Why Physical State Matters in a Hess-Law Calculation

A student is given the equation

\[
\ce{H2(g) + 1/2O2(g) -> H2O(l)}
\]

with an enthalpy change of \(-285.8\ \mathrm{kJ\,mol^{-1}}\). They then use the same value for the reaction that produces water vapour:

\[
\ce{H2(g) + 1/2O2(g) -> H2O(g)}
\]

The atoms are identical. The bonds formed are identical. Surely the energy change should be the same?

It isn't. The second reaction has an enthalpy change of approximately \(-241.8\ \mathrm{kJ\,mol^{-1}}\). There is a difference of \(44.0\ \mathrm{kJ\,mol^{-1}}\).

**Physical state matters in a Hess's Law calculation because enthalpy depends on the exact initial and final states of the substances, not just their chemical formulas.** Liquid water and water vapour contain the same molecules, but they have different enthalpies. Changing between them involves an energy transfer that must be included in the calculation.

This is why state symbols such as \((s)\), \((l)\), \((g)\), and \((aq)\) are part of a thermochemical equation, not optional decorations.

## Why the same substance can have different enthalpies

Picture a glass of water sitting on a bench. Now imagine heating it until the water becomes vapour.

The chemical formula hasn't changed. Each molecule is still \(\ce{H2O}\). However, the molecules in the gas are much farther apart than those in the liquid.

In liquid water, neighbouring molecules attract one another through intermolecular forces, particularly hydrogen bonding. Energy must be supplied to separate the molecules during vaporisation.

The reverse process, condensation, releases energy.

[IMAGE: Two scientifically accurate molecular views side by side: liquid water molecules close together in an irregular arrangement with dotted intermolecular hydrogen bonds, and water vapour molecules widely separated. Show that the individual H2O molecules and their internal covalent bonds remain unchanged while their intermolecular arrangement changes.]

The important distinction is between two kinds of interaction:

- Intramolecular bonds hold the atoms together within each water molecule.
- Intermolecular attractions act between neighbouring water molecules.

A bond-enthalpy calculation considers the first type. A physical state change involves the second.

This is one reason a calculation that counts covalent bonds can give a different answer from an experimentally determined reaction enthalpy.

### The energy difference between liquid water and water vapour

At \(298.15\ \mathrm{K}\) (approximately \(25^\circ\mathrm{C}\)) and a standard pressure of \(1\ \mathrm{bar}\), the approximate standard enthalpies of formation of water are:

| Formation reaction | Standard enthalpy change |
|---|---:|
| \(\ce{H2(g) + 1/2O2(g) -> H2O(l)}\) | \(-285.83\ \mathrm{kJ\,mol^{-1}}\) |
| \(\ce{H2(g) + 1/2O2(g) -> H2O(g)}\) | \(-241.83\ \mathrm{kJ\,mol^{-1}}\) |

These are rounded reference values from the NIST Chemistry WebBook.

The difference is:

\[
\begin{aligned}
\Delta H_{\mathrm{vap}}
&= -241.83-(-285.83)\\
&=+44.00\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Therefore:

\[
\ce{H2O(l) -> H2O(g)}
\qquad
\Delta H=+44.00\ \mathrm{kJ\,mol^{-1}}
\]

Vaporisation is endothermic because energy is absorbed to separate the water molecules. Condensation is the reverse process:

\[
\ce{H2O(g) -> H2O(l)}
\qquad
\Delta H=-44.00\ \mathrm{kJ\,mol^{-1}}
\]

Notice the sign change. The two equations describe opposite processes, so their enthalpy changes must have opposite signs.

[IMAGE: Annotated Hess energy-level diagram for water formation. Show H2(g) + 1/2O2(g) at the top reference level, H2O(g) at an intermediate level with formation enthalpy -241.83 kJ mol-1, and H2O(l) at a lower level with formation enthalpy -285.83 kJ mol-1. Label the downward gas-to-liquid difference -44.00 kJ mol-1 and the reverse liquid-to-gas difference +44.00 kJ mol-1. Make the two alternative formation pathways clear.]

The lower enthalpy of liquid water is not caused by different covalent bonds inside its molecules. It comes from the attractive interactions between neighbouring molecules.

This is the physical reason the state symbol changes the enthalpy.

## Seeing the energy of a state change experimentally

There is a useful way to investigate this energy difference without carrying out a chemical reaction.

Imagine measuring the energy needed to turn liquid water into steam. This is a physical change, but it gives us a direct way to investigate why the same chemical substance can have different enthalpies in different states.

