Kirchhoff-Type Heat-Capacity Corrections

Introduction

A reaction enthalpy measured or tabulated near room temperature need not have exactly the same value at a higher process temperature. Reactants and products warm by different enthalpy amounts because their heat capacities differ. Kirchhoff's relation corrects a known reaction enthalpy using the difference between product and reactant heat capacities over the temperature interval.

Core explanation

For a balanced reaction at compatible pressure and phases, define Δ rC p = Σν products C p,products − Σν reactants C p,reactants. Each C p is a molar constant-pressure heat capacity, for example J mol⁻¹ K⁻¹, and coefficients multiply it just as they multiply formation enthalpies. Then Δ rH(T₂) = Δ rH(T₁) + ∫ from T₁ to T₂ of Δ rC p(T)dT. If Δ rC p is treated as constant across the interval, this becomes Δ rH(T₂) ≈ Δ rH(T₁) + Δ rC p(T₂ − T₁).

The correction's units must match the original enthalpy. If Δ rC p is in J mol⁻¹ K⁻¹ and ΔT is in K, their product is J mol⁻¹; divide by 1000 before adding to a value in kJ mol⁻¹. A negative Δ rC p means the product side's summed heat capacity is smaller than the reactant side's. Increasing temperature then makes Δ rH more negative or less positive, according to the formula. This is a slope relationship, not a statement that the reaction necessarily releases more measured heat in every reactor configuration.

Take a hypothetical reaction with Δ rH(298 K) = −92.0 kJ mol⁻¹ extent and approximately constant Δ rC p = −20.0 J mol⁻¹ K⁻¹ from 298 to 500 K. The temperature difference is 202 K, so the correction is (−20.0)(202) = −4040 J mol⁻¹ = −4.04 kJ mol⁻¹. Therefore Δ rH(500 K) ≈ −96.0 kJ mol⁻¹ extent. If one forgot the J-to-kJ conversion, the answer would be wrong by a factor of 1000.

If individual heat capacities are given, build Δ rC p with the balanced equation. For A + 2B → C, Δ rC p = C p(C) − [C p(A) + 2C p(B)]. The negative sign applies to the entire reactant bracket. A supplied list of mass-specific heat capacities must first be converted to molar heat capacities or used with a matching mass-basis reaction calculation; one cannot simply insert J g⁻¹ K⁻¹ into a molar enthalpy formula.

Phase changes make the simple one-interval approximation unsuitable unless their latent enthalpies are included. For example, if a product changes from liquid to gas between T₁ and T₂, heating liquid to boiling, vaporising, and heating gas are distinct contributions. Heat capacities also vary with temperature, especially over a wide range, so integrating a temperature-dependent model is more accurate than a constant Δ rC p. The university-hosted thermodynamics treatment of reaction enthalpy states the integral and constant-ΔC p forms.

Step-by-step reasoning

1. Write the balanced reaction with phases at the reference and target temperatures. 2. Obtain Δ rH(T₁) and compute coefficient-weighted Δ rC p from compatible molar heat capacities. 3. Check whether Δ rC p can reasonably be treated as constant and whether phases remain the same. 4. Multiply Δ rC p by T₂ − T₁, converting J to kJ if necessary. 5. Add the signed correction to Δ rH(T₁) and state the approximation.

Visual explanation

Draw two sloping enthalpy lines against temperature: one for summed reactants and one for summed products. Their vertical separation at T₁ is Δ rH(T₁). Their slope difference is Δ rC p, so the vertical separation changes by Δ rC pΔT by T₂. Mark the same physical phases along both lines.

Real-world analogy

Two runners start with a fixed distance between them but move at different speeds. Their separation later equals initial separation plus the speed difference times elapsed time. Reaction enthalpy at a new temperature similarly equals its starting difference plus the heat-capacity-slope difference times the temperature interval.

Real-world example

A process calculation may know a reaction enthalpy at 298 K but operate near 500 K. If reactant and product heat-capacity data are available, applying a temperature correction improves the predicted energy balance. A plant-scale calculation would also include feed heating, product cooling, phase changes and heat losses as separate terms.

Why?

Why do product and reactant heat capacities appear as a difference? Reaction enthalpy is H products − H reactants. On warming, each side gains enthalpy according to its heat capacity. Subtracting their gains leaves the product-side warming contribution minus the reactant-side warming contribution.

Common misconception

“Add every species heat capacity together.” Reactants are subtracted, and coefficients matter. Another error is adding a correction in joules to a reaction enthalpy in kilojoules. Dimensional analysis should be done before signed arithmetic.

Worked example

For A(g) + 2B(g) → C(g), take molar C p values of A = 30.0, B = 25.0 and C = 70.0 J mol⁻¹ K⁻¹, treated as constant from 300 to 400 K. Δ rC p = 70.0 − [30.0 + 2(25.0)] = −10.0 J mol⁻¹ K⁻¹. If Δ rH(300 K) = +50.0 kJ mol⁻¹ extent, the correction over +100 K is −1000 J mol⁻¹ = −1.00 kJ mol⁻¹. Thus Δ rH(400 K) ≈ +49.0 kJ mol⁻¹. The reaction remains endothermic on this approximate basis.

Quick check

1. If Δ rC p = +5.0 J mol⁻¹ K⁻¹ over a 200 K rise, what is the enthalpy correction in kJ mol⁻¹? Answer: +5.0 × 200 = +1000 J mol⁻¹ = +1.0 kJ mol⁻¹.

Exam focus

Use product minus reactant heat-capacity sums, each multiplied by stoichiometric coefficients. Convert energy units before adding to Δ rH, and check that no unaccounted phase change crosses the temperature interval. Label the result as an approximation if constant C p is assumed.

Advanced insight

When heat capacity is represented by a polynomial in T, integrate each term analytically before taking the coefficient-weighted product-minus-reactant difference. This preserves the same Kirchhoff principle while allowing a temperature-varying correction over a broader range.

Summary

Reaction enthalpy changes with temperature according to the integrated difference in product and reactant heat capacities. With approximately constant Δ rC p, add Δ rC pΔT to the known reference enthalpy. Coefficients, phases, signs and joule-to-kilojoule conversion determine a reliable result.

Practice questions

1. For A + B → C, C p values are 20, 30 and 60 J mol⁻¹ K⁻¹ respectively. Find Δ rC p. Answer: 60 − (20 + 30) = +10 J mol⁻¹ K⁻¹. 2. If Δ rH(300 K) = −40.0 kJ mol⁻¹, estimate Δ rH(400 K) with that constant Δ rC p. Answer: Correction is +10 × 100 = +1000 J = +1.0 kJ mol⁻¹, so Δ rH(400 K) ≈ −39.0 kJ mol⁻¹. 3. Why is one constant-C p correction insufficient if a liquid product boils in the interval? Answer: Vaporisation adds latent enthalpy and changes the relevant heat capacity; the path needs separate heating and phase-change stages.