Why Reversing a Thermochemical Equation Changes the Sign of ΔH

A strip of magnesium reacts with acid. Bubbles form, the solution warms, and a thermometer records a temperature rise. The reaction has an enthalpy change of approximately \(-470\ \mathrm{kJ\,mol^{-1}}\) in an illustrative calorimetry experiment.

Now reverse the chemical equation. Does the enthalpy change remain negative because the same chemicals are involved?

No. It becomes \(+470\ \mathrm{kJ\,mol^{-1}}\).

Reversing a thermochemical equation changes the sign of \(\Delta H\) because it swaps the initial and final states of the reaction. The enthalpy change is calculated by subtracting the enthalpy of the reactants from the enthalpy of the products. Reversing the equation swaps those two groups, so the subtraction runs in the opposite direction.

Let’s work through why this happens, how experiments support the rule, and how to avoid losing marks when reversing and scaling complete thermochemical equations.

01Why reversing a reaction changes the sign of ΔH

First, think about what an enthalpy change actually represents.

Enthalpy, represented by \(H\), is a thermodynamic property of a chemical system. The enthalpy change, \(\Delta H\), describes the difference between the final and initial states.

For a forward reaction:

\[ \Delta H_{\mathrm{forward}} = H_{\mathrm{products}}-H_{\mathrm{reactants}} \]

For the reverse reaction, the products become the reactants and the reactants become the products:

\[ \Delta H_{\mathrm{reverse}} = H_{\mathrm{reactants}}-H_{\mathrm{products}} \]

Therefore:

\[ \boxed{\Delta H_{\mathrm{reverse}} =-\Delta H_{\mathrm{forward}}} \]

This is why the sign must change.

It isn’t an arbitrary rule to memorise. It follows directly from how an enthalpy change is calculated.

Think about a change in height

Imagine walking down a hill from a lookout to a valley. Your elevation decreases by 120 metres.

If you walk back up the same hill, your elevation increases by 120 metres.

The two changes have equal magnitudes but opposite signs because you have swapped your starting and finishing points.

Enthalpy behaves similarly. The analogy is about the difference between two endpoints, not the physical nature of energy. Chemical enthalpy is not literally a height, and the analogy doesn’t describe activation energy or reaction rates.

The important idea is that the enthalpy change depends on the initial and final states, not on the direction you happen to write the equation.

What happens to exothermic and endothermic reactions?

Remember the sign convention:

An exothermic reaction releases energy to its surroundings. Its enthalpy decreases.

An endothermic reaction absorbs energy from its surroundings. Its enthalpy increases.

When a reaction is reversed, these descriptions swap.

For example:

\[ \ce{Mg(s) + 2H+(aq) -> Mg^2+(aq) + H2(g)} \]

Suppose:

\[ \Delta H=-470\ \mathrm{kJ\,mol^{-1}} \]

The forward reaction is exothermic. Reversing it gives:

\[ \ce{Mg^2+(aq) + H2(g) -> Mg(s) + 2H+(aq)} \]

\[ \Delta H=+470\ \mathrm{kJ\,mol^{-1}} \]

The reverse reaction is endothermic. It requires an energy input of 470 kJ per mole of reaction under the same specified conditions.

Notice something important: reversing a reaction does not mean it will automatically happen in the opposite direction.

The reverse magnesium reaction, for example, requires energy input and a suitable process. The thermochemical equation tells us its enthalpy change if that transformation occurs. It does not tell us whether the reaction is spontaneous or how quickly it happens.

02A thermochemical equation includes more than its chemical formula

A complete thermochemical equation contains three important pieces of information:

• A balanced chemical equation.
• The physical states of all reactants and products.
• The enthalpy change for the reaction as written.

All three matter when manipulating an equation.

Consider the formation of liquid water:

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

\[ \Delta H=-286\ \mathrm{kJ\,mol^{-1}} \]

This value describes the formation of one mole of liquid water from hydrogen and oxygen gases.

If we reverse the equation:

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

then:

\[ \Delta H=+286\ \mathrm{kJ\,mol^{-1}} \]

The physical states have stayed exactly the same. Only the direction has changed.

