Carbon burns to carbon dioxide with a standard enthalpy of about −393.5 kJ per mole, while carbon monoxide burns onward to carbon dioxide with about −283.0 kJ per mole. Subtracting the second thermochemical reaction from the first gives an enthalpy of −110.5 kJ for C(graphite)+½O₂(g)→CO(g). The arithmetic is simple only after the chemical equations have been reversed, scaled and cancelled correctly.
Hess’s law uses the state-function nature of enthalpy to calculate an otherwise unspecified reaction from known thermochemical steps. This guide works the carbon–carbon-monoxide–carbon-dioxide cycle by two routes, explains where sign and coefficient errors occur, and shows why a valid standard-state calculation does not automatically describe the measured heat or feasibility of a real process.
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Define the Hess-law cycle
Chapters 1–3. Start here.
Reverse, add and cross-check
Chapters 4–6. Start here.
Audit coefficients and physical states
Chapters 7–10. Start here.
Practise and report
Chapters 11–12. Start here.
Full chapter index · Worked exercises · Thermochemistry subject root
Chapter index
Define the Hess-law cycle · Chapters 1–3
Reverse, add and cross-check · Chapters 4–6
Audit coefficients and physical states · Chapters 7–10
Practise and report · Chapters 11–12
Suppose the desired reaction is C(graphite,s)+½O₂(g)→CO(g). We want the standard reaction enthalpy for the balanced equation as written. That phrase is important: it refers to one mole of carbon converting to one mole of carbon monoxide with half a mole of oxygen in the indicated states.
A known reference reaction is complete carbon combustion, C(graphite,s)+O₂(g)→CO₂(g), with illustrative standard ΔH_A≈−393.5 kJ per mole of reaction. A second known reaction is CO(g)+½O₂(g)→CO₂(g), with ΔH_B≈−283.0 kJ per mole of reaction.
Both known reactions make carbon dioxide, but the target makes carbon monoxide. The task is not to decide which number ‘looks nearer’ the target; it is to construct a reaction pathway whose net reactants and products match the target exactly.
We use mutually compatible rounded standard enthalpies near the same reference temperature, with graphite, oxygen and the two gas molecules in the stated phases. A high-precision table must use its own stated conditions and uncertainties.
Enthalpy H is a state function, so the enthalpy change between specified initial and final states does not depend on the intermediate reaction pathway used to connect them. If an overall reaction is the sum of steps, its ΔH is the sum of the corresponding step ΔH values.
This permits a reaction with an inconvenient or difficult direct measurement to be evaluated from suitable reference reactions. But the resulting sum is valid only when the chemical species, stoichiometric amounts, states and reference conditions form a consistent path.
A catalyst may change the mechanism or speed of a reaction without changing the enthalpy difference between the same initial and final states. This is an important distinction: Hess’s law says nothing about reaction rate, activation barrier or whether the desired process occurs rapidly.
Treat the chemical equations like signed vectors. Reversing an equation reverses the sign of its enthalpy. Multiplying all stoichiometric coefficients by a factor multiplies its energy change by that same factor. Adding equations cancels species that appear on both sides.
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Label complete graphite combustion as equation A: C(s)+O₂(g)→CO₂(g), ΔH_A=−393.5 kJ. The carbon is the graphite reference form; oxygen is molecular O₂ gas; carbon dioxide is CO₂ gas.
Label carbon monoxide combustion as equation B: CO(g)+½O₂(g)→CO₂(g), ΔH_B=−283.0 kJ. Each equation is balanced in carbon and oxygen. It is acceptable to write a half-molecule coefficient in a thermochemical equation because the coefficients represent stoichiometric mole proportions.
Record the sign beside each equation. Negative enthalpies indicate exothermic changes in the standard convention. Neither negative value should be read as a positive ‘amount of heat contained in carbon dioxide’; these are changes between specified reaction states.
Before any arithmetic, identify what must disappear from the net equation. Our target contains no CO₂, so one equation must produce it and another must consume it. That immediately suggests reversing equation B.
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Equation B was CO+½O₂→CO₂ with ΔH_B=−283.0 kJ. Reverse it: CO₂(g)→CO(g)+½O₂(g). The reversed enthalpy is +283.0 kJ for that written direction.
