A hypothetical reaction releases 50.0 kJ per mole of reaction at 298 K, but its products have a greater heat capacity than its reactants. At 398 K, the same standard-state reaction can have an enthalpy change of −48.5 kJ/mol instead. The change is not an error in the original calorimeter measurement; the reactant and product enthalpies respond differently as temperature changes.
Kirchhoff’s law connects the temperature dependence of reaction enthalpy to the difference between product and reactant heat capacities. This guide derives the relation, calculates fixed-Cp and variable-Cp examples, shows why reaction coefficients and units matter, and separates a thermochemical state-function correction from the temperature history of a real reactor.
Find the question you need
Work all twelve chapters in order, or use the reading routes below to find a particular formula, worked table or checking method.
Build Kirchhoff’s law
Chapters 1–3. Begin this route
Calculate constant heat-capacity examples
Chapters 4–6. Begin this route
Integrate changing heat capacities
Chapters 7–8. Begin this route
Check physical limits
Chapters 9–10. Begin this route
Practise and report
Chapters 11–12. Begin this route
Full chapter index · Applied thermochemistry lab · Exercises · Thermochemistry subject root
Chapter index
Build Kirchhoff’s law · Chapters 1–3
Calculate constant heat-capacity examples · Chapters 4–6
Integrate changing heat capacities · Chapters 7–8
Check physical limits · Chapters 9–10
Practise and report · Chapters 11–12
A reaction enthalpy ΔHrxn(T) compares the enthalpies of products and reactants in a specified balanced reaction at a chosen temperature, under specified pressure and phase conventions. When the reference temperature changes, the enthalpy of each participating substance generally changes, so the reaction enthalpy can change.
That does not mean an earlier laboratory result has become invalid. A reference value reported near 298 K and a calculated value near 398 K answer different state comparisons. The connection between them depends on the heat capacities of all substances in the written reaction.
Our numerical examples use invented species A, B and C so the calculation can be audited without pretending to provide real chemical reference data. Whenever a real chemical is used, heat capacities and phase behaviour should come from an appropriate source for the temperature and pressure range.
Always attach the reporting basis. The phrase ‘−50 kJ/mol’ is incomplete unless it says per mole of what reaction, written with what coefficients and phase labels.
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At fixed pressure, the heat capacity Cp of a substance describes how its enthalpy changes with temperature under the relevant state conditions. For reaction stoichiometric coefficients νᵢ, taken positive for products and negative for reactants, the reaction enthalpy is ΔHrxn=ΣνᵢHᵢ.
Differentiating the reaction enthalpy with respect to temperature at the specified reference conditions gives d(ΔHrxn)/dT=ΔCp,rxn, where ΔCp,rxn=ΣνᵢCp,i. This is the differential form of Kirchhoff’s law.
The product-minus-reactant direction is essential. A larger total product heat capacity makes ΔCp positive; a larger total reactant heat capacity makes it negative. Individual substances have positive heat capacities in many ordinary stable regimes, yet their stoichiometric difference may have either sign.
These equations connect defined thermodynamic states; they do not require the reacting material in a real vessel to follow one specific heating path. That is why Kirchhoff’s law can be used as a reference correction in a state-function calculation.
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Integrate from reference temperature T₁ to target temperature T₂. The exact relation under the appropriate definitions is ΔHrxn(T₂)=ΔHrxn(T₁)+∫ from T₁ to T₂ ΔCp,rxn(T)dT.
If the heat-capacity difference is approximately constant over the interval, the integral simplifies to ΔCp,rxn(T₂−T₁). This is a convenient linear temperature correction, but it is an approximation when the heat capacities depend significantly on temperature.
Pay attention to units: a heat-capacity difference of 15 J/(mol·K) times a 100 K temperature interval gives 1500 J/mol = 1.5 kJ/mol. A mismatch between joules and kilojoules produces an error of a factor of one thousand even when the formula is conceptually correct.
Both T₁ and T₂ must refer to absolute thermodynamic temperatures in the underlying thermodynamic functions. Their difference has the same numerical magnitude in degrees Celsius and kelvin, but an expression using T itself, such as a polynomial Cp(T), must use the units for which the coefficients were defined.
