Bond breaking and bond making

Definition

Bond breaking

An endothermic process in which energy is taken in to overcome the attraction between two bonded atoms and separate them.

Definition

Bond making

An exothermic process in which energy is released as two atoms are joined by a new bond.

During a chemical reaction, the bonds in the reactant particles break, the atoms rearrange, and new bonds form to make the products.
In a covalent bond, the two atoms are held together by the strong attraction between their nuclei and the shared pair of electrons.
Bond breaking is an endothermic process, because energy must be taken in to overcome this attraction and pull the atoms apart.
Bond making is an exothermic process, because energy is released when two atoms are drawn together and a new bond forms.
Forming a particular bond releases exactly the same amount of energy as is needed to break that same bond.
Every reaction involves both processes, so the overall energy change depends on which of the two involves more energy.

Explaining the enthalpy change

Exothermic reactions
1. The energy released when bonds form in the products is greater than the energy taken in to break the bonds in the reactants.
2. The extra energy is transferred to the surroundings, so $ΔH\Delta H$ is negative.
3. Burning a fuel is exothermic because the bonds formed in carbon dioxide and water are very strong, so making them releases more energy than breaking the bonds in the fuel and oxygen takes in.
Endothermic reactions
1. The energy taken in to break the bonds in the reactants is greater than the energy released when bonds form in the products.
2. The shortfall is taken in from the surroundings, so $ΔH\Delta H$ is positive.
The enthalpy change is therefore the balance between two energy transfers, one into the chemicals and one out of them.

Calculating enthalpy change from bond energies

Definition

Bond energy

The energy needed to break one mole of a particular covalent bond, measured in $kJ/mol\text{kJ/mol}$ .

A bond energy is the energy needed to break one mole of a particular covalent bond, and the same amount is released when one mole of that bond forms.
Bond energies are given in $kJ/mol\text{kJ/mol}$ and are always listed as positive values, because breaking a bond always takes energy in.
Stronger bonds have larger bond energies, so a triple bond such as $N≡N\text{N}\equiv\text{N}$ at $kJ/mol945\ \text{kJ/mol}$ is much stronger than a single bond such as $H-H\text{H-H}$ at $kJ/mol436\ \text{kJ/mol}$ .
The enthalpy change of a reaction is calculated with $form\Delta H = \text{energy taken in to break bonds} - \text{energy released when bonds form}$ .
Work through every calculation in the same order.
1. Write the balanced equation and draw the displayed formula of every reactant and product, so that each bond can be seen.
2. List every bond broken in the reactants, multiplying by the number of molecules shown in front of each formula.
3. Add the bond energies to find the total energy taken in to break bonds.
4. List every bond formed in the products in the same way, and add their bond energies to find the total energy released when bonds form.
5. Subtract the bond-making total from the bond-breaking total, and give the answer with its sign and the unit $kJ/mol\text{kJ/mol}$ .
A negative answer shows an exothermic reaction, and a positive answer shows an endothermic reaction.

Example

Hydrogen reacting with chlorine

The equation is $H-Cl\text{H-H} + \text{Cl-Cl} \rightarrow 2\,\text{H-Cl}$ , with bond energies of $kJ/mol436\ \text{kJ/mol}$ for $H-H\text{H-H}$ , $kJ/mol242\ \text{kJ/mol}$ for $Cl-Cl\text{Cl-Cl}$ and $kJ/mol431\ \text{kJ/mol}$ for $H-Cl\text{H-Cl}$ .
Bonds broken: one $H-H\text{H-H}$ bond and one $Cl-Cl\text{Cl-Cl}$ bond.

\text{energy taken in} = 436 + 242 = 678\ \text{kJ/mol}

Bonds formed: two $H-Cl\text{H-Cl}$ bonds, because the equation makes two molecules of hydrogen chloride.

\text{energy released} = 2 \times 431 = 862\ \text{kJ/mol}

\Delta H = 678 - 862 = -184\ \text{kJ/mol}

The value is negative, so the reaction is exothermic.

Example

Combustion of methane

In $CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ , methane has four $C-H\text{C-H}$ bonds, each oxygen molecule has one $O=O\text{O=O}$ bond, carbon dioxide has two $C=O\text{C=O}$ bonds and each water molecule has two $O-H\text{O-H}$ bonds.
The bond energies used are $C-H=413\text{C-H} = 413$ , $O=O=498\text{O=O} = 498$ , $C=O=805\text{C=O} = 805$ and $O-H=463\text{O-H} = 463$ , all in $kJ/mol\text{kJ/mol}$ .
Bonds broken: four $C-H\text{C-H}$ bonds and two $O=O\text{O=O}$ bonds.

