becomes more ordered.

\Delta S^\circ=-40
Concept assessment exercise 6.9
For which of the following reactions is \Delta S^\circ positive or negative? (a) \ce{CO2(g) -> CO2(s)}; (b) \ce{2H2(g) + O2(g) -> 2H2O(l)}; (c) \ce{H2O(l) -> H2O(g)}.
  1. (a) A gas becomes a solid. The molecules go from moving freely to a fixed lattice, so the disorder falls.
    \Delta S^\circ<0\quad\text{(negative)}
  2. (b) Three moles of gas (2\,\ce{H2}+1\,\ce{O2}) become two moles of liquid. The number of gas molecules falls to zero and the products are a liquid, so the disorder falls.
    \Delta S^\circ<0\quad\text{(negative)}
  3. (c) A liquid becomes a gas. The molecules gain freedom of movement, so the disorder rises.
    \Delta S^\circ>0\quad\text{(positive)}
\text{(a) negative; (b) negative; (c) positive}

6.7Gibbs free energyOften examined

Gibbs free energy is a measure of the amount of usable energy of a system at constant temperature and pressure. It is defined as the enthalpy of the system minus the product of its temperature and entropy. It is also known as the 'available energy'. It is denoted by the symbol G, after its discoverer, the American scientist Josiah Willard Gibbs, who introduced it in 1876.

G=H-TS
Symbols in G=H-TS
SymbolMeaning
GGibbs free energy
HEnthalpy of the reaction (of the system)
SEntropy
TTemperature in kelvin

Gibbs free energy is a state function: it does not depend on the path followed by a reaction. So the change in Gibbs free energy is given by

\Delta G^\circ=\Delta H^\circ-\Delta(TS)

When the reaction is carried out at constant temperature, \Delta T=0, so \Delta(TS)=T\,\Delta S and

\Delta G^\circ=\Delta H^\circ-T\,\Delta S^\circ

This equation is called the Gibbs–Helmholtz equation. \Delta G^\circ is particularly useful in deciding whether a chemical reaction is spontaneous or at equilibrium.

The sign and size of \Delta G^\circ show whether the process is spontaneous or at equilibrium. There are three possibilities.

Sign of \Delta G^\circ and spontaneity
Value of \Delta G^\circMeaning
\Delta G^\circ<0 (negative)The process is spontaneous in the forward direction. The system releases energy and tends to move to a lower energy state.
\Delta G^\circ>0 (positive)The process is non-spontaneous in the forward direction and requires energy to be put in.
\Delta G^\circ=0The system is at equilibrium.
How the signs of \Delta H and \Delta S decide the sign of \Delta G (from the equation)
\Delta H\Delta S\Delta G=\Delta H-T\Delta SSpontaneous?
NegativePositiveAlways negativeAt all temperatures
PositiveNegativeAlways positiveAt no temperature
NegativeNegativeNegative only when T is low (|\Delta H|>T|\Delta S|)At low temperatures
PositivePositiveNegative only when T is high (T\Delta S>\Delta H)At high temperatures

Gibbs free energy is also useful for understanding phase transitions, such as from liquid to solid or gas and back. Consider the melting of ice:

\ce{H2O(s) -> H2O(l)}

At the melting point, \Delta G^\circ=0, so the solid and liquid phases are in equilibrium.

Worked example: is ammonia formation spontaneous, and at what temperature?
For \ce{N2(g) + 3H2(g) -> 2NH3(g)}, \Delta H^\circ=-92.2\,\mathrm{kJ} and \Delta S^\circ=-198.7\,\mathrm{J\,K^{-1}}. Find \Delta G^\circ at 298\,\mathrm{K} and at 500\,\mathrm{K} and the temperature at which the reaction stops being spontaneous.
Given
  • \Delta H^\circ = -92.2\,\mathrm{kJ}
  • \Delta S^\circ = -198.7\,\mathrm{J\,K^{-1}}=-0.1987\,\mathrm{kJ\,K^{-1}} convert to kJ
To find
  • \Delta G^\circ at 298\,\mathrm{K} and 500\,\mathrm{K}
  • temperature where \Delta G^\circ=0
  1. Both quantities must be in kJ, so change \Delta S^\circ from J to kJ (divide by 1000). Use \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ.
  2. At 298\,\mathrm{K}.
    \Delta G^\circ=-92.2-298(-0.1987)=-92.2+59.2=-33.0\,\mathrm{kJ}
  3. \Delta G^\circ<0, so the reaction is spontaneous at 298\,\mathrm{K}.
  4. At 500\,\mathrm{K}.
    \Delta G^\circ=-92.2+500(0.1987)=-92.2+99.35=+7.2\,\mathrm{kJ}
  5. \Delta G^\circ>0, so it is non-spontaneous at 500\,\mathrm{K}. Here \Delta H and \Delta S are both negative, so the reaction is spontaneous only at low temperature, as in the table of signs.
  6. Equilibrium is where \Delta G^\circ=0, that is \Delta H^\circ=T\Delta S^\circ.
    T=\dfrac{\Delta H^\circ}{\Delta S^\circ}=\dfrac{-92.2}{-0.1987}=464\,\mathrm{K}
Free energy change of ammonia formation against temperature
\Delta G^\circ=-92.2+0.1987\,T (in kJ). The line crosses zero at 464\,\mathrm{K}: below it the reaction is spontaneous, above it is not.
\Delta G^\circ_{298}=-33.0\,\mathrm{kJ};\; \Delta G^\circ_{500}=+7.2\,\mathrm{kJ};\; T=464\,\mathrm{K}

6.7Enthalpy and calorie contents of food

The amount of energy in a food or drink is measured in calories. The calories in food refer to the amount of energy stored in the food, which is released through the metabolic processes in our body. The enthalpy change linked with these processes corresponds to the amount of energy released or absorbed during digestion and metabolism.

The energy content of food is measured in kilocalories, and is often just called calories. One kilocalorie (kcal) is approximately equal to 4.184\,\mathrm{kJ}. This conversion factor relates the energy released in a chemical reaction to the calorie content of food.

1\ \text{kcal}\approx4.184\,\mathrm{kJ}

Formula sheet

Heat of reaction as enthalpy change
\Delta H=H_\text{products}-H_\text{reactants}
\Delta H<0 for an exothermic reaction and \Delta H>0 for an endothermic reaction.
Units: \mathrm{kJ} or \mathrm{kJ\,mol^{-1}}
Enthalpy
H=E+PV
E is the internal energy, P the pressure and V the volume of the system.
Unit: \mathrm{kJ}
Calories and joules
1\ \text{cal}=4.18\ \mathrm{J},\qquad 1\ \text{kcal}\approx4.184\ \mathrm{kJ}
The calorie is the heat needed to raise 1\,\mathrm{g} of water from 14.5^\circ\mathrm{C} to 15.5^\circ\mathrm{C}.
Units: \mathrm{J}, \mathrm{kJ}
Enthalpy change from bond energies
\Delta H=\sum H_\text{(bonds broken)}-\sum H_\text{(bonds formed)}
Bond energies are positive numbers; breaking bonds absorbs energy, forming bonds releases it.
Unit: \mathrm{kJ}
Heat capacity
\text{heat capacity}=\dfrac{q}{\Delta T}
q is the heat absorbed and \Delta T the temperature change.
Unit: \mathrm{J\,K^{-1}}
Heat absorbed or released
q=m\,c\,\Delta T
m is the mass, c the specific heat capacity and \Delta T the change in temperature.
Units: \mathrm{J} (with c in \mathrm{J\,g^{-1}\,K^{-1}} and m in \mathrm{g})
Molar heat capacity
C_\text{molar}=c\times M
c is the specific heat capacity and M the molar mass.
Unit: \mathrm{J\,mol^{-1}\,K^{-1}}
Enthalpy change from calorimetry
\Delta H=-\dfrac{q}{n}=-\dfrac{m\,c\,\Delta T}{n}
n is the number of moles of the reaction (for neutralization, moles of water formed).
Unit: \mathrm{kJ\,mol^{-1}}
Hess's law
\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\cdots,\qquad \sum\Delta H(\text{cycle})=0
The enthalpy change is the same whether the reaction takes place in one step or in several steps.
Unit: \mathrm{kJ}
Enthalpy of reaction from enthalpies of formation
\Delta H^\circ=\sum\text{coeff}\,\Delta H^\circ_\mathrm{f}(\text{products})-\sum\text{coeff}\,\Delta H^\circ_\mathrm{f}(\text{reactants})
coeff is the coefficient in the balanced equation; \Delta H^\circ_\mathrm{f} of an element in its standard state is zero.
Unit: \mathrm{kJ}
Born–Haber cycle (NaCl type)
\Delta H^\circ_\mathrm{f}=\Delta H_\mathrm{s}+\Delta H_\mathrm{IE}+\Delta H_\mathrm{at}+\Delta H_\mathrm{EA}+\Delta H_\mathrm{l}
Sublimation, ionization energy, atomization, electron affinity and lattice energy; the lattice energy is found by difference.
Unit: \mathrm{kJ\,mol^{-1}}
Charge density
\text{charge density}=\dfrac{\text{ionic charge}}{\text{ionic radius}}
Larger charge density gives larger lattice energy and larger enthalpy of hydration.
charge per unit radius
Enthalpy of solution
\Delta H_\text{solution}=\Delta H_\text{reverse lattice}+\Delta H_\text{hydration}
\Delta H_\text{reverse lattice}=-\Delta H_\text{lattice}; \Delta H_\text{hydration} is the sum for the cation and the anion.
Unit: \mathrm{kJ\,mol^{-1}}
Entropy change of a reaction
\Delta S^\circ=\sum S^\circ_\text{products}-\sum S^\circ_\text{reactants}
Sum of the standard entropies, each times its coefficient.
Unit: \mathrm{J\,K^{-1}\,mol^{-1}}
Gibbs free energy
G=H-TS
H is the enthalpy, T the temperature in kelvin and S the entropy.
Unit: \mathrm{kJ}
Gibbs–Helmholtz equation
\Delta G^\circ=\Delta H^\circ-T\,\Delta S^\circ
\Delta G^\circ<0 spontaneous, \Delta G^\circ>0 non-spontaneous, \Delta G^\circ=0 equilibrium.
Unit: \mathrm{kJ}

