Bond Energy Revision Notes for HSC SSCE Chemistry

Bond Energy

What is bond energy?

Bond energy is the amount of energy needed to break a chemical bond. When we break bonds apart, we must supply energy to overcome the forces holding the atoms together. This concept helps us understand why some compounds are more stable than others and why certain reactions release or absorb energy.

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The symbol commonly used for bond energy is $B$ . For example, we write $B_{\text{Cl-Cl}}$ for the chlorine-chlorine bond energy, $B_{\text{C-H}}$ for the carbon-hydrogen bond energy, and $B_{\text{C-C}}$ for the carbon-carbon bond energy.

Examples of bond energies:

Chlorine molecule: The $\text{Cl-Cl}$ bond energy is the enthalpy change when a chlorine molecule breaks into two separate chlorine atoms:

$\text{Cl}_2(g) \rightarrow 2\text{Cl}(g)$

Methane molecule: The $\text{C-H}$ bond energy is one quarter of the enthalpy change needed to break methane into completely separated atoms. We divide by four because methane contains four $\text{C-H}$ bonds:

$\text{CH}_4(g) \rightarrow \text{C}(g) + 4\text{H}(g)$

Ethane molecule: The $\text{C-C}$ bond energy is the enthalpy change when ethane ( $\text{C}_2\text{H}_6$ or $\text{H}_3\text{C-CH}_3$ ) breaks into two $\text{CH}_3$ fragments:

$\text{H}_3\text{C-CH}_3(g) \rightarrow 2\text{CH}_3(g)$

Bond energies can be calculated using standard enthalpies of formation and Hess's law. However, we sometimes need an additional quantity called the heat of atomisation.

Heat of atomisation

The heat of atomisation ( $\Delta H_{\text{at}}^{\circ}$ ) of an element is the enthalpy change required to produce gaseous atoms from the element in its standard state. Standard state means the element at a pressure of $100.0 \text{ kPa}$ and at the specified temperature (usually $298 \text{ K}$ or $25°\text{C}$ ). This value is expressed per mole of atoms produced.

Example for carbon: The heat of atomisation of carbon at $25°\text{C}$ is the $\Delta H^{\circ}$ per mole for:

$\text{C}(s) \rightarrow \text{C}(g)$

This represents the energy needed to convert solid carbon (graphite) into individual gaseous carbon atoms.

Example for chlorine: The heat of atomisation of chlorine at $25°\text{C}$ is the $\Delta H^{\circ}$ per mole of chlorine atoms. Since chlorine naturally exists as $\text{Cl}_2$ molecules, we use half a mole of molecules to get one mole of atoms:

$\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g)$

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Important relationship: The heat of atomisation of an element equals the standard heat of formation of gaseous atoms of that element.

Table of heats of atomisation

Atom	$\Delta H_{\text{at}}^{\circ}$ (kJ mol⁻¹)	Atom	$\Delta H_{\text{at}}^{\circ}$ (kJ mol⁻¹)	Atom	$\Delta H_{\text{at}}^{\circ}$ (kJ mol⁻¹)
H	218	Br	112	Ca	177
O	249	I	107	Zn	131
N	473	C	717	P	315
Cl	122	Na	109	S	279

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Key observations from the table:

Nitrogen has a very high heat of atomisation ( $473 \text{ kJ mol}^{-1}$ ) because the $\text{N}\equiv\text{N}$ triple bond is very strong
Carbon also has a high value ( $717 \text{ kJ mol}^{-1}$ ) because it must be converted from solid graphite to gaseous atoms
Iodine has a relatively low value ( $107 \text{ kJ mol}^{-1}$ ) because the $\text{I-I}$ bond is weaker than other halogen bonds

Calculating bond energies using Hess's law

We can calculate bond energies by combining standard enthalpies of formation with heats of atomisation, using Hess's law to connect different reaction pathways. The key principle is that the total enthalpy change is the same regardless of which path we take.

lightbulbExample

Worked Example: Calculating the O-H bond energy in water

Let's calculate the bond energy of the $\text{O-H}$ bond in water using an enthalpy diagram approach.

