How to Check a Hess Cycle by Following Two Reaction Paths

Two calculations give different answers for the same reaction: one gives \(-392\ \text{kJ mol}^{-1}\), while another gives \(-393.5\ \text{kJ mol}^{-1}\). Has someone made a mistake, or can both results be correct?

This is exactly what checking a Hess cycle is designed to establish. Calculate the total enthalpy change along each complete path, making sure both paths have identical starting and finishing states. The totals should agree within experimental uncertainty and rounding.

The important part is knowing what you are comparing. A calculated sum of tabulated enthalpy changes, an experimental measurement, and an estimate using average bond enthalpies are not necessarily equally precise.

01Why two paths must give the same enthalpy change

Imagine walking from the ground floor of a building to the third floor. You could take the stairs or the lift. The route is different, but your overall change in height is identical.

Enthalpy works in a similar way. It is a state function, meaning its change depends on the initial and final states of a system, not the route taken between them.

Hess’s Law applies this idea to chemical reactions.

If a reaction occurs directly or through several intermediate reactions, the overall enthalpy change must be the same, provided the initial and final states are identical.

Mathematically:

\[ \Delta H_{\text{direct}} = \Delta H_1+\Delta H_2+\cdots+\Delta H_n \]

Here, each \(\Delta H\) represents the enthalpy change of one reaction step, in kJ mol\(^{-1}\).

For a complete cycle, returning to the original chemical state means the total enthalpy change must be zero:

\[ \sum \Delta H_{\text{cycle}}=0 \]

There is one important limitation to the building analogy. Enthalpy is not the distance travelled or the amount of energy consumed along a physical route. A Hess cycle is an energy-accounting model, not necessarily a description of the actual mechanism of a reaction.

02First check: are the starting and finishing states identical?

Before adding any numbers, compare the chemical equations at the beginning and end of each path.

This is one of the most easily overlooked parts of Hess cycle questions.

Consider these substances:

These are not interchangeable.

For example, producing carbon dioxide gas and producing carbon dioxide dissolved in water are different thermochemical processes. The dissolution of carbon dioxide involves an additional enthalpy change.

Similarly, forming liquid water is not the same reaction as forming water vapour.

Check three things before comparing paths:

1. Are the same reactants present, in the same physical states?
2. Are the same products present, in the same physical states?
3. Do the equations represent the same quantities of substances?

If one path produces two moles of a substance while another produces one mole, the equations must be scaled appropriately.

A Hess cycle cannot be checked properly until these conditions are satisfied.

03Second check: have the complete equations been manipulated correctly?

The most important calculation skill is manipulating thermochemical equations.

Every time you change a chemical equation, you must make the corresponding change to its enthalpy value.

There are three basic operations.

1. Reversing an equation

If a reaction is reversed, its enthalpy change changes sign.

For example:

\[ \ce{A -> B} \qquad \Delta H=-50\ \text{kJ mol}^{-1} \]

Reversing the reaction gives:

\[ \ce{B -> A} \qquad \Delta H=+50\ \text{kJ mol}^{-1} \]

The reverse reaction requires the same amount of energy that the forward reaction releases.

2. Multiplying an equation

If every coefficient in an equation is multiplied by a number, its enthalpy change must also be multiplied by that number.

For example:

\[ \ce{A + B -> C} \qquad \Delta H=-40 \]

Multiplying the entire equation by two gives:

\[ \ce{2A + 2B -> 2C} \qquad \Delta H=-80 \]

The enthalpy change doubles because twice as much reaction occurs.

3. Adding equations

When you add thermochemical equations, add their enthalpy changes as well.

Any substances appearing on both sides can be cancelled, provided their chemical identities and physical states are identical.