Key terms

Energetics
The study of the energy changes associated with all types of physical and chemical changes.
Thermochemistry
The branch of chemistry which deals with the heat or thermal energy changes in chemical reactions.
Thermal energy
The energy in an object or system due to the movement of its molecules and atoms.
Calorie
The heat required to raise the temperature of one gram of water from 14.5^\circ\mathrm{C} to 15.5^\circ\mathrm{C}; 1\,\text{cal}=4.18\,\mathrm{J}.
Thermochemical equation
A balanced chemical equation which also shows the heat change of the reaction.
Exothermic reaction
A reaction that proceeds with the evolution of heat; \Delta H is negative.
Endothermic reaction
A reaction that proceeds with the absorption of heat; \Delta H is positive.
Heat of reaction
The heat evolved or absorbed in a reaction when the molar quantities of reactants and products are the same as in the chemical equation.
Enthalpy change, \Delta H
The heat of reaction at constant pressure: H_\text{products}-H_\text{reactants}.
Thermodynamics
The study of all types of energy changes associated with chemical and physical changes.
State function
A property determined by the state of the system regardless of how that state was reached.
Internal energy
The sum of all kinds of energies of the particles of a system.
Enthalpy, H
The internal energy of a system plus the product of its pressure and volume: H=E+PV.
Standard states
1 atm for a gas; the most stable physical state at 1\,\mathrm{atm} and 25^\circ\mathrm{C} for an element or compound; 1\,\mathrm{M} for a substance in aqueous solution.
Standard enthalpy of reaction
The enthalpy change when reactants and products are in their standard states and in the molar quantities of the balanced equation.
Standard enthalpy of formation
The enthalpy change when one mole of a compound is formed from its elements, all in their standard states.
Standard enthalpy of combustion
The enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions.
Standard enthalpy of atomization
The enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions.
Standard enthalpy of neutralization
The heat evolved when one mole of \ce{H+} from an acid combines with one mole of \ce{OH-} from a base to form one mole of water.
Standard enthalpy of solution
The enthalpy change when one mole of a substance is dissolved in so much solvent that further dilution gives no detectable heat change.
First electron affinity
The energy released when one mole of gaseous atoms gains one electron each to form gaseous ions with a single negative charge.
Lattice energy
The energy released when one mole of an ionic solid is formed from its gaseous ions.
Bond dissociation energy
The energy required to break one mole of a particular bond to form neutral atoms.
Bond energy
The energy released when one mole of a particular bond forms from neutral atoms.
Heat capacity
The heat required to raise the temperature of a given amount of a substance by 1 kelvin (\mathrm{J\,K^{-1}}).
Specific heat capacity
The heat required to raise the temperature of one gram of a substance by 1 kelvin (\mathrm{J\,g^{-1}\,K^{-1}}).
Calorimeter
A device that measures heat flow, at constant pressure or at constant volume.
Hess's law
The enthalpy change in a chemical reaction is the same whether the reaction takes place in one step or in several steps.
Born–Haber cycle
An application of Hess's law to ionic compounds that is used to find lattice energies.
Charge density
The ratio of the ionic charge to the ionic radius of an ion.
Enthalpy of hydration
The energy released when gaseous ions bind to water molecules.
Entropy, S
The degree of disorder and randomness of a system; the number of possible arrangements of its particles and their energy.
Gibbs free energy, G
The usable energy of a system at constant temperature and pressure: G=H-TS.
Spontaneous process
A process with \Delta G^\circ<0, which goes ahead by itself in the forward direction.

Multiple choice questionsp. 140–141

i
Which of the following substances have zero value for their standard enthalpy of formation?
  1. A\ce{O3}
  2. B\ce{H2O}
  3. C\ce{ZnO}
  4. DNone of these.Answer
Why

Only an element in its standard state has a standard enthalpy of formation of zero. The standard state of oxygen is \ce{O2(g)}, so \ce{O3}, a different form of oxygen, is not zero.

\ce{H2O} and \ce{ZnO} are compounds, not elements, so their enthalpies of formation are not zero (Table 6.1 gives -285.8\,\mathrm{kJ\,mol^{-1}} for liquid water).

Note. The book prints no key for this question; the answer was worked out.

ii
Calorie is equivalent to:
  1. A4.18\,\mathrm{J}Answer
  2. B4.18\,\mathrm{kJ}
  3. C0.418\,\mathrm{J}
  4. D0.418\,\mathrm{kJ}
Why

The book defines the calorie as the heat needed to raise the temperature of 1 g of water from 14.5^\circ\mathrm{C} to 15.5^\circ\mathrm{C}, and 1\,\mathrm{calorie}=4.18\,\mathrm{J}.

Option B is 1000 times too large, option D is 100 times too large and option C is 10 times too small.

Note. The book prints no key for this question; the answer was worked out.

iii
Enthalpy of neutralization of all the strong acids and strong bases has the same value due to:
  1. AThe formation of salt and water
  2. BThe formation of salt
  3. CThe complete ionization of acids and bases
  4. DThe combination of \ce{H+} and \ce{OH-} ionsAnswer
Why

Strong acids and strong bases are fully ionized, so in every case the only real change is \ce{H+(aq) + OH-(aq) -> H2O(l)}. The spectator ions, such as \ce{Na+} and \ce{Cl-}, do not change, and the heat comes from the formation of water: \Delta H^\circ_n=-57.4\,\mathrm{kJ\,mol^{-1}} in each case, as the book states.

Options A and B are wrong because the salt depends on the acid and base used. Option C is true of strong acids and bases, but complete ionization is not the source of the constant heat value.

Note. The book prints no key for this question; the answer was worked out.

iv
Total heat content of a system is called:
  1. AEnthalpyAnswer
  2. BInternal energy
  3. CHeat
  4. DState function
Why

The book defines enthalpy as H=E+PV, the internal energy plus the product of pressure and volume. It is the total heat content of the system.

Note. The book prints no key for this question; the answer was worked out.

v
Heat of ______ of a substance is always negative.
  1. AFormation
  2. BCombustionAnswer
  3. CDecomposition
  4. DSolution
Why

Combustion is always exothermic, so the standard enthalpy of combustion is always negative (for example -393.5\,\mathrm{kJ\,mol^{-1}} for carbon).

Formation, decomposition and solution can be positive: benzene has \Delta H^\circ_f=+49.0\,\mathrm{kJ\,mol^{-1}} (Table 6.1), and the solution of \ce{NH4Cl} has \Delta H^\circ_\text{sol}=+15.1\,\mathrm{kJ\,mol^{-1}}.

Note. The book prints no key for this question; the answer was worked out.

vi
A balloon filled with oxygen is placed in a freezer. Identify the system:
  1. ABalloon
  2. BOxygenAnswer
  3. CFreezer
  4. DAll of these
Why

The system is the part of the universe under study, here the oxygen gas. The balloon wall is the boundary between the system and its surroundings, and the freezer is part of the surroundings, so options A, C and D are wrong.

Note. The book prints no key for this question; the answer was worked out.

vii
A bomb calorimeter is used in ______ calorimetry.
  1. AConstant volumeAnswer
  2. BConstant pressure
  3. CBoth a and b
  4. DConstant temperature
Why

A bomb calorimeter is a sealed, rigid vessel, so the reaction takes place at constant volume and the heat measured is at constant volume.

The coffee-cup calorimeter is the constant-pressure type (section 6.3.1), so option B is not the bomb's use and option C is wrong. Constant temperature is not a type of calorimetry in the book.

Note. The book prints no key for this question; the answer was worked out.

viii
Born Haber cycle is used to determine lattice energies of:
  1. AMolecular solids
  2. BIonic solidsAnswer
  3. CCovalent solids
  4. DMetallic solids
Why

Lattice energy is the energy released when gaseous ions form one mole of an ionic solid. The Born–Haber cycle is a special application of Hess's law to binary ionic compounds, so it gives lattice energies of ionic solids.

Note. The book prints no key for this question; the answer was worked out.

ix
Enthalpy of combustion for C is -393.5\,\mathrm{kJ\,mol^{-1}}. Enthalpy of formation of \ce{CO2} would be:
\ce{C(s) + O2(g) -> CO2(g)}\qquad \Delta H^\circ_\text{combustion}=-393.5\,\mathrm{kJ\,mol^{-1}}
  1. A+393.5\,\mathrm{kJ}
  2. B-393.5\,\mathrm{kJ}Answer
  3. CZero
  4. DCannot be predicted from the given equation.
Why

The equation \ce{C(s) + O2(g) -> CO2(g)} forms one mole of \ce{CO2} from its elements, both in their standard states, so it is also the formation reaction of \ce{CO2}.

