Enthalpy Definition and Motivation

Boil a kettle open to the atmosphere and the water does more than heat up. It also pushes back the air above it as steam expands outward. Some of the energy you supply never raises the temperature at all. It goes into shoving the atmosphere out of the way. Chemists almost always work at constant pressure, open to the lab air, not in a sealed rigid box. The First Law tracks internal energy $U$ , which counts heat and work together, but at constant pressure the "useful" heat measured by a calorimeter is not $\Delta U$ . It is a different quantity. This article defines that quantity, enthalpy, and shows why it is the natural bookkeeping tool for constant-pressure chemistry. You will see where enthalpy comes from. You will see why it equals heat at constant pressure. You will work two full numerical examples. Later chapters build on this to define heat capacity at constant pressure, $C_p$ , and to compute reaction enthalpies from tabulated data.

Key takeaways

Enthalpy is defined as $H = U + pV$ , a state function built from other state functions.
At constant pressure with only expansion work, $\Delta H$ equals the heat absorbed, $q_p$ .
The correction term $\Delta(pV)$ accounts for work done pushing back the atmosphere.
$H$ is preferred over $U$ in chemistry because most reactions run open to constant atmospheric pressure.
Enthalpy changes are exact and path-independent, just like $\Delta U$ , because $H$ is built from state functions.
Key idea: enthalpy is internal energy plus a bookkeeping term that automatically subtracts atmospheric expansion work, so $\Delta H$ directly equals the heat you measure at constant pressure.

The core insight

Enthalpy, $H = U + pV$ , is engineered so that its change under constant pressure equals the heat flow, $\Delta H = q_p$ , because the $pV$ term exactly cancels the expansion work that a constant-pressure process must perform against the surroundings.

The physical picture

Consider a reaction run in an open beaker on a lab bench, at constant atmospheric pressure $p$ . Suppose the reaction produces gas, like zinc dissolving in hydrochloric acid to release H₂. As gas forms, the system must expand, pushing the atmosphere back. This costs energy: work equal to $p\,\Delta V$ is done on the surroundings, energy that leaves the system but never shows up as a temperature rise or as measurable "chemical heat." The First Law states that internal energy changes as

\Delta U = q + w

where $w$ is the work done on the system. For expansion against a constant external pressure, $w = -p\,\Delta V$ . So the heat absorbed by the system is

q = \Delta U + p\,\Delta V

This is exactly the combination that would arise if we tracked a new quantity equal to $U + pV$ . Since $p$ is constant, $p\,\Delta V = \Delta(pV)$ , and we get $q_p = \Delta U + \Delta(pV) = \Delta(U + pV)$ . Define that combination as $H$ , the enthalpy. Then heat at constant pressure is simply $\Delta H$ : no extra correction term needed, because it is already built in. Crucially, $H$ is a state function. It is constructed entirely from $U$ , $p$ , and $V$ , all of which are themselves state functions. That means $\Delta H$ between two states does not depend on the path taken between them, even though heat and work individually are path-dependent quantities. This is what makes enthalpy so useful: calorimetry measures a path-dependent quantity, $q_p$ , but under the constant-pressure constraint that quantity happens to equal a path-independent one, $\Delta H$ . This lets chemists tabulate enthalpy values and combine them algebraically (as in Hess's Law) regardless of how a reaction is actually carried out.

Figure

Energy partitioning during constant-pressure expansion of a gas-producing reaction

The core relations

H = U + pV

where: $H$ is enthalpy, $U$ is internal energy, $p$ is pressure, $V$ is volume — all state functions, so $H$ is a state function.

\Delta H = \Delta U + \Delta(pV)

where: for a process between two states at possibly different $p, V$ ; simplifies to $\Delta H = \Delta U + p\Delta V$ if $p$ is constant.

q_p = \Delta H \quad (\text{constant } p,\ \text{only } pV\text{-work})

where: $q_p$ is heat absorbed at constant external pressure with no non-expansion work.

\Delta H = \Delta U + p\,\Delta V \quad (\text{constant } p)

where: $p\,\Delta V$ is the expansion work magnitude; this is the correction separating $\Delta H$ from $\Delta U$ .

Derivation AFrom the First Law to Δ H = q_p

Goal. Show that at constant pressure, with only $pV$ -expansion work allowed, the heat absorbed equals the change in the quantity $H \equiv U + pV$ .

Step 1: Start from the First Law. For a closed system undergoing any process,

\Delta U = q + w

where $w$ is work done on the system by the surroundings.