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Consider the fictional reaction A(g)→B(g). At T₁=298 K, suppose its reaction enthalpy is ΔH₁=−50.00 kJ/mol reaction. Let Cp,A=30 J/(mol·K) and Cp,B=45 J/(mol·K), both treated as constant from 298 to 398 K.
The heat-capacity difference is ΔCp=45−30=+15 J/(mol·K). Raising the reference temperature by 100 K therefore changes the reaction enthalpy by +1500 J/mol=+1.50 kJ/mol.
So at 398 K, ΔH₂=−50.00+1.50=−48.50 kJ/mol reaction. The reaction remains exothermic under the stipulated model, but its enthalpy change is less negative than at 298 K.
The plus sign is not optional. A positive ΔCp makes ΔH increase as temperature rises; increasing a negative number makes it less negative. Some students incorrectly subtract the correction because the original reaction was exothermic, mixing the sign of ΔH with the sign of its derivative.
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Under the constant ΔCp=15 J/(mol·K) approximation, the reaction enthalpy is linear in T. At 348 K, 50 K above reference, ΔH≈−49.25 kJ/mol. At 398 K, it is −48.50 kJ/mol. At 498 K, 200 K above reference, it is −47.00 kJ/mol.
This table is a consistency check, not a licence to extrapolate the same heat capacities to any temperature. Far beyond a modest interval, real species can have changing rotational, vibrational and phase behaviour that alters heat capacities, and chemical stability itself may differ.
A quick estimate can help catch a misplaced factor of ten: a 15 J/(mol·K) difference over 100 K is about one or two kilojoules per mole, not 150 kJ/mol. Compare the correction’s scale with the reference enthalpy before accepting a surprising result.
A careful calculation explicitly shows T₂−T₁ in kelvin, the signed ΔCp value, the resulting energy correction and the updated enthalpy with its reaction basis.
| Temperature | ΔT from 298 K | ΔH correction | Illustrative ΔH |
|---|---|---|---|
| 298 K | 0 K | 0 kJ/mol | −50.00 kJ/mol |
| 348 K | 50 K | +0.75 kJ/mol | −49.25 kJ/mol |
| 398 K | 100 K | +1.50 kJ/mol | −48.50 kJ/mol |
| 498 K | 200 K | +3.00 kJ/mol | −47.00 kJ/mol |
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Now take a different fictional reaction 2A(g)+B(g)→3C(g). Let Cp,A=30, Cp,B=40 and Cp,C=25 J/(mol·K). These are made-up constants for the exercise.
The product contribution is 3×25=75 J/(mol reaction·K). The reactant contribution is 2×30+40=100 J/(mol reaction·K). Their difference is ΔCp,rxn=−25 J/(mol reaction·K).
If the reaction enthalpy at 300 K is −80.0 kJ per mole of the equation as written, raising T to 400 K changes it by (−25)(100)/1000=−2.5 kJ, giving −82.5 kJ per mole of reaction.
The wrong shortcut ‘there are three product molecules and two reactant species’ gives a misleading answer. Multiply each species’ heat capacity by its stoichiometric coefficient. The count of different chemical names is not the same as the total stoichiometric amount.
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Suppose the heat-capacity difference for a third fictional reaction varies with temperature as ΔCp(T)=10+0.020(T−300) J/(mol·K), with T in kelvin and the formula intended for 300–500 K. The slope coefficient therefore carries units consistent with J/(mol·K²).
At 300 K the difference is 10 J/(mol·K), at 400 K it is 12, and at 500 K it is 14. Using a single constant value of 10 across the full 200 K interval would give only 2.00 kJ/mol of enthalpy correction.
Integrate the actual function: ∫ from 300 to 500 [10+0.020(T−300)]dT = 10×200+0.010×200² = 2400 J/mol = 2.40 kJ/mol.
If the reaction enthalpy at 300 K is −100.0 kJ/mol, it becomes −97.60 kJ/mol at 500 K under the stated model. A positive average ΔCp again moves a negative enthalpy toward zero, though no real chemical identity is implied by our constructed numbers.
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Suppose no closed-form polynomial is supplied, only tabulated ΔCp values: 10 at 300 K, 12 at 400 K and 14 at 500 K. The trapezoidal rule on the first 100 K interval contributes (10+12)/2×100=1100 J/mol.