This is an exact reversal.

Physical states cannot be ignored

Now consider a different equation:

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

This is vaporisation. It requires energy, so its enthalpy change is positive.

At approximately 298 K, the molar enthalpy of vaporisation of water is about \(+44.0\ \mathrm{kJ\,mol^{-1}}\).

Reversing this physical change gives condensation:

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

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

The chemical formula hasn’t changed, but the physical state has.

Liquid water and water vapour have different enthalpies because energy is required to overcome intermolecular attractions and separate the water molecules.

This distinction becomes important when using experimental data or constructing enthalpy calculations. You cannot simply cancel \(\ce{H2O(l)}\) against \(\ce{H2O(g)}\).

They are different thermodynamic states.

03Three rules for manipulating thermochemical equations

There are three operations you need to recognise.

These rules follow from the same principle: the enthalpy change belongs to the reaction in the quantities and direction specified by the equation.

If you double the amount of reaction, you double the total enthalpy change. If you halve the equation, you halve the enthalpy change.

04How a calorimetry experiment demonstrates the sign change

So far, we’ve treated enthalpy changes as supplied values.

But where do those numbers come from?

One way to determine an experimental reaction enthalpy is through calorimetry.

The basic question is straightforward: when a chemical reaction occurs, does the reacting system gain or lose energy?

Start with a temperature prediction

Consider magnesium reacting with hydrochloric acid.

The net ionic equation is:

\[ \ce{Mg(s) + 2H+(aq) -> Mg^2+(aq) + H2(g)} \]

If the reaction is exothermic, what should happen to the temperature of the solution?

It should increase.

The reacting chemicals transfer energy to their surroundings. In this experiment, the solution and calorimeter absorb some of that energy.

How the apparatus works

A simple experimental arrangement uses a polystyrene cup calorimeter.

The apparatus consists of:

• A polystyrene cup containing a measured volume of hydrochloric acid.
• A supporting beaker around the cup.
• A lid to reduce heat exchange with the surroundings.
• A thermometer to record the solution temperature.
• A stirrer to distribute the heat throughout the solution.

A small strip of magnesium is added to the acid. As the reaction proceeds, hydrogen gas bubbles form and the magnesium gradually disappears.

The thermometer records the temperature increase.

Why the first experimental design needs improving

Imagine using an ordinary open glass beaker instead of an insulated cup.

You would probably observe a temperature increase, but the measurement would have limitations.

Some energy would:

• Heat the glass beaker.
• Escape into the surrounding air.
• Be transferred to the thermometer and other apparatus.

If you only calculated the energy absorbed by the solution, you would underestimate the energy released by the reaction.

A better design uses an insulated cup, a lid, and a calibrated calorimeter.

The calorimeter’s heat capacity can be measured beforehand. This allows us to account for energy absorbed by the apparatus as well as the solution.

Stirring also helps distribute energy evenly, while recording the temperature at regular intervals can help account for heat loss during the measurement.

The observation is still a temperature rise, but the improved apparatus gives a more reliable estimate of the reaction enthalpy.

From temperature change to reaction enthalpy

At constant pressure, assuming only pressure-volume work, the heat transferred to the reacting system is related to its enthalpy change:

\[ q_p=\Delta H \]

For the surroundings, the heat absorbed by the solution is:

\[ q_{\mathrm{solution}}=mc\Delta T \]

where:

• \(m\) is the mass of the solution in grams.
• \(c\) is its specific heat capacity in \(\mathrm{J\,g^{-1}\,K^{-1}}\).
• \(\Delta T\) is its temperature change in kelvin.

The calorimeter also absorbs energy:

\[ q_{\mathrm{cal}}=C_{\mathrm{cal}}\Delta T \]

where \(C_{\mathrm{cal}}\) is the calorimeter heat capacity in \(\mathrm{J\,K^{-1}}\).

The total energy gained by the surroundings is therefore:

\[ q_{\mathrm{surroundings}} = q_{\mathrm{solution}}+q_{\mathrm{cal}} \]

By conservation of energy:

\[ q_{\mathrm{reaction}} = -q_{\mathrm{surroundings}} \]

That negative sign is important.

