Kirchhoff's Law

A blast furnace runs near 1900 K, yet standard enthalpies of reaction are tabulated at 298 K. Can you still use those tables? Kirchhoff's law says yes, provided you correct for the temperature dependence of heat capacity. This correction matters everywhere reactions run hot: combustion, metallurgy, atmospheric chemistry, industrial synthesis. Without it, engineers would badly misjudge the heat released or absorbed at operating temperature. This article derives Kirchhoff's equation from the First Law. It shows how to integrate it using constant, linear, and empirical heat capacity data. Then it applies the result to two fully worked numerical examples.

Key takeaways

Kirchhoff's law gives the temperature dependence of a reaction enthalpy: $d(\Delta_r H)/dT = \Delta_r C_p$ .
It follows directly from differentiating Hess's law summations with respect to temperature at constant pressure.
Integration from a reference temperature (usually 298.15 K) gives $\Delta_r H(T_2)$ from $\Delta_r H(T_1)$ and $\Delta_r C_p$ .
If $\Delta_r C_p$ is treated as constant, the correction is simply $\Delta_r C_p \times \Delta T$ .
Real heat capacities vary with $T$ , so accurate work uses polynomial forms like $C_p = a + bT + c/T^2$ .
Key idea: the reaction enthalpy changes with temperature at a rate set by the difference in heat capacities between products and reactants.

The core insight

Because enthalpy is a state function whose temperature slope at constant pressure is heat capacity, the temperature slope of a reaction enthalpy must be the difference in heat capacities between products and reactants, $\Delta_r C_p$ ; integrating this difference over a temperature range converts an enthalpy known at one temperature into the enthalpy at any other.

The physical picture

Imagine heating the reactants of a reaction from $T_1$ to $T_2$ , letting them react at $T_2$ , then imagine instead letting them react first at $T_1$ and heating the products up to $T_2$ . Because enthalpy is a state function, both paths must give the same total enthalpy change between the same initial and final states. The difference between "heat then react" and "react then heat" is exactly the difference in heat capacity between products and reactants, integrated over the temperature interval. Each substance's enthalpy rises with temperature according to its own heat capacity at constant pressure,

\left(\frac{\partial H}{\partial T}\right)_p = C_p .

For a reaction, $\Delta_r H$ is a weighted sum of the enthalpies of products minus reactants, each weighted by its stoichiometric coefficient. Differentiating that sum with respect to $T$ simply replaces each $H$ by its own $C_p$ , giving the analogous relation for the whole reaction. If products have a larger total heat capacity than reactants, $\Delta_r H$ becomes more positive (or less negative) as temperature rises; if reactants dominate, the reverse happens.

Figure

Two equivalent paths between the same initial and final states, showing why the enthalpy change is path-independent

Figure

Reaction enthalpy versus temperature for constant ΔrCp of −41 J K⁻¹ mol⁻¹, the ammonia synthesis example below

The core relations

\left(\frac{\partial \Delta_r H}{\partial T}\right)_p = \Delta_r C_p

where $\Delta_r C_p = \sum_i \nu_i C_{p,m}(i)$ , products positive, reactants negative, stoichiometric coefficients $\nu_i$ .

\Delta_r H(T_2) = \Delta_r H(T_1) + \int_{T_1}^{T_2} \Delta_r C_p \, dT

where $T_1$ is usually 298.15 K (standard tables) and $T_2$ is the temperature of interest.

\Delta_r H(T_2) = \Delta_r H(T_1) + \Delta_r C_p \,(T_2 - T_1) \quad \text{(if } \Delta_r C_p \text{ is constant)}

where the correction term is a single number once $\Delta_r C_p$ is known.

\Delta_r C_p(T) = \Delta_r a + \Delta_r b\, T + \frac{\Delta_r c}{T^2}

where $\Delta_r a, \Delta_r b, \Delta_r c$ are the differences of the empirical Shomate/NIST-style coefficients between products and reactants.

Derivation AKirchhoff's equation from the First Law and Hess's law

Goal. Show that the temperature derivative of a reaction enthalpy at constant pressure equals $\Delta_r C_p$ , and integrate the result.

Step 1. Write the reaction enthalpy as a sum of molar enthalpies. For a reaction $\sum_i \nu_i \, \text{X}_i = 0$ (products positive, reactants negative), Hess's law gives

\Delta_r H(T) = \sum_i \nu_i H_{m,i}(T).

Each $H_{m,i}(T)$ is the molar enthalpy of species $i$ at temperature $T$ and constant pressure.

Step 2. Differentiate with respect to temperature at constant pressure. Since differentiation is linear and each $\nu_i$ is a temperature-independent stoichiometric number,

\left(\frac{\partial \Delta_r H}{\partial T}\right)_p = \sum_i \nu_i \left(\frac{\partial H_{m,i}}{\partial T}\right)_p.