Step 2: Specify the work term. Restrict to systems that only do expansion (pressure–volume) work against a constant external pressure $p_{\text{ext}} = p$ . Mechanically reversible or not, the work done on the system is

w = -p\,\Delta V

The negative sign reflects that if the system expands ( $\Delta V > 0$ ), it does work on the surroundings, so energy leaves the system.

Step 3: Substitute into the First Law.

\Delta U = q_p - p\,\Delta V

Solve for the heat absorbed at constant pressure:

q_p = \Delta U + p\,\Delta V

Step 4: Recognize $p\,\Delta V$ as $\Delta(pV)$ when $p$ is constant. Since $p$ does not change during the process,

p\,\Delta V = p\,(V_2 - V_1) = (pV_2) - (pV_1) = \Delta(pV)

q_p = \Delta U + \Delta(pV)

Step 5: Define enthalpy so this combination becomes a single state-function difference. Define $H \equiv U + pV$ . Because $U$ , $p$ , and $V$ are each functions of the state alone, so is $H$ . Then

\Delta H = H_2 - H_1 = (U_2 + p_2V_2) - (U_1 + p_1V_1)

At constant pressure, $p_1 = p_2 = p$ , so

\Delta H = (U_2 - U_1) + p(V_2 - V_1) = \Delta U + p\,\Delta V

This is identical to the expression for $q_p$ found in Step 4.

Step 6: Conclude.

\boxed{q_p = \Delta H \quad \text{(constant pressure, expansion work only)}}

Enthalpy change is therefore directly measurable: run the reaction in an open vessel at constant pressure (e.g., atmospheric), measure the heat exchanged, and that heat is $\Delta H$ , independent of how the process actually unfolded microscopically.

Worked examples

Example 1 — Foundational: melting ice at constant pressure. One mole of ice melts at 0 °C (273.15 K) and 1 atm (101325 Pa). The molar enthalpy of fusion is $\Delta H_{fus} = 6.01\ \text{kJ mol}^{-1}$ . The molar volume of ice is $V_{ice} = 19.6\ \text{cm}^3\ \text{mol}^{-1}$ and of liquid water is $V_{liq} = 18.0\ \text{cm}^3\ \text{mol}^{-1}$ at this temperature. Find $q_p$ , $w$ , and $\Delta U$ for melting 1 mol of ice. Step 1. Since melting occurs at constant pressure with only $pV$ -work,

q_p = \Delta H = 6.01\ \text{kJ}

Step 2. Compute the volume change.

\Delta V = V_{liq} - V_{ice} = 18.0 - 19.6 = -1.6\ \text{cm}^3 = -1.6\times10^{-6}\ \text{m}^3

(Water is denser than ice, so volume decreases on melting — the system contracts slightly.) Step 3. Compute the work done on the system.

w = -p\,\Delta V = -(101325\ \text{Pa})(-1.6\times10^{-6}\ \text{m}^3) = +1.62\times10^{-1}\ \text{J} = 0.000162\ \text{kJ}

This is positive: the shrinking system has work done on it by the atmosphere (a tiny push inward). Step 4. Compute $\Delta U$ from the First Law.

\Delta U = q_p + w = 6.01\ \text{kJ} + 0.000162\ \text{kJ} = 6.010\ \text{kJ}

Sanity check: the $pV$ correction is minuscule (~0.003% of $\Delta H$ ) because liquids and solids have tiny molar volumes, so $\Delta U \approx \Delta H$ for condensed-phase transitions. This confirms the well-known rule that $\Delta H \approx \Delta U$ whenever no gas is produced or consumed.

Example 2 — Applied: combustion producing gas. Consider the combustion of 1.00 mol of solid naphthalene, $\text{C}_{10}\text{H}_8(s)$ , in excess oxygen at 298 K and 1 bar (100000 Pa):

\text{C}_{10}\text{H}_8(s) + 12\,\text{O}_2(g) \rightarrow 10\,\text{CO}_2(g) + 4\,\text{H}_2\text{O}(l)

The measured constant-pressure heat released is $q_p = -5157\ \text{kJ mol}^{-1}$ (so $\Delta H = -5157\ \text{kJ mol}^{-1}$ , exothermic). Find $\Delta U$ for the reaction, treating gases as ideal and neglecting the volumes of solids and liquids. Step 1. Count moles of gas before and after.

n_{gas,\ reactants} = 12\ \text{mol O}_2, \qquad n_{gas,\ products} = 10\ \text{mol CO}_2

\Delta n_{gas} = 10 - 12 = -2\ \text{mol}

(Naphthalene is solid and water is liquid, so they contribute negligible volume by comparison.) Step 2. Use the ideal gas law to convert $\Delta n_{gas}$ into a volume/pressure work term.

