How to Scale Enthalpy in Thermochemical Equations

A thermochemical equation tells you two things: which substances react and how much energy is transferred. But what happens when you multiply the equation by 2, 3, or even 0.5?

Suppose you are given the decomposition of calcium carbonate:

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

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

This equation describes the decomposition of one mole of calcium carbonate. Now suppose a question asks you to calculate the enthalpy change when 3.00 mol of calcium carbonate decomposes.

Should you leave the enthalpy at +178 kJ because the chemical reaction is unchanged, or multiply it by 3?

You must multiply it by 3. The new thermochemical equation is:

\[
\ce{3CaCO3(s) -> 3CaO(s) + 3CO2(g)}
\]

\[
\Delta H=3(+178)=+534\text{ kJ}
\]

The rule is simple: whenever you multiply a complete chemical equation by a factor, multiply its enthalpy change by exactly the same factor.

The sign stays the same when you multiply by a positive number. The reaction is still endothermic, but three times as much reaction now absorbs three times as much energy.

01Why does enthalpy change when you multiply an equation?

Think about a chemical equation as a recipe for a particular amount of reaction.

If you triple a recipe, you triple every ingredient. You cannot triple the flour but leave the other ingredients unchanged and expect to make three times as many identical cakes.

Chemical equations work similarly. Their coefficients describe the relative amounts of each substance involved in a reaction.

For example:

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

This equation represents:

• 2 mol of hydrogen gas reacting.
• 1 mol of oxygen gas reacting.
• 2 mol of liquid water forming.

If you multiply the equation by 2, every coefficient doubles:

\[
\ce{4H2(g) + 2O2(g) -> 4H2O(l)}
\]

The reaction now involves twice as much of every substance.

Because enthalpy measures the energy change associated with the reaction, doubling the amount of reaction doubles the energy transferred.

The recipe analogy has one limitation: chemical equations describe mole ratios, not individual kitchen ingredients. The energy change also depends on the physical states and conditions of the substances.

The mathematical rule

Suppose a thermochemical equation has an enthalpy change of \(\Delta H\).

If you multiply every coefficient by a positive factor \(k\), the new enthalpy change is:

\[
\boxed{\Delta H_{\text{new}}=k\Delta H_{\text{original}}}
\]

Here:

• \(k\) is the multiplying factor.
• \(\Delta H_{\text{original}}\) is the original enthalpy change.
• \(\Delta H_{\text{new}}\) is the enthalpy change after scaling.

The factor \(k\) applies to every coefficient, including fractional coefficients.

For example, consider:

\[
\ce{2A + B -> 2C}
\]

\[
\Delta H=-300\text{ kJ}
\]

Multiplying the equation by 3 gives:

\[
\ce{6A + 3B -> 6C}
\]

\[
\Delta H=3(-300)=-900\text{ kJ}
\]

Notice that the negative sign remains. The reaction is still exothermic.

What happens when you divide an equation?

Dividing an equation by a number works in exactly the same way.

Suppose you divide the original equation by 2:

\[
\ce{A + 1/2B -> C}
\]

The enthalpy change must also be divided by 2:

\[
\Delta H=\frac{-300}{2}=-150\text{ kJ}
\]

You are now describing half the original amount of reaction, so half as much energy is transferred.

The following table summarises the different possibilities.

02Understand what the enthalpy value actually represents

A common mistake is assuming that the enthalpy value always refers to one mole of the reactant.

That is not necessarily true.

The enthalpy change belongs to the complete thermochemical equation as written.

Consider:

\[
\ce{2NO(g) + O2(g) -> 2NO2(g)}
\]

\[
\Delta H=-114\text{ kJ}
\]

The reaction releases 114 kJ when the stoichiometric amounts shown react.

That means:

• 2 mol of NO react.
• 1 mol of oxygen reacts.
• 2 mol of nitrogen dioxide form.
• 114 kJ of energy is released.

The value is not automatically the energy released per mole of nitrogen dioxide.

