Why Boiling Water Does Not Break Its Covalent Bonds

At 100°C, water boils. Bubbles rise through the liquid, the water level drops, and vapour escapes into the air. But if boiling breaks the covalent bonds inside water molecules, what exactly is escaping from the beaker? Hydrogen and oxygen atoms? Hydrogen gas and oxygen gas?

Neither. Under ordinary boiling conditions, the escaping particles are still water molecules, \(\ce{H2O}\).

Boiling separates whole molecules. It does not break the covalent bonds holding the atoms within each molecule together. The energy supplied during boiling is used to overcome intermolecular attractions, particularly hydrogen bonding between water molecules.

The difference becomes much easier to understand when you look at what is happening inside a molecule compared with what is happening between neighbouring molecules.

01Two different types of attraction exist in molecular substances

Consider a single water molecule.

Its formula is \(\ce{H2O}\), meaning it contains two hydrogen atoms covalently bonded to one oxygen atom.

The two O-H covalent bonds hold the atoms together. They are part of the molecule’s internal structure.

Now place another water molecule nearby. The two molecules attract one another through intermolecular forces, including hydrogen bonding.

These are two completely different types of attraction.

Covalent bonds: within a molecule

A covalent bond forms when atoms share electrons.

In water, each oxygen atom shares electrons with two hydrogen atoms:

\[ \ce{H-O-H} \]

The O-H bonds are strong. They hold the atoms together as one water molecule.

These bonds are called intramolecular bonds because they exist within a molecule.

Intermolecular forces: between molecules

Water molecules also attract neighbouring water molecules.

Oxygen is more electronegative than hydrogen, so the shared electrons in each O-H bond are attracted more strongly towards oxygen.

This produces:

• A partially negative oxygen atom, \(\delta^-\).
• Two partially positive hydrogen atoms, \(\delta^+\).

The partially positive hydrogen of one water molecule is attracted to the partially negative oxygen of another.

This attraction is called a hydrogen bond.

A hydrogen bond is an intermolecular attraction, not another covalent O-H bond.

The distinction is:

If you want to understand exactly how the O-H bonds form, revisit Lewis Dot Diagrams for Atoms, Ions and Covalent Molecules.

02What actually happens when water boils?

Let’s compare liquid water with water vapour.

In liquid water, molecules are relatively close together. They move around, but intermolecular attractions keep them near one another.

There is a constantly changing network of hydrogen-bond attractions between neighbouring molecules.

When water is heated, its particles gain kinetic energy and move more rapidly.

Eventually, at the boiling point for the surrounding pressure, bubbles of water vapour form throughout the liquid.

What happens to the molecules inside those bubbles?

They move much further apart.

The energy supplied during boiling allows molecules to overcome enough of the intermolecular attractions holding them in the liquid. They can then move independently through the gas.

Crucially, each molecule still contains two hydrogen atoms covalently bonded to one oxygen atom.

The change can be represented as:

\[ \ce{H2O(l) -> H2O(g)} \]

Notice what has changed:

• The physical state changes from liquid to gas.
• The distance between molecules increases.
• Intermolecular attractions are overcome sufficiently for molecules to separate.
• The covalent O-H bonds remain intact.
• The chemical identity remains \(\ce{H2O}\).

An important detail: intermolecular attractions do not suddenly become nonexistent in the gas. Water molecules can still interact during collisions, and brief associations can occur. The key difference is that these attractions no longer hold the molecules together in a liquid structure.

Why doesn’t the temperature keep increasing during boiling?

At a fixed external pressure, a pure substance undergoing boiling stays approximately at its boiling temperature while liquid and vapour coexist.

The extra energy is being used for the phase change rather than increasing the average kinetic energy of the particles.

For water at approximately 101.3 kPa, this temperature is 100°C.

The energy required to convert one mole of liquid into gas at its boiling point is called the molar enthalpy of vaporisation, \(\Delta H_{\mathrm{vap}}\).