### How the experiment works

A simple electrical heating experiment uses:

- An insulated vessel containing water.
- An electrical heating element immersed in the water.
- An ammeter connected in series with the heater.
- A voltmeter connected in parallel across the heater.
- A thermometer to monitor the water temperature.
- A balance to measure the mass of water before and after heating.
- A stopwatch to measure the heating time.

The vessel has a loose, vented lid to reduce heat loss while allowing steam to escape safely. It must never be sealed.

[IMAGE: Clearly labelled experimental apparatus for measuring the specific latent heat of vaporisation of water. Show an insulated vessel containing boiling water, an immersed electrical heating element, thermometer, ammeter in series, voltmeter in parallel, a vented lid, and steam escaping through the vent. Include a balance beneath the vessel as a separate measurement step. Show correct electrical connections.]

First, the water is heated until it reaches its boiling point. Once boiling is established, the temperature remains approximately constant while water continues to turn into steam.

Before reading further, make a prediction: if the temperature stays constant, is the heater still transferring energy into the water?

Yes. The mass of liquid water decreases as steam escapes, even though the temperature does not rise.

The energy is being used to change the physical state rather than increase the average kinetic energy of the molecules.

This is called the latent heat of vaporisation: the energy required to convert a unit mass of liquid into gas at a specified temperature and pressure.

### A small experimental calculation

Suppose an illustrative experiment produces the following measurements:

| Measurement | Value |
|---|---:|
| Voltage across heater | \(12.0\ \mathrm{V}\) |
| Current through heater | \(4.00\ \mathrm{A}\) |
| Heating time | \(300\ \mathrm{s}\) |
| Mass of water vaporised | \(6.40\ \mathrm{g}\) |

Assume all electrical energy supplied goes into vaporising the water. In a real experiment, heat losses and energy absorbed by the apparatus would need to be considered.

Electrical energy supplied is:

\[
E=VIt
\]

where \(E\) is energy in joules, \(V\) is voltage in volts, \(I\) is current in amperes, and \(t\) is time in seconds.

Substituting:

\[
\begin{aligned}
E&=12.0\times4.00\times300\\
&=14400\ \mathrm{J}\\
&=14.4\ \mathrm{kJ}
\end{aligned}
\]

The energy required per gram of water is:

\[
L_v=\frac{14.4}{6.40}
=2.25\ \mathrm{kJ\,g^{-1}}
\]

Converting to a molar quantity using the molar mass of water, \(18.0\ \mathrm{g\,mol^{-1}}\):

\[
\begin{aligned}
\Delta H_{\mathrm{vap}}
&=2.25\times18.0\\
&=40.5\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The result is approximately \(40.5\ \mathrm{kJ\,mol^{-1}}\) at the boiling temperature, close to the accepted value of approximately \(40.7\ \mathrm{kJ\,mol^{-1}}\) at \(100^\circ\mathrm{C}\).

The experiment demonstrates that changing liquid water into water vapour requires a substantial energy input, even though the chemical formula stays the same.

There is an important limitation: this measurement is made at approximately \(100^\circ\mathrm{C}\), whereas the standard enthalpy values in the earlier table refer to \(298.15\ \mathrm{K}\). They cannot be treated as identical without accounting for temperature-dependent enthalpy changes.

### What the experimental design must control

An initial, poorly insulated setup would lose some energy to the surroundings. Some energy would also heat the vessel and the electrical element rather than vaporise water.

That would make the calculated latent heat too low.

A better design uses insulation, a vented lid, a consistent boiling temperature, and measurements taken after steady boiling begins. A more accurate investigation would also account for energy absorbed by the apparatus and heat lost to the surroundings.

The key experimental lesson is that an enthalpy value belongs to a specified initial and final state under specified conditions. It is not simply a number attached to a chemical formula.

## Complete thermochemical equations: what must match?

A thermochemical equation gives you more information than an ordinary balanced chemical equation.

It specifies:
- The substances reacting and forming.
- The mole ratio between them.
- The physical state of every substance.
- The enthalpy change for the reaction as written.

Consider:

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

\[
\Delta H=-571.6\ \mathrm{kJ}
\]

This equation describes the formation of two moles of liquid water. If you divide every coefficient by two, you must also divide the enthalpy change by two:

\[
\ce{H2(g) + 1/2O2(g) -> H2O(l)}
\]

\[
\Delta H=-285.8\ \mathrm{kJ}
\]

The same rule applies when multiplying or reversing an equation.

However, changing the physical state is not just another way of rewriting the equation. It describes an additional physical process with its own enthalpy change.