The positive value is expected. Decomposing carbon dioxide into carbon monoxide and oxygen represents the reverse chemical transformation under the same states, so it requires the opposite enthalpy change. It is not the same as asserting that a reactor can perform this reverse reaction simply by adding that exact amount of heat.
Keep all species intact while reversing; reverse both sides of the whole equation, not merely the CO₂ formula. Retaining the original negative sign on the reversed step would give −676.5 kJ for the sum, an algebraically inconsistent result.
Write ‘+283.0’ explicitly beside the reverse reaction before combining equations. That small discipline prevents one of the most common Hess-law errors, especially when students manipulate more than two reference steps.
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Add equation A and the reversed equation B. On the left are C+O₂+CO₂. On the right are CO₂+CO+½O₂. Cancel CO₂ because one mole appears on both sides of the summed equations.
Next cancel oxygen: one mole of O₂ on the left minus half a mole on the right leaves half a mole on the left. The remaining net reaction is exactly C(s)+½O₂(g)→CO(g).
Now add enthalpies with their manipulated signs: ΔH_target=−393.5+(+283.0)=−110.5 kJ per mole of the target reaction as written. The negative sign means this ideal standard transformation is exothermic.
The species cancellation is the proof that the numerical addition applies to the desired equation. Without that cancellation check, an apparently plausible energy answer can belong to a different reaction.
| Step | Balanced thermochemical reaction | ΔH° for the written equation |
|---|---|---|
| A | C(s) + O₂(g) → CO₂(g) | −393.5 kJ |
| B | CO(g) + ½O₂(g) → CO₂(g) | −283.0 kJ |
| Reverse B | CO₂(g) → CO(g) + ½O₂(g) | +283.0 kJ |
| A plus reverse B | C(s) + ½O₂(g) → CO(g) | −110.5 kJ |
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Suppose instead we name the unknown formation reaction C+½O₂→CO as step F. Follow it with CO+½O₂→CO₂, known to release 283.0 kJ. Adding these two steps gives C+O₂→CO₂, known to release 393.5 kJ.
Thus ΔH_F+(−283.0)=−393.5, and algebra gives ΔH_F=−393.5+283.0=−110.5 kJ. This route avoids explicitly writing the reverse step but uses the same state-function relationship.
The consistency equation is also a way to audit the result. Substitute −110.5 back into the sum: −110.5−283.0=−393.5, exactly the given complete-combustion enthalpy at the selected rounding.
Independent checks are valuable in thermochemistry because a sign mistake can otherwise survive for many lines. Showing both a reaction-cancellation route and a scalar enthalpy equation makes the logic accessible to readers who are less confident with symbolic chemistry.
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Double the target reaction to 2C(s)+O₂(g)→2CO(g). Its enthalpy becomes −221.0 kJ for the doubled reaction as written, not −110.5 kJ. The energy change is extensive in reaction amount under the idealised conditions.
Similarly, 2CO+O₂→2CO₂ has ΔH=2(−283.0)=−566.0 kJ, while 2C+2O₂→2CO₂ has ΔH=2(−393.5)=−787.0 kJ. Subtracting these two doubled equations gives the same doubled target and −787−(−566)=−221.0 kJ.
A reader who obtains −221.0 kJ but reports ‘−221.0 kJ per mole of CO formed’ has a basis error: the doubled equation makes two moles of CO. Dividing by two returns −110.5 kJ per mole of CO produced.
Always attach a specific amount or written reaction to an energy value. The symbols kJ/mol are insufficient without the mole basis and coefficients, particularly when comparing engineering tables with classroom equations.
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The reference reaction explicitly starts from graphite, the conventional standard-state carbon form under ordinary reference conditions. Replacing graphite with diamond changes the initial chemical state and therefore changes the enthalpy difference. Both are carbon, but state functions distinguish their physical forms.
Likewise CO(g), CO₂(g) and oxygen gas cannot be replaced silently with alternative phases or solvated species. Formation enthalpies and Hess-law paths rely on consistent phases and temperatures. The same chemical formula can carry a different enthalpy in a different state.
An equation omitting the oxygen coefficient can also violate atom conservation. C+O₂→CO, as written, has two oxygen atoms on the left and one on the right; it is not a valid chemical target until corrected to ½O₂.