The next interval contributes (12+14)/2×100=1300 J/mol. Add them to obtain 2400 J/mol, matching the exact integral of our linear test function.
The agreement here is exact for this synthetic linear heat-capacity function; do not infer that two trapezoids exactly integrate every real material’s nonlinear Cp(T). For non-linear data, grid spacing and numerical integration error require attention.
A clear table of temperature, reactant and product heat capacities, reaction ΔCp, and the interval contribution helps readers reproduce the calculation. It also reveals whether interpolation has crossed a phase transition, where a smooth heat-capacity treatment alone may be inadequate.
| Temperature | ΔCₚ | Correction from previous row |
|---|---|---|
| 300 K | 10 J/(mol·K) | Start |
| 400 K | 12 J/(mol·K) | +1.10 kJ/mol |
| 500 K | 14 J/(mol·K) | +1.30 kJ/mol |
| Total 300–500 K | Linear function | +2.40 kJ/mol |
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Kirchhoff’s law in the simple integrated form assumes the reference substance states and phases are handled consistently over the temperature range. If a reactant or product melts, boils or changes crystal form, the enthalpy change for that phase transition must be included appropriately.
A liquid water product at one reference temperature and water vapour at another does not follow the same heat-capacity path as one fixed-phase gas. Attempting to correct a liquid-water reaction enthalpy using only gas-phase heat capacities omits the phase-change contribution.
The same warning applies if a compound dissociates significantly or changes composition over the chosen temperature interval. An enthalpy of a hypothetical stoichiometric transformation can still be defined, but it must not be confused with the heat generated by an equilibrium mixture or an incompletely reacting real system.
Always identify phase labels in the balanced equation and state which heat-capacity dataset applies to those phases. Checking for phase transitions is a substantive thermochemical step, not a small metadata detail.
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A standard reaction enthalpy at temperature T compares substances in specified standard states. It is not automatically the heat measured in a working reactor that has feed streams entering at different temperatures, products leaving hot, incomplete conversion, external heat loss and possibly shaft or pressure–volume work.
A real process energy balance may include sensible enthalpy carried by incoming and outgoing streams, latent heats, formation enthalpies, nonideal mixing and reaction extent. Kirchhoff’s relation helps calculate temperature-dependent reaction enthalpy but does not eliminate those other terms.
For example, calculating ΔHrxn(500 K) from a 300 K reference does not mean a reactor held at 500 K must release exactly that much energy per mole of feed. You must know how many moles reacted and account for the thermodynamic states of the inlet and outlet materials.
This distinction ties the present guide to Limiting Reagents and the Correct Mole in Calorimetry, which asks a separate but indispensable question: what amount of reaction actually occurred?
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Problem A: At 300 K, an abstract reaction has ΔH=+20.0 kJ/mol and ΔCp=−10 J/(mol·K). Find ΔH at 450 K. The correction is −10×150/1000=−1.50 kJ/mol, giving +18.50 kJ/mol. The reaction remains endothermic under this model.
Problem B: A student computes ΔCp=+20 J/(mol·K), raises temperature by 50 K, and adds 1000 kJ/mol. The multiplication should give 1000 J/mol=1.00 kJ/mol. The mistake is the unit conversion, not the sign.
Problem C: If ΔCp(T)=10+0.020(T−300), integrate from 300 to 400 K. The correction is 10×100+0.010×100²=1100 J/mol. This equals the first trapezoidal interval of the table example.
Problem D: Suppose one reactant boils between T₁ and T₂. Can you use its gas heat capacity from T₁ all the way to T₂ without accounting for its liquid phase and enthalpy of vaporisation? No. Break the path into appropriate phases and include the phase transition.
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Write the balanced reaction with states and a clear reaction-normalisation basis. Record a trustworthy reference enthalpy at T₁. Obtain appropriate heat capacities for every species, in the proper phases and units, and calculate the signed product-minus-reactant ΔCp.
Choose the integration method based on what is known. Use a constant ΔCp approximation only when its accuracy is defensible; otherwise integrate a validated heat-capacity function or tabulated data, documenting interpolation and phase transitions. Check units at each stage.
Report ΔH(T₂) together with T₂, reaction as written, the model and the uncertainty or approximation. A table that only states a corrected number without its temperature basis invites readers to compare incompatible values.