### When can substances cancel?

This is where students often make a costly mistake.

Suppose you have:

\[
\ce{H2O(l)}
\]

on one side of an equation and:

\[
\ce{H2O(g)}
\]

on the other.

Can you cancel them?

No. They have the same chemical formula, but they represent different physical states and different enthalpies.

For cancellation, the chemical species and physical states must match exactly.

| Species | Can it cancel with...? |
|---|---|
| \(\ce{H2O(l)}\) | \(\ce{H2O(l)}\) |
| \(\ce{H2O(g)}\) | \(\ce{H2O(g)}\) |
| \(\ce{H2O(l)}\) | Not \(\ce{H2O(g)}\) |
| \(\ce{NaCl(s)}\) | Not \(\ce{NaCl(aq)}\) |

The last comparison matters too. Dissolving a substance involves interactions between its particles and the surrounding water molecules. The aqueous state cannot automatically be treated as equivalent to the pure substance.

[IMAGE: A chemical equation comparison showing H2O(l) and H2O(g) as separate species, with a labelled phase-change equation connecting them. Include a second comparison of NaCl(s) and NaCl(aq), illustrating the difference between a solid ionic lattice and separated hydrated ions.]

### A reliable decision process

Before adding any thermochemical equations, use this sequence:

1. **Write the target reaction completely.** Include every state symbol and coefficient.
2. **Compare the given equations with the target.** Check both the formulas and the physical states.
3. **Adjust the equations.** Reverse reactions or multiply coefficients where necessary. Apply the same changes to their enthalpy values.
4. **Add any missing phase-change reactions.** For example, convert liquid water into water vapour if the target requires gaseous water.
5. **Cancel identical species.** Do not cancel substances that have different state symbols.
6. **Add the enthalpy changes.** Check that the final equation matches the target exactly.

For phase changes, remember the direction of energy transfer:

| Physical change | Enthalpy change |
|---|---|
| Solid → liquid (melting) | Positive |
| Liquid → solid (freezing) | Negative |
| Liquid → gas (vaporisation) | Positive |
| Gas → liquid (condensation) | Negative |
| Solid → gas (sublimation) | Positive |

The magnitude depends on the substance and the temperature and pressure. Never assume every phase change has the same enthalpy.

## Worked example: Combustion of propane producing water vapour

A student is given the following thermochemical equation:

\[
\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}
\]

\[
\Delta H=-2220\ \mathrm{kJ\,mol^{-1}}
\]

Calculate the enthalpy change when propane undergoes complete combustion to produce water vapour instead.

Use the illustrative phase-change data:

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

\[
\Delta H=+44.0\ \mathrm{kJ\,mol^{-1}}
\]

Assume all values refer to the same temperature and pressure, and that the supplied combustion value is for one mole of propane.

**Step 1**: Identify the difference between the supplied and target equations.

The original reaction produces four moles of liquid water.

The target reaction must produce four moles of gaseous water.

We therefore need to convert four moles of liquid water into four moles of water vapour.

**Step 2**: Multiply the vaporisation equation by four.

\[
\ce{4H2O(l) -> 4H2O(g)}
\]

\[
\Delta H=4(+44.0)
\]

\[
\Delta H=+176.0\ \mathrm{kJ}
\]

Notice that the coefficient and enthalpy change both increase by a factor of four.

**Step 3**: Add the two thermochemical equations.

The liquid water appears on opposite sides, so it cancels.

The resulting equation is:

\[
\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)}
\]

**Step 4**: Add the enthalpy changes.

\[
\begin{aligned}
\Delta H&=-2220+176\\
&=-2044\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The combustion reaction producing water vapour has an enthalpy change of:

\[
\boxed{\Delta H=-2044\ \mathrm{kJ\,mol^{-1}}}
\]

The reaction is still exothermic, but it releases less energy than combustion producing liquid water.

Why? The gaseous water retains more enthalpy than liquid water. The additional energy that would have been released during condensation is not included in the gas-product reaction.

[IMAGE: A Hess cycle for propane combustion. Show propane and oxygen forming carbon dioxide and four liquid water molecules by the direct route, and an alternative route that continues through the vaporisation of four water molecules. Label the combustion enthalpy -2220 kJ mol-1, the phase-change enthalpy +176 kJ, and the overall gas-product enthalpy -2044 kJ mol-1.]

