Enthalpy as U + PV

Introduction

Many laboratory reactions occur near constant atmospheric pressure. When gases expand or contract, pressure-volume work affects the heat observed. Enthalpy, defined by H = U + PV, packages internal energy with a pressure-volume term so that its change is especially useful for constant-pressure processes.

Core explanation

Internal energy U is a state function. Pressure P and volume V also describe a system's state, so their product PV has a state-determined value. Therefore H = U + PV is itself a state function. Between two specified states, ΔH = ΔU + Δ(PV). Do not simplify Δ(PV) to PΔV unless pressure is the same at both endpoints; in general Δ(PV) = P fV f − P iV i.

For a closed system doing only pressure-volume work against a constant external pressure equal to the maintained system pressure, the first law gives ΔU = q p − PΔV. Then ΔH = ΔU + PΔV = q p. This is why the measured heat at constant pressure equals the enthalpy change under the stated conditions. If significant electrical or other non-P–V work occurs, or pressure is not maintained appropriately, the simple equality needs qualification.

Reaction enthalpy refers to the enthalpy difference between products and reactants in a specified balanced equation. An exothermic reaction commonly has Δ rH < 0: products have lower enthalpy than reactants under the same reference conditions. An endothermic reaction has Δ rH > 0. The sign refers to the chosen forward reaction; reversing the equation reverses the sign. Scaling the entire reaction by a factor scales Δ rH by that factor.

Enthalpy is not “heat stored in a substance.” Heat is energy crossing a boundary during a process, whereas H is a state property constructed from U, P and V. The equality q p = ΔH is conditional, not an identity between the concepts. A substance can have an enthalpy relative to an arbitrary reference, but only differences are normally needed for chemical calculations.

The PV term is especially transparent for an ideal gas: PV = nRT. A gas-generating reaction at constant temperature changes PV partly because gas mole number changes. This contributes to the difference between ΔH and ΔU. For liquids and solids, volume changes are often relatively small, so the distinction may be modest, but it should not be discarded without considering conditions.

Standard enthalpy changes use specified reference states, commonly pure substances at a reference pressure and stated temperature. A reported reaction enthalpy is therefore incomplete without a balanced equation, physical-state labels and units such as kJ per mole of reaction as written. Formation, combustion and neutralisation enthalpies are special cases with carefully defined reaction extents.

Step-by-step reasoning

1. State H = U + PV and define initial and final states. 2. Calculate ΔH as ΔU + P fV f − P iV i. 3. If pressure is constant, simplify to ΔH = ΔU + PΔV. 4. Check whether only P–V work occurs before setting ΔH = q p. 5. Tie the sign to the reaction direction and equation scale.

Visual explanation

Draw two energy boxes labelled U i and U f, each with a separate PV strip attached. The total heights are H i and H f. Next show a constant-pressure piston where incoming heat both changes U and makes room for expansion; the combined change equals ΔH.

Real-world analogy

The price of delivering a package may combine the item cost with a required shipping charge. Enthalpy combines internal energy with a pressure-volume term useful for a common experimental condition. The analogy should not imply PV is always an external fee or that heat is a stored quantity.

Real-world example

Combustion in an open vessel occurs near atmospheric pressure, and a calorimeter can infer heat released at approximately constant pressure. Under suitable conditions the reaction's heat is linked to ΔH. A sealed rigid bomb calorimeter instead more directly measures a quantity related to ΔU.

Why?

Why does adding PV help? At constant pressure, the PΔV term accounts for expansion work in the first-law balance, leaving a state-function change equal to the heat transfer under ordinary P–V-only conditions.

Common misconception

“Enthalpy is exactly the heat inside a system.” H is U + PV, a state property. Heat q is transferred during a process. They are numerically equal only as changes or transfers under specified constant-pressure conditions.

Worked example

A closed gas system undergoes a constant-pressure change with ΔU = +1.5 kJ and performs 0.4 kJ of expansion work. For the gas, w = −0.4 kJ. Since w = −PΔV, PΔV = +0.4 kJ, so ΔH = ΔU + PΔV = +1.9 kJ. The first law also gives q p = ΔU − w = 1.5 − (−0.4) = +1.9 kJ, consistent with the enthalpy result under the stated conditions.

Quick check

1. Is H a state function or a path function? Answer: A state function, because U, P and V are state properties.

Exam focus

Write H = U + PV before using special formulas. State conditions for q p = ΔH, include physical states in reaction equations and reverse the enthalpy sign when reversing a reaction.

Advanced insight

In flowing systems, enthalpy naturally includes the pressure-flow work associated with moving material across a control-volume boundary. This is another reason it is widely used in chemical engineering, though the open-system energy balance also includes kinetic, potential, heat and shaft-work terms when relevant.

Summary

Enthalpy is the state function H = U + PV. At constant pressure with only P–V work under suitable conditions, ΔH equals heat transferred. Reaction enthalpy is an endpoint difference for a specified equation, not heat stored in a substance.

Practice questions

1. If ΔU = −80 kJ and PΔV = +5 kJ at constant pressure, find ΔH. Answer: ΔH = ΔU + PΔV = −75 kJ. 2. Does q p = ΔH always hold when electrical work also crosses the boundary? Answer: No. Other work modes require additional terms in the first-law balance. 3. What happens to ΔH when a balanced reaction is reversed? Answer: Its sign reverses while the magnitude is unchanged for the same states and conditions.