Continue with The Gold Standard of Thermochemistry and Gas Moles, Internal Energy and the Enthalpy Correction. Kirchhoff’s law changes the reference temperature; the ΔngasRT relation compares ΔH and ΔU under ideal-gas assumptions. They solve different thermodynamic questions.
Frequently asked questions
Does a positive ΔCₚ make an exothermic reaction endothermic?
Not necessarily. It makes ΔH increase as temperature increases, but the reaction may remain exothermic throughout the interval. Only a sufficiently large correction could change the sign under the model.
Is Kirchhoff’s law the same as ΔH=ΔU+ΔnRT?
No. Kirchhoff’s law relates reaction enthalpy values at different temperatures. The ΔnRT expression relates reaction enthalpy and internal-energy changes for ideal gases under appropriate state assumptions.
What if heat capacity changes during heating?
Integrate the relevant ΔCₚ(T) rather than multiplying by one constant. If phase changes occur, include the enthalpy of those transitions and use phase-appropriate data.
Sources and related articles
The numerical exercises are original thermodynamic models. No invented heat capacity is attributed to a real substance.
- Chemistry LibreTexts — DeVoe, Molar Reaction Enthalpy and Kirchhoff’s Equation
- OpenStax Chemistry 2e — Enthalpy
- The Gold Standard of Thermochemistry | Gas Moles and Enthalpy
- The Gold Standard of Thermochemistry | Limiting Reagents
Applied thermochemistry lab: reaction cycles, changing Cₚ and uncertainty
Build the heat-capacity difference from species data
Rather than start from a supplied ΔCₚ, calculate it from a balanced fictional reaction: 2A(g)+B(g)→3C(g). Suppose the temperature-dependent heat capacities are Cₚ,A=30+0.010(T−300), Cₚ,B=40+0.020(T−300), and Cₚ,C=25+0.015(T−300), all in J/(mol·K) with T in kelvin.
The product contribution is three times Cₚ,C; the reactant contribution is twice Cₚ,A plus Cₚ,B. Subtracting gives ΔCₚ(T)=[75+0.045(T−300)]−[100+0.040(T−300)]=−25+0.005(T−300) J/(mol reaction·K).
Do not silently discard the stoichiometric multipliers when temperature-dependent functions are used. Every coefficient multiplies the entire expression for its species, including both constant and temperature-dependent parts.
Check the result at 300 K: ΔCₚ=−25 J/(mol·K), consistent with the constant-data example. At 500 K it is −24 J/(mol·K), slightly less negative. A table of endpoint values provides a simple way to catch a sign or multiplication mistake before integrating.
Integrate a changing negative heat-capacity difference
Use the ΔCₚ(T)=−25+0.005(T−300) model from 300 to 500 K. Let u=T−300 so u runs from zero to 200 K. The correction is ∫[−25+0.005u]du, evaluated over that interval.
The constant component contributes −25×200=−5000 J per mole of reaction. The linear component contributes 0.5×0.005×200²=+100 J per mole. Together they give −4900 J/mol=−4.90 kJ/mol.
If ΔH(300 K)=−80.00 kJ/mol reaction as written, then ΔH(500 K)=−80.00−4.90=−84.90 kJ/mol. The difference is slightly less negative than the −5.00 kJ/mol that a fixed ΔCₚ of −25 would predict over the same interval.
The distinction is modest in this hypothetical example but mathematically clear. A changing heat capacity alters the correction by the area between the variable ΔCₚ curve and the horizontal constant approximation. The sign of that area determines which estimate is larger.
Check Kirchhoff’s law with a Hess-style state pathway
Because enthalpy is a state function, the enthalpy difference between reactants at T₂ and products at T₂ can be assembled from a reference reaction at T₁ plus the heating of products and minus the heating of reactants. This is the thermodynamic-cycle interpretation of Kirchhoff’s law.
Return to fictional A(g)→B(g), with constant Cₚ,A=30 and Cₚ,B=45 J/(mol·K). Heating A through 100 K raises its enthalpy by 3.00 kJ/mol; heating B through the same 100 K raises its enthalpy by 4.50 kJ/mol. The difference between those heating increments is +1.50 kJ/mol.
Therefore a reference reaction enthalpy of −50.00 kJ/mol at 298 K becomes −48.50 kJ/mol at 398 K, matching the direct formula ΔH₂=ΔH₁+ΔCₚΔT. This independent route is useful because it makes the product-minus-reactant signs physically transparent.