For example:

\[ \ce{A -> B} \]

\[ \ce{B -> C} \]

Adding these equations cancels the intermediate substance B:

\[ \ce{A -> C} \]

The overall enthalpy change is:

\[ \Delta H_{\text{overall}}=\Delta H_1+\Delta H_2 \]

You cannot cancel \(\ce{H2O(l)}\) against \(\ce{H2O(g)}\). They represent different states.

A useful habit is to write out the complete equations before doing any arithmetic. It is much easier to spot a missing coefficient or an incorrect cancellation when the chemistry is visible.

04What is measured, and what is calculated?

This distinction matters when you are checking whether experimental results support a Hess cycle.

A temperature change measured in a laboratory is an observation. An enthalpy change calculated from that observation is a derived value.

For example, a simple constant-pressure solution calorimeter contains an insulated cup holding a measured quantity of solution. A thermometer or temperature probe passes through the lid, alongside a stirrer.

The experimental process follows a logical sequence:

1. Measure the initial temperature of the solution.
2. Add known quantities of the reactants.
3. Stir the mixture and record its temperature change.
4. Use the temperature change to calculate the heat transferred.
5. Convert the heat transferred into an enthalpy change per mole of reaction.

The prediction is straightforward. If the reaction is exothermic, the solution should warm up. If it is endothermic, the solution should cool down.

For a simplified solution calorimeter, the heat absorbed by the solution is:

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

Where:

• \(q\) is heat energy in joules (J).
• \(m\) is the mass of the solution in grams (g).
• \(c\) is its specific heat capacity in J g\(^{-1}\) °C\(^{-1}\).
• \(\Delta T\) is the temperature change in °C.

Assuming the calorimeter absorbs negligible heat:

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

At constant pressure, the reaction heat corresponds to the enthalpy change for the amount of reaction that occurred.

A quick measurement calculation

Suppose a reaction warms 100.0 g of solution by 0.68 °C.

Use:

• \(c=4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}\)
• \(n=0.00500\ \text{mol}\) of reaction

First, calculate the heat absorbed by the solution:

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=100.0\times4.18\times0.68\\ &=284.2\ \text{J}\\ &=0.2842\ \text{kJ} \end{aligned} \]

The reaction released this energy:

\[ q_{\text{reaction}}=-0.2842\ \text{kJ} \]

Now calculate the molar enthalpy change:

\[ \begin{aligned} \Delta H &=\frac{q_{\text{reaction}}}{n}\\ &=\frac{-0.2842}{0.00500}\\ &=-56.8\ \text{kJ mol}^{-1} \end{aligned} \]

The negative sign indicates an exothermic reaction.

This is a calculated experimental result, not a direct reading from the thermometer.

In a real investigation, the cup and thermometer can also absorb heat. Heat can escape to the surroundings. A better design uses a lid, additional insulation, stirring, and a calibrated calorimeter heat capacity to account for these effects.

This matters because an experimental result will have uncertainty. You should not expect two independent measurements to agree to every decimal place.

Three types of enthalpy information

Standard enthalpies of formation are particularly useful in Hess cycles.

The standard enthalpy of formation, \(\Delta H_f^\circ\), is the enthalpy change when one mole of a substance forms from its constituent elements in their standard states.

Elements in their standard states have a standard enthalpy of formation of zero.

For a reaction, the formation-enthalpy calculation is:

\[ \Delta H^\circ_{\text{reaction}} = \sum n\Delta H_f^\circ(\text{products}) - \sum n\Delta H_f^\circ(\text{reactants}) \]

The coefficient \(n\) is the number of moles represented by each substance in the balanced equation.

This is another way to follow an indirect thermochemical route.

Why bond enthalpies give estimates rather than exact values

Suppose you want to estimate the enthalpy change for:

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

Illustrative average bond enthalpy data:

The reaction breaks one H-H bond and one Cl-Cl bond. It forms two H-Cl bonds.