Step 1: Set up the enthalpy pathways

We have two pathways from $\text{H}_2\text{O}(g)$ to separated atoms $2\text{H}(g) + \text{O}(g)$ :

Direct pathway: $\Delta H_4$
Indirect pathway: $\Delta H_1 + \Delta H_2 + \Delta H_3$

By Hess's law, these must be equal:

$\Delta H_4 = \Delta H_1 + \Delta H_2 + \Delta H_3 \quad \text{...(equation 1)}$

Step 2: Identify what each $\Delta H$ represents

$\Delta H_1$ is the negative of $\Delta H_{\text{f}}^{\circ}(\text{H}_2\text{O}, g)$ (going backwards from water to elements)
$\Delta H_2$ is twice $\Delta H_{\text{at}}(\text{H})$ (converting one mole of $\text{H}_2$ to two moles of H atoms)
$\Delta H_3$ is $\Delta H_{\text{at}}(\text{O})$ (converting half a mole of $\text{O}_2$ to one mole of O atoms)
$\Delta H_4 = 2B_{\text{O-H}}$ (breaking two $\text{O-H}$ bonds in water)

Step 3: Substitute into equation 1

$2B_{\text{O-H}} = -\Delta H_{\text{f}}^{\circ}(\text{H}_2\text{O}, g) + 2\Delta H_{\text{at}}(\text{H}) + \Delta H_{\text{at}}(\text{O})$

Step 4: Insert numerical values

From Table 1: $\Delta H_{\text{f}}^{\circ}(\text{H}_2\text{O}, g) = -242 \text{ kJ mol}^{-1}$

From Table 2: $\Delta H_{\text{at}}(\text{H}) = 218 \text{ kJ mol}^{-1}$ and $\Delta H_{\text{at}}(\text{O}) = 249 \text{ kJ mol}^{-1}$

$2B_{\text{O-H}} = -(-242) + 2 \times 218 + 249$

$2B_{\text{O-H}} = 242 + 436 + 249 = 927 \text{ kJ mol}^{-1}$

$\boxed{B_{\text{O-H}} = 463 \text{ kJ mol}^{-1}}$

Common bond energies

The table below shows bond energies for common chemical bonds at $298 \text{ K}$ . These values are averages measured across many different compounds containing each type of bond.

Bond	Bond energy (kJ mol⁻¹)	Bond	Bond energy (kJ mol⁻¹)	Bond	Bond energy (kJ mol⁻¹)
H—Br	366	I—I	151	C≡O in CO₂	805
H—Cl	432	N—N	163	C≡O in CO	1080
H—I	299	O—O	146	O=O in O₂	498
H—O	463	C—C	348	O—O in O₃	302
H—N	391	C—H	413	N≡N in N₂	945
H—S	360	C—O	358	C=C in C₂H₄	610
H—H	436	C—Cl	339	C≡C in C₂H₂	835
Br—Br	193	C—Br	285
Cl—Cl	242	C—F	485

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Key patterns to notice:

Single bonds have lower energies than double bonds, which have lower energies than triple bonds
For hydrogen-halogen bonds: $\text{H-Cl}$ (432) $>$ $\text{H-Br}$ (366) $>$ $\text{H-I}$ (299)
The $\text{C-F}$ bond is particularly strong (485 kJ mol⁻¹)
Multiple bonds: $\text{N}\equiv\text{N}$ (945) is one of the strongest bonds

Bond energy, stability and reactivity

There is a direct relationship between bond energy and the stability and reactivity of compounds: the greater the bond energy, the more stable the compound and the less reactive it is. This principle helps us predict and explain chemical behaviour.

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Example 1: Oxygen vs ozone

Oxygen ( $\text{O}_2$ ) is a very stable gas that we breathe safely. Its allotrope ozone ( $\text{O}_3$ ) is notoriously unstable - samples of liquid ozone can spontaneously explode! This dramatic difference arises from their bond energies:

Bond energy in $\text{O}_2$ : 498 kJ mol⁻¹
Bond energy in $\text{O}_3$ : 302 kJ mol⁻¹

The much lower bond energy in ozone makes it highly unstable and very reactive compared to normal oxygen.