## Worked example: Combining three equations with different reaction directions and states

A chemist wants to calculate the enthalpy change for:

\[
\ce{CO(g) + H2O(g) -> CO2(g) + H2(g)}
\]

The following illustrative experimental thermochemical data are available:

**Reaction 1:**

\[
\ce{CO(g) + 1/2O2(g) -> CO2(g)}
\]

\[
\Delta H=-283.0\ \mathrm{kJ}
\]

**Reaction 2:**

\[
\ce{H2(g) + 1/2O2(g) -> H2O(l)}
\]

\[
\Delta H=-285.8\ \mathrm{kJ}
\]

**Reaction 3:**

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

\[
\Delta H=+44.0\ \mathrm{kJ\,mol^{-1}}
\]

Calculate the enthalpy change for the target reaction.

**Step 1**: Compare the target equation with Reaction 1.

Reaction 1 already has carbon monoxide as a reactant and carbon dioxide as a product.

It is in the correct direction, so no adjustment is needed.

**Step 2**: Reverse Reaction 2.

The target requires hydrogen gas as a product, but Reaction 2 consumes hydrogen gas.

Reverse the equation:

\[
\ce{H2O(l) -> H2(g) + 1/2O2(g)}
\]

Reversing a reaction changes the sign of its enthalpy change:

\[
\Delta H=+285.8\ \mathrm{kJ}
\]

**Step 3**: Reverse Reaction 3.

The target requires gaseous water as a reactant. The reversed Reaction 2 produces liquid water.

We therefore need to convert gaseous water into liquid water:

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

\[
\Delta H=-44.0\ \mathrm{kJ\,mol^{-1}}
\]

The negative sign is essential because condensation releases energy.

**Step 4**: Add all three adjusted equations.

\[
\begin{aligned}
&\ce{CO(g) + 1/2O2(g) -> CO2(g)}\\
&\ce{H2O(l) -> H2(g) + 1/2O2(g)}\\
&\ce{H2O(g) -> H2O(l)}
\end{aligned}
\]

Cancel the intermediate substances:

- \(\ce{1/2O2(g)}\) appears on both sides.
- \(\ce{H2O(l)}\) appears on both sides.

The remaining equation is:

\[
\ce{CO(g) + H2O(g) -> CO2(g) + H2(g)}
\]

It matches the target exactly.

**Step 5**: Calculate the total enthalpy change.

\[
\begin{aligned}
\Delta H&=-283.0+285.8-44.0\\
&=-41.2\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta H=-41.2\ \mathrm{kJ\,mol^{-1}}}
\]

The reaction is exothermic.

The important point is that the liquid-to-gas correction cannot be ignored just because the formula \(\ce{H2O}\) appears in both equations. The state change is a separate step in the energy pathway.

[IMAGE: An annotated three-reaction Hess cycle for CO(g) + H2O(g) forming CO2(g) + H2(g). Show the CO oxidation reaction in the forward direction, hydrogen combustion reversed to produce H2, and water condensation reversed to produce liquid water as an intermediate. Clearly show the cancellation of O2(g) and H2O(l).]

## Bond enthalpy estimates are not the same as measured reaction enthalpies

There is another reason physical state matters: bond enthalpy calculations use a particular model.

You may have used the equation:

\[
\Delta H\approx
\sum E_{\mathrm{broken}}
-
\sum E_{\mathrm{formed}}
\]

This estimates the energy change by adding the energy required to break bonds and subtracting the energy released when new bonds form.

The approximation symbol matters.

Mean bond enthalpies are averages based on bonds in gaseous molecules. They do not automatically include the energy changes involved in converting a substance between solid, liquid, and gas states.

They also cannot perfectly represent every bond in every molecular environment.

### Applying the model to hydrogen combustion

Consider the reaction:

\[
\ce{H2(g) + 1/2O2(g) -> H2O(g)}
\]

Suppose the following illustrative mean bond enthalpies are supplied:

| Bond | Mean bond enthalpy |
|---|---:|
| H-H | \(436\ \mathrm{kJ\,mol^{-1}}\) |
| O=O | \(498\ \mathrm{kJ\,mol^{-1}}\) |
| O-H | \(463\ \mathrm{kJ\,mol^{-1}}\) |

**Bonds broken:**

- One H-H bond.
- Half an O=O bond.

The energy absorbed is:

\[
436+\frac{498}{2}=685\ \mathrm{kJ}
\]

**Bonds formed:**

- Two O-H bonds.

The energy released is:

\[
2(463)=926\ \mathrm{kJ}
\]

Therefore:

\[
\begin{aligned}
\Delta H&\approx685-926\\
&\approx-241\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

This is an estimate for forming gaseous water.

Compare it with the reference value of approximately \(-241.83\ \mathrm{kJ\,mol^{-1}}\) for the gaseous product.