Therefore:

\[ \begin{aligned} \Delta H_{\text{estimate}} &=\text{bonds broken}-\text{bonds formed}\\ &=(436+243)-2(432)\\ &=-185\ \text{kJ mol}^{-1} \end{aligned} \]

Suppose an independent experimental determination gives:

\[ \Delta H=-184.6\pm1.0\ \text{kJ mol}^{-1} \]

The two values agree within the experimental uncertainty.

However, the bond enthalpy result is still an estimate. Average bond enthalpies are not exact values for every individual chemical environment.

This is why you must not treat a bond-energy calculation as an exact experimental measurement when assessing whether a Hess cycle closes.

07What to check when the two paths do not agree

If your calculated values differ substantially, do not immediately assume Hess’s Law has failed.

Work through the possible sources of error.

One final rule: never average two conflicting values simply to make a cycle appear consistent. First establish whether the equations describe the same overall reaction.

08Questions and solutions

The following questions progress from straightforward equation manipulation to checking competing thermochemical models. All numerical data identified as illustrative are supplied for these questions.

Question 1

The following thermochemical equation is provided:

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

\[ \Delta H=-92.4\ \text{kJ mol}^{-1} \]

Determine the enthalpy change for the reverse reaction:

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

Solution 1

The enthalpy change is \(+92.4\ \text{kJ mol}^{-1}\).

Reversing a thermochemical equation changes the sign of its enthalpy change.

The original reaction is exothermic, releasing 92.4 kJ per mole of reaction as written. The reverse reaction requires that energy to be absorbed.

Therefore:

\[ \boxed{\Delta H=+92.4\ \text{kJ mol}^{-1}} \]

The coefficients remain unchanged because the amount of each substance has not changed.

Question 2

Consider the following illustrative thermochemical equations:

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

\[ \Delta H=-221.0\ \text{kJ mol}^{-1} \]

And:

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

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

Use Hess’s Law to calculate the enthalpy change for:

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

Show how the complete equations are manipulated.

Solution 2

The enthalpy change is \(-566.0\ \text{kJ mol}^{-1}\).

The first equation must be reversed because carbon monoxide is a reactant in the target reaction.

The second equation must be doubled because the target reaction produces two moles of carbon dioxide.

Step 1: Double the second equation.

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

Its enthalpy change becomes:

\[ \Delta H=2(-393.5)=-787.0\ \text{kJ mol}^{-1} \]

Step 2: Reverse the first equation.

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

Its enthalpy change changes sign:

\[ \Delta H=+221.0\ \text{kJ mol}^{-1} \]

Step 3: Add the equations.

\[ \begin{aligned} \ce{2C + 2O2 -> 2CO2}\\ \ce{2CO -> 2C + O2} \end{aligned} \]

Cancel the two moles of carbon on both sides.

Cancel two moles of oxygen against one mole on the opposite side.

The resulting equation is:

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

Step 4: Add the enthalpy changes.

\[ \begin{aligned} \Delta H &=-787.0+221.0\\ &=-566.0\ \text{kJ mol}^{-1} \end{aligned} \]

The negative result indicates that the oxidation of carbon monoxide to carbon dioxide is exothermic.

Question 3

A student is checking a Hess cycle involving carbon dioxide.

The supplied illustrative data are:

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

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

And:

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

\[ \Delta H=-19.0\ \text{kJ mol}^{-1} \]

Calculate the enthalpy change for:

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

Explain why the answer differs from the enthalpy change for producing carbon dioxide gas.

Solution 3

The enthalpy change for producing aqueous carbon dioxide is \(-412.5\ \text{kJ mol}^{-1}\).

The two reactions must be added because the second reaction converts gaseous carbon dioxide into aqueous carbon dioxide.

Step 1: Write the two supplied equations together.

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

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

Step 2: Cancel the intermediate substance.

Carbon dioxide gas appears as a product in the first equation and a reactant in the second.