Therefore \Delta H^\circ_f(\ce{CO2})=-393.5\,\mathrm{kJ\,mol^{-1}}. Option A has the wrong sign, and options C and D are wrong because the value is known.

Note. The book prints no key for this question; the answer was worked out.

x
Which of the following is not a state function of a system?
  1. AThermal energy at constant pressure
  2. BEnthalpy
  3. CInternal energy
  4. DWork doneAnswer
Why

A state function depends only on the state of the system, not on the path taken. Enthalpy and internal energy are state functions, and the heat at constant pressure equals the enthalpy change, so option A is also a state function.

Work done depends on the path followed (for example, how the pressure changes during an expansion), so it is not a state function.

Note. The book prints no key for this question; the answer was worked out.

xi
For writing a thermochemical equation for enthalpy of combustion of an element requires: (a) 1 mole of element as reactant, (b) 1 mole of oxide of element as product, (c) standard states of all the substances, (d) balanced equation of 1 mole of element.
  1. A(a) only
  2. B(a) and (b)
  3. C(a), (c) and (d)
  4. D(a), (b), (c) and (d)Answer
Why

The book's method for the enthalpy of combustion (section 6.1.6) is: write 1 mole of the element and oxygen as reactants, write the oxides of the element as products, show the standard states of all substances and balance the atoms.

Each of (a) to (d) is part of that method, so all four are required.

Note. The book prints its options as numbers 1 to 4 (a, b, c, d combinations); they are relabelled here as A to D. The book prints no key for this question; the answer was worked out.

Short questionsp. 141–143

23 marks
Name and define units of thermal energy.
Solution

The SI unit of heat, and of all forms of energy, is the joule (J). Thermal energy is the energy in an object or system due to the movement of its molecules and atoms.

The calorie is the other common unit of heat. One calorie is the heat (thermal energy) needed to raise the temperature of 1 g of water from 14.5^\circ\mathrm{C} to 15.5^\circ\mathrm{C}.

1\,\mathrm{calorie}=4.18\,\mathrm{J}\qquad 1\,\mathrm{kilocalorie}\approx4.184\,\mathrm{kJ}

The kilocalorie is the unit in which the energy content of food is measured (section 6.7).

33 marks
Define the terms system, surroundings, boundary, state function, heat capacity, internal energy, enthalpy of a substance.
Solution
System
The part of the universe chosen for study, such as the reactants and products in a reaction.
Surroundings
Everything else in the universe that can exchange energy with the system.
Boundary
The real or imaginary surface that separates the system from its surroundings.
State function
A property that depends only on the present state of the system (its temperature, pressure, volume and amount), not on how that state was reached. Enthalpy and internal energy are state functions.
Heat capacity
The heat required to raise the temperature of a given amount of a substance by 1 K, in \mathrm{J\,K^{-1}}. The heat capacity of 1 g of a substance is its specific heat capacity.
Internal energy
The sum of all kinds of energy of the particles of the system.
Enthalpy of a substance
H=E+PV: the internal energy of the system plus the product of its pressure and volume. The enthalpy change of a reaction, \Delta H=H_\text{products}-H_\text{reactants}, is the heat exchanged at constant pressure.
43 marks
Classify reactions as exothermic or endothermic.
Solution

A reaction is classified by whether heat is released or absorbed, which is shown by the sign of its enthalpy change at constant pressure, \Delta H^\circ.

Exothermic reactionEndothermic reaction
Heat is evolved: the system transfers energy to the surroundings.Heat is absorbed: heat is transferred from the surroundings to the system.
\Delta H^\circ is negative.\Delta H^\circ is positive.
Burning of fuels, for example \ce{C(s) + O2(g) -> CO2(g)}, \Delta H^\circ=-393.5\,\mathrm{kJ}.For example \ce{H2(g) + I2(g) -> 2HI(g)}, \Delta H^\circ=+53.8\,\mathrm{kJ}; and \ce{C(s) + H2O(g) -> CO(g) + H2(g)}, \Delta H^\circ=+131.4\,\mathrm{kJ}.

Reversing a reaction changes the sign of \Delta H^\circ but not its size, so a reaction that is exothermic in one direction is endothermic in the reverse direction.

53 marks
Define bond dissociation energy.
Solution

The bond dissociation energy is the energy required to break one mole of a particular bond so as to form neutral atoms.

\ce{H2(g) -> 2H(g)}\qquad \Delta H=+436\,\mathrm{kJ\,mol^{-1}}

Breaking bonds always absorbs energy, so bond dissociation energy is positive. Bond energy is the energy released when one mole of neutral atoms forms the bond, and it has the same magnitude.

\Delta H=\Sigma H_\text{(bonds broken)}-\Sigma H_\text{(bonds formed)}
143 marks
Write the balanced equation for the formation reaction of each of the following substances.
Solution
(a) Butanol, \ce{C4H9OH}.

Write the elements as reactants and one mole of butanol, a liquid, as product. Four carbon atoms, ten hydrogen atoms and one oxygen atom must be balanced.

\ce{4C(s) + 5H2(g) + 1/2 O2(g) -> C4H9OH(l)}
(b) Rust, \ce{Fe3O4}.

The book gives the formula \ce{Fe3O4} for rust, so the equation is balanced from three iron atoms and four oxygen atoms.

\ce{3Fe(s) + 2O2(g) -> Fe3O4(s)}
(c) Acetic acid, \ce{CH3CO2H}.

Acetic acid is a liquid under standard conditions. Two carbon, four hydrogen and two oxygen atoms are needed on the product side.

\ce{2C(s) + 2H2(g) + O2(g) -> CH3COOH(l)}
(d) Urea, \ce{(NH2)2CO}.

Urea is a solid. It needs two nitrogen atoms, four hydrogen atoms, one carbon atom and one oxygen atom, so the reactants are \ce{N2}, \ce{H2}, \ce{C} and half a mole of \ce{O2}.

\ce{N2(g) + 2H2(g) + C(s) + 1/2 O2(g) -> CO(NH2)2(s)}
203 marks
Arrange the following compounds in order of increasing entropy. Justify your order: \ce{CH3OH(s)}, \ce{CH3OH(g)}, \ce{CH3OH(l)}.
Solution

Entropy measures the disorder of a system and the number of ways its particles can be arranged. The particles of a solid have a fixed, ordered arrangement and the least energy. In a liquid they move more freely, and in a gas they move freely through the whole container, so a gas has the most arrangements.

\ce{CH3OH(s)} < \ce{CH3OH(l)} < \ce{CH3OH(g)}

Melting increases the entropy (\Delta S>0), as the book shows for ice melting, and vaporization increases it further. Any change to a less ordered state gives a positive entropy change, so the order of increasing entropy is solid, liquid, gas.

Long questionsp. 143

196 marks
Develop a hypothetical reaction and calculate the Gibbs free energy. Discuss conditions under which the reaction would be spontaneous or non-spontaneous.
Solution

Hypothetical reaction

Take the gas-phase reaction \ce{A(g) + B(g) -> C(g)} with \Delta H^\circ=-40\,\mathrm{kJ\,mol^{-1}} and \Delta S^\circ=-100\,\mathrm{J\,K^{-1}\,mol^{-1}}, which is -0.100\,\mathrm{kJ\,K^{-1}\,mol^{-1}}. Two moles of gas become one, so a fall in entropy is expected.

Gibbs free energy at 298 K

  1. Gibbs–Helmholtz equation at constant temperature.
    \Delta G^\circ=\Delta H^\circ-T\,\Delta S^\circ
  2. Substitute T=298\,\mathrm{K} and express both terms in kJ.
    \Delta G^\circ=-40\,\mathrm{kJ\,mol^{-1}}-(298\,\mathrm{K})(-0.100\,\mathrm{kJ\,K^{-1}\,mol^{-1}})
  3. Multiply out and add.
    \Delta G^\circ=-40+29.8=-10.2\,\mathrm{kJ\,mol^{-1}}
\Delta G^\circ_{298}=-10.2\,\mathrm{kJ\,mol^{-1}}

The value is negative, so the reaction is spontaneous in the forward direction at 298 K.

Conditions for spontaneity

A reaction is spontaneous when \Delta G^\circ is negative, non-spontaneous when it is positive, and at equilibrium when it is zero. Because the entropy term is multiplied by the temperature, the sign of \Delta G^\circ depends on temperature. The reaction reaches equilibrium where \Delta G^\circ=0:

T=\dfrac{\Delta H^\circ}{\Delta S^\circ}=\dfrac{-40\,\mathrm{kJ\,mol^{-1}}}{-0.100\,\mathrm{kJ\,K^{-1}\,mol^{-1}}}=400\,\mathrm{K}
Gibbs free energy against temperature
\Delta G^\circ in \mathrm{kJ\,mol^{-1}} is negative (spontaneous) below 400 K and positive (non-spontaneous) above it.
Sign of \Delta H^\circSign of \Delta S^\circBehaviour
NegativePositiveSpontaneous at all temperatures
NegativeNegativeSpontaneous below T=\Delta H^\circ/\Delta S^\circ (the hypothetical reaction: below 400 K)
PositivePositiveSpontaneous only above T=\Delta H^\circ/\Delta S^\circ, for example melting of ice above its melting point
PositiveNegativeNever spontaneous

So the reaction is spontaneous at 298 K and below 400 K, and it becomes non-spontaneous above 400 K, where the entropy term outweighs the enthalpy term.