The agreement is close.

But what if the question asks for liquid water?

You must include condensation:

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

\[
\Delta H=-44.0\ \mathrm{kJ\,mol^{-1}}
\]

The estimated enthalpy change becomes:

\[
\begin{aligned}
\Delta H&\approx-241-44.0\\
&\approx-285\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

This is much closer to the reference value for liquid water, approximately \(-285.83\ \mathrm{kJ\,mol^{-1}}\).

[IMAGE: A two-part chemistry illustration. On the left, show gaseous H2 and O2 molecules with their covalent bonds being broken and new O-H bonds forming in gaseous H2O. On the right, show water vapour molecules condensing into closely packed liquid water molecules. Distinguish the intramolecular bond changes from the intermolecular attractions involved in condensation.]

### How do you distinguish the two types of data?

| Feature | Experimental/reference enthalpy | Mean bond enthalpy estimate |
|---|---|---|
| Source | Calorimetry and evaluated thermochemical data | Average bond-energy model |
| Molecular environment | Actual substances and specified states | Average bond environments |
| Physical state | Included in the thermochemical equation | Assumes gaseous molecules |
| Accuracy | Depends on measurement and data quality | Approximate |
| Best use | Calculations where reliable reaction or formation data are available | Estimating reactions when suitable thermochemical data are unavailable |

One important distinction: a published standard enthalpy value is not necessarily the result of directly measuring that exact formation reaction. It may have been determined indirectly using calorimetry and Hess's Law, then critically evaluated.

Hess's Law does not make approximate input data exact. It preserves the energy relationship between the stated endpoints. The reliability of your final result still depends on the quality and suitability of the values supplied.

For the formal definition of mean bond enthalpy, see the [IUPAC Gold Book](https://goldbook.iupac.org/terms/view/08147). The liquid and gas formation enthalpies used earlier can also be checked against the [NIST Chemistry WebBook](https://webbook.nist.gov/cgi/cbook.cgi?Name=water&cTC=on&cTG=on).

## Questions and solutions

The following questions progress from a straightforward phase correction to problems involving different physical states, model estimates, and temperature-dependent enthalpy changes.

All additional numerical data described as illustrative are supplied for the questions and are not claimed to be experimental measurements.

### Question 1

Hydrogen reacts with oxygen to form liquid water:

\[
\ce{H2(g) + 1/2O2(g) -> H2O(l)}
\]

\[
\Delta H=-285.8\ \mathrm{kJ\,mol^{-1}}
\]

The enthalpy change for vaporising one mole of liquid water is \(+44.0\ \mathrm{kJ\,mol^{-1}}\).

Calculate the enthalpy change for forming one mole of gaseous water from hydrogen and oxygen.

### Solution 1

The enthalpy change is \(-241.8\ \mathrm{kJ\,mol^{-1}}\).

The original equation forms liquid water, whereas the target equation forms water vapour. We must include the energy absorbed during vaporisation.

Write the phase-change equation:

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

\[
\Delta H=+44.0\ \mathrm{kJ\,mol^{-1}}
\]

Add this to the original formation reaction:

\[
\begin{aligned}
\Delta H&=-285.8+44.0\\
&=-241.8\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The final reaction is:

\[
\ce{H2(g) + 1/2O2(g) -> H2O(g)}
\]

The result is less negative because gaseous water has a higher enthalpy than liquid water. The difference is the energy required to vaporise the product.

### Question 2

Ammonia can be formed as a gas or dissolved in water.

A chemist is given the following illustrative thermochemical data:

**Formation of gaseous ammonia:**

\[
\ce{N2(g) + 3H2(g) -> 2NH3(g)}
\]

\[
\Delta H=-92.2\ \mathrm{kJ}
\]

**Dissolution of ammonia:**

\[
\ce{NH3(g) -> NH3(aq)}
\]

\[
\Delta H=-34.0\ \mathrm{kJ\,mol^{-1}}
\]

Calculate the enthalpy change for:

\[
\ce{N2(g) + 3H2(g) -> 2NH3(aq)}
\]

A student claims that the answer must be \(-126.2\ \mathrm{kJ}\), because they simply added the two supplied enthalpy values.

Identify the student's error and calculate the correct value.

### Solution 2

The correct enthalpy change is \(-160.2\ \mathrm{kJ}\).

The student's mistake is forgetting that the formation reaction produces two moles of ammonia, while the dissolution equation describes only one mole.

