Topics

Bond enthalpy calculations

Breaking and making bonds

Two facts drive every calculation on this page:

Breaking bonds requires energy — you must pull the atoms apart. Endothermic, energy in.
Making bonds releases energy — the atoms settle into a lower-energy arrangement. Exothermic, energy out.

What decides the sign of a reaction

Every reaction breaks the reactants' bonds and makes new ones in the products. Whether it is exothermic or endothermic depends on which is bigger:

More energy released making bonds than absorbed breaking them → exothermic, $\Delta_rH$ negative.
More energy absorbed breaking bonds than released making them → endothermic, $\Delta_rH$ positive.

The relationship

$\Delta_rH = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$

$\sum(\text{bonds broken})$ — total bond enthalpy of all bonds in the reactants.
$\sum(\text{bonds formed})$ — total bond enthalpy of all bonds in the products.
A useful way to remember the order: "broken minus formed" — reactants first, products second, same order as the equation.

Bond enthalpy

Bond enthalpy is the energy needed to break one mole of a particular bond.
It is always quoted as a positive number, because breaking always takes energy in.
Typical values (in $kJ\ mol^{-1}$ ): C–H 413, C–C 347, C=C 614, O=O 498, C=O 805, O–H 464, H–H 436.
Stronger bonds have larger values, so a C=C takes more energy to break than a C–C.

The method

Draw or list every bond in the reactants, and every bond in the products.
Count them carefully, including the coefficients in the equation.
Add up the bond enthalpies for the reactants — this is bonds broken.
Add up the bond enthalpies for the products — this is bonds formed.
Subtract: broken − formed.
Check the sign and give units.

Counting bonds without missing any

This is where nearly all the marks are lost, so be systematic:

Write out the structure of each molecule rather than working from the formula.
Multiply by the coefficient: $2H_2O$ contains 4 O–H bonds, not 2.
Count every bond, including all the C–H bonds in an organic molecule.
Tally them in a list before you add anything up.

Tips and Tricks

Get the order right: broken minus formed. Reversing it gives an answer with exactly the right number and exactly the wrong sign, which loses the marks for both the sign and the conclusion.

Count bonds from structures, not formulae. Methane's formula $CH_4$ makes the four C–H bonds obvious, but $C_2H_6$ hides the fact that there are six C–H bonds and one C–C. Sketch the structure every time.

Do not forget the coefficients. In $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ there are two O=O bonds broken and four O–H bonds formed. Missing a coefficient is the most common arithmetic error in this topic.

Finally, remember bond enthalpies are average values, so a calculated $\Delta_rH$ is an estimate and may differ slightly from the measured value. Saying so is often worth a mark in an evaluation question.

Worked ExampleA full bond enthalpy calculation

Use the bond enthalpies below to calculate $\Delta_rH$ for the combustion of methane, and state whether the reaction is exothermic or endothermic.

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

C–H = 413, O=O = 498, C=O = 805, O–H = 464 (all $kJ\ mol^{-1}$ )

Step 1 — List the bonds in the reactants

$CH_4$ — the carbon is bonded to four hydrogens: 4 × C–H

$2O_2$ — each oxygen molecule has one double bond, and there are two molecules: 2 × O=O

Step 2 — List the bonds in the products

$CO_2$ — the structure is $O{=}C{=}O$ , so there are 2 × C=O

$2H_2O$ — each water molecule has two O–H bonds, and there are two molecules: $2 \times 2 =$ 4 × O–H

Step 3 — Add up the bonds broken (reactants)

$\sum(\text{broken}) = (4 \times 413) + (2 \times 498)$ $= 1652 + 996$ $= 2648\ kJ\ mol^{-1}$

Step 4 — Add up the bonds formed (products)

$\sum(\text{formed}) = (2 \times 805) + (4 \times 464)$ $= 1610 + 1856$ $= 3466\ kJ\ mol^{-1}$

Step 5 — Subtract, in the right order

$\Delta_rH = \sum(\text{broken}) - \sum(\text{formed})$ $\Delta_rH = 2648 - 3466$ $\Delta_rH = -818\ kJ\ mol^{-1}$

Step 6 — Interpret the result

The value is negative, so the reaction is exothermic.