\text{energy taken in} = (4 \times 413) + (2 \times 498) = 2648\ \text{kJ/mol}

Bonds formed: two $C=O\text{C=O}$ bonds and four $O-H\text{O-H}$ bonds, because two water molecules each contain two $O-H\text{O-H}$ bonds.

\text{energy released} = (2 \times 805) + (4 \times 463) = 3462\ \text{kJ/mol}

\Delta H = 2648 - 3462 = -814\ \text{kJ/mol}

The value is negative, so the combustion of methane is exothermic.

Example

Decomposition of hydrogen bromide

The equation is $H-Br→H-H+Br-Br2\,\text{H-Br} \rightarrow \text{H-H} + \text{Br-Br}$ , with bond energies of $kJ/mol366\ \text{kJ/mol}$ for $H-Br\text{H-Br}$ , $kJ/mol436\ \text{kJ/mol}$ for $H-H\text{H-H}$ and $kJ/mol193\ \text{kJ/mol}$ for $Br-Br\text{Br-Br}$ .
Bonds broken: two $H-Br\text{H-Br}$ bonds.

\text{energy taken in} = 2 \times 366 = 732\ \text{kJ/mol}

Bonds formed: one $H-H\text{H-H}$ bond and one $Br-Br\text{Br-Br}$ bond.

\text{energy released} = 436 + 193 = 629\ \text{kJ/mol}

\Delta H = 732 - 629 = +103\ \text{kJ/mol}

The value is positive, so the reaction is endothermic, because more energy is taken in to break the bonds than is released when the new bonds form.

Common Mistake

Do not forget the numbers in front of formulae, because two molecules of water contain four $O-H\text{O-H}$ bonds, not two.
Do not subtract the wrong way round, because only bonds broken minus bonds formed gives the correct sign.
Do not say that breaking bonds releases energy, because energy is always needed to break a bond.
Do not leave the sign off your final answer, because a value without a sign does not show whether the reaction is exothermic or endothermic.

Exam technique

Bond energy questions

Set out your working in three lines, giving the total for bonds broken, the total for bonds formed and the subtraction, because each line can earn credit even if a later step goes wrong.
To explain why a reaction is exothermic, make two points: bond breaking takes in energy and bond making releases energy, and more energy is released making bonds than is taken in breaking them.
For an endothermic reaction, keep the first point and reverse the second, so more energy is taken in breaking bonds than is released making them.
Finish every calculation with the sign and the unit, as in $kJ/mol-184\ \text{kJ/mol}$ .
Sketch the displayed formulae before you count, because double bonds such as $C=O\text{C=O}$ and $O=O\text{O=O}$ have their own bond energies and are easy to miss.

Self review

State whether bond breaking is an endothermic or an exothermic process.
Explain, in terms of bond breaking and bond making, why a reaction is exothermic.
Write the expression used to calculate $ΔH\Delta H$ from bond energies.
State how many $O-H\text{O-H}$ bonds are formed when two molecules of water are made.
Calculate $ΔH\Delta H$ for $N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , using $N≡N=945\text{N}\equiv\text{N} = 945$ , $H-H=436\text{H-H} = 436$ and $N-H=391\text{N-H} = 391$ , all in $kJ/mol\text{kJ/mol}$ .

Bond breaking and bond making

Definition

Bond breaking

An endothermic process in which energy is taken in to overcome the attraction between two bonded atoms and separate them.

Definition

Bond making

An exothermic process in which energy is released as two atoms are joined by a new bond.

During a chemical reaction, the bonds in the reactant particles break, the atoms rearrange, and new bonds form to make the products.
In a covalent bond, the two atoms are held together by the strong attraction between their nuclei and the shared pair of electrons.
Bond breaking is an endothermic process, because energy must be taken in to overcome this attraction and pull the atoms apart.
Bond making is an exothermic process, because energy is released when two atoms are drawn together and a new bond forms.
Forming a particular bond releases exactly the same amount of energy as is needed to break that same bond.
Every reaction involves both processes, so the overall energy change depends on which of the two involves more energy.

