Connecting Reaction ΔH and ΔU

Introduction

Reaction enthalpy and internal-energy change are related but need not be equal. For ideal-gas reactants and products at a common temperature, their difference follows the change in gaseous mole count: ΔH = ΔU + Δn gRT. The equation is a useful correction when comparing constant-pressure and constant-volume measurements.

Core explanation

Begin with H = U + PV. For a reaction between specified initial and final states at the same temperature, ΔH = ΔU + Δ(PV). If gas phases behave ideally, PV = n gRT for each side's gas contribution. At common T, Δ(PV) = (n g,products − n g,reactants)RT = Δn gRT. Thus ΔH = ΔU + Δn gRT for the reaction as written under these assumptions. Condensed-phase PV contributions are usually small enough to neglect in this introductory correction; the precise relation retains them if needed.

Count only gaseous stoichiometric coefficients when computing Δn g. For N₂(g) + 3H₂(g) → 2NH₃(g), product gas moles are 2 and reactant gas moles 1 + 3 = 4, so Δn g = 2 − 4 = −2. Therefore ΔH = ΔU − 2RT for one mole of reaction as written. The negative correction reflects a reduction in the gas PV term. Scaling the reaction by two doubles Δn g and both energy changes.

For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), the gas count is 1 on the product side and 3 on the reactant side, so Δn g = −2. Liquid water is excluded. If the product is H₂O(g) instead, product gas count becomes 3 and Δn g = 0. The reaction enthalpy also changes because the water phase changes; one cannot replace liquid with gas only in the mole count while keeping the same reaction ΔH.

Use R in units consistent with ΔH and ΔU. At 298 K, RT ≈ (8.314 J mol⁻¹ K⁻¹)(298 K) ≈ 2478 J mol⁻¹, or 2.48 kJ per mole of gas-count change. For Δn g = −2, the correction is about −4.96 kJ per mole of reaction. This can be modest compared with a large combustion enthalpy but is not identically zero.

The formula assumes a common temperature for reactant and product reference states, ideal-gas behavior for gases and consistent stoichiometric scaling. It is not a general shortcut for arbitrary heating or cooling processes where T changes. It also does not directly yield the heat of a process with electrical work or non-equilibrium material flow. Keep the relationship between state functions distinct from the path-dependent heat that a particular calorimeter measures.

If Δn g = 0, the ideal-gas correction vanishes and ΔH ≈ ΔU within the stated condensed-phase approximation. A zero correction does not mean the reaction releases no heat; it means the two energy changes are approximately equal.

Step-by-step reasoning

1. Write a balanced reaction with physical states. 2. Add gaseous stoichiometric coefficients on each side. 3. Compute Δn g = products minus reactants. 4. Convert RT to the same energy units as ΔU or ΔH. 5. Substitute into ΔH = ΔU + Δn gRT and check the sign.

Visual explanation

Draw two boxes labelled gas reactants and gas products, each containing a number of molecule icons proportional to stoichiometric coefficients. Put liquid or solid species below a separate line. An arrow marked Δn g links gas counts, and a PV strip changes height accordingly.

Real-world analogy

A bill may include a core cost plus a charge per package shipped. Changing the number of packages changes the second term even if the core cost also changes. In the analogy, U is the core term and PV is the gas-count-related term; the analogy does not replace the actual thermodynamic derivation.

Real-world example

A bomb calorimeter measures an energy change related to ΔU for combustion. To compare it with a tabulated constant-pressure combustion enthalpy, a chemist can apply the gas-mole correction at the specified temperature, while also ensuring the same product phases and reaction extent.

Why?

Why do solids and liquids usually not enter Δn g? Their PV terms are much smaller than gas PV terms under ordinary conditions, and the simplified formula isolates the ideal-gas contribution. Exact high-pressure work may require retaining condensed-phase changes.

Common misconception

“Δn g counts every species in the equation.” It counts only gaseous stoichiometric coefficients for this approximation. Including liquid water or solid carbon gives the wrong correction.

Worked example

For N₂(g) + 3H₂(g) → 2NH₃(g), let ΔU = −87.0 kJ per reaction as written at 298 K. Δn g = −2 and RT ≈ 2.478 kJ mol⁻¹. Then ΔH = −87.0 + (−2)(2.478) = −91.956 kJ, about −92.0 kJ per reaction as written. The enthalpy is more negative because gas mole count falls.

Quick check

1. What is Δn g for H₂(g) + Cl₂(g) → 2HCl(g)? Answer: 2 − (1 + 1) = 0.

Exam focus

Balance first, use state labels and count only gas coefficients. Keep the reaction scale fixed and convert J to kJ if needed. State assumptions before applying the ideal-gas correction.

Advanced insight

For nonideal gases, PV is not exactly nRT, and Δ(PV) must be obtained from an equation of state or measured state data. The compact Δn gRT result is thus a model-dependent consequence of H = U + PV, not the definition of enthalpy.

Summary

For specified ideal-gas reactions at common temperature, ΔH − ΔU ≈ Δn gRT, with gaseous product coefficients minus gaseous reactant coefficients. Physical phases, stoichiometric scale and units determine the correct result. Zero Δn g means the two changes are approximately equal, not zero.

Practice questions

1. Find Δn g for CaCO₃(s) → CaO(s) + CO₂(g). Answer: 1 − 0 = +1, counting only CO₂(g). 2. At 300 K, estimate ΔH − ΔU for a reaction with Δn g = +2. Answer: 2RT ≈ 2(8.314)(300) J ≈ +4.99 kJ per reaction as written. 3. Why does changing H₂O(l) to H₂O(g) require more than recounting gas moles? Answer: The product state and its enthalpy change too, so the reaction itself is different and its ΔH must be updated.