For water:

\[ \Delta H_{\mathrm{vap}} \approx 40.7\ \mathrm{kJ\,mol^{-1}} \]

This energy describes the liquid-to-gas transition. It is not the energy required to break the O-H covalent bonds.

03How much stronger are covalent bonds than intermolecular attractions?

It helps to compare the energy scales involved.

The molar enthalpy of vaporisation of water is approximately 40.7 kJ mol\(^{-1}\).

An average O-H covalent bond enthalpy is approximately 463 kJ mol\(^{-1}\).

These values describe different processes, but their sizes reveal why ordinary boiling does not break the O-H bonds.

04How can an experiment show that water molecules remain intact?

Watching water boil tells us that a gas is being produced. It does not, by itself, tell us the molecular identity of that gas.

To investigate the question properly, we need to collect or analyse the vapour.

The first experimental design: Boil and condense the water

Imagine an apparatus containing:

• A flask of liquid water connected to a heat source.
• A delivery tube leading from the flask.
• A water-cooled condenser connected to the delivery tube.
• A receiving flask at the other end to collect the condensed liquid.

As the water boils, vapour leaves the heated flask and travels through the delivery tube.

It then passes through the condenser, where cooling water circulates around the inner vapour tube. The vapour loses energy, condenses, and flows into the receiving flask.

What should this experiment show?

If boiling produces intact water molecules, cooling the vapour should produce liquid water again.

If the water had decomposed substantially into hydrogen and oxygen, we would expect different gaseous products rather than simply recovering the original water.

The collection of water supports the idea that water is being vaporised and condensed.

However, there is a limitation.

The receiving flask only tells us what condensed and was collected. It does not directly identify every substance that passed through the hot apparatus. Some gaseous products could escape without condensing, and a small amount of decomposition might go undetected.

We need a more direct investigation.

The revised design: Analyse the hot water vapour

A stronger approach is to examine the vapour before it cools.

One method uses infrared spectroscopy.

The apparatus consists of:

1. A heated flask containing water.
2. A heated sampling tube that carries the water vapour away from the flask.
3. A heated gas cell through which the vapour passes.
4. An infrared source on one side of the cell.
5. A detector on the other side that measures the radiation passing through.

The cell is kept hot enough to prevent the water vapour from condensing.

Infrared radiation passes through the sample. Water molecules absorb particular frequencies associated with vibrations of their O-H bonds and their bent molecular structure.

An absorption pattern characteristic of gaseous \(\ce{H2O}\) provides evidence that intact water molecules are present in the vapour.

What does the revised experiment establish?

A gas-phase spectrum characteristic of water supports the conclusion that the vapour contains intact \(\ce{H2O}\) molecules.

This is more direct evidence than simply collecting the condensed liquid.

There is still an experimental limitation. Infrared spectroscopy does not detect every gas equally well. In particular, \(\ce{H2}\) and \(\ce{O2}\) do not produce strong characteristic infrared absorption in the same way that water does. A separate gas-analysis method would be needed to test for those products directly.

Also, no measurement proves that absolutely zero molecules undergo bond-breaking events. Experiments have detection limits.

The important conclusion is that ordinary boiling produces predominantly intact water molecules, not significant quantities of hydrogen and oxygen through covalent bond cleavage.

05When can heating actually break covalent bonds?

Boiling is a physical change, but heating can also cause chemical changes.

The important distinction is whether the substance simply changes physical state or whether its molecular structure changes.

Contrast 1: Boiling water versus electrolysis

During boiling:

\[ \ce{H2O(l) -> H2O(g)} \]

The water molecules remain chemically unchanged.

During electrolysis, electrical energy drives a chemical reaction that decomposes water into hydrogen and oxygen:

\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]

The original O-H covalent bonds are disrupted, and new H-H and O=O bonds form.

The products are chemically different substances.