Step 3. Recognize each term as a heat capacity. By the defining property of $C_p$ established for pure substances,

\left(\frac{\partial H_{m,i}}{\partial T}\right)_p = C_{p,m}(i).

Substituting,

\left(\frac{\partial \Delta_r H}{\partial T}\right)_p = \sum_i \nu_i C_{p,m}(i) \equiv \Delta_r C_p.

This is Kirchhoff's equation in differential form.

Step 4. Integrate between two temperatures. Separating variables and integrating from $T_1$ to $T_2$ at constant pressure,

\int_{\Delta_r H(T_1)}^{\Delta_r H(T_2)} d(\Delta_r H) = \int_{T_1}^{T_2} \Delta_r C_p \, dT.

The left side telescopes trivially since it is an exact differential of a state function.

Step 5. Evaluate the right-hand integral for the two common cases. If $\Delta_r C_p$ is treated as temperature-independent over the interval, it comes out of the integral:

\Delta_r H(T_2) - \Delta_r H(T_1) = \Delta_r C_p (T_2 - T_1).

If $\Delta_r C_p(T) = \Delta_r a + \Delta_r b T + \Delta_r c/T^2$ , term-by-term integration gives

\Delta_r H(T_2) - \Delta_r H(T_1) = \Delta_r a (T_2-T_1) + \frac{\Delta_r b}{2}(T_2^2 - T_1^2) - \Delta_r c\left(\frac{1}{T_2}-\frac{1}{T_1}\right).

Boxed result.

\boxed{\Delta_r H(T_2) = \Delta_r H(T_1) + \int_{T_1}^{T_2} \Delta_r C_p(T)\, dT}

Worked examples

Example 1 — Ammonia synthesis, constant ΔrCp (foundational). Reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ , with $\Delta_r H^\circ(298\ \text{K}) = -92.2\ \text{kJ mol}^{-1}$ . Standard molar heat capacities (constant, in J K⁻¹ mol⁻¹): $C_{p,m}(\text{NH}_3) = 35.1$ , $C_{p,m}(\text{N}_2) = 29.1$ , $C_{p,m}(\text{H}_2) = 28.8$ . Step 1: compute $\Delta_r C_p$ .

\Delta_r C_p = 2(35.1) - 1(29.1) - 3(28.8) = 70.2 - 29.1 - 86.4 = -45.3\ \text{J K}^{-1}\text{mol}^{-1}.

(A commonly quoted rounded literature value near $-41\ \text{J K}^{-1}\text{mol}^{-1}$ appears when slightly different tabulated $C_p$ values are used; here we carry through our own numbers consistently.) Step 2: apply Kirchhoff's law from 298 K to 700 K, $\Delta T = 402\ \text{K}$ .

\Delta_r H(700\ \text{K}) = -92\,200\ \text{J mol}^{-1} + (-45.3\ \text{J K}^{-1}\text{mol}^{-1})(402\ \text{K})

= -92\,200 - 18\,210.6 = -110\,410.6\ \text{J mol}^{-1} \approx -110.4\ \text{kJ mol}^{-1}.

Sanity check: since products (NH₃) have a larger total heat capacity contribution relative to reactants is not the deciding factor here — reactants dominate ( $115.5$ vs $70.2\ \text{J K}^{-1}\text{mol}^{-1}$ ), so $\Delta_r C_p$ is negative, and the reaction becomes more exothermic at higher temperature. The magnitude of the shift, about 18 kJ mol⁻¹ over 400 K, is a modest but non-negligible fraction of the room-temperature value, confirming the correction is significant for high-temperature industrial design (the Haber process runs near 700 K).

Example 2 — Combustion of methane with temperature-dependent Cp (applied, harder). Reaction: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$ , $\Delta_r H^\circ(298\ \text{K}) = -802.3\ \text{kJ mol}^{-1}$ (using gaseous water). Use empirical heat capacities of the form $C_{p,m} = a + bT$ (J K⁻¹ mol⁻¹, $T$ in K), with representative coefficients:

Species	$a$	$b$ (×10³)
CH₄	23.6	47.9
O₂	25.7	12.98
CO₂	26.0	43.5
H₂O(g)	30.0	10.7

Step 1: form $\Delta_r a$ and $\Delta_r b$ using $\nu_{\text{CO}_2}=1,\ \nu_{\text{H}_2\text{O}}=2,\ \nu_{\text{CH}_4}=-1,\ \nu_{\text{O}_2}=-2$ .

\Delta_r a = [1(26.0) + 2(30.0)] - [1(23.6) + 2(25.7)] = 86.0 - 75.0 = 11.0\ \text{J K}^{-1}\text{mol}^{-1}

\Delta_r b = [1(43.5) + 2(10.7) - 1(47.9) - 2(12.98)]\times 10^{-3} = (43.5+21.4-47.9-25.96)\times10^{-3} = -8.96\times10^{-3}\ \text{J K}^{-2}\text{mol}^{-1}

Step 2: integrate from $T_1 = 298\ \text{K}$ to $T_2 = 1000\ \text{K}$ .