The dissolution equation must therefore be multiplied by two:

\[
\ce{2NH3(g) -> 2NH3(aq)}
\]

\[
\Delta H=2(-34.0)=-68.0\ \mathrm{kJ}
\]

Now add the formation and dissolution reactions:

\[
\begin{aligned}
\Delta H&=-92.2-68.0\\
&=-160.2\ \mathrm{kJ}
\end{aligned}
\]

The gaseous ammonia cancels, leaving:

\[
\ce{N2(g) + 3H2(g) -> 2NH3(aq)}
\]

The more negative value makes sense because dissolving ammonia in water releases additional energy.

The important lesson is that the state correction must be applied to the correct number of moles. A phase-change or dissolution enthalpy is not automatically the value for an entire reaction.

### Question 3

A student uses mean bond enthalpies to estimate the combustion of methane:

\[
\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}
\]

Their calculated value is:

\[
\Delta H_{\mathrm{model}}=-815\ \mathrm{kJ\,mol^{-1}}
\]

A thermochemical reference value for the reaction producing liquid water is:

\[
\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}
\]

\[
\Delta H_{\mathrm{reference}}=-890\ \mathrm{kJ\,mol^{-1}}
\]

Use a water vaporisation enthalpy of \(+44.0\ \mathrm{kJ\,mol^{-1}}\) to:

**(a)** Calculate the enthalpy change corresponding to the reference reaction if it produces gaseous water instead.

**(b)** Calculate the difference between this adjusted reference value and the bond-enthalpy estimate.

**(c)** Explain why comparing \(-815\) directly with \(-890\) does not provide a fair assessment of the bond-enthalpy model.

### Solution 3

The adjusted reference value for gaseous water is \(-802\ \mathrm{kJ\,mol^{-1}}\), which differs from the bond-enthalpy estimate by \(13\ \mathrm{kJ\,mol^{-1}}\).

**(a) Convert the reference reaction to gaseous water.**

The original reaction produces two moles of liquid water.

To produce gaseous water, vaporise both moles:

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

\[
\Delta H=2(+44.0)=+88.0\ \mathrm{kJ}
\]

Add the phase-change enthalpy to the reference reaction:

\[
\begin{aligned}
\Delta H_{\mathrm{gas}}
&=-890+88.0\\
&=-802\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

**(b) Compare the adjusted reference value with the model.**

The difference is:

\[
\begin{aligned}
\text{Difference}
&=|-815-(-802)|\\
&=13\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The bond-enthalpy estimate is \(13\ \mathrm{kJ\,mol^{-1}}\) more negative than the adjusted reference value.

**(c) Explain the discrepancy.**

The tempting approach is to subtract the two original numbers:

\[
|-815-(-890)|=75\ \mathrm{kJ\,mol^{-1}}
\]

and conclude that the bond-enthalpy model is inaccurate by \(75\ \mathrm{kJ\,mol^{-1}}\).

That comparison is invalid because the two equations describe different final states.

The model produces gaseous water, while the reference reaction produces liquid water. The difference of \(88\ \mathrm{kJ\,mol^{-1}}\) comes from condensing two moles of water vapour.

After correcting for the physical state, the remaining discrepancy is only \(13\ \mathrm{kJ\,mol^{-1}}\).

That remaining difference is consistent with the limitations of mean bond enthalpies. They are averaged values and cannot perfectly represent the actual molecular environment of every bond.

The correct approach is to match the reaction endpoints before evaluating the accuracy of a model.

### Question 4

Carbon exists in different solid forms, including graphite and diamond.

A chemist has the following illustrative rounded thermochemical data:

**Combustion of graphite:**

\[
\ce{C(s, graphite) + O2(g) -> CO2(g)}
\]

\[
\Delta H=-393.5\ \mathrm{kJ\,mol^{-1}}
\]

**Combustion of diamond:**

\[
\ce{C(s, diamond) + O2(g) -> CO2(g)}
\]

\[
\Delta H=-395.4\ \mathrm{kJ\,mol^{-1}}
\]

Calculate the enthalpy change for:

\[
\ce{C(s, graphite) -> C(s, diamond)}
\]

Explain why writing both substances simply as \(\ce{C(s)}\) could lead to an incorrect Hess's Law calculation.

[IMAGE: Side-by-side scientific representations of graphite and diamond crystal structures. Show graphite as stacked hexagonal sheets of carbon atoms with strong covalent bonds within each layer and weaker attractions between layers. Show diamond as a three-dimensional tetrahedral covalent network. Clearly label the two different carbon allotropes.]

### Solution 4

The conversion of graphite into diamond has an enthalpy change of \(+1.9\ \mathrm{kJ\,mol^{-1}}\).