The element’s standard formation enthalpy is conventionally zero only for its specified reference state. Graphite C and O₂(g) can therefore be zero in a compatible standard formation-enthalpy table, while CO’s formation enthalpy is the nonzero result we have derived.
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The standard enthalpy of formation of CO(g) corresponds to forming one mole of CO from C(graphite) and half a mole of O₂(g) in their standard reference states. That is exactly the target reaction.
The products-minus-reactants formula gives ΔH°_f(CO)=ΔH°_f(CO)−[ΔH°_f(Cgraphite)+½ΔH°_f(O₂)]. The elemental reference contributions are zero by convention, leaving the tabulated CO formation enthalpy directly as the target result.
By our Hess cycle, this is approximately −110.5 kJ/mol, consistent with typical rounded thermochemical tables. The agreement serves as a third check that the reversed-equation process produced the intended state comparison.
In real reporting, use formation-enthalpy data from one consistent temperature and reference framework, not a patchwork of numbers chosen merely because their difference yields a preferred answer.
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CHAPTER 10 OF 12
10. Apply the result to partial reaction extent without overclaiming heat
If 0.20 mol of carbon undergoes the ideal target formation reaction completely, the corresponding standard reaction enthalpy is 0.20×(−110.5)=−22.1 kJ under the stated reference-state assumptions.
That scaled amount is not automatically the heat a thermometer would register. A real reactor can have incomplete conversion, feed and product streams at different temperatures, heat losses, other reactions and work transfers.
The earlier article on limiting reagents and calorimetry explains why a correct molar enthalpy must be multiplied by actual reaction extent rather than the total amount of one reagent initially charged.
One should also distinguish the target CO-formation reaction from complete combustion of carbon to CO₂. Producing CO and then oxidising it further are separate thermochemical transformations, with different chemical products and potentially different practical process outcomes.
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The numbers −393.5 and −283.0 are rounded reference values. Their subtraction justifies −110.5 at that reporting precision, not an arbitrarily long series of decimals. A spreadsheet may preserve more digits internally, but the source measurements and conventions determine meaningful reporting.
For a purely illustrative uncertainty example, suppose two independent reference enthalpies have standard uncertainties of 0.2 and 0.3 kJ respectively. The standard uncertainty in their difference would be √(0.2²+0.3²)≈0.36 kJ, under the independence assumption.
If the reference enthalpies share systematic calibration or data sources, their errors may be correlated. The uncertainty of A−B then involves covariance: Var(A−B)=Var(A)+Var(B)−2Cov(A,B). The simple square-root sum is not universally valid.
The goal of an educational calculation is to show the dependence of the result on source values and consistent states. A defensible publication lists data sources, rounding and any relevant uncertainty rather than presenting a precise-looking number without its provenance.
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Exercise A: Given C+O₂→CO₂, ΔH=−393.5 kJ and CO+½O₂→CO₂, ΔH=−283.0 kJ, find C+½O₂→CO. Reverse the second equation, add to the first, and obtain −110.5 kJ.
Exercise B: What is ΔH for 2CO₂→2CO+O₂, the reverse of the doubled CO combustion? Start from 2CO+O₂→2CO₂, ΔH=−566.0 kJ, then reverse. The answer is +566.0 kJ for the doubled reverse equation.
Exercise C: A student reports −676.5 kJ for carbon monoxide formation after adding −393.5 and −283.0. What went wrong? The CO combustion equation was not reversed before addition, so the chemical cancellation and energy sign do not match the target.
Exercise D: A solution gets −221.0 kJ for producing two moles of CO but labels it per mole CO. Correct the normalisation: −110.5 kJ per mole CO produced in the chosen standard states.
Applied thermochemical reaction lab
A formal species-balance audit
List species in the order C(s), O₂(g), CO(g), CO₂(g). The coefficient-change vector for C+O₂→CO₂ is (−1,−1,0,+1), while that for CO+½O₂→CO₂ is (0,−0.5,−1,+1).
Subtract the second vector from the first: (−1,−0.5,+1,0). It describes exactly C+½O₂→CO. This algebraic representation shows why Hess’s law can be implemented as a linear combination of stoichiometric reactions.
Students do not need matrix theory to solve the basic case, but the vector check becomes valuable when a reaction network contains many intermediates and potential cancellation errors.