Note. The reaction is a hypothetical example chosen for this answer; the book prints no key.

218 marks
Strontium fluoride is an ionic solid. It dissolves in water and can be used as an optical coating for lenses.
Solution
a. Draw Born–Haber cycle for the enthalpy change of solution of \ce{SrF2(s)}.

The cycle connects the solid, the gaseous ions and the aqueous ions. The solid is broken into gaseous ions (the reverse of lattice energy), the gaseous ions are hydrated, and the enthalpy of solution is the direct route from the solid to the aqueous ions.

Energy cycle for the solution of \ce{SrF2(s)}, in the form of Figure 6.6: the reverse lattice enthalpy plus the hydration enthalpy equals the enthalpy of solution.
b. Label each enthalpy change.
  • \Delta H^\circ_\text{solution}: from \ce{SrF2(s)} directly to the aqueous ions, calculated in part c.
  • \Delta H^\circ_\text{reverse lattice}: the reverse of the lattice energy, from \ce{SrF2(s)} up to the gaseous ions, +2492\,\mathrm{kJ\,mol^{-1}}.
  • \Delta H^\circ_\text{hydration}: from the gaseous ions to the hydrated ions, \Delta H^\circ_\text{hyd}(\ce{Sr^2+})+2\,\Delta H^\circ_\text{hyd}(\ce{F-})=-2492\,\mathrm{kJ\,mol^{-1}}.
c. Calculate the enthalpy change of solution of \ce{SrF2}, using the following data.
Enthalpy change\Delta H^\circ, kJ mol^{-1}
Lattice energy of \ce{SrF2}-2492
Enthalpy of hydration for \ce{Sr^2+} ion-1480
Enthalpy of hydration for \ce{F-} ion-506
  1. Hess's law: the enthalpy of solution is the reverse lattice enthalpy plus the hydration enthalpies of the ions. The fluoride value is counted twice because there are two \ce{F-} ions.
    \Delta H^\circ_\text{solution}=\Delta H^\circ_\text{reverse lattice}+\Delta H^\circ_\text{hyd}(\ce{Sr^2+})+2\,\Delta H^\circ_\text{hyd}(\ce{F-})
  2. Reverse the sign of the lattice energy.
    \Delta H^\circ_\text{reverse lattice}=-(-2492)=+2492\,\mathrm{kJ\,mol^{-1}}
  3. Substitute the values.
    \Delta H^\circ_\text{solution}=+2492+(-1480)+2(-506)
  4. Add the terms.
    =+2492-1480-1012=0\,\mathrm{kJ\,mol^{-1}}

The hydration enthalpies of the two ions (total -2492\,\mathrm{kJ\,mol^{-1}}) exactly cancel the lattice energy, so the enthalpy of solution is zero under these data.

\Delta H^\circ_\text{solution}=0\,\mathrm{kJ\,mol^{-1}}
d. Write the names of the factors that affect enthalpy of hydration.

The magnitude of the enthalpy of hydration depends on the charge density of the ion, which is the ionic charge divided by the ionic radius. The two factors are:

  1. Ionic charge: a higher charge attracts water molecules more strongly and makes the enthalpy of hydration more exothermic (more negative).
  2. Ionic radius: a smaller ion has a higher charge density, attracts water molecules more strongly and makes the enthalpy of hydration more exothermic.
e. Differentiate enthalpy of hydration from lattice energy.
Lattice energyEnthalpy of hydration
Energy released when one mole of an ionic solid forms from its gaseous ions: \ce{M^{n+}(g) + X^{n-}(g) -> MX(s)}.Energy released when gaseous ions are surrounded by water molecules: \ce{M^{n+}(g) ->[H2O] M^{n+}(aq)}.
Concerns the solid: the ions are packed into a crystal.Concerns the solution: the ions are bonded to water molecules (ion–dipole attraction).
For \ce{SrF2} it is -2492\,\mathrm{kJ\,mol^{-1}}.For \ce{Sr^2+} it is -1480\,\mathrm{kJ\,mol^{-1}} and for \ce{F-} it is -506\,\mathrm{kJ\,mol^{-1}}.

Lattice energy is the energy to build the solid from ions, while hydration is the energy released when the separated ions are solvated. The enthalpy of solution depends on both, as shown in part c.

Note. Part a asks for a Born–Haber cycle, but the cycle that links the solid, the gaseous ions and the hydrated ions for a solution is the energy cycle of Figure 6.6, which is what is drawn. The book prints no answers for this question.

Numerical problemsp. 141–142

Method: enthalpies of reaction from formation or combustion data with Hess's law, and q=mc\,\Delta T for calorimetry.

63 marks
When ethanol burns in oxygen, carbon dioxide and water are formed.
Solution
(a) Write the equation which describes this reaction.

Ethanol is a liquid under standard conditions. Balance the carbon, hydrogen and oxygen atoms: two carbon atoms, six hydrogen atoms and seven oxygen atoms on the reactant side.

\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)}
(b) Using the following data, calculate the enthalpy of combustion for ethanol, \ce{C2H5OH}.
  • \Delta H^\circ_f for ethanol (l) =-277.0\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_f for \ce{CO2(g)}=-393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_f for \ce{H2O(l)}=-285.8\,\mathrm{kJ\,mol^{-1}}
Given
  • \Delta H^\circ_f(\ce{C2H5OH,l}) = -277.0\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_f(\ce{CO2,g}) = -393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_f(\ce{H2O,l}) = -285.8\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_f(\ce{O2,g}) = 0\,\mathrm{kJ\,mol^{-1}} element in its standard state
To find
  • \Delta H^\circ_c(\ce{C2H5OH})
  1. Enthalpy change of reaction from formation enthalpies: products minus reactants, each multiplied by its coefficient.
    \Delta H^\circ_c=\sum\text{coeff}\,\Delta H^\circ_f(\text{products})-\sum\text{coeff}\,\Delta H^\circ_f(\text{reactants})
  2. Two \ce{CO2} and three \ce{H2O} are products; one ethanol and three \ce{O2} are reactants.
    \Delta H^\circ_c=[2(-393.5)+3(-285.8)]-[(-277.0)+3(0)]
  3. Multiply out the products.
    =-787.0-857.4-(-277.0)
  4. Add and subtract.
    \Delta H^\circ_c=-1644.4+277.0=-1367.4\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_c(\ce{C2H5OH})=-1367.4\,\mathrm{kJ\,mol^{-1}}

Note. The book prints no key for this question. Table 6.1 lists ethanol (l) as -277.7\,\mathrm{kJ\,mol^{-1}}, which would give -1366.7\,\mathrm{kJ\,mol^{-1}}; the question's -277.0 gives -1367.4, in line with Table 6.2 (-1367).

73 marks
Calculate from the data in table 11.1 and 11.2, the enthalpy changes of the following reaction at standard conditions.
\ce{C6H6(l) -> 6C(s) + 3H2(g)}
Solution
Given
  • \Delta H^\circ_c(\ce{C6H6,l}) = -3268\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\text{C, graphite}) = -393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{H2,g}) = -285.8\,\mathrm{kJ\,mol^{-1}}
To find
  • \Delta H^\circ \text{ of } \ce{C6H6(l) -> 6C(s) + 3H2(g)}
  1. Each substance is burnt to the same products, \ce{CO2} and \ce{H2O}. By Hess's law the reaction enthalpy is the combustion enthalpies of the reactants minus those of the products.
    \Delta H^\circ=\sum\Delta H^\circ_c(\text{reactants})-\sum\text{coeff}\,\Delta H^\circ_c(\text{products})
  2. Reactant: one mole of liquid benzene.
    \Delta H^\circ_c(\ce{C6H6})=-3268\,\mathrm{kJ\,mol^{-1}}
  3. Products: six moles of carbon and three moles of hydrogen.
    6(-393.5)+3(-285.8)=-2361.0-857.4=-3218.4\,\mathrm{kJ}
  4. Subtract.
    \Delta H^\circ=-3268-(-3218.4)=-49.6\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ=-49.6\,\mathrm{kJ\,mol^{-1}}

Book answer. -49.6 kJ/mole

Note. The question cites 'table 11.1 and 11.2'; the chapter's tables are 6.1 (formation) and 6.2 (combustion). The book's answer uses the combustion data. With the formation data of Table 6.1 (benzene +49.0\,\mathrm{kJ\,mol^{-1}}) the result would be -49.0\,\mathrm{kJ\,mol^{-1}}.