The two combustion reactions both produce carbon dioxide. We can use this common product to construct a Hess cycle.

**Step 1**: Write the graphite combustion reaction in the forward direction.

\[
\ce{C(s, graphite) + O2(g) -> CO2(g)}
\]

\[
\Delta H=-393.5\ \mathrm{kJ}
\]

**Step 2**: Reverse the diamond combustion reaction.

We need diamond as a product, so reverse its combustion equation:

\[
\ce{CO2(g) -> C(s, diamond) + O2(g)}
\]

Reversing the equation changes the sign:

\[
\Delta H=+395.4\ \mathrm{kJ}
\]

**Step 3**: Add the equations.

\[
\begin{aligned}
&\ce{C(s, graphite) + O2(g) -> CO2(g)}\\
&\ce{CO2(g) -> C(s, diamond) + O2(g)}
\end{aligned}
\]

Carbon dioxide and oxygen cancel.

The remaining equation is:

\[
\ce{C(s, graphite) -> C(s, diamond)}
\]

**Step 4**: Calculate the enthalpy change.

\[
\begin{aligned}
\Delta H&=-393.5+395.4\\
&=+1.9\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The conversion is endothermic.

The important distinction is that physical state alone does not always identify a substance completely. Graphite and diamond are both solids, but they have different crystal structures and different enthalpies.

They are allotropes: different structural forms of the same element.

Writing only \(\ce{C(s)}\) hides this distinction. A complete thermochemical equation should identify the particular form of carbon when it matters.

This question extends the same principle used for liquid and gaseous water: the exact starting and finishing states determine the enthalpy change.

### Question 5

**Extension: Temperature-dependent phase changes**

A student compares two sets of thermochemical data for water.

The first comes from reference formation enthalpies at \(298.15\ \mathrm{K}\):

\[
\Delta_fH^\circ[\ce{H2O(l)}]
=-285.83\ \mathrm{kJ\,mol^{-1}}
\]

\[
\Delta_fH^\circ[\ce{H2O(g)}]
=-241.83\ \mathrm{kJ\,mol^{-1}}
\]

The second comes from an experimental investigation of water vaporisation at \(373.15\ \mathrm{K}\):

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

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

The student argues:

"The enthalpy difference at room temperature is \(44.0\ \mathrm{kJ\,mol^{-1}}\), but the experimental value is only \(40.7\ \mathrm{kJ\,mol^{-1}}\). Hess's Law must have failed, or one of the measurements must be wrong."

You are also given the following approximate molar heat capacities:

| Substance | Molar heat capacity |
|---|---:|
| Liquid water | \(75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}\) |
| Water vapour | \(33.6\ \mathrm{J\,mol^{-1}\,K^{-1}}\) |

Assume these heat capacities remain constant between \(298.15\ \mathrm{K}\) and \(373.15\ \mathrm{K}\).

**(a)** Explain why the two vaporisation enthalpies cannot be compared directly.

**(b)** Use a Hess cycle to calculate the approximate enthalpy change for vaporising water at \(298.15\ \mathrm{K}\), using the experimental value at \(373.15\ \mathrm{K}\).

**(c)** Compare your result with the value obtained from the standard enthalpies of formation. Explain why any small difference does not mean Hess's Law has failed.

### Solution 5

The two values do not contradict each other. The approximate vaporisation enthalpy at \(298.15\ \mathrm{K}\), calculated using the supplied heat capacities, is \(+43.8\ \mathrm{kJ\,mol^{-1}}\).

The crucial issue is that the measurements describe different temperatures. We must account for heating the liquid and cooling the vapour before comparing them.

**(a) Explain the difference in conditions.**

The reference formation enthalpies describe water in its liquid and gaseous standard states at \(298.15\ \mathrm{K}\).

The experimental vaporisation enthalpy describes the change from liquid to gas at \(373.15\ \mathrm{K}\).

These are different thermodynamic conditions.

The tempting approach is to treat \(40.7\ \mathrm{kJ\,mol^{-1}}\) as the vaporisation enthalpy at every temperature.

However, enthalpy depends on temperature as well as physical state. The energy required to heat liquid water is not the same as the energy required to heat water vapour.

We need to account for both changes.

**(b) Construct the Hess cycle.**

Start with one mole of liquid water at \(298.15\ \mathrm{K}\).

We want to finish with one mole of water vapour at the same temperature.