A complete two-mole consistency check
Double both given reactions: 2C+2O₂→2CO₂, ΔH=−787.0 kJ, and 2CO+O₂→2CO₂, ΔH=−566.0 kJ. Reverse the second doubled equation and add it to the first.
The cancelled net is 2C+O₂→2CO, with ΔH=−787.0+566.0=−221.0 kJ. Divide by two to return the one-mole result −110.5 kJ.
This exercise reinforces two independent operations: reversing changes the sign, whereas scaling changes both coefficients and energy magnitude. Neither operation may be applied to only part of a thermochemical equation.
Explain why a negative ΔH does not give a reaction rate
Exothermicity means the products have lower enthalpy than the reactants in the stated comparison. It does not tell us how quickly the pathway proceeds, which reaction intermediates occur or what activation energy must be overcome.
A catalyst can change the reaction mechanism and rate without changing ΔH between the same initial and final states. Hess’s law remains valid even when a real process travels through many intermediate reactions.
Teach this distinction by asking whether the calculated −110.5 kJ guarantees that a piece of graphite instantly forms CO under ambient conditions. It does not; kinetic, transport and chemical-environment constraints are separate questions.
Handle a change of temperature or phase
If one reference reaction’s enthalpy is tabulated at 298 K and another at a different temperature, they cannot always be subtracted as though their state comparisons were identical. Appropriate heat-capacity corrections can be required to bring them to a common temperature.
Likewise, changing the starting carbon allotrope or the product’s physical state changes the enthalpy comparison. The reaction vector may look superficially similar, but the thermodynamic species are not identical.
This connects the Hess cycle to the wider thermochemistry series. The correct equation is necessary for the algebra, while common reference conditions are necessary for the energy numbers to form a defensible sum.
Frequently asked questions
Why does reversing the CO reaction make ΔH positive?
The reversed reaction swaps its initial and final chemical states, so the enthalpy difference changes sign from −283.0 to +283.0 kJ.
Why does doubling the equation double ΔH?
It describes twice the amount of reaction. The doubled formation of two moles of CO has ΔH≈−221.0 kJ for that written equation.
Does an exothermic ΔH prove a rapid reaction?
No. Reaction kinetics and activation barriers are separate from the state-function enthalpy difference.
Sources and connected reading
The numerical reaction data are representative rounded standard values. The worked calculations demonstrate exact reaction cancellation at that precision.
- OpenStax Chemistry: Atoms First 2e — Hess’s law, carbon and carbon monoxide example
- Chemistry LibreTexts — Hess’s Law worked reactions and signs
- The Gold Standard of Thermochemistry — subject root
Advanced Hess-law lab: cycle closure, actual extent and correlated uncertainty
Use a closed thermochemical cycle as an error detector
Call complete carbon combustion A, carbon monoxide combustion B, and formation of carbon monoxide F. Their balanced equations obey F+B=A. Therefore a closed cycle of reactions satisfies ΔH_F+ΔH_B−ΔH_A=0 when all data refer to the same states and conditions.
Insert the rounded values: (−110.5)+(−283.0)−(−393.5)=0 kJ. The cycle closes. This is an excellent check because it combines the chemical stoichiometry with the enthalpy bookkeeping; a wrong reversal sign or doubled coefficient would usually leave a nonzero residual.
Suppose a hypothetical dataset instead claimed ΔH_F=−108.0 kJ while keeping the other two entries unchanged. Its closure residual would be +2.5 kJ. That discrepancy shows the three reference values cannot all satisfy one ideal internally consistent cycle at the claimed precision, but it does not by itself identify which measurement, convention or calculation caused the mismatch.
Real thermochemical databases reconcile measurements with uncertainties and shared standards. A cycle-closure audit is useful precisely because it reveals inconsistency without pretending the direct experiment or one favourite table must automatically be correct.
Reconcile the result with an actual amount of carbon
For the target C(graphite)+½O₂→CO, the illustrative standard enthalpy is −110.5 kJ per mole of reaction. If the reaction extent is 0.25 mol, its standard-state enthalpy change scales to 0.25×(−110.5)=−27.625 kJ, usually reported as about −27.6 kJ at the available source precision.