83 marks
The heat of combustion of liquid benzene, \ce{C6H6}, to form \ce{H2O} and \ce{CO2} at 1 atm and 25^\circ\mathrm{C} is -3268\,\mathrm{kJ\,mol^{-1}} of benzene. What is the heat of formation of liquid benzene under these conditions?
Solution
Given
  • \Delta H^\circ_c(\ce{C6H6,l}) = -3268\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\text{C, graphite}) = -393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{H2,g}) = -285.8\,\mathrm{kJ\,mol^{-1}}
To find
  • \Delta H^\circ_f(\ce{C6H6,l})
  1. Formation reaction: six moles of carbon and three moles of hydrogen form one mole of liquid benzene.
    \ce{6C(s) + 3H2(g) -> C6H6(l)}
  2. Combustion route: combustion enthalpies of the reactants minus that of the product.
    \Delta H^\circ_f=[6(-393.5)+3(-285.8)]-(-3268)
  3. Evaluate.
    =-3218.4+3268=+49.6\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_f(\ce{C6H6,l})=+49.6\,\mathrm{kJ\,mol^{-1}}

Book answer. +49.6 kJ/mole

Note. Table 6.1 lists +49.0\,\mathrm{kJ\,mol^{-1}} for liquid benzene. The book's answer +49.6 comes from the combustion data (Table 6.2), as in the previous question.

104 marks
Camphor (\ce{C10H16O}) has a heat of combustion of 5903.6\,\mathrm{kJ\,mol^{-1}}. A sample of camphor having mass 0.1204\,\mathrm{g} is burned in a bomb calorimeter. The temperature increases by 2.28^\circ\mathrm{C}. Calculate the heat capacity of the calorimeter.
Solution
Given
  • m = 0.1204\,\mathrm{g}
  • M_{\ce{C10H16O}} = 152\,\mathrm{g\,mol^{-1}} from C 12, H 1, O 16
  • \Delta H^\circ_c = -5903.6\,\mathrm{kJ\,mol^{-1}}
  • \Delta T = 2.28\,\mathrm{K}
To find
  • C_\text{cal}, the heat capacity of the calorimeter
  1. Moles of camphor burned.
    n=\dfrac{m}{M}=\dfrac{0.1204}{152}=7.92\times10^{-4}\,\mathrm{mol}
  2. Heat released by burning this sample, which the calorimeter absorbs.
    q=n\,|\Delta H^\circ_c|=7.92\times10^{-4}\times5903.6=4.676\,\mathrm{kJ}
  3. Heat capacity is the heat absorbed for each kelvin rise, C=q/\Delta T.
    C_\text{cal}=\dfrac{4.676\,\mathrm{kJ}}{2.28\,\mathrm{K}}=2.051\,\mathrm{kJ\,K^{-1}}
C_\text{cal}\approx2.05\,\mathrm{kJ\,K^{-1}}

Book answer. 2.04531 kJ/K

Note. With a molar mass of 152\,\mathrm{g\,mol^{-1}} the value is 2.051\,\mathrm{kJ\,K^{-1}}. The book's 2.04531\,\mathrm{kJ\,K^{-1}} corresponds to a molar mass of about 152.4\,\mathrm{g\,mol^{-1}}, so the difference is rounding of the molar mass. The method is the same.

113 marks
In a coffee-cup calorimeter 100\,\mathrm{cm^3} of 1.0\,\mathrm{M} \ce{HCl} and 100\,\mathrm{cm^3} of 1.0\,\mathrm{M} \ce{NaOH} are mixed at 24.6^\circ\mathrm{C} and the temperature rises by 6.9^\circ\mathrm{C}. Calculate the enthalpy of neutralization of \ce{HCl} by \ce{NaOH} from the given data. Heat capacity of water is 4.18\,\mathrm{J\,g^{-1}\,K^{-1}}.
Solution
Given
  • V = 100\,\mathrm{cm^3}\text{ of }1.0\,\mathrm{mol\,dm^{-3}}\text{ HCl and NaOH each}
  • c = 4.18\,\mathrm{J\,g^{-1}\,K^{-1}}
  • \Delta T = 6.9\,\mathrm{K}
To find
  • \Delta H^\circ_n
  1. Moles: the reaction is 1:1, so \ce{HCl} and \ce{NaOH} are both 0.100 mol, and 0.100 mol of water forms.
    n=1.0\,\mathrm{mol\,dm^{-3}}\times0.100\,\mathrm{dm^3}=0.100\,\mathrm{mol}
  2. Mass of the mixture: 200 cm^3 at 1 g cm^{-3}.
    m=200\,\mathrm{cm^3}\times1\,\mathrm{g\,cm^{-3}}=200\,\mathrm{g}
  3. Heat absorbed by the solution.
    q=mc\,\Delta T=200\times4.18\times6.9=5768.4\,\mathrm{J}
  4. The reaction gives out this heat, so its sign is negative. Divide by the moles of water formed.
    \Delta H^\circ_n=-\dfrac{5768.4\,\mathrm{J}}{0.100\,\mathrm{mol}}=-57\,684\,\mathrm{J\,mol^{-1}}
\Delta H^\circ_n=-57.684\,\mathrm{kJ\,mol^{-1}}

Book answer. -57.684 kJ/mole

Note. Density of the mixture is taken as 1\,\mathrm{g\,cm^{-3}}, as in the book's Example 6.4; the question does not state it.

122026 Annual I, Set B, Q.2 (iv)3 marks
Calculate \Delta H^\circ for the reaction \ce{S(s) + O2(g) -> SO2(g)} from the following data.
\ce{S(s) + 3/2 O2(g) -> SO3(g)}\qquad \Delta H^\circ=-395.2\,\mathrm{kJ}
\ce{2SO2(g) + O2(g) -> 2SO3(g)}\qquad \Delta H^\circ=-198.2\,\mathrm{kJ}
Solution
  1. The first equation already has \ce{S} and \ce{O2} as reactants and \ce{SO3} as product, so it is kept as written.
    \ce{S(s) + 3/2 O2(g) -> SO3(g)}\qquad \Delta H^\circ_1=-395.2\,\mathrm{kJ}
  2. Halve the second equation (its enthalpy halves too). It has \ce{SO2} as reactant, which the target needs as product.
    \ce{SO2(g) + 1/2 O2(g) -> SO3(g)}\qquad \Delta H^\circ=-99.1\,\mathrm{kJ}
  3. Reverse the halved equation to remove \ce{SO3} from the products; the sign of \Delta H^\circ changes.
    \ce{SO3(g) -> SO2(g) + 1/2 O2(g)}\qquad \Delta H^\circ=+99.1\,\mathrm{kJ}
  4. Add the first equation to the reversed equation. \ce{SO3} cancels, and \tfrac12\ce{O2} cancels half of the \ce{O2} needed.
    \ce{S(s) + O2(g) -> SO2(g)}\qquad \Delta H^\circ=-395.2+99.1=-296.1\,\mathrm{kJ}
\Delta H^\circ=-296.1\,\mathrm{kJ\,mol^{-1}}

Note. The book prints no key. The second equation is printed as 2\,\ce{SO2(g)O2(g)} with the plus sign missing; it is read as 2\ce{SO2(g)} + \ce{O2(g)}. The result is close to the formation enthalpy of \ce{SO2} in Table 6.1 (-296.8\,\mathrm{kJ\,mol^{-1}}).

133 marks
An aluminium frying pan weighs 745\,\mathrm{g} and is heated on a stove from 25^\circ\mathrm{C} to 205^\circ\mathrm{C}. What is q for the frying pan? C_P for Al is 24.35\,\mathrm{J\,mol^{-1}\,K^{-1}}.
Solution
Given
  • m = 745\,\mathrm{g}
  • M_\ce{Al} = 27\,\mathrm{g\,mol^{-1}}
  • C_P = 24.35\,\mathrm{J\,mol^{-1}\,K^{-1}}
  • \Delta T = 205-25=180\,\mathrm{K}
To find
  • q
  1. Moles of aluminium in the pan.
    n=\dfrac{745}{27}=27.5926\,\mathrm{mol}
  2. Temperature change (a change of 1 °C equals a change of 1 K).
    \Delta T=205-25=180\,\mathrm{K}
  3. Heat absorbed, q=n\,C_P\,\Delta T.
    q=27.5926\times24.35\times180=120\,938\,\mathrm{J}
q=120\,938\,\mathrm{J}\approx1.21\times10^{5}\,\mathrm{J}

Book answer. 120938.3 J

Note. The book's answer is reproduced. The molar heat capacity is used with the number of moles, as the question gives it.