Notice that both processes can involve energy transfer, but energy transfer alone does not tell us whether a chemical reaction has occurred.

The products and changes in bonding tell us that.

Contrast 2: Molecular substances versus covalent network structures

There is another important distinction.

So far, we have discussed molecular substances such as water. These consist of separate molecules held near one another by intermolecular forces.

However, not every covalent substance contains separate molecules.

Consider silicon dioxide, \(\ce{SiO2}\), which occurs naturally in substances such as quartz.

Although its formula resembles that of carbon dioxide, \(\ce{CO2}\), ordinary solid silicon dioxide does not consist of individual \(\ce{SiO2}\) molecules.

Instead, its atoms form an extended covalent network.

Each silicon atom is bonded to four oxygen atoms, and each oxygen atom connects two silicon atoms. The covalent bonding extends throughout the structure.

This changes how we think about heating the substance.

For molecular water, we can separate intact molecules by overcoming intermolecular attractions.

For a covalent network, there are no separate molecules to pull apart. Heating and melting involve disruption and rearrangement of covalent bonds within the extended structure.

This is why giant covalent structures generally have much higher melting temperatures than ordinary molecular substances.

For a more detailed comparison of diamond, silicon dioxide, and molecular covalent substances, see Covalent Network Structures: Diamond, SiO2 and Molecules.

06A reliable way to answer HSC questions about boiling

When a question asks you to explain why boiling does not break covalent bonds, work through three ideas.

First: Identify the structural units.

Does the substance consist of separate molecules, or is it an extended covalent network?

Second: Identify what holds those units together.

For a molecular substance, distinguish the covalent bonds within molecules from the intermolecular attractions between molecules.

Third: Identify what changes during the process.

If the substance simply changes from liquid to gas, its molecular identity remains unchanged. The intermolecular attractions are overcome sufficiently for the molecules to separate.

A strong HSC explanation would read:

During boiling, energy is supplied to overcome the intermolecular forces between neighbouring molecules. This allows the molecules to move further apart and enter the gaseous state. The covalent bonds within each molecule remain intact, so the chemical identity of the substance does not change.

Avoid writing that boiling breaks covalent bonds because the particles gain enough energy to move apart.

The ability to move independently is not the same as the ability to separate the atoms within a molecule.

07Questions and solutions

Question 1

Water is heated until it reaches its normal boiling point of 100°C.

A student claims that the O-H covalent bonds break because the water molecules are moving further apart.

Explain why this claim is incorrect. Identify the type of attraction that is overcome during boiling.

Solution 1

The O-H covalent bonds remain intact. Boiling overcomes intermolecular attractions between neighbouring water molecules, rather than breaking the bonds within each molecule.

Each water molecule contains two hydrogen atoms covalently bonded to one oxygen atom.

In liquid water, hydrogen bonding and other intermolecular attractions keep neighbouring molecules relatively close together.

When water boils, sufficient energy is supplied to overcome these attractions, allowing whole water molecules to enter the gas phase.

The process is:

\[ \ce{H2O(l) -> H2O(g)} \]

The formula remains \(\ce{H2O}\), so the chemical identity is unchanged.

The student’s mistake is confusing the separation of molecules with the separation of atoms within a molecule. These are different processes.

Question 2

A sample of 0.350 mol of ethanol, \(\ce{CH3CH2OH}\), is heated to its normal boiling point and completely vaporised.

The molar enthalpy of vaporisation of ethanol is \(38.6\ \mathrm{kJ\,mol^{-1}}\).

(a) Calculate the energy required to vaporise the sample.

(b) Explain what happens to the covalent bonds within each ethanol molecule during vaporisation.

(c) Identify the main intermolecular attraction that must be overcome between ethanol molecules.

Solution 2

(a) The energy required is 13.5 kJ. (b) The covalent bonds remain intact. (c) Intermolecular hydrogen-bond attractions are particularly important.