To calculate the enthalpy change per mole of nitrogen dioxide:

\[
\frac{-114}{2}=-57.0\text{ kJ mol}^{-1}
\]

This distinction matters whenever a question gives you an enthalpy value alongside a complete equation.

Equation enthalpy versus molar enthalpy

Always check what the word per mole refers to.

For example, an enthalpy of combustion of \(-800\text{ kJ mol}^{-1}\) normally refers to the combustion of one mole of the specified fuel.

If the thermochemical equation shows two moles of fuel, the total enthalpy change for that equation is twice the molar value.

Do not multiply twice by the same amount of substance.

What about physical states?

Physical states must remain unchanged when you simply scale an equation.

For example:

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

is not thermochemically identical to:

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

Both equations describe water formation, but the water is in a different physical state. The enthalpy changes are therefore different.

Multiplying an equation does not give you permission to change its state symbols.

04Measured enthalpy values versus model estimates

There is another distinction worth understanding before you start more complicated enthalpy calculations.

An enthalpy value can come from an experiment or from a theoretical model.

Both can be scaled when you multiply a chemical equation. However, they do not have the same scientific status.

Experimental values: Calorimetry

Calorimetry is used to determine energy changes by measuring temperature changes.

Imagine mixing hydrochloric acid and sodium hydroxide in an insulated cup.

The apparatus consists of:

• A polystyrene cup nested inside a supporting beaker.
• A measured volume of hydrochloric acid.
• A measured volume of sodium hydroxide.
• A thermometer to record temperature.
• A lid and stirring device to reduce heat exchange and maintain an even temperature.

The two solutions are combined, stirred, and monitored.

If the reaction is exothermic, the temperature of the solution increases.

But what does the thermometer actually measure?

It measures temperature, not enthalpy directly.

You use the temperature change to calculate how much energy the surroundings absorb.

For the solution:

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

Where:

• \(q_{\text{solution}}\) is the heat absorbed by the solution, in joules.
• \(m\) is the solution mass, in grams.
• \(c\) is the specific heat capacity, in J g\(^{-1}\) °C\(^{-1}\).
• \(\Delta T\) is the temperature change, in °C.

If the calorimeter itself absorbs heat, include that contribution:

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

Here, \(C_{\text{cal}}\) is the heat capacity of the calorimeter, measured in J °C\(^{-1}\) or kJ °C\(^{-1}\).

The reaction and its surroundings exchange energy, so:

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

For an exothermic reaction, the surroundings gain heat and the reaction loses heat. This is why the reaction enthalpy is negative.

Why the experimental design matters

A simple, uncovered cup can lose heat to the surrounding air. The cup itself can also absorb energy.

If you calculate heat using only the solution’s temperature change, you may underestimate the total energy released by the reaction.

A better design uses nested insulated cups, a lid, and a calibrated calorimeter heat capacity. You then account for the heat absorbed by both the solution and the apparatus.

Even an improved setup has limitations. Heat loss, incomplete reaction, and measurement uncertainty can affect the calculated result.

The value obtained is therefore an experimental determination based on measurements and assumptions, not a perfect direct reading of enthalpy.

Model estimates: Mean bond enthalpies

You can also estimate reaction enthalpy using mean bond enthalpies.

This method calculates the energy required to break bonds and the energy released when new bonds form.

The relationship is:

\[
\Delta H_{\text{estimate}}\approx
\sum E_{\text{bonds broken}}
–
\sum E_{\text{bonds formed}}
\]

Mean bond enthalpies are average energy values for bonds in gaseous molecules.

Consider:

\[
\ce{H2(g) + Cl2(g) -> 2HCl(g)}
\]

The following rounded bond enthalpies are illustrative supplied data.

To estimate the enthalpy change, identify the bonds broken and formed.