If the solution and calorimeter gain energy, the reacting system must lose energy.

To calculate the molar enthalpy change, divide the reaction heat by the number of moles that reacted:

\[ \Delta H=\frac{q_{\mathrm{reaction}}}{n} \]

Here, \(n\) is the amount of the limiting reactant in moles, provided the equation is written with one mole of that reactant per mole of reaction.

05Measured enthalpy values versus model estimates

There is another distinction worth understanding.

Not every reaction enthalpy comes from a calorimetry experiment.

Sometimes chemists calculate an estimated reaction enthalpy using average bond enthalpies. These are not the same type of result.

For example, a bond-enthalpy calculation uses:

\[ \Delta H_{\mathrm{reaction}} \approx \sum D_{\mathrm{broken}} - \sum D_{\mathrm{formed}} \]

Here, \(D\) represents the average enthalpy required to break a particular type of bond.

The result is an estimate because the same type of bond can have slightly different enthalpies in different molecular environments.

This method also does not automatically include energy changes associated with dissolving substances or changing physical states.

You can explore the distinction between bond-energy estimates and reaction enthalpies in How to Decide Which Bonds Are Broken and Formed in a Reaction.

Does reversing a model estimate also change its sign?

Yes.

Suppose a bond-enthalpy calculation estimates a forward reaction enthalpy of:

\[ \Delta H_{\mathrm{model}}=-455\ \mathrm{kJ\,mol^{-1}} \]

The reverse model estimate is:

\[ \Delta H_{\mathrm{model,reverse}} =+455\ \mathrm{kJ\,mol^{-1}} \]

The same applies to an experimental value.

If calorimetry gives \(-470\ \mathrm{kJ\,mol^{-1}}\), reversing the exact reaction gives \(+470\ \mathrm{kJ\,mol^{-1}}\).

Do not reverse a model estimate and replace it with an experimental value, or vice versa. Each value must be manipulated consistently with the method that produced it.

The sign reversal rule is exact for the same specified initial and final states. The numerical agreement between a model and an experiment is a separate question.

06Questions and solutions

The following questions progress from simple equation reversal to problems involving calorimetry, equation manipulation, and the difference between measured and estimated enthalpy changes.

Question 1

Calcium carbonate decomposes according to the thermochemical equation:

\[ \ce{CaCO3(s) -> CaO(s) + CO2(g)} \]

\[ \Delta H=+178\ \mathrm{kJ} \]

Write the complete thermochemical equation for the reverse reaction and determine its enthalpy change.

Explain why the reaction changes from endothermic to exothermic.

Solution 1

The reverse reaction has an enthalpy change of \(-178\ \mathrm{kJ}\) and is exothermic.

Start by swapping the reactants and products:

\[ \ce{CaO(s) + CO2(g) -> CaCO3(s)} \]

The original reaction absorbed 178 kJ as calcium carbonate decomposed.

The reverse reaction forms calcium carbonate from calcium oxide and carbon dioxide. It releases the same amount of energy under the same conditions.

Therefore:

\[ \boxed{\Delta H=-178\ \mathrm{kJ}} \]

The tempting mistake is to keep the positive sign because the same three substances are involved. However, the initial and final states have swapped, so the enthalpy change must also reverse.

Question 2

Sulfur dioxide reacts with oxygen to form sulfur trioxide:

\[ \ce{2SO2(g) + O2(g) -> 2SO3(g)} \]

The enthalpy change is:

\[ \Delta H=-198\ \mathrm{kJ} \]

Determine the enthalpy change for the decomposition of three moles of sulfur trioxide:

\[ \ce{3SO3(g) -> 3SO2(g) + 3/2O2(g)} \]

Show how both the direction and coefficients affect your answer.

Solution 2

The decomposition of three moles of sulfur trioxide has an enthalpy change of \(+297\ \mathrm{kJ}\).

Both reversing and scaling the original equation are necessary.

Step 1: Reverse the original equation.

\[ \ce{2SO3(g) -> 2SO2(g) + O2(g)} \]

Change the sign:

\[ \Delta H=+198\ \mathrm{kJ} \]

Step 2: Compare the coefficients with the required equation.