154 marks
The human body burns glucose for energy. Burning 1.0 g of glucose produces 15.65 kJ of heat.
Solution
(a) Write the balanced equation for the combustion of glucose.
\ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(l)}
(b) Determine the molar heat of combustion of glucose.
Given
  • q = 15.65\,\mathrm{kJ\ per\ 1.0\,g}
  • M_{\ce{C6H12O6}} = 6(12)+12(1)+6(16)=180\,\mathrm{g\,mol^{-1}}
To find
  • \Delta H^\circ_c
  1. Scale the heat from 1.0 g to one mole (180 g). Burning releases heat, so the sign is negative.
    \Delta H^\circ_c=-\dfrac{15.65\,\mathrm{kJ}}{1.0\,\mathrm{g}}\times180\,\mathrm{g\,mol^{-1}}
  2. Evaluate.
    \Delta H^\circ_c=-2817\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_c=-2817\,\mathrm{kJ\,mol^{-1}}
(c) Heats of combustions of C and \ce{H2} are -393.5\,\mathrm{kJ\,mol^{-1}} and -285.8\,\mathrm{kJ\,mol^{-1}} respectively. Determine the heat of formation of glucose.
Given
  • \Delta H^\circ_c(\text{C}) = -393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{H2}) = -285.8\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{C6H12O6}) = -2817\,\mathrm{kJ\,mol^{-1}} from part b
To find
  • \Delta H^\circ_f(\ce{C6H12O6})
  1. Formation reaction: six moles of carbon and six moles of hydrogen, with three moles of oxygen, give one mole of glucose.
    \ce{6C(s) + 6H2(g) + 3O2(g) -> C6H12O6(s)}
  2. Combustion route: combustion enthalpies of the reactants minus that of glucose.
    \Delta H^\circ_f=[6(-393.5)+6(-285.8)]-(-2817)
  3. Evaluate.
    =-2361.0-1714.8+2817=-1258.8\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_f(\ce{C6H12O6})=-1258.8\,\mathrm{kJ\,mol^{-1}}

Book answer. (b) -2817 kJ/mole; (c) -1258.8 kJ/mole

162025 Annual I, Set D, Q.6 OR3 marks
The standard combustion enthalpies of carbon, hydrogen and acetic acid are -393.5\,\mathrm{kJ\,mol^{-1}}, -285.8\,\mathrm{kJ\,mol^{-1}} and -875\,\mathrm{kJ\,mol^{-1}} respectively. Deduce the value of the standard enthalpy of formation of acetic acid, \ce{CH3COOH}.
Solution
Given
  • \Delta H^\circ_c(\text{C}) = -393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{H2}) = -285.8\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{CH3COOH,l}) = -875\,\mathrm{kJ\,mol^{-1}}
To find
  • \Delta H^\circ_f(\ce{CH3COOH})
  1. Formation reaction: two carbon atoms, two moles of hydrogen and one mole of oxygen form one mole of acetic acid.
    \ce{2C(s) + 2H2(g) + O2(g) -> CH3COOH(l)}
  2. Combustion route: combustion enthalpies of the reactants minus that of the product.
    \Delta H^\circ_f=[2(-393.5)+2(-285.8)]-(-875)
  3. Evaluate.
    =-1358.6+875=-483.6\,\mathrm{kJ\,mol^{-1}}
\Delta H^\circ_f(\ce{CH3COOH})=-483.6\,\mathrm{kJ\,mol^{-1}}

Book answer. -483.6 kJ/mole

174 marks
Is the conversion of magnetite, \ce{Fe3O4}, to hematite, \ce{Fe2O3}, by oxygen endothermic or exothermic? Justify your answer.
\ce{3Fe(s) + 2O2(g) -> Fe3O4(s)}\qquad \Delta H^\circ=-1118\,\mathrm{kJ}
\ce{2Fe(s) + 3/2 O2(g) -> Fe2O3(s)}\qquad \Delta H^\circ=-824.2\,\mathrm{kJ}
Solution
  1. Balance the conversion: two moles of magnetite and half a mole of oxygen give three moles of hematite.
    \ce{2Fe3O4(s) + 1/2 O2(g) -> 3Fe2O3(s)}
  2. Multiply the hematite equation by 3, and reverse the magnetite equation and multiply it by 2 (the sign of its \Delta H^\circ changes).
    3(-824.2)=-2472.6\,\mathrm{kJ}\qquad -2(-1118)=+2236\,\mathrm{kJ}
  3. Add the two equations. Iron atoms cancel, and the oxygen left over is \tfrac12\ce{O2}.
    \begin{aligned}&\ce{2Fe3O4(s) + 1/2 O2(g) -> 3Fe2O3(s)}\\ &\Delta H^\circ=-2472.6+2236=-236.6\,\mathrm{kJ}\end{aligned}
  4. Per mole of magnetite the change is -236.6/2=-118.3\,\mathrm{kJ}. The enthalpy change is negative, so heat is released.
\Delta H^\circ=-236.6\,\mathrm{kJ}\ \text{(exothermic)}

Note. The book prints no key for this question; the answer was worked out.

183 marks
Calculate \Delta H for the conversion of graphite to diamond. Heats of combustion of graphite and diamond measured at 25^\circ\mathrm{C} and 1 atm pressure are -393.5\,\mathrm{kJ} and -395.4\,\mathrm{kJ} respectively.
Solution
Given
  • \Delta H_c(\text{graphite}) = -393.5\,\mathrm{kJ}
  • \Delta H_c(\text{diamond}) = -395.4\,\mathrm{kJ}
To find
  • \Delta H \text{ for } \ce{C(graphite) -> C(diamond)}
  1. Both forms are burnt to the same product, \ce{CO2}. Hess's law gives the change as the combustion enthalpy of the reactant minus that of the product.
    \Delta H=\Delta H_c(\text{graphite})-\Delta H_c(\text{diamond})
  2. Substitute.
    \Delta H=-393.5-(-395.4)=+1.9\,\mathrm{kJ\,mol^{-1}}
\Delta H=+1.9\,\mathrm{kJ\,mol^{-1}}

Book answer. +1.9 kJ/mole

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 51 mark
Exothermic reaction has \Delta H:
  1. APositive
  2. BZero
  3. CNegativeAnswer
  4. DInfinite
Why

An exothermic reaction gives out heat to the surroundings, so the enthalpy of the products is lower than that of the reactants. \Delta H=H_\text{products}-H_\text{reactants} is therefore negative. A positive \Delta H (A) belongs to an endothermic reaction, where heat is absorbed.

2025 Annual I, Set B, MCQ 51 mark
What will be the enthalpy of solution of \ce{NaCl} if \Delta H (reverse lattice) is +786\ \mathrm{kJ\,mol^{-1}} and \Delta H (hyd) is -783\ \mathrm{kJ\,mol^{-1}}?
  1. A+3\ \mathrm{kJ\,mol^{-1}}Answer
  2. B-3\ \mathrm{kJ\,mol^{-1}}
  3. C+1569\ \mathrm{kJ\,mol^{-1}}
  4. D-1569\ \mathrm{kJ\,mol^{-1}}
Why

The energy cycle for dissolving gives \Delta H_\text{solution}=\Delta H_\text{reverse lattice}+\Delta H_\text{hydration}.

\Delta H_\text{solution}=+786+(-783)=+3\,\mathrm{kJ\,mol^{-1}}

Option B (-3) comes from getting the sign backwards. Option D (-1569) adds the two sizes with negative signs and C (+1569) adds them with positive signs; but the lattice must be broken (energy absorbed, +786) and the ions hydrated (energy released, -783), so the two terms nearly cancel.

2026 Annual I, Set B, Q.2 (iv)3 marks
Calculate \Delta H^{\circ} of the reaction from the following data: \ce{S(s) + O2(g) -> SO2(g)}; \ce{S(s) + 3/2 O2(g) -> SO3(g)} \Delta H^{\circ}=-395.2\ \mathrm{kJ}; \ce{2SO2(g) + O2(g) -> 2SO3(g)} \Delta H^{\circ}=-198.2\ \mathrm{kJ}
Solution

Use Hess's law: the enthalpy change of a reaction is the same whether it takes place in one step or in several steps, so the given equations can be added after reversing or multiplying them, and their \Delta H^\circ values are treated in the same way.

Given
  • \ce{S(s)+3/2O2(g)->SO3(g)} = \Delta H^\circ=-395.2\,\mathrm{kJ}
  • \ce{2SO2(g)+O2(g)->2SO3(g)} = \Delta H^\circ=-198.2\,\mathrm{kJ}
To find
  • \Delta H^\circ for \ce{S(s) + O2(g) -> SO2(g)}
  1. The target equation has \ce{S(s)} and \ce{O2(g)} as reactants and \ce{SO2(g)} as product. The first given equation forms \ce{SO3} from the elements; keep it as it is.
    \ce{S(s) + 3/2 O2(g) -> SO3(g)}\qquad \Delta H^\circ_1=-395.2\,\mathrm{kJ}
  2. The target needs one \ce{SO2}, but the second equation is written for 2\ce{SO2}. Divide it by 2 (its \Delta H^\circ is halved too).
    \ce{SO2(g) + 1/2 O2(g) -> SO3(g)}\qquad \Delta H^\circ=\dfrac{-198.2}{2}=-99.1\,\mathrm{kJ}
  3. \ce{SO2} is a reactant here but is needed as a product, so reverse this equation and change the sign of \Delta H^\circ.
    \ce{SO3(g) -> SO2(g) + 1/2 O2(g)}\qquad \Delta H^\circ_2=+99.1\,\mathrm{kJ}
  4. Add the first equation and the reversed equation. \ce{SO3} cancels, and \tfrac12\ce{O2} cancels from the \tfrac32\ce{O2}, leaving \ce{O2}.
    \ce{S(s) + O2(g) -> SO2(g)}
  5. Add the enthalpy changes (Hess's law).
    \Delta H^\circ=\Delta H^\circ_1+\Delta H^\circ_2=-395.2+99.1=-296.1\,\mathrm{kJ}
\Delta H^\circ=-296.1\,\mathrm{kJ}
2025 Annual I, Set B, Q.2 (vi)1.5+1.5 = 3 marks
Why enthalpy of solution for \ce{NH4NO3} is +25.7\ \mathrm{kJ\,mol^{-1}} (endothermic process) whereas that of \ce{CaCl2} is -83.0\ \mathrm{kJ\,mol^{-1}} (exothermic process)? Explain briefly.
Solution
Basis (1.5 marks)

When an ionic solid dissolves, the lattice must be broken (energy absorbed, the reverse of the lattice energy) and the gaseous ions are then hydrated by water molecules (energy released). The enthalpy of solution is the net result.