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Example 2: Water vs hydrogen peroxide

Water ( $\text{H-O-H}$ ) is very stable and relatively unreactive - which is fortunate since it covers most of Earth's surface! Hydrogen peroxide ( $\text{H-O-O-H}$ ) is much less stable and considerably more reactive:

$\text{O-O}$ bond energy in hydrogen peroxide: 146 kJ mol⁻¹
$\text{O-H}$ bond energy: 463 kJ mol⁻¹

The two $\text{H-O}$ bonds in hydrogen peroxide have similar energies to those in water. However, the strength of the weakest bond in a molecule determines its stability and reactivity. The weak $\text{O-O}$ bond makes hydrogen peroxide much more prone to decomposition and chemical reactions than water.

Estimating ΔH from bond energies

We can use bond energies to estimate the enthalpy change ( $\Delta H^{\circ}$ ) for a reaction. The method involves imagining that all bonds in the reactants break first, then new bonds form to create the products.

Key equation:

$\Delta H^{\circ} = \{\text{energy required to break bonds}\} - \{\text{energy released when new bonds form}\}$

Alternatively:

$\Delta H^{\circ} = \{\text{sum of bond energies of bonds broken}\} - \{\text{sum of bond energies of bonds formed}\} \quad \text{...(15.7)}$

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Remember: breaking bonds requires energy (positive value), while forming bonds releases energy (negative value).

lightbulbExample

Worked Example: Ethane reacting with bromine

Calculate $\Delta H^{\circ}$ for the reaction between ethane and bromine to form bromoethane ( $\text{C}_2\text{H}_5\text{Br}$ ) and hydrogen bromide, with all substances in the gas phase.

Step 1: Write the balanced equation

$\text{C}_2\text{H}_6(g) + \text{Br}_2(g) \rightarrow \text{C}_2\text{H}_5\text{Br}(g) + \text{HBr}(g)$

Step 2: Identify bonds broken and formed

Bonds broken:

One $\text{C-H}$ bond in ethane
One $\text{Br-Br}$ bond in bromine

Bonds formed:

One $\text{C-Br}$ bond in bromoethane
One $\text{H-Br}$ bond in hydrogen bromide

Step 3: Apply equation 15.7

$\Delta H^{\circ} = B_{\text{C-H}} + B_{\text{Br-Br}} - B_{\text{C-Br}} - B_{\text{H-Br}}$

Step 4: Substitute values from Table 3

$\Delta H^{\circ} = 413 + 193 - 285 - 366$

$\boxed{\Delta H^{\circ} = -45 \text{ kJ mol}^{-1}}$

The negative value indicates this reaction is exothermic - it releases energy.

The energy diagram above shows the step-by-step process. The reactants first gain energy as bonds break, then release more energy as new bonds form, resulting in a net energy release of $45 \text{ kJ mol}^{-1}$ .

Accuracy and limitations

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Important consideration about accuracy:

Enthalpy changes calculated from bond energies are not particularly accurate. This is because bond energies vary slightly depending on the molecule. For example:

The $\text{C-Cl}$ bond energy in chloromethane is not exactly the same as the $\text{C-Cl}$ bond energy in dichloroethane
Values in Table 15.3 are averages across many compounds

When to use bond energies:

For discussing stability and reactivity of compounds
For estimating $\Delta H^{\circ}$ when standard enthalpies of formation are not available
For getting a quick approximation of reaction energetics

For accurate calculations: Use standard enthalpies of formation with Hess's law whenever possible.

bookmarkSummary

Key Points to Remember:

Bond energy is the energy needed to break a chemical bond. The symbol $B$ is used, such as $B_{\text{C-H}}$ or $B_{\text{O-H}}$ .
Heat of atomisation ( $\Delta H_{\text{at}}^{\circ}$ ) is the enthalpy change to convert an element in its standard state into gaseous atoms, expressed per mole of atoms.
Strong bonds mean stable molecules: Higher bond energies indicate more stable, less reactive compounds. Weak bonds make molecules unstable and reactive.
To estimate ΔH for reactions: Use the formula $\Delta H^{\circ} = \text{(bonds broken)} - \text{(bonds formed)}$ . Breaking bonds requires energy, forming bonds releases energy.
Bond energies are averages: Values vary slightly between different molecules, so calculations give estimates rather than exact values. Use standard enthalpies of formation for accurate results.

Bond Energy (HSC SSCE Chemistry): Revision Notes

Bond Energy

What is bond energy?

Heat of atomisation

Table of heats of atomisation

Calculating bond energies using Hess's law

Common bond energies

Bond energy, stability and reactivity

Estimating ΔH from bond energies

Accuracy and limitations

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