The reversed equation represents the decomposition of two moles of sulfur trioxide.

The question requires three moles, so multiply every coefficient by \(1.5\).

\[ \ce{3SO3(g) -> 3SO2(g) + 3/2O2(g)} \]

Step 3: Multiply the enthalpy change by the same factor.

\[ \Delta H=1.5\times198 \]

\[ \Delta H=+297\ \mathrm{kJ} \]

The positive sign means that decomposing three moles of sulfur trioxide absorbs 297 kJ.

A common error is to reverse the equation correctly but forget to scale the enthalpy change. The coefficients and \(\Delta H\) must always change together.

Question 3

Calcium oxide reacts with water to form calcium hydroxide:

\[ \ce{CaO(s) + H2O(l) -> Ca(OH)2(s)} \]

In an insulated calorimeter, \(0.560\ \mathrm{g}\) of calcium oxide is added to \(50.0\ \mathrm{g}\) of water.

The temperature increases by \(2.8\ \mathrm{K}\).

The following data are supplied:

• Molar mass of calcium oxide: \(56.1\ \mathrm{g\,mol^{-1}}\).
• Mass of the reaction mixture: \(50.56\ \mathrm{g}\).
• Specific heat capacity of the mixture: \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).
• Calorimeter heat capacity: \(20.0\ \mathrm{J\,K^{-1}}\).

Assume complete reaction, excess water, negligible heat loss, and constant pressure.

a. Calculate the experimental molar enthalpy change for the reaction as written.

b. Write the reverse thermochemical equation and calculate its enthalpy change.

c. Explain why the temperature increase does not mean the reaction has a positive enthalpy change.

Solution 3

The forward reaction has an experimental enthalpy change of approximately \(-64.9\ \mathrm{kJ\,mol^{-1}}\). Its reverse reaction has an enthalpy change of \(+64.9\ \mathrm{kJ\,mol^{-1}}\).

Step 1: Calculate the amount of calcium oxide.

Calcium oxide is the limiting reactant.

\[ n=\frac{m}{M} \]

\[ n(\ce{CaO})=\frac{0.560}{56.1} \]

\[ n(\ce{CaO})=0.00998\ \mathrm{mol} \]

Step 2: Calculate the heat absorbed by the reaction mixture.

Use:

\[ q_{\mathrm{mixture}}=mc\Delta T \]

Substitute:

\[ q_{\mathrm{mixture}} = 50.56\times4.18\times2.8 \]

\[ q_{\mathrm{mixture}}\approx592\ \mathrm{J} \]

Step 3: Calculate the heat absorbed by the calorimeter.

\[ q_{\mathrm{cal}}=C_{\mathrm{cal}}\Delta T \]

\[ q_{\mathrm{cal}}=20.0\times2.8 \]

\[ q_{\mathrm{cal}}=56.0\ \mathrm{J} \]

Step 4: Determine the heat released by the reaction.

The surroundings have absorbed:

\[ q_{\mathrm{surroundings}}=592+56 \]

\[ q_{\mathrm{surroundings}}=648\ \mathrm{J} \]

Therefore:

\[ q_{\mathrm{reaction}}=-648\ \mathrm{J} \]

Convert to kilojoules:

\[ q_{\mathrm{reaction}}=-0.648\ \mathrm{kJ} \]

Step 5: Calculate the molar enthalpy change.

\[ \Delta H=\frac{q_{\mathrm{reaction}}}{n} \]

\[ \Delta H=\frac{-0.648}{0.00998} \]

\[ \Delta H\approx-64.9\ \mathrm{kJ\,mol^{-1}} \]

The negative sign indicates that the reaction is exothermic.

Step 6: Write the reverse equation.

\[ \ce{Ca(OH)2(s) -> CaO(s) + H2O(l)} \]

Reverse the sign:

\[ \Delta H=+64.9\ \mathrm{kJ\,mol^{-1}} \]

The reverse reaction requires energy input.

Step 7: Explain the temperature increase.

The thermometer measures the temperature of the surroundings, not the enthalpy change of the reacting chemicals.