We have an experimental value for vaporisation at \(373.15\ \mathrm{K}\), so use this three-step pathway:

1. Heat liquid water from \(298.15\ \mathrm{K}\) to \(373.15\ \mathrm{K}\).
2. Vaporise the water at \(373.15\ \mathrm{K}\).
3. Cool the water vapour from \(373.15\ \mathrm{K}\) back to \(298.15\ \mathrm{K}\).

[IMAGE: A three-stage Hess cycle for vaporising one mole of water at 298.15 K using an alternative route through 373.15 K. Show the starting state H2O(l) at 298.15 K, heating to H2O(l) at 373.15 K, vaporisation to H2O(g) at 373.15 K, and cooling to H2O(g) at 298.15 K. Label the three arrows with the supplied heat-capacity and vaporisation data, without showing a final calculated result.]

For a temperature change, use:

\[
q=nC_m\Delta T
\]

where:
- \(q\) is the enthalpy change in joules.
- \(n\) is the amount of substance in moles.
- \(C_m\) is the molar heat capacity in \(\mathrm{J\,mol^{-1}\,K^{-1}}\).
- \(\Delta T\) is the temperature change in kelvin.

The temperature interval is:

\[
\Delta T=373.15-298.15=75.00\ \mathrm{K}
\]

**Step 1: Heat the liquid water.**

For one mole of liquid water:

\[
\begin{aligned}
q_{\mathrm{liquid}}
&=1\times75.3\times75.00\\
&=5647.5\ \mathrm{J}\\
&=+5.65\ \mathrm{kJ}
\end{aligned}
\]

The value is positive because heating the liquid absorbs energy.

**Step 2: Vaporise the water at \(373.15\ \mathrm{K}\).**

The supplied experimental value is:

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

This is positive because vaporisation is endothermic.

**Step 3: Cool the water vapour.**

The gas cools from \(373.15\ \mathrm{K}\) to \(298.15\ \mathrm{K}\).

Therefore:

\[
\Delta T=-75.00\ \mathrm{K}
\]

Using the molar heat capacity of water vapour:

\[
\begin{aligned}
q_{\mathrm{gas}}
&=1\times33.6\times(-75.00)\\
&=-2520\ \mathrm{J}\\
&=-2.52\ \mathrm{kJ}
\end{aligned}
\]

The value is negative because cooling the gas releases energy.

**Step 4: Add the three enthalpy changes.**

\[
\begin{aligned}
\Delta H_{298}
&=+5.65+40.7-2.52\\
&=+43.83\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Therefore, to the appropriate precision:

\[
\boxed{\Delta H_{\mathrm{vap},298}
\approx+43.8\ \mathrm{kJ\,mol^{-1}}}
\]

**(c) Compare with the reference value.**

Using the standard enthalpies of formation:

\[
\begin{aligned}
\Delta H_{\mathrm{vap},298}
&=-241.83-(-285.83)\\
&=+44.00\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

The difference between the two results is:

\[
44.00-43.83=0.17\ \mathrm{kJ\,mol^{-1}}
\]

This small difference is consistent with the rounded experimental data and the assumption that the molar heat capacities remain constant across the temperature interval.

The student's original conclusion was incorrect.

Hess's Law has not failed. The two original values referred to different temperatures, so the student was comparing different initial and final conditions.

The deeper principle is that a valid Hess cycle must connect the same thermodynamic states, including both physical state and temperature.

This temperature correction is an extension of the basic HSC Hess-cycle method, rather than a claim that students must memorise these particular heat-capacity values.

## Where to take this next

The most useful habit to develop is to treat every thermochemical equation as a complete description of a reaction, rather than a formula with an energy number attached.

Before combining equations, check three things:

- Are the chemical substances identical?
- Are their physical states identical?
- Are the reaction conditions consistent?

If any of these differ, you may need an additional reaction or correction before the equations can be combined.

For more practice constructing complete energy pathways, read [Hess's Law Energy Cycles: HSC Chemistry Calculations Explained](https://tutorgum.com/learn/resources/hsc-chemistry-hess-s-law-energy-cycles/).

To strengthen your understanding of why substances have different physical states, see [State of Matter Trends in the Periodic Table for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-state-of-matter-trends-in-the-periodic-table/).

**NSW syllabus context:** Hess's Law and enthalpy are covered in the Chemistry Stage 6 Syllabus (2017). That syllabus remains in use until implementation of the Chemistry 11-12 Syllabus (2025), beginning with Year 11 in 2028. The first HSC examination under the new syllabus is scheduled for 2029. Check the [official NESA Chemistry syllabus information](https://www.nsw.gov.au/education-and-training/nesa/curriculum/science/chemistry-stage-6-2017) for the course applying to your year group.