Step 1: Identify the relevant equation.

\[ q=n\Delta H_{\mathrm{vap}} \]

Step 2: Substitute the given values.

\[ q=0.350\times38.6 \]

\[ q=13.51\ \mathrm{kJ} \]

Therefore:

\[ \boxed{q=13.5\ \mathrm{kJ}} \]

Step 3: Explain what happens to the ethanol molecules.

Ethanol contains C-C, C-H, C-O, and O-H covalent bonds.

These bonds remain intact during ordinary vaporisation. Each ethanol molecule enters the gas phase without losing its molecular structure.

Step 4: Identify the intermolecular attractions.

Ethanol molecules can form hydrogen bonds because they contain an O-H group.

They also experience dipole-dipole and London dispersion attractions.

The supplied energy allows the molecules to separate by overcoming these intermolecular attractions sufficiently.

The important point is that ethanol does not need to break its O-H or C-C covalent bonds to become a gas.

Question 3

A student investigates whether boiling water breaks covalent bonds.

They heat 12.0 g of liquid water, direct the vapour through a condenser, and collect 11.4 g of liquid in a receiving flask.

The student concludes:

“The experiment proves that all water molecules remained intact during boiling because the collected substance is still water.”

(a) Calculate the percentage recovery of water.

(b) Explain what the result supports.

(c) Identify two possible reasons why 0.6 g of water was not recovered.

(d) Explain why this experiment cannot conclusively establish the molecular identity of every particle in the hot vapour. Propose a more direct experimental method.

Solution 3

(a) The percentage recovery is 95.0%. (b) The result supports the presence of water in the condensed vapour, but it does not prove that every particle in the hot vapour was an intact water molecule.

(a) Calculate the percentage recovery.

Use:

\[ \text{Recovery}= \frac{\text{mass recovered}} {\text{initial mass}}\times100 \]

Substitute:

\[ \text{Recovery}= \frac{11.4}{12.0}\times100 \]

\[ \boxed{\text{Recovery}=95.0\%} \]

(b) Interpret the observation.

The experiment successfully collected 11.4 g of liquid after the vapour was cooled.

This supports the conclusion that water was present in the vapour and condensed back into liquid water.

However, the observation does not identify every substance that travelled through the apparatus.

(c) Explain the missing mass.

The missing mass is:

\[ 12.0-11.4=0.6\ \mathrm{g} \]

Possible explanations include:

• Some water vapour escaped through connections or was not condensed.
• Some liquid remained inside the delivery tube or condenser.

These are experimental losses, not evidence that covalent bonds were broken.

(d) Propose a better investigation.

The main limitation is that the experiment analyses the condensed liquid rather than directly examining the hot vapour.

A more direct method would use a heated gas-phase infrared spectroscopy cell.

The vapour would travel through a heated sampling tube into the cell, where infrared radiation would pass through it.

A characteristic water-vapour absorption spectrum would provide evidence that intact \(\ce{H2O}\) molecules are present in the gas.

A separate gas-analysis method could test for possible hydrogen and oxygen products.

This approach provides more direct evidence about the molecular identity of the vapour. It still has detection limits, so it cannot establish that absolutely no decomposition occurs.

Question 4

Carbon dioxide and silicon dioxide have the formulas \(\ce{CO2}\) and \(\ce{SiO2}\), respectively.

A student argues that both substances must consist of separate molecules because their formulas contain one central atom and two oxygen atoms.

Explain why this reasoning is incorrect.

In your answer, compare:

• The structural units in each substance.
• The bonding within and between those units.
• What must happen to the bonding when each substance is heated and converted into a gaseous form.

Solution 4

Carbon dioxide consists of separate molecules, whereas ordinary solid silicon dioxide forms an extended covalent network. Their similar-looking formulas do not mean they have the same structure.

Carbon dioxide:

Carbon dioxide consists of individual linear molecules:

\[ \ce{O=C=O} \]

Within each molecule, carbon and oxygen are held together by strong covalent bonds.