\int_{298}^{1000}\Delta_r C_p\,dT = \Delta_r a\,(T_2-T_1) + \frac{\Delta_r b}{2}(T_2^2 - T_1^2)

= 11.0(1000-298) + \frac{-8.96\times10^{-3}}{2}(1\,000\,000 - 88\,804)

= 11.0(702) + (-4.48\times10^{-3})(911\,196)

= 7722.0 - 4082.2 = 3639.8\ \text{J mol}^{-1} \approx 3.64\ \text{kJ mol}^{-1}.

Step 3: apply Kirchhoff's law.

\Delta_r H(1000\ \text{K}) = -802.3\ \text{kJ mol}^{-1} + 3.64\ \text{kJ mol}^{-1} = -798.7\ \text{kJ mol}^{-1}.

Sanity check: the correction is small (< 1% of $\Delta_r H$ ) because $\Delta_r a$ is modest and the quadratic term partly cancels it — a common feature for combustions where product and reactant heat capacities are numerically similar. This also shows why constant- $C_p$ estimates are often adequate unless $T_2$ is far from 298 K or $\Delta_r C_p$ is unusually large.

Limiting cases & common traps

Situation	What to do
$T_2$ close to 298 K, modest $\Delta_r C_p$	Constant- $\Delta_r C_p$ approximation is fine; error is second order in $\Delta T$
Large $\Delta T$ (hundreds of K) or high precision needed	Use empirical $C_p(T)$ polynomials and integrate term by term
Phase change occurs between $T_1$ and $T_2$	Must add the enthalpy of transition explicitly; Kirchhoff's law alone does not capture latent heat
$\Delta_r C_p \approx 0$	$\Delta_r H$ is nearly temperature-independent, common for isomerization reactions with similar bonding

The chemist behind it

Gustav Kirchhoff is best known to physical chemists for two contributions: the spectroscopic laws bearing his name (with Robert Bunsen) and this thermochemical relation, published in 1858. Working in an era before enthalpy was formally named, Kirchhoff reasoned from calorimetric heat measurements and Hess's law — itself barely a decade old — to show that the heat of a chemical reaction must depend on temperature through the heat capacities of the substances involved. The result was a natural consequence of energy conservation applied rigorously to chemical change, and it remains one of the cleanest applications of the First Law to thermochemistry.

Origins · 1 minGustav Robert Kirchhoff and how this was discovered

Gustav Robert Kirchhoff

1824–1887 · German physicist who unified circuit laws, spectroscopy, and thermochemistry through rigorous application of conservation principles

Ask the tutor: How would you modify Kirchhoff's law to also track the temperature dependence of ΔrS and ΔrG?

Why does Kirchhoff's law use ΔrCp rather than the individual heat capacities separately

Because $\Delta_r H$ is defined as a weighted sum of product and reactant enthalpies, its temperature derivative is automatically the same weighted sum of heat capacities. Only the difference, $\Delta_r C_p = \sum \nu_i C_{p,m}(i)$ , survives in the final relation. Individual heat capacities matter only insofar as they combine into this single effective quantity, which is why tables often report $\Delta_r C_p$ directly for common reactions rather than requiring you to look up every species separately.

Can Kirchhoff's law be applied at constant volume instead of constant pressure

Yes, with the analogous substitution: $(\partial \Delta_r U/\partial T)_V = \Delta_r C_V$ , since internal energy's temperature derivative at constant volume is $C_V$ by definition. The two forms are physically parallel, but chemists overwhelmingly work at constant pressure (open vessels, atmospheric conditions), so the enthalpy version with $C_p$ is the one used in essentially all practical thermochemistry.

Does Kirchhoff's law tell us anything about whether a reaction becomes spontaneous at high temperature

No — spontaneity is governed by $\Delta_r G = \Delta_r H - T\Delta_r S$ , which requires both enthalpy and entropy at the temperature of interest. Kirchhoff's law only supplies the corrected $\Delta_r H(T)$ ; you would need an analogous temperature correction for $\Delta_r S$ (via $\Delta_r C_p/T$ ) before assembling $\Delta_r G(T)$ and drawing conclusions about spontaneity.

Why is 298.15 K almost always chosen as the reference temperature T1

Because essentially all standard thermochemical data — standard enthalpies of formation, combustion, and reaction — are tabulated at 298.15 K by convention. Using this as $T_1$ lets you plug tabulated $\Delta_r H^\circ$ values directly into Kirchhoff's equation without any additional lookup, then integrate $\Delta_r C_p$ up or down to whatever process temperature $T_2$ you actually need.