This makes sense: more energy was released forming bonds (3466 kJ) than was absorbed breaking them (2648 kJ), so there is a net release of energy — which is exactly what you expect from burning a fuel.

Bond enthalpy calculations

Covers AS91164

Breaking and making bonds

Two facts drive every calculation on this page:

Breaking bonds requires energy — you must pull the atoms apart. Endothermic, energy in.
Making bonds releases energy — the atoms settle into a lower-energy arrangement. Exothermic, energy out.

What decides the sign of a reaction

Every reaction breaks the reactants' bonds and makes new ones in the products. Whether it is exothermic or endothermic depends on which is bigger:

More energy released making bonds than absorbed breaking them → exothermic, $\Delta_rH$ negative.
More energy absorbed breaking bonds than released making them → endothermic, $\Delta_rH$ positive.

The relationship

$\Delta_rH = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$

$\sum(\text{bonds broken})$ — total bond enthalpy of all bonds in the reactants.
$\sum(\text{bonds formed})$ — total bond enthalpy of all bonds in the products.
A useful way to remember the order: "broken minus formed" — reactants first, products second, same order as the equation.

Bond enthalpy

Bond enthalpy is the energy needed to break one mole of a particular bond.
It is always quoted as a positive number, because breaking always takes energy in.
Typical values (in $kJ\ mol^{-1}$ ): C–H 413, C–C 347, C=C 614, O=O 498, C=O 805, O–H 464, H–H 436.
Stronger bonds have larger values, so a C=C takes more energy to break than a C–C.

The method

Draw or list every bond in the reactants, and every bond in the products.
Count them carefully, including the coefficients in the equation.
Add up the bond enthalpies for the reactants — this is bonds broken.
Add up the bond enthalpies for the products — this is bonds formed.
Subtract: broken − formed.
Check the sign and give units.

Counting bonds without missing any

This is where nearly all the marks are lost, so be systematic:

Write out the structure of each molecule rather than working from the formula.
Multiply by the coefficient: $2H_2O$ contains 4 O–H bonds, not 2.
Count every bond, including all the C–H bonds in an organic molecule.
Tally them in a list before you add anything up.

Tips and Tricks

Get the order right: broken minus formed. Reversing it gives an answer with exactly the right number and exactly the wrong sign, which loses the marks for both the sign and the conclusion.

Worked ExampleA full bond enthalpy calculation

Use the bond enthalpies below to calculate $\Delta_rH$ for the combustion of methane, and state whether the reaction is exothermic or endothermic.

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

C–H = 413, O=O = 498, C=O = 805, O–H = 464 (all $kJ\ mol^{-1}$ )

Step 1 — List the bonds in the reactants

$CH_4$ — the carbon is bonded to four hydrogens: 4 × C–H

$2O_2$ — each oxygen molecule has one double bond, and there are two molecules: 2 × O=O

Step 2 — List the bonds in the products

$CO_2$ — the structure is $O{=}C{=}O$ , so there are 2 × C=O

$2H_2O$ — each water molecule has two O–H bonds, and there are two molecules: $2 \times 2 =$ 4 × O–H

Step 3 — Add up the bonds broken (reactants)

$\sum(\text{broken}) = (4 \times 413) + (2 \times 498)$ $= 1652 + 996$ $= 2648\ kJ\ mol^{-1}$

Step 4 — Add up the bonds formed (products)

$\sum(\text{formed}) = (2 \times 805) + (4 \times 464)$ $= 1610 + 1856$ $= 3466\ kJ\ mol^{-1}$

Step 5 — Subtract, in the right order

$\Delta_rH = \sum(\text{broken}) - \sum(\text{formed})$ $\Delta_rH = 2648 - 3466$ $\Delta_rH = -818\ kJ\ mol^{-1}$

Step 6 — Interpret the result

The value is negative, so the reaction is exothermic.