Between separate carbon dioxide molecules, much weaker intermolecular attractions operate, mainly London dispersion forces.

When solid carbon dioxide sublimes at normal atmospheric pressure, it produces gaseous \(\ce{CO2}\) molecules.

The C=O covalent bonds remain intact.

Silicon dioxide:

Ordinary solid silicon dioxide does not consist of individual \(\ce{SiO2}\) molecules.

Instead, it forms a continuous three-dimensional covalent network.

Each silicon atom is bonded to four oxygen atoms, and each oxygen atom bridges two silicon atoms.

The formula \(\ce{SiO2}\) represents the overall ratio of silicon to oxygen atoms, not the composition of a separate molecule.

Heating silicon dioxide therefore does not simply separate intact \(\ce{SiO2}\) molecules by overcoming intermolecular attractions.

Its extended covalent network must undergo bond disruption and rearrangement as the material melts or undergoes further high-temperature changes.

The student’s mistake is assuming that a chemical formula always represents an individual molecule.

The correct approach is to identify the actual structural units before predicting how a substance behaves when heated.

Question 5

A chemistry investigation is designed to distinguish between two possible explanations for the behaviour of boiling water.

A sample containing 0.100 mol of liquid water is heated to 100°C at a pressure of 101.3 kPa.

The vapour is collected in a heated gas syringe maintained at the same temperature and pressure, preventing condensation before the volume is measured.

The following information is recorded:

• Initial amount of water: 0.100 mol.
• Temperature of collected gas: 373 K.
• Pressure: 101.3 kPa.
• Measured gas volume: \(3.06\pm0.02\ \mathrm{L}\).
• A gas-phase infrared spectrum shows strong absorption features characteristic of water vapour.

Assume ideal gas behaviour, no loss of material, and that any decomposition follows:

\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]

All products remain gaseous and are collected.

(a) Calculate the expected volume if all water molecules remain intact during vaporisation.

(b) Calculate the expected volume if all the water decomposes completely according to the given equation.

(c) Suppose a fraction \(f\) of the original water molecules decomposes, while the remainder vaporises without decomposition. Derive an expression for the total gas volume in terms of \(f\) and the volume calculated in part (a).

(d) Use the upper limit of the measured volume to estimate the maximum percentage of water molecules that could have decomposed, under the stated assumptions.

(e) A student argues that the large volume of gas proves that boiling has created additional molecules. Evaluate this argument using the volume measurements, reaction stoichiometry, and infrared evidence. State what the experiment establishes and what it cannot prove.

Solution 5

The results strongly support vaporisation of intact water molecules. Under the stated assumptions, the gas-volume measurement limits decomposition to approximately 1.2% or less. The larger gas volume does not mean that boiling creates additional chemical molecules.

The key is to distinguish the number of water molecules originally present from the number of gas particles produced by a possible chemical reaction.

(a) Calculate the volume if water molecules remain intact.

Use the ideal gas equation:

\[ PV=nRT \]

Rearrange to make volume the subject:

\[ V=\frac{nRT}{P} \]

Use:

• \(n=0.100\ \mathrm{mol}\)
• \(R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\)
• \(T=373\ \mathrm{K}\)
• \(P=101.3\ \mathrm{kPa}\)

Substitute:

\[ V=\frac{0.100\times8.314\times373}{101.3} \]

\[ \boxed{V=3.06\ \mathrm{L}} \]

This is the expected volume of 0.100 mol of intact water molecules in the gaseous state.

(b) Calculate the volume if all water decomposes.

Start with the balanced equation:

\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]

The equation tells us that:

• 2 mol of water produces 2 mol of hydrogen gas.
• 2 mol of water also produces 1 mol of oxygen gas.

Therefore, 2 mol of water produces 3 mol of gaseous particles.