Do not sum the heating increments as 3.00+4.50. Reactant enthalpy appears on the opposite side of the reaction enthalpy difference from product enthalpy. A Hess-style diagram can prevent a sign error that is easy to conceal in a single line of arithmetic.
How approximation uncertainty enters the corrected enthalpy
Suppose a classroom calculation treats ΔCₚ=15 J/(mol·K) as constant from 298 to 398 K, but that value is only known to within about ±5 J/(mol·K). Holding the reference enthalpy exact just for this illustration, the 100 K temperature correction is 1.50±0.50 kJ/mol.
The exact formal uncertainty depends on how heat-capacity values were obtained, their correlation, the temperature interval and the integration model. The ±0.50 kJ/mol example is simply the direct effect of multiplying the stated heat-capacity uncertainty by 100 K.
An independent uncertainty in the reference reaction enthalpy must also be considered. It is misleading to print a corrected value such as −48.500000 kJ/mol when the heat-capacity information is uncertain by tenths of a kilojoule per mole.
The same lesson applies to a fitted polynomial. A smooth-looking curve does not prove that its coefficients are exact or valid outside the fitting interval. Report the provenance and intended temperature range of the functions used.
Distinguish heating of streams from chemical reaction heat
Consider a continuous reactor with feed entering at 300 K and product leaving at 500 K. A standard-state enthalpy correction does not by itself tell you the amount of external heating or cooling the reactor needs. That depends on flow rates, which reactants actually convert, product composition and the sensible enthalpy carried in each stream.
For the fictional reaction A→B, the reference ΔH at 300 K might be negative while an actual process also requires energy to heat the incoming or outgoing matter. Whether heat must be supplied or removed from the equipment depends on the complete first-law balance.
Kirchhoff’s law is useful within that balance because it supplies a consistent reaction enthalpy at the chosen temperature. It does not replace calculating inlet and outlet enthalpy contributions, reaction extent, nonideal mixing or any work transferred across the system boundary.
A parent or student new to thermochemistry can use a simple rule: the chemical reaction enthalpy compares states defined by an equation, whereas actual apparatus heat asks where energy crossed the measured boundary. Do not use one as the answer to the other without stating assumptions.
A more demanding practice set with answer explanations
Practice A: The reaction 2A+B→3C has ΔCₚ(T)=−25+0.005(T−300) J/(mol·K). Estimate the correction from 300 to 500 K. Answer: −4.90 kJ/mol reaction, found by integrating the constant and linear terms separately.
Practice B: A reference reaction enthalpy at 298 K is +10.0 kJ/mol, and the constant reaction heat-capacity difference is +20 J/(mol·K). At 348 K, the correction is +1.0 kJ/mol, giving +11.0 kJ/mol. The reaction remains endothermic and becomes more endothermic under the stated model.
Practice C: The same +20 J/(mol·K) difference is used to extrapolate from 298 K to 1200 K, crossing a known phase change. Should the simple constant-Cₚ calculation be accepted? No. Find phase-appropriate data, include transition enthalpy and check whether reactant and product states remain chemically meaningful.
Practice D: A calculation predicts ΔH(398 K)=−48.50 kJ/mol, but measured heat from a real vessel is much smaller. Does that prove Kirchhoff’s law is wrong? No. Check actual reaction extent, losses, vessel heat capacity, feed and product temperatures, pressure conditions, side reactions and measurement uncertainty before attributing a discrepancy to the state-function relation.
Why a temperature-corrected enthalpy cannot determine equilibrium by itself
Thermochemistry also distinguishes enthalpy from Gibbs energy. A reaction enthalpy correction tells us how the enthalpy difference changes with temperature under the model; it does not directly specify the equilibrium constant or whether a reaction is thermodynamically favoured under real conditions. Those questions involve entropy, Gibbs energy and the activities of the participating substances.
Students often overinterpret a negative ΔH as meaning a reaction must proceed completely. It does not. Exothermicity describes an enthalpy change, while equilibrium composition and reaction progress depend on additional thermodynamic and kinetic factors. Kirchhoff’s law is therefore an important part of an energy analysis, not a standalone prediction of reactor conversion, speed or safety.