Energy required to break the original bonds:

\[
436+243=679\text{ kJ mol}^{-1}
\]

Energy released when two H-Cl bonds form:

\[
2(432)=864\text{ kJ mol}^{-1}
\]

Therefore:

\[
\begin{aligned}
\Delta H_{\text{estimate}}
&=679-864\\
&=-185\text{ kJ mol}^{-1}
\end{aligned}
\]

The model estimates that the reaction releases 185 kJ per mole of reaction as written.

Now suppose you multiply the equation by 3:

\[
\ce{3H2(g) + 3Cl2(g) -> 6HCl(g)}
\]

The estimated enthalpy becomes:

\[
\Delta H_{\text{estimate}}=3(-185)
\]

\[
\boxed{\Delta H_{\text{estimate}}=-555\text{ kJ}}
\]

The scaling rule still applies. However, the result remains a model estimate.

Why?

Mean bond enthalpies are averages. The actual energy of a particular bond depends on its molecular environment. The method also uses gas-phase bond enthalpies, so it does not automatically account for the energy changes associated with liquid and solid states.

Consequently, bond enthalpy calculations provide estimates rather than direct experimental measurements.

Comparing the two sources of enthalpy data

There is one important point: multiplying an equation does not make an estimate more accurate or turn it into an experimental result.

It simply changes the amount of reaction represented by the equation.

If an experimental value has uncertainty, that uncertainty must also be considered when you scale the value.

05A reliable method for scaling thermochemical equations

When you encounter a question involving a multiplied or divided thermochemical equation, follow this process.

1. Check that the original equation is balanced.

The coefficients must represent the correct mole ratios. If you need to revise your balancing skills, see Balancing Chemical Equations for HSC Chemistry.

2. Identify the amount of substance in the original equation.

Pay attention to the coefficient of the substance mentioned in the question.

3. Calculate the multiplying factor.

Use:

\[
k=\frac{\text{required amount}}{\text{original amount}}
\]

Make sure both amounts refer to the same substance.

4. Multiply every coefficient by the factor.

Do not change chemical formula subscripts or physical states.

5. Multiply the enthalpy change by the same factor.

Use:

\[
\Delta H_{\text{new}}=k\Delta H_{\text{original}}
\]

6. Check the sign.

Multiplying by a positive number preserves the sign. Reversing the equation changes the sign.

7. Check what the enthalpy value represents.

Determine whether you are working with an experimental result, a model estimate, a total reaction enthalpy, or a molar enthalpy.

A common mistake is multiplying an already scaled enthalpy by the number of moles a second time. Always check the basis of the supplied value before doing another calculation.

06Questions and solutions

The following questions progress from straightforward equation scaling to limiting reactants, model estimates, and the interpretation of experimental uncertainty.

All numerical reaction data are illustrative teaching data unless identified as experimental results within a question.

Question 1

The following thermochemical equation describes the formation of sulfur trioxide:

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

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

Write the complete thermochemical equation, including its enthalpy change, when 6.00 mol of sulfur trioxide is produced.

Solution 1

The enthalpy change is \(-594\text{ kJ}\), and the complete equation must be multiplied by 3.

The original equation produces 2 mol of sulfur trioxide. The required amount is 6.00 mol.

Step 1: Calculate the multiplying factor.

\[
k=\frac{6.00}{2.00}=3
\]

Step 2: Multiply every coefficient by 3.

\[
\ce{6SO2(g) + 3O2(g) -> 6SO3(g)}
\]

Step 3: Multiply the enthalpy change by 3.

\[
\begin{aligned}
\Delta H_{\text{new}}&=3(-198)\\
&=-594\text{ kJ}
\end{aligned}
\]

The final equation is:

\[
\boxed{\ce{6SO2(g) + 3O2(g) -> 6SO3(g)}}
\]

\[
\boxed{\Delta H=-594\text{ kJ}}
\]

The negative sign indicates that the reaction releases energy. Producing three times as much sulfur trioxide requires three times the energy transfer.

Question 2

Consider the following thermochemical equation:

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

\[
\Delta H=-566\text{ kJ}
\]

Write the thermochemical equation for the decomposition of 0.500 mol of carbon dioxide into carbon monoxide and oxygen.