For 0.100 mol of water:

\[ n(\ce{H2})=0.100\ \mathrm{mol} \]

\[ n(\ce{O2})=0.050\ \mathrm{mol} \]

Total gas:

\[ n_{\mathrm{total}}=0.150\ \mathrm{mol} \]

At the same temperature and pressure, gas volume is proportional to the number of moles.

Therefore:

\[ V_{\mathrm{decomp}}=1.5\times3.06 \]

\[ \boxed{V_{\mathrm{decomp}}=4.59\ \mathrm{L}} \]

Complete decomposition would produce a substantially larger gas volume than intact vaporisation.

(c) Derive an expression for partial decomposition.

Let \(f\) represent the fraction of the original water molecules that decomposes.

The fraction that remains intact is \(1-f\).

The amount of intact water vapour is:

\[ n_{\mathrm{H_2O}}=0.100(1-f) \]

From the balanced decomposition equation, each mole of decomposed water produces 1.5 mol of gaseous products.

Therefore:

\[ n_{\mathrm{products}}=1.5(0.100f) \]

The total number of moles of gas is:

\[ n_{\mathrm{gas}}=0.100(1-f)+0.150f \]

Simplifying:

\[ n_{\mathrm{gas}}=0.100(1+0.5f) \]

At constant temperature and pressure, volume is proportional to the total number of gas particles.

If \(V_0\) is the volume when no decomposition occurs:

\[ \boxed{V=V_0(1+0.5f)} \]

This relationship is important.

If \(f=0\), all molecules remain intact and \(V=V_0\).

If \(f=1\), all molecules decompose and \(V=1.5V_0\).

(d) Calculate the maximum possible decomposition.

The measured volume is:

\[ 3.06\pm0.02\ \mathrm{L} \]

The upper limit is:

\[ V_{\mathrm{max}}=3.08\ \mathrm{L} \]

Using the expression derived in part (c):

\[ 3.08=3.061(1+0.5f) \]

Divide both sides by 3.061:

\[ \frac{3.08}{3.061}=1+0.5f \]

Rearrange:

\[ f=2\left(\frac{3.08}{3.061}-1\right) \]

\[ f\approx0.012 \]

Convert to a percentage:

\[ \boxed{f\approx1.2\%} \]

Therefore, under the stated assumptions, no more than approximately 1.2% of the original water molecules could have decomposed without exceeding the upper measured volume.

(e) Evaluate the student’s argument.

The student’s argument is incorrect because a large gas volume does not necessarily indicate that additional molecules have been chemically produced.

Consider the two possible processes.

During ordinary vaporisation:

\[ \ce{H2O(l) -> H2O(g)} \]

One water molecule becomes one water molecule in the gas phase.

The total amount remains 0.100 mol.

The molecules simply occupy a much larger volume because gas particles are much further apart than particles in a liquid.

During complete decomposition:

\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]

Two water molecules produce three gaseous molecules.

The number of gas particles increases because a chemical reaction has occurred.

The measured volume of \(3.06\pm0.02\ \mathrm{L}\) agrees with the prediction for intact water vapour and is far below the 4.59 L predicted for complete decomposition.

The infrared spectrum provides additional evidence that intact \(\ce{H2O}\) molecules are present in the hot vapour.

Together, the measurements strongly support ordinary vaporisation rather than substantial chemical decomposition.

However, the results do not establish that absolutely no decomposition occurs. The volume calculation places an approximate upper bound under the specified reaction model, while the infrared measurement has its own detection limits.

The deeper lesson is that a change in the physical arrangement of particles is not the same as a change in their chemical identity.

08Where to take this concept next

Once you can distinguish covalent bonds within molecules from intermolecular forces between molecules, you can use the same reasoning to explain why different molecular substances have different boiling points.

The next useful step is comparing hydrogen bonding, dipole-dipole attractions, and London dispersion forces, then predicting how their relative strengths affect boiling points.

That is where understanding the structure of a molecule starts becoming a powerful tool for predicting the physical properties of an entire substance.