The reaction releases energy, which heats the mixture and calorimeter. Therefore, the surroundings gain energy while the reacting system loses energy.

This is why a temperature increase corresponds to a negative reaction enthalpy in this experiment.

Question 4

2017 Chemistry Stage 6 syllabus: Year 11, Module 4 – Drivers of Reactions.

Consider the following thermochemical equations:

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

\[ \Delta H_1=-393.5\ \mathrm{kJ} \]

And:

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

\[ \Delta H_2=-283.0\ \mathrm{kJ} \]

Calculate the enthalpy change for:

\[ \ce{2CO(g) -> 2C(s, graphite) + O2(g)} \]

Show which equation must be reversed, and explain why simply adding the two supplied equations does not produce the required reaction.

Solution 4

The required reaction has an enthalpy change of \(+221.0\ \mathrm{kJ}\).

The key is to reverse the carbon monoxide combustion equation so that carbon dioxide can cancel when the equations are added.

Step 1: Identify the required direction.

The target equation decomposes carbon monoxide into carbon and oxygen.

However, the supplied carbon combustion equation produces carbon dioxide, while the carbon monoxide combustion equation also produces carbon dioxide.

We need to reverse the second equation to allow carbon dioxide to cancel.

Step 2: Reverse the carbon monoxide combustion equation.

Original:

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

Reversed:

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

The enthalpy change becomes:

\[ \Delta H_2=+283.0\ \mathrm{kJ} \]

Step 3: Add the two equations.

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

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

Carbon dioxide appears on both sides, so it cancels.

The resulting equation is:

\[ \ce{C(s,graphite) + 1/2O2(g) -> CO(g)} \]

Add the enthalpy changes:

\[ \Delta H=-393.5+283.0 \]

\[ \Delta H=-110.5\ \mathrm{kJ} \]

Step 4: Reverse the resulting equation.

The target reaction requires carbon monoxide to decompose, not form.

Therefore:

\[ \ce{CO(g) -> C(s,graphite) + 1/2O2(g)} \]

Its enthalpy change is:

\[ \Delta H=+110.5\ \mathrm{kJ} \]

Step 5: Double the equation.

The required equation produces two moles of carbon and one mole of oxygen.

Multiply every coefficient and the enthalpy change by two:

\[ \ce{2CO(g) -> 2C(s,graphite) + O2(g)} \]

\[ \Delta H=2(110.5) \]

\[ \Delta H=+221.0\ \mathrm{kJ} \]

The reaction is endothermic.

The important trap is reversing an equation but retaining its original negative sign. The second equation had to be reversed before cancellation, so its enthalpy change had to become positive.

Question 5

A student is investigating the decomposition of liquid water:

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

Two different sources provide information about the energy changes.

Source A: Calorimetry

The experimentally determined enthalpy change for water formation is:

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

\[ \Delta H_{\mathrm{experimental}}=-572\ \mathrm{kJ} \]

Source B: Average bond enthalpies

The following illustrative average bond enthalpies are supplied:

Also given:

\[ \Delta H_{\mathrm{vap}}(\ce{H2O})= +44.0\ \mathrm{kJ\,mol^{-1}} \]

A student uses the bond enthalpies to estimate the formation of gaseous water:

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

They obtain an approximate value of \(-482\ \mathrm{kJ}\).

The student reverses this value and claims that the decomposition of two moles of liquid water must therefore have an enthalpy change of \(+482\ \mathrm{kJ}\).

Another student argues that the experimental result must be \(+572\ \mathrm{kJ}\).

Resolve the disagreement.

Your answer must:

a. Verify the bond-enthalpy estimate for gaseous water formation.

b. Adjust the model estimate for the formation of liquid water.

c. Determine both the experimental and model-based enthalpy changes for the decomposition of liquid water.

d. Explain precisely why the first student’s \(+482\ \mathrm{kJ}\) answer does not describe the target reaction, and why the two remaining values do not have to be identical.

Solution 5

For the decomposition of two moles of liquid water, the experimental enthalpy change is \(+572\ \mathrm{kJ}\), while the adjusted bond-enthalpy model gives approximately \(+570\ \mathrm{kJ}\).