Calculate the corresponding enthalpy change.

Solution 2

The decomposition is endothermic, and the required enthalpy change is \(+141.5\text{ kJ}\).

There are two separate manipulations: reversing the reaction and scaling it down.

Step 1: Reverse the original equation.

Swap the reactants and products:

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

Reversing the reaction changes the sign of the enthalpy:

\[
\Delta H=+566\text{ kJ}
\]

Step 2: Calculate the multiplying factor.

The reversed equation decomposes 2 mol of carbon dioxide.

The question requires 0.500 mol.

\[
k=\frac{0.500}{2.00}=0.250
\]

Step 3: Multiply every coefficient by 0.250.

\[
\ce{0.500CO2(g) -> 0.500CO(g)}
\]

\[
\ce{+ 0.250O2(g)}
\]

Step 4: Multiply the reversed enthalpy change by 0.250.

\[
\begin{aligned}
\Delta H_{\text{new}}&=0.250(+566)\\
&=+141.5\text{ kJ}
\end{aligned}
\]

Therefore:

\[
\boxed{\ce{0.500CO2(g) -> 0.500CO(g) + 0.250O2(g)}}
\]

\[
\boxed{\Delta H=+141.5\text{ kJ}}
\]

The reaction is endothermic because energy must be absorbed to decompose carbon dioxide.

The key is to reverse the equation first, changing the sign, and then scale both the coefficients and enthalpy by 0.250.

Question 3

Magnesium reacts with oxygen according to the following thermochemical equation:

\[
\ce{2Mg(s) + O2(g) -> 2MgO(s)}
\]

\[
\Delta H=-1204\text{ kJ}
\]

A reaction vessel contains:

• 0.300 mol of magnesium.
• 0.100 mol of oxygen.

Assume the reaction proceeds until one reactant is completely consumed.

Calculate:

(a) The limiting reactant.

(b) The amount of magnesium oxide produced.

(c) The enthalpy change for the reaction.

(d) The complete thermochemical equation for the actual amount of reaction.

Solution 3

Oxygen is the limiting reactant. The reaction produces 0.200 mol of magnesium oxide and has an enthalpy change of \(-120.4\text{ kJ}\).

The important detail is that you cannot automatically multiply the equation by the amount of magnesium available. You must first determine which reactant controls the amount of reaction.

Step 1: Identify the mole ratio.

The balanced equation is:

\[
\ce{2Mg + O2 -> 2MgO}
\]

The mole ratio is:

\[
2:1:2
\]

This means 2 mol of magnesium require 1 mol of oxygen.

Step 2: Determine the limiting reactant.

For 0.100 mol of oxygen, the required magnesium is:

\[
n(\ce{Mg})=2\times0.100
\]

\[
n(\ce{Mg})=0.200\text{ mol}
\]

We have 0.300 mol of magnesium available.

Therefore, magnesium is in excess and oxygen is the limiting reactant.

The magnesium remaining is:

\[
0.300-0.200=0.100\text{ mol}
\]

Step 3: Calculate the amount of magnesium oxide formed.

The mole ratio of oxygen to magnesium oxide is 1:2.

Therefore:

\[
n(\ce{MgO})=2\times0.100
\]

\[
\boxed{n(\ce{MgO})=0.200\text{ mol}}
\]

Step 4: Determine the multiplying factor.

The original equation uses 1 mol of oxygen.

The actual reaction uses 0.100 mol of oxygen.

\[
k=\frac{0.100}{1.00}=0.100
\]

Step 5: Calculate the enthalpy change.

\[
\begin{aligned}
\Delta H_{\text{new}}&=0.100(-1204)\\
&=-120.4\text{ kJ}
\end{aligned}
\]

Step 6: Write the complete scaled equation.

\[
\boxed{\ce{0.200Mg(s) + 0.100O2(g) -> 0.200MgO(s)}}
\]

\[
\boxed{\Delta H=-120.4\text{ kJ}}
\]

The reaction releases 120.4 kJ of energy.