If 0.25 mol of carbon was charged but only 0.10 mol actually converted to CO, scaling by the charged amount would be wrong. The standard reaction enthalpy for 0.10 mol conversion would instead be −11.05 kJ, before an actual vessel’s temperature and heat-loss corrections are considered.
By contrast, if the same carbon goes all the way to CO₂, the chemical reaction is different and the complete-combustion standard enthalpy applies. This is not a difference caused by rounding; it reflects different final chemical products.
Students should always state which species is being formed and how much reaction occurred. A reliable molar reference value does not determine reaction extent or measured useful heat without additional evidence.
Check uncertainties when two reference values share calibration
Assume solely for illustration that the carbon-to-CO₂ enthalpy A has standard uncertainty 0.2 kJ and the CO-to-CO₂ enthalpy B has standard uncertainty 0.3 kJ. If they are independent, F=A−B has uncertainty √(0.2²+0.3²)≈0.36 kJ.
If the two reference values have positive error correlation ρ=+0.5, their covariance is 0.5×0.2×0.3=0.03 kJ². The variance of A−B is 0.04+0.09−2(0.03)=0.07 kJ², giving standard uncertainty about 0.265 kJ.
With negative correlation ρ=−0.5, the difference variance is 0.04+0.09+0.06=0.19 kJ², giving uncertainty about 0.436 kJ. Positive shared errors partly cancel in a difference, while negative correlations can make the difference more uncertain.
These numbers are not claimed uncertainties for the published formation enthalpies. They demonstrate why precision statements require provenance and covariance information rather than simply appending extra digits to a Hess-law result.
Explain how thermochemical states differ from reaction kinetics
Under Hess’s law, a hypothetical indirect path through carbon monoxide and a direct reference-state transformation to carbon dioxide yield the same net enthalpy change when they share initial and final states. That mathematical equality does not establish that each reaction pathway can be carried out readily under ambient laboratory conditions.
For example, a catalyst can affect a chemical rate or activation barrier without altering the standard enthalpy of the same final transformation. A reaction may be exothermic but kinetically slow, or it may require special conditions to achieve a measurable conversion.
A calorimeter also measures an energy balance that may include apparatus heat capacity, unreacted material and exchange with surroundings. The standard Hess-law enthalpy is a state difference; the observed temperature response needs its own measurement model.
This distinction explains why the correct −110.5 kJ result should not be described as a recipe for obtaining that quantity of usable heat from an arbitrary carbon sample. It is the reference enthalpy for a specified thermochemical conversion.
A final Hess-law quality-control worksheet
First write the target with correct atoms, phases and coefficients. Then list each known reaction with its direction and signed enthalpy. Mark every reversal and multiplier visibly before forming the reaction sum. If CO₂ remains in the net equation when the target contains none, the cancellation is not complete.
Second calculate the sum in two ways where possible: a manipulated-reaction pathway and a formation-enthalpy table or cycle closure. If the numerical results disagree, inspect reference states, reaction amounts, source rounding and copied signs before trying to ‘average’ the two calculations.
Third distinguish the reference state function from experimental heat and reaction feasibility. Record temperature, pressure or standard-state convention, reaction extent and any corrections needed to compare an actual measurement with the derived enthalpy.
Finally ask the learner to explain the sign in plain language: reversing the CO combustion step changes −283.0 into +283.0 because the product and reactant states swap. A result should make that causal reasoning apparent, not merely display a correct arithmetic answer.
A final check of the reaction’s chemical meaning
Carbon monoxide and carbon dioxide are distinct final products, not interchangeable names for partially or completely burnt carbon. Their standard formation and oxidation enthalpies cannot be compared without keeping the oxygen amount and reaction extent consistent. That is exactly why the Hess cycle explicitly cancels carbon dioxide and leaves half a mole of oxygen on the target’s reactant side.
The −110.5 kJ result is meaningful only for the reaction C(graphite)+½O₂(g)→CO(g) at the chosen reference conditions and per mole of reaction as written. An actual process involving excess oxygen, elevated temperatures, incomplete conversion or further oxidation to CO₂ requires a different overall reaction and a full energy balance.
This distinction is especially important in educational material: the correctness of the enthalpy arithmetic does not make carbon monoxide a safe laboratory product or imply a recommended practical experiment. Carbon monoxide is toxic, and real reactions require appropriate qualified safety procedures.