The \(+482\ \mathrm{kJ}\) value describes the decomposition of water vapour, not liquid water.

This distinction is the key to the entire question.

Step 1: Calculate the energy required to break the reactant bonds.

The gaseous reactants are:

\[ \ce{2H2(g) + O2(g)} \]

There are two H-H bonds and one O=O bond.

Therefore:

\[ E_{\mathrm{broken}} = 2(436)+498 \]

\[ E_{\mathrm{broken}}=1370\ \mathrm{kJ} \]

Step 2: Calculate the energy released when product bonds form.

Each water molecule contains two O-H bonds.

The equation produces two water molecules, so four O-H bonds form.

\[ E_{\mathrm{formed}}=4(463) \]

\[ E_{\mathrm{formed}}=1852\ \mathrm{kJ} \]

Step 3: Calculate the estimated enthalpy change for gaseous water formation.

Use the bond-enthalpy relationship:

\[ \Delta H_{\mathrm{model}} \approx E_{\mathrm{broken}}-E_{\mathrm{formed}} \]

Substitute:

\[ \Delta H_{\mathrm{model}} \approx1370-1852 \]

\[ \Delta H_{\mathrm{model}}\approx-482\ \mathrm{kJ} \]

This is the estimated enthalpy change for:

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

It is not yet the enthalpy change for liquid water formation.

Step 4: Account for the condensation of water vapour.

The required product is liquid water.

The enthalpy of vaporisation is:

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

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

Condensation is the reverse process:

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

Therefore:

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

Two moles of water condense, so:

\[ \Delta H_{\mathrm{condensation}} =2(-44.0) \]

\[ \Delta H_{\mathrm{condensation}}=-88.0\ \mathrm{kJ} \]

Add this to the gaseous water formation estimate:

\[ \Delta H_{\mathrm{model,liquid}} =-482-88 \]

\[ \Delta H_{\mathrm{model,liquid}} \approx-570\ \mathrm{kJ} \]

This is now the model estimate for the formation of liquid water.

Step 5: Reverse the model-based thermochemical equation.

The model estimate for liquid water formation is:

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

\[ \Delta H_{\mathrm{model}}\approx-570\ \mathrm{kJ} \]

Reverse the equation:

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

The model estimate becomes:

\[ \Delta H_{\mathrm{model,reverse}} \approx+570\ \mathrm{kJ} \]

Step 6: Reverse the experimental thermochemical equation.

The experimental value for liquid water formation is:

\[ \Delta H_{\mathrm{experimental}}=-572\ \mathrm{kJ} \]

Reversing the complete equation gives:

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

Therefore:

\[ \Delta H_{\mathrm{experimental,reverse}} =+572\ \mathrm{kJ} \]

Step 7: Explain the difference between the two results.

The two remaining values are:

The difference is approximately \(2\ \mathrm{kJ}\).

This is reasonable because average bond enthalpies are approximate values. They simplify the behaviour of individual molecular environments, and the supplied values are rounded.

The tempting answer of \(+482\ \mathrm{kJ}\) came from reversing the gaseous water equation:

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

But the question asks about liquid water.

The missing energy is the energy required to vaporise two moles of liquid water:

\[ 2\times44.0=88.0\ \mathrm{kJ} \]

Adding that energy gives the correct model estimate for the liquid-water decomposition.

The deeper lesson is that reversing an equation always changes the sign for the same initial and final states. It does not make two different thermochemical models produce identical numerical values, and it does not allow you to ignore physical states.

07Where this leads next

Once you can reverse and scale thermochemical equations reliably, you can use the same rules to combine several reactions into a single target equation.

This is the foundation of Hess’s Law: different reaction pathways between the same initial and final states must have the same overall enthalpy change.

For students following the Chemistry Stage 6 Syllabus (2017), Hess’s Law is covered in Year 11 Module 4. The Chemistry 11-12 Syllabus (2025) is implemented from 2028; the Hess-law material here should not be assumed to be a required 2025-syllabus outcome.

The next useful skill is deciding which supplied equations to reverse, multiply, and combine so that unwanted substances cancel without losing their enthalpy signs. See How to Build a Hess Cycle from Combustion Data.