Explaining the enthalpy change

Exothermic reactions
1. The energy released when bonds form in the products is greater than the energy taken in to break the bonds in the reactants.
2. The extra energy is transferred to the surroundings, so $ΔH\Delta H$ is negative.
3. Burning a fuel is exothermic because the bonds formed in carbon dioxide and water are very strong, so making them releases more energy than breaking the bonds in the fuel and oxygen takes in.
Endothermic reactions
1. The energy taken in to break the bonds in the reactants is greater than the energy released when bonds form in the products.
2. The shortfall is taken in from the surroundings, so $ΔH\Delta H$ is positive.
The enthalpy change is therefore the balance between two energy transfers, one into the chemicals and one out of them.

Calculating enthalpy change from bond energies

Definition

Bond energy

The energy needed to break one mole of a particular covalent bond, measured in $kJ/mol\text{kJ/mol}$ .

A bond energy is the energy needed to break one mole of a particular covalent bond, and the same amount is released when one mole of that bond forms.
Bond energies are given in $kJ/mol\text{kJ/mol}$ and are always listed as positive values, because breaking a bond always takes energy in.
Stronger bonds have larger bond energies, so a triple bond such as $N≡N\text{N}\equiv\text{N}$ at $kJ/mol945\ \text{kJ/mol}$ is much stronger than a single bond such as $H-H\text{H-H}$ at $kJ/mol436\ \text{kJ/mol}$ .
The enthalpy change of a reaction is calculated with $form\Delta H = \text{energy taken in to break bonds} - \text{energy released when bonds form}$ .
Work through every calculation in the same order.
1. Write the balanced equation and draw the displayed formula of every reactant and product, so that each bond can be seen.
2. List every bond broken in the reactants, multiplying by the number of molecules shown in front of each formula.
3. Add the bond energies to find the total energy taken in to break bonds.
4. List every bond formed in the products in the same way, and add their bond energies to find the total energy released when bonds form.
5. Subtract the bond-making total from the bond-breaking total, and give the answer with its sign and the unit $kJ/mol\text{kJ/mol}$ .
A negative answer shows an exothermic reaction, and a positive answer shows an endothermic reaction.

Example

Hydrogen reacting with chlorine

The equation is $H-Cl\text{H-H} + \text{Cl-Cl} \rightarrow 2\,\text{H-Cl}$ , with bond energies of $kJ/mol436\ \text{kJ/mol}$ for $H-H\text{H-H}$ , $kJ/mol242\ \text{kJ/mol}$ for $Cl-Cl\text{Cl-Cl}$ and $kJ/mol431\ \text{kJ/mol}$ for $H-Cl\text{H-Cl}$ .
Bonds broken: one $H-H\text{H-H}$ bond and one $Cl-Cl\text{Cl-Cl}$ bond.

\text{energy taken in} = 436 + 242 = 678\ \text{kJ/mol}

Bonds formed: two $H-Cl\text{H-Cl}$ bonds, because the equation makes two molecules of hydrogen chloride.

\text{energy released} = 2 \times 431 = 862\ \text{kJ/mol}

\Delta H = 678 - 862 = -184\ \text{kJ/mol}

The value is negative, so the reaction is exothermic.

Example

Combustion of methane

In $CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ , methane has four $C-H\text{C-H}$ bonds, each oxygen molecule has one $O=O\text{O=O}$ bond, carbon dioxide has two $C=O\text{C=O}$ bonds and each water molecule has two $O-H\text{O-H}$ bonds.
The bond energies used are $C-H=413\text{C-H} = 413$ , $O=O=498\text{O=O} = 498$ , $C=O=805\text{C=O} = 805$ and $O-H=463\text{O-H} = 463$ , all in $kJ/mol\text{kJ/mol}$ .
Bonds broken: four $C-H\text{C-H}$ bonds and two $O=O\text{O=O}$ bonds.

\text{energy taken in} = (4 \times 413) + (2 \times 498) = 2648\ \text{kJ/mol}

Bonds formed: two $C=O\text{C=O}$ bonds and four $O-H\text{O-H}$ bonds, because two water molecules each contain two $O-H\text{O-H}$ bonds.

\text{energy released} = (2 \times 805) + (4 \times 463) = 3462\ \text{kJ/mol}

\Delta H = 2648 - 3462 = -814\ \text{kJ/mol}

The value is negative, so the combustion of methane is exothermic.