It cancels, leaving:

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

Step 3: Add the enthalpy changes.

\[ \begin{aligned} \Delta H &=-393.5+(-19.0)\\ &=-412.5\ \text{kJ mol}^{-1} \end{aligned} \]

The result is more negative because dissolving carbon dioxide releases additional energy.

The tempting mistake is to assume that \(\ce{CO2(g)}\) and \(\ce{CO2(aq)}\) are equivalent final states. They are not. The dissolution step must be included to make the two pathways comparable.

Question 4

Nitrogen monoxide reacts with oxygen to produce nitrogen dioxide:

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

The following illustrative standard enthalpies of formation are supplied:

An independent experimental determination gives:

\[ \Delta H_{\text{measured}}=-113\pm2\ \text{kJ mol}^{-1} \]

Calculate the reaction enthalpy using Hess’s Law and determine whether the experimental result is consistent with the calculated value.

Solution 4

The Hess cycle gives \(-114.2\ \text{kJ mol}^{-1}\), which is consistent with the experimental result.

The key is to construct formation equations for both products and reactants, then reverse the formation of nitrogen monoxide.

Step 1: Write the formation equation for nitrogen monoxide, multiplied by two.

\[ \ce{N2(g) + O2(g) -> 2NO(g)} \]

\[ \Delta H=2(90.3)=+180.6\ \text{kJ mol}^{-1} \]

Step 2: Write the formation equation for nitrogen dioxide, multiplied by two.

\[ \ce{N2(g) + 2O2(g) -> 2NO2(g)} \]

\[ \Delta H=2(33.2)=+66.4\ \text{kJ mol}^{-1} \]

Step 3: Reverse the nitrogen monoxide equation.

\[ \ce{2NO(g) -> N2(g) + O2(g)} \]

Its enthalpy change becomes:

\[ \Delta H=-180.6\ \text{kJ mol}^{-1} \]

Step 4: Add the equations.

\[ \begin{aligned} \ce{2NO(g) -> N2(g) + O2(g)}\\ \ce{N2(g) + 2O2(g) -> 2NO2(g)} \end{aligned} \]

Cancel nitrogen gas and one mole of oxygen gas.

The resulting equation is:

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

Step 5: Calculate the overall enthalpy change.

\[ \begin{aligned} \Delta H &=-180.6+66.4\\ &=-114.2\ \text{kJ mol}^{-1} \end{aligned} \]

The experimental uncertainty range is:

\[ -115\text{ to }-111\ \text{kJ mol}^{-1} \]

The calculated result of \(-114.2\ \text{kJ mol}^{-1}\) falls within this range.

Therefore, the experimental and calculated values are consistent.

This question demonstrates why using complete formation equations is useful. You can see exactly which substances cancel and why the reaction enthalpy has its particular sign.

Question 5

A student is investigating the enthalpy change for forming liquid water:

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

Three different approaches produce the following information.

Path A: Direct experimental measurement

An illustrative calorimetry result gives:

\[ \Delta H_A=-286.0\pm0.8\ \text{kJ mol}^{-1} \]

Path B: Using reference enthalpy data

The standard enthalpy change for forming water vapour is:

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

\[ \Delta H=-241.8\ \text{kJ mol}^{-1} \]

The enthalpy change for condensation is:

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

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

Path C: Using average bond enthalpies

The following illustrative average bond enthalpies are provided:

A student makes three claims:

• The gas-phase pathway should use \(+44.0\ \text{kJ mol}^{-1}\) for condensation because vaporisation is endothermic.
• The reference-data pathway proves that Hess’s Law is incorrect because its calculated result is not exactly \(-286.0\).
• The bond-energy pathway must produce exactly the same result as the experimental measurement because all three approaches describe the same overall reaction.

Evaluate each claim. Calculate the enthalpy change using Paths B and C, and explain which results can be treated as exact thermochemical pathway calculations and which are estimates.

Solution 5

The reference-data pathway gives \(-285.8\ \text{kJ mol}^{-1}\), while the bond-energy pathway estimates \(-285.0\ \text{kJ mol}^{-1}\).