The tempting approach is to use all 0.300 mol of magnesium to calculate the energy change. However, there is insufficient oxygen to react with all the magnesium.

This is why limiting reactants must be considered before scaling a thermochemical equation.

Question 4

Consider the reaction:

\[
\ce{H2(g) + Cl2(g) -> 2HCl(g)}
\]

The following illustrative mean bond enthalpies are supplied:

(a) Calculate the estimated enthalpy change for the reaction as written.

(b) Calculate the estimated enthalpy change when 6.00 mol of hydrogen chloride forms.

(c) Explain why the calculated value is an estimate rather than an experimentally measured enthalpy change.

Solution 4

The estimated enthalpy change is \(-185\text{ kJ mol}^{-1}\) for the original reaction and \(-555\text{ kJ}\) when 6.00 mol of hydrogen chloride forms.

Step 1: Identify the bonds broken.

One H-H bond and one Cl-Cl bond must be broken.

The energy required is:

\[
436+243=679\text{ kJ mol}^{-1}
\]

Step 2: Identify the bonds formed.

Two H-Cl bonds form.

The energy released is:

\[
2\times432=864\text{ kJ mol}^{-1}
\]

Step 3: Calculate the estimated enthalpy change.

Use:

\[
\Delta H_{\text{estimate}}=
E_{\text{broken}}-E_{\text{formed}}
\]

Substitute:

\[
\begin{aligned}
\Delta H_{\text{estimate}}
&=679-864\\
&=-185\text{ kJ mol}^{-1}
\end{aligned}
\]

The negative sign indicates an exothermic reaction.

Step 4: Calculate the multiplying factor for 6.00 mol of hydrogen chloride.

The original equation produces 2 mol of hydrogen chloride.

\[
k=\frac{6.00}{2.00}=3
\]

Step 5: Multiply the equation and estimated enthalpy by 3.

The scaled equation is:

\[
\ce{3H2(g) + 3Cl2(g) -> 6HCl(g)}
\]

The estimated enthalpy change is:

\[
\begin{aligned}
\Delta H_{\text{estimate,new}}
&=3(-185)\\
&=-555\text{ kJ}
\end{aligned}
\]

Step 6: Explain the limitation of the model.

The bond enthalpies are average values for gaseous molecules. They do not necessarily represent the exact bond energies in the particular molecules involved.

Therefore, the calculated enthalpy is a theoretical estimate rather than a direct experimental measurement.

Scaling the estimate by 3 is mathematically correct, but it does not make the underlying model more precise.

Question 5

A chemist investigates the reaction:

\[
\ce{H2(g) + Cl2(g) -> 2HCl(g)}
\]

Using mean bond enthalpies, a theoretical model estimates the enthalpy change as:

\[
\Delta H_{\text{model}}=-185\text{ kJ mol}^{-1}
\]

An independent experimental determination gives:

\[
\Delta H_{\text{experimental}}
=-184\pm2\text{ kJ mol}^{-1}
\]

The experimental uncertainty refers to the enthalpy change per mole of reaction as written.

A student multiplies the chemical equation by 4:

\[
\ce{4H2(g) + 4Cl2(g) -> 8HCl(g)}
\]

They calculate:

• Model estimate: \(-740\text{ kJ}\).
• Experimental value: \(-736\text{ kJ}\).

The student then argues:

“The model was originally only 1 kJ away from the experimental value. After multiplying the equation by 4, it is 4 kJ away. Therefore, the model has become less accurate.”

Evaluate this conclusion.

In your answer:

(a) Explain whether the student’s conclusion is correct.

(b) Calculate the uncertainty in the scaled experimental value.

(c) Determine whether the model estimate can be distinguished from the experimental result using the supplied data.

(d) Explain why multiplying an equation does not improve the reliability of an enthalpy estimate.