\Delta H_\text{solution}=\Delta H_\text{reverse lattice}+\Delta H_\text{hydration}
\ce{NH4NO3} and \ce{CaCl2} (1.5 marks)
\ce{NH4NO3}\ce{CaCl2}
\Delta H_\text{solution}+25.7\,\mathrm{kJ\,mol^{-1}} (endothermic)-83.0\,\mathrm{kJ\,mol^{-1}} (exothermic)
Energy of hydration compared with the lattice energySmaller: the crystal binds the ions more tightly than water doesLarger: water binds the ions more tightly than the crystal does
IonsSingly charged, large ions, low charge density\ce{Ca^2+} is small and doubly charged, high charge density
ResultSolution coolsSolution becomes warm

For \ce{NH4NO3} the energy needed to break up the lattice is greater than the energy released on hydration, so heat is absorbed from the surroundings and the process is endothermic. For \ce{CaCl2} the doubly charged \ce{Ca^2+} ion of high charge density attracts water molecules strongly (ion–dipole attraction), so the hydration energy exceeds the lattice energy and the process is exothermic.

2026 Annual I, Set B, Q.2 (xi) OR1+2 = 3 marks
Briefly explain the term enthalpy change of atomization with one example.
Solution
Definition (1 mark)

The standard enthalpy of atomization, \Delta H^\circ_\mathrm{at}, is the enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions.

Example (2 marks)
\ce{1/2 Cl2(g) -> Cl(g)}\qquad \Delta H^\circ_\mathrm{at}=+121\,\mathrm{kJ\,mol^{-1}}
\ce{1/2 H2(g) -> H(g)}\qquad \Delta H^\circ_\mathrm{at}=+218\,\mathrm{kJ\,mol^{-1}}

The product is one mole of atoms, which is why only half a mole of a diatomic molecule is written on the left. Atomization breaks bonds (here the Cl–Cl or H–H bond), so it always needs energy and \Delta H^\circ_\mathrm{at} is positive.

2025 Annual I, Set B, Q.2 (xiii) OR3 marks
Calculate total entropy change when \ce{NaCl} dissolves in water under standard conditions: \ce{NaCl(s) -> Na+(aq) + Cl-(aq)}, with S^{\circ}(\ce{NaCl})=72.1\ \mathrm{J\,mol^{-1}\,K^{-1}}, S^{\circ}(\ce{Na+})=321\ \mathrm{J\,mol^{-1}\,K^{-1}}, S^{\circ}(\ce{Cl-})=56.5\ \mathrm{J\,mol^{-1}\,K^{-1}}
Solution
Given
  • S^\circ(\ce{NaCl,s}) = 72.1\,\mathrm{J\,mol^{-1}\,K^{-1}}
  • S^\circ(\ce{Na+,aq}) = 321\,\mathrm{J\,mol^{-1}\,K^{-1}}
  • S^\circ(\ce{Cl-,aq}) = 56.5\,\mathrm{J\,mol^{-1}\,K^{-1}}
To find
  • \Delta S^\circ for \ce{NaCl(s) -> Na+(aq) + Cl-(aq)}
  1. The entropy change of a reaction is the sum of the standard entropies of the products minus that of the reactants, each multiplied by its coefficient.
    \Delta S^\circ=\sum S^\circ_\text{products}-\sum S^\circ_\text{reactants}
  2. Substitute the data (all coefficients are 1).
    \Delta S^\circ=\left[S^\circ(\ce{Na+})+S^\circ(\ce{Cl-})\right]-S^\circ(\ce{NaCl})=(321+56.5)-72.1
  3. Evaluate.
    \Delta S^\circ=377.5-72.1=+305.4\,\mathrm{J\,mol^{-1}\,K^{-1}}

\Delta S^\circ is positive: an ordered crystal lattice breaks up into free, hydrated ions in solution, so the disorder of the system increases.

\Delta S^\circ=+305.4\,\mathrm{J\,mol^{-1}\,K^{-1}}
2025 Annual I, Set B, Q.2 (xiv)3 marks
Draw Born-Haber cycle for the formation of \ce{KCl}. (Calculation is NOT required.)
Solution

The Born–Haber cycle is the application of Hess's law to an ionic compound. The formation of \ce{KCl(s)} from its elements is broken into five steps, the last of which is the formation of the lattice.

Born–Haber cycle for the formation of \ce{KCl(s)} (levels not to scale). Upward arrows are endothermic steps, downward arrows are exothermic steps; the accent arrow is the lattice energy.
StepWhat happensEquation
ISublimation of solid potassium, \Delta H_\mathrm{s} (endothermic)\ce{K(s) -> K(g)}
IIFirst ionization energy of potassium, \Delta H_\mathrm{IE} (endothermic)\ce{K(g) -> K+(g) + e-}
IIIAtomization of chlorine, \Delta H_\mathrm{at} (endothermic)\ce{1/2 Cl2(g) -> Cl(g)}
IVElectron affinity of chlorine, \Delta H_\mathrm{EA} (exothermic)\ce{Cl(g) + e- -> Cl-(g)}
VLattice energy, \Delta H_\mathrm{l} (exothermic)\ce{K+(g) + Cl-(g) -> KCl(s)}

By Hess's law the sum of the five steps equals the enthalpy of formation of \ce{KCl}, \ce{K(s) + 1/2 Cl2(g) -> KCl(s)}:

\Delta H^\circ_\mathrm{f}=\Delta H_\mathrm{s}+\Delta H_\mathrm{IE}+\Delta H_\mathrm{at}+\Delta H_\mathrm{EA}+\Delta H_\mathrm{l}
2026 Annual I, Set D, MCQ 51 mark
Combustion is a/an ______ process.
  1. AExothermicAnswer
  2. BNeutral
  3. CEndothermic
  4. DReversible
Why

Combustion (burning in oxygen) always gives out heat, so it is an exothermic process and the standard enthalpy of combustion is always negative (for example -393.5\,\mathrm{kJ\,mol^{-1}} for carbon). It is not endothermic (C), and it is not neutral (B): it has a definite, negative \Delta H.

Reversible (D) describes whether a reaction can go back, not its heat change.

2025 Annual I, Set D, MCQ 51 mark
Which of the following is the correct relationship for Gibbs free energy?
  1. AG = H + TS
  2. BG = H \times TS
  3. CG = H - TSAnswer
  4. DG = \dfrac{H}{TS}
Why

Gibbs free energy is defined as the enthalpy of the system minus the product of its temperature and entropy: G=H-TS. At constant temperature this gives the Gibbs–Helmholtz equation \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ.

The signs matter: option A (H+TS) has the wrong sign, and options B and D combine H and TS by multiplication and division, which is not dimensionally possible because H and TS are both energies and can only be added or subtracted.

2025 Annual I, Set D, Q.2 (i) OR3 marks
Briefly explain the terms with examples: (a) enthalpy of atomization (b) lattice energy (c) electron affinity
Solution
(a) Enthalpy of atomization (1 mark)

The standard enthalpy of atomization is the enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions. It is always positive.

\ce{1/2 Cl2(g) -> Cl(g)}\qquad \Delta H^\circ_\mathrm{at}=+121\,\mathrm{kJ\,mol^{-1}}
(b) Lattice energy (1 mark)

The energy released when one mole of an ionic solid is formed from its gaseous ions. It is negative as an enthalpy change.

\ce{Na+(g) + Cl-(g) -> NaCl(s)}\qquad \Delta H=-787\,\mathrm{kJ\,mol^{-1}}
(c) Electron affinity (1 mark)

The standard enthalpy of first electron affinity is the energy released when one mole of gaseous atoms gains an electron each to form one mole of gaseous ions with a single negative charge.

\ce{Cl(g) + e- -> Cl-(g)}\qquad \Delta H^\circ_\mathrm{EA}=-349\,\mathrm{kJ\,mol^{-1}}
2026 Annual I, Set D, Q.2 (iv)3 marks
Calculate bond energy of \ce{HCl}, when: bond dissociation energy of \ce{H2} =436\ \mathrm{kJ/mol}, bond dissociation energy of \ce{Cl2} =243\ \mathrm{kJ/mol}
Solution

The book's method (Example 6.3) takes the energy absorbed in breaking the bonds of \ce{H2} and \ce{Cl2} as equal to the energy released in forming the bonds of \ce{HCl}.