Example

Decomposition of hydrogen bromide

The equation is $H-Br→H-H+Br-Br2\,\text{H-Br} \rightarrow \text{H-H} + \text{Br-Br}$ , with bond energies of $kJ/mol366\ \text{kJ/mol}$ for $H-Br\text{H-Br}$ , $kJ/mol436\ \text{kJ/mol}$ for $H-H\text{H-H}$ and $kJ/mol193\ \text{kJ/mol}$ for $Br-Br\text{Br-Br}$ .
Bonds broken: two $H-Br\text{H-Br}$ bonds.

\text{energy taken in} = 2 \times 366 = 732\ \text{kJ/mol}

Bonds formed: one $H-H\text{H-H}$ bond and one $Br-Br\text{Br-Br}$ bond.

\text{energy released} = 436 + 193 = 629\ \text{kJ/mol}

\Delta H = 732 - 629 = +103\ \text{kJ/mol}

The value is positive, so the reaction is endothermic, because more energy is taken in to break the bonds than is released when the new bonds form.

Common Mistake

Do not forget the numbers in front of formulae, because two molecules of water contain four $O-H\text{O-H}$ bonds, not two.
Do not subtract the wrong way round, because only bonds broken minus bonds formed gives the correct sign.
Do not say that breaking bonds releases energy, because energy is always needed to break a bond.
Do not leave the sign off your final answer, because a value without a sign does not show whether the reaction is exothermic or endothermic.

Exam technique

Bond energy questions

Set out your working in three lines, giving the total for bonds broken, the total for bonds formed and the subtraction, because each line can earn credit even if a later step goes wrong.
To explain why a reaction is exothermic, make two points: bond breaking takes in energy and bond making releases energy, and more energy is released making bonds than is taken in breaking them.
For an endothermic reaction, keep the first point and reverse the second, so more energy is taken in breaking bonds than is released making them.
Finish every calculation with the sign and the unit, as in $kJ/mol-184\ \text{kJ/mol}$ .
Sketch the displayed formulae before you count, because double bonds such as $C=O\text{C=O}$ and $O=O\text{O=O}$ have their own bond energies and are easy to miss.

Self review

State whether bond breaking is an endothermic or an exothermic process.
Explain, in terms of bond breaking and bond making, why a reaction is exothermic.
Write the expression used to calculate $ΔH\Delta H$ from bond energies.
State how many $O-H\text{O-H}$ bonds are formed when two molecules of water are made.
Calculate $ΔH\Delta H$ for $N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , using $N≡N=945\text{N}\equiv\text{N} = 945$ , $H-H=436\text{H-H} = 436$ and $N-H=391\text{N-H} = 391$ , all in $kJ/mol\text{kJ/mol}$ .

5.1.3 Bond breaking, bond making and bond energies

Bond breaking and bond making

Explaining the enthalpy change

Calculating enthalpy change from bond energies

Hydrogen reacting with chlorine

Combustion of methane

Decomposition of hydrogen bromide

Bond energy questions

Review 5.1.3 Bond breaking, bond making and bond energies by teaching Genie

5.1.3 Bond breaking, bond making and bond energies

Bond breaking and bond making

Explaining the enthalpy change

Calculating enthalpy change from bond energies

Hydrogen reacting with chlorine

Combustion of methane

Decomposition of hydrogen bromide

Bond energy questions

Review 5.1.3 Bond breaking, bond making and bond energies by teaching Genie

5.1.3 Bond breaking, bond making and bond energies

Bond breaking and bond making

Explaining the enthalpy change

Calculating enthalpy change from bond energies

Hydrogen reacting with chlorine

Combustion of methane

Decomposition of hydrogen bromide

Bond energy questions

Review 5.1.3 Bond breaking, bond making and bond energies by teaching Genie

5.1.3 Bond breaking, bond making and bond energies Revision Guide

5.1.3 Bond breaking, bond making and bond energies

Bond breaking and bond making

Explaining the enthalpy change

Calculating enthalpy change from bond energies

Hydrogen reacting with chlorine

Combustion of methane

Decomposition of hydrogen bromide

Bond energy questions

Review 5.1.3 Bond breaking, bond making and bond energies by teaching Genie

5.1.3 Bond breaking, bond making and bond energies Revision Guide