The reference-data result is consistent with the direct measurement. The bond-energy result is an approximation, so it is not expected to match the experimental value exactly.

Step 1: Correct the sign of the condensation enthalpy.

The student has confused condensation with vaporisation.

Vaporisation is:

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

It requires energy, so its enthalpy change is positive:

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

Condensation is the reverse process:

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

Therefore:

\[ \Delta H_{\text{cond}}=-44.0\ \text{kJ mol}^{-1} \]

The first claim is incorrect because the reaction direction determines the sign.

Step 2: Calculate Path B using reference enthalpy data.

The first step forms gaseous water:

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

\[ \Delta H_1=-241.8\ \text{kJ mol}^{-1} \]

The second step condenses the water:

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

\[ \Delta H_2=-44.0\ \text{kJ mol}^{-1} \]

Adding the equations cancels gaseous water.

The net equation is:

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

The total enthalpy change is:

\[ \begin{aligned} \Delta H_B &=-241.8+(-44.0)\\ &=-285.8\ \text{kJ mol}^{-1} \end{aligned} \]

Compare this with the direct measurement:

\[ -286.0\pm0.8\ \text{kJ mol}^{-1} \]

The experimental range is:

\[ -286.8\text{ to }-285.2\ \text{kJ mol}^{-1} \]

The calculated value of \(-285.8\) lies within this range.

Therefore, the reference-data pathway and direct experimental measurement agree within uncertainty.

The second claim is incorrect. Hess’s Law does not require rounded experimental and reference values to agree to every decimal place.

Step 3: Calculate the bond-energy estimate for forming gaseous water.

First, identify the bonds broken.

One H-H bond is broken, and half an O=O bond is required per mole of water formed.

Energy absorbed:

\[ 436+\frac{498}{2}=685\ \text{kJ mol}^{-1} \]

Next, identify the bonds formed.

Two O-H bonds are formed.

Energy released:

\[ 2(463)=926\ \text{kJ mol}^{-1} \]

The estimated enthalpy change is:

\[ \begin{aligned} \Delta H_{\text{estimate}} &=\text{bonds broken}-\text{bonds formed}\\ &=685-926\\ &=-241.0\ \text{kJ mol}^{-1} \end{aligned} \]

This estimates the formation of gaseous water, not liquid water.

Step 4: Complete the estimated pathway.

Add the condensation enthalpy:

\[ \begin{aligned} \Delta H_C &=-241.0+(-44.0)\\ &=-285.0\ \text{kJ mol}^{-1} \end{aligned} \]

The bond-energy estimate differs from the direct experimental result by:

\[ |-285.0-(-286.0)|=1.0\ \text{kJ mol}^{-1} \]

This does not demonstrate a failure of Hess’s Law.

The average bond enthalpies are approximate values. They describe average bond energies rather than the exact enthalpy changes for every individual chemical environment.

Consequently, this pathway is a model estimate, not an exact independent thermochemical determination.

Step 5: Evaluate the three claims.

The deeper lesson is that Hess’s Law applies to the actual enthalpy changes between identical initial and final states. Differences between experimental results, rounded reference values, and approximate models do not automatically indicate a contradiction.

09Where to go next

Once you can check whether two paths agree, the next useful skill is constructing a Hess cycle from unfamiliar reaction data.

Practise identifying which equations need reversing, which need multiplying, and which intermediate substances must cancel before attempting the arithmetic.

For a broader explanation of constructing and calculating enthalpy cycles, see Hess’s Law Energy Cycles: HSC Chemistry Calculations Explained.

Course scope: This explanation follows the Chemistry Stage 6 Syllabus (2017), which includes Hess’s Law in Year 11 Drivers of Reactions. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. Students entering the new course should check its specific content and their school’s program rather than assume that every detail on this page is a named examinable requirement.