Given
  • E(\ce{H-H}) = 436\,\mathrm{kJ\,mol^{-1}} bond dissociation energy of \ce{H2}
  • E(\ce{Cl-Cl}) = 243\,\mathrm{kJ\,mol^{-1}} bond dissociation energy of \ce{Cl2}
To find
  • bond energy of \ce{H-Cl}
  1. Write the reaction. One mole of \ce{H-H} and one mole of \ce{Cl-Cl} bonds break, and two moles of \ce{H-Cl} bonds form.
    \ce{H2 + Cl2 -> 2HCl}
  2. Energy absorbed in breaking the bonds equals energy released in forming the new bonds.
    E(\ce{H-H})+E(\ce{Cl-Cl})=2\,E(\ce{H-Cl})
  3. Put in the data.
    436+243=2\,E(\ce{H-Cl})
  4. Add and divide by two.
    E(\ce{H-Cl})=\dfrac{679}{2}=339.5\,\mathrm{kJ\,mol^{-1}}

The experimentally measured bond energy of \ce{HCl} is 431.5\,\mathrm{kJ\,mol^{-1}}. The calculated value differs because it assumes no net energy change and uses averages; the real reaction is exothermic and the unequal sharing of the electron pair (electronegativity difference) makes the \ce{H-Cl} bond stronger than the estimate.

E(\ce{H-Cl})\approx339.5\,\mathrm{kJ\,mol^{-1}}
2025 Annual I, Set D, Q.2 (vii)1+1+1 = 3 marks
What is entropy? Describe the increase or decrease in entropy of system with sign in the following reactions: (a) \ce{N2O4(g) -> 2NO2(s)} (b) \ce{NH3(g) + HCl(g) -> NH4Cl(s)}
Solution
Entropy (1 mark)

Entropy, symbol S, is the degree of disorder or randomness of a system (the number of possible arrangements of its particles and their energy). When a system goes from a more ordered to a less ordered state its entropy increases and \Delta S is positive (\Delta S>0); when it becomes more ordered, \Delta S is negative (\Delta S<0).

(a) \ce{N2O4(g) -> 2NO2(g)} (1 mark)

One mole of gas gives two moles of gas, so the number of gaseous molecules increases and the system becomes more disordered. Entropy increases, \Delta S>0 (positive).

(b) \ce{NH3(g) + HCl(g) -> NH4Cl(s)} (1 mark)

Two moles of gas combine to give a solid. The particles go from free movement to a fixed lattice, so the disorder falls. Entropy decreases, \Delta S<0 (negative).

2026 Annual I, Set D, Q.2 (xi) OR1.5+1.5 = 3 marks
Briefly explain standard enthalpy of electron affinity and standard enthalpy of Lattice (\Delta H_{Lattice}) with examples.
Solution
Standard enthalpy of first electron affinity (1.5 marks)

It is the energy released when one mole of gaseous atoms gains an electron each to form one mole of gaseous ions with a single negative charge, under standard conditions.

\ce{Cl(g) + e- -> Cl-(g)}\qquad \Delta H^\circ_\mathrm{EA}=-349\,\mathrm{kJ\,mol^{-1}}

The sign is negative because energy is released when the incoming electron is attracted by the nucleus of the atom.

Standard enthalpy of lattice energy (1.5 marks)

The lattice energy is the energy released when one mole of an ionic solid is formed from its constituent gaseous ions.

\ce{Na+(g) + Cl-(g) -> NaCl(s)}\qquad \Delta H_\mathrm{lattice}=-787\,\mathrm{kJ\,mol^{-1}}

It is released when the lattice forms, so it carries a negative sign as an enthalpy change. Its size is 787\,\mathrm{kJ\,mol^{-1}}, and it cannot be measured directly but is found from the Born–Haber cycle.

2025 Annual I, Set D, Q.6 OR2+5 = 7 marks
State Hess's law with an example. Apply Hess's law to calculate enthalpy of formation of \ce{CH3COOH}. The standard enthalpies of combustion of carbon, hydrogen and acetic acid are -393.5, -285.8 and -875\ \mathrm{kJ\,mol^{-1}} respectively.
Solution
Hess's law with an example (2 marks)

Hess's law states that the enthalpy change in a chemical reaction is the same whether the reaction takes place in a single step or in several steps. It follows because enthalpy is a state function.

Example: carbon burns to \ce{CO2} directly, or in two steps through \ce{CO}. The sum of the two steps equals the one-step value:

\begin{aligned}&\ce{C(s) + O2(g) -> CO2(g)} &&\Delta H^\circ=-393.5\,\mathrm{kJ}\\&\ce{C(s) + 1/2 O2(g) -> CO(g)} &&\Delta H^\circ_1=-110.52\,\mathrm{kJ}\\&\ce{CO(g) + 1/2 O2(g) -> CO2(g)} &&\Delta H^\circ_2=-282.98\,\mathrm{kJ}\\&\Delta H^\circ_1+\Delta H^\circ_2=-393.5\,\mathrm{kJ}\end{aligned}
Enthalpy of formation of \ce{CH3COOH} (5 marks)
Given
  • \Delta H^\circ_c(\ce{C}) = -393.5\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{H2}) = -285.8\,\mathrm{kJ\,mol^{-1}}
  • \Delta H^\circ_c(\ce{CH3COOH}) = -875\,\mathrm{kJ\,mol^{-1}}
To find
  • \Delta H^\circ_f(\ce{CH3COOH})
  1. Write the formation reaction: elements in their standard states give one mole of acetic acid (a liquid).
    \ce{2C(s) + 2H2(g) + O2(g) -> CH3COOH(l)}
  2. Write the combustion equations given. Carbon and hydrogen burn to \ce{CO2} and \ce{H2O}; acetic acid burns to the same products.
    \begin{aligned}\text{(i)}\;&\ce{C(s) + O2(g) -> CO2(g)} &&-393.5\\ \text{(ii)}\;&\ce{H2(g) + 1/2 O2(g) -> H2O(l)} &&-285.8\\ \text{(iii)}\;&\ce{CH3COOH(l) + 2O2(g) -> 2CO2(g) + 2H2O(l)} &&-875\end{aligned}
  3. Multiply (i) by 2 and (ii) by 2, so that 2 C and 2 \ce{H2} appear on the left. Reverse (iii), because \ce{CH3COOH} must be a product, and change the sign of its \Delta H^\circ.
    2\times\text{(i)}:\ -787.0\qquad 2\times\text{(ii)}:\ -571.6\qquad -\text{(iii)}:\ +875
  4. Add the three equations and cancel \ce{2CO2} and \ce{2H2O}, which appear on both sides. Oxygen: 2+1=3\ \ce{O2} on the left and 2\ \ce{O2} on the right, so 1\ \ce{O2} is left on the left.
    \ce{2C(s) + 2H2(g) + O2(g) -> CH3COOH(l)}
  5. Add the enthalpy changes.
    \Delta H^\circ_f=-787.0-571.6+875=-483.6\,\mathrm{kJ\,mol^{-1}}

The same result follows from \Delta H^\circ_f=[2(-393.5)+2(-285.8)]-(-875)=-1358.6+875=-483.6\,\mathrm{kJ\,mol^{-1}}.

\Delta H^\circ_f(\ce{CH3COOH})=-483.6\,\mathrm{kJ\,mol^{-1}}

Exam focus

  1. Apply Hess's law to find an enthalpy of reaction or of formation (for example \ce{S + O2 -> SO2}, and \ce{CH3COOH} from combustion data). (numerical)
    A 7-mark Section C question in 2025 Set D Q.6 OR and 3 marks in 2026 Set B Q.2 (iv) and in Set D (same idea). They are book numerical questions 12 and 16.
    Go to answer
  2. Draw and explain a Born–Haber cycle for an ionic solid, and use it to find the lattice energy. (long)
    Asked in 2025 Set B Q.2 (xiv) (draw the cycle for \ce{KCl}). The book's cycle for \ce{NaCl} and the exercise for \ce{SrF2} use the same five steps.
    Go to answer
  3. Define enthalpy of atomization, lattice energy and first electron affinity with examples. (short)
    Asked in three of the four sets: 2026 Set B Q.2 (xi) OR, 2026 Set D Q.2 (xi) OR and 2025 Set D Q.2 (i) OR. Learn the definition and one equation with its sign for each.
    Go to answer
  4. Explain the enthalpy of solution through lattice energy and enthalpy of hydration (why \ce{NH4NO3} is endothermic and \ce{CaCl2} exothermic). (short)
    Asked in 2025 Set B (MCQ 5, +786-783=+3\,\mathrm{kJ}, and Q.2 (vi)). It is also book long question 21 (\ce{SrF2}).
    Go to answer
  5. Define entropy, predict the sign of \Delta S, and use \Delta G=\Delta H-T\Delta S. (short)
    Asked in 2025: Set D Q.2 (vii) (sign of \Delta S), Set D MCQ 5 (G=H-TS) and Set B Q.2 (xiii) OR (calculation of \Delta S^\circ). Book short question 20 and long question 19.
    Go to answer
  6. Calculate a bond energy or the enthalpy change of a reaction from bond energies. (numerical)
    Asked in 2026 Set D Q.2 (iv) (the \ce{H-Cl} bond energy, book Example 6.3).
    Go to answer
  7. Know that exothermic reactions and combustion have a negative \Delta H. (short)
    Two MCQs in 2026 (Set B MCQ 5 and Set D MCQ 5). Compare book exercise MCQ v on heat of combustion.
    Go to answer
  8. Calorimetry numericals with q=mc\Delta T and \Delta H=-q/n (neutralization, bomb calorimeter, heat of combustion). (numerical)
    Not in the 2025 or 2026 papers, but book numerical questions 5, 7 and 10 use it, and the SLOs list it.
    Go to answer

Mark this unit as done to fill the Class 11 Chemistry ring.

Next, unit 7
Chemical kinetics

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