Solution 5

The student’s conclusion is incorrect. The model has not become less accurate. Both the enthalpy difference and the experimental uncertainty increase by the same factor of 4.

This question tests an important distinction between the size of an absolute error and the reliability of a result.

Step 1: Calculate the original difference between the model and experimental values.

The model predicts:

\[
-185\text{ kJ mol}^{-1}
\]

The experimental result is:

\[
-184\pm2\text{ kJ mol}^{-1}
\]

The difference between the central values is:

\[
|-185-(-184)|=1\text{ kJ mol}^{-1}
\]

The model is 1 kJ mol\(^{-1}\) more exothermic than the experimental central value.

Step 2: Scale the thermochemical equation.

Multiplying by 4 gives:

\[
\boxed{\ce{4H2(g) + 4Cl2(g) -> 8HCl(g)}}
\]

The model estimate becomes:

\[
\begin{aligned}
\Delta H_{\text{model,new}}&=4(-185)\\
&=-740\text{ kJ}
\end{aligned}
\]

The experimental central value becomes:

\[
\begin{aligned}
\Delta H_{\text{experimental,new}}&=4(-184)\\
&=-736\text{ kJ}
\end{aligned}
\]

Step 3: Scale the experimental uncertainty.

The original uncertainty is:

\[
\pm2\text{ kJ mol}^{-1}
\]

Because the equation has been multiplied by 4, the absolute uncertainty must also be multiplied by 4.

\[
\text{New uncertainty}=4\times2
\]

\[
\boxed{\pm8\text{ kJ}}
\]

The scaled experimental result is therefore:

\[
\boxed{\Delta H_{\text{experimental}}=-736\pm8\text{ kJ}}
\]

Step 4: Compare the model with the experimental range.

The experimental range is:

\[
-736-8=-744\text{ kJ}
\]

to:

\[
-736+8=-728\text{ kJ}
\]

Therefore, the experimental interval is:

\[
-744\text{ kJ}\leq\Delta H\leq-728\text{ kJ}
\]

The model estimate is:

\[
-740\text{ kJ}
\]

This value falls inside the experimental interval.

The model is therefore consistent with the experimental result within the supplied uncertainty.

Step 5: Explain why the student’s reasoning is incorrect.

The student has compared absolute differences without considering that the uncertainty also scales.

Originally:

• Absolute difference: 1 kJ mol\(^{-1}\).
• Experimental uncertainty: ±2 kJ mol\(^{-1}\).

After multiplying by 4:

• Absolute difference: 4 kJ.
• Experimental uncertainty: ±8 kJ.

The relative disagreement has not changed.

For example, the original difference relative to the experimental central value is:

\[
\frac{1}{184}\times100=0.543\%
\]

After scaling:

\[
\frac{4}{736}\times100=0.543\%
\]

The relative difference is identical.

Multiplying an equation does not generate new experimental evidence, reduce uncertainty, or improve the theoretical model.

It simply describes a larger amount of the same reaction.

There is also an important qualification: agreement within the experimental uncertainty does not prove that the model is exact. It means the supplied experiment cannot distinguish the model prediction from the measured result at the stated uncertainty.

Final conclusion: Scaling a thermochemical equation multiplies the enthalpy change and its absolute uncertainty by the same factor. The relative uncertainty and relative disagreement remain unchanged.

07Where to go next

Once you can scale complete thermochemical equations, you can use the same rules to reverse reactions and combine multiple equations.

This is the foundation of Hess’s Law calculations: when equations are added, their enthalpy changes must also be added, after applying the correct scaling and sign changes.

For students following the Chemistry Stage 6 Syllabus (2017), this leads directly into Hess’s Law cycles.

Students following the Chemistry 11-12 Syllabus (2025) should check their prescribed course content and treat the full Hess-cycle method as extension unless it is specified in their program.

Continue with How to Build a Hess Cycle from Combustion Data to see how scaled and reversed thermochemical equations can be combined to calculate an unknown reaction enthalpy.