\Delta(pV)_{gas} = \Delta n_{gas}\,RT

\Delta(pV)_{gas} = (-2\ \text{mol})(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K}) = -4955\ \text{J} = -4.955\ \text{kJ}

Step 3. Since $\Delta H = \Delta U + \Delta(pV)$ at constant $T,p$ for ideal gases,

\Delta U = \Delta H - \Delta(pV) = -5157\ \text{kJ} - (-4.955\ \text{kJ})

\Delta U = -5157 + 4.955 = -5152\ \text{kJ mol}^{-1}

Sanity check: because the gas moles decrease ( $\Delta n_{gas} < 0$ ), the system contracts overall, so the surroundings do a small amount of work on the system. That makes $\Delta U$ slightly less negative than $\Delta H$ — consistent with our result ( $-5152$ vs. $-5157$ kJ/mol), a difference of only about 0.1%, small but not negligible for precise thermochemical tables.

Limiting cases & common traps

Situation	Relationship	Why
Condensed phases only (no gas involved)	$\Delta H \approx \Delta U$	Molar volumes of solids/liquids are tiny, so $p\Delta V$ is negligible
Reaction with $\Delta n_{gas} = 0$	$\Delta H = \Delta U$ exactly (ideal gas)	$\Delta(pV) = \Delta n_{gas}RT = 0$
Reaction produces net gas ( $\Delta n_{gas} > 0$ )	$\Delta H > \Delta U$	System must do extra work expanding against atmosphere
Reaction consumes net gas ( $\Delta n_{gas} < 0$ )	$\Delta H < \Delta U$	Surroundings do work compressing the system
Constant-volume calorimetry (bomb calorimeter)	measures $q_v = \Delta U$ , not $\Delta H$	No expansion work possible; must convert using $\Delta H = \Delta U + \Delta n_{gas}RT$

The chemist behind it

The term "enthalpy" (from Greek enthalpein, "to warm in") was coined in the early twentieth century, but the underlying quantity $U + pV$ traces to nineteenth-century engineering thermodynamics developed to analyze steam engines, where constant-pressure heat exchange is the norm. The formalization of enthalpy as a distinct state function for chemical thermodynamics is closely associated with the broader program of Gibbs, Clausius, and their successors who built the state-function framework that lets chemists tabulate and combine thermochemical data reliably.

Origins · 1 minHeike Kamerlingh Onnes and how this was discovered

Heike Kamerlingh Onnes

1853–1926 · Dutch physicist credited with popularizing the term "enthalpy" around 1909, building on the state-function formalism developed for engineering thermodynamics of gases and refrigeration

Ask the tutor: How does enthalpy connect to heat capacity at constant pressure, $C_p$ , and why is $C_p$ defined as a derivative of $H$ rather than of $U$ ?

Is enthalpy only defined for chemical reactions?

No. Enthalpy is a state function defined for any system in a well-defined thermodynamic state — a pure gas, a mixture, a single phase, anything with definable $U$ , $p$ , and $V$ . It does not require a reaction. Chemists use it heavily for reactions and phase changes because those are commonly studied at constant pressure, but $H$ is just as meaningful for, say, a fixed sample of nitrogen gas sitting in a cylinder. The reaction context is where $\Delta H$ becomes most useful, not where $H$ is defined.

Why not just always use ΔU and skip enthalpy entirely?

You could, in principle, but it would be inconvenient. Constant-pressure conditions are the default in real chemistry: open beakers, flasks vented to the atmosphere, biological systems under roughly constant pressure. Under these conditions $\Delta U$ requires correcting for expansion work every single time, while $\Delta H$ already has that correction built in. Enthalpy saves a step in the overwhelmingly common experimental situation, which is exactly why thermochemical tables report $\Delta H$ values rather than $\Delta U$ values.

Does ΔH = q_p hold if the reaction does non-expansion work, like electrical work in a battery?

No. The derivation assumes the only work is $pV$ -expansion work against the surroundings. If the system also performs additional (non-expansion) work — electrical work from a galvanic cell, for example — then $q_p \neq \Delta H$ in general. Instead, $\Delta H$ still equals $\Delta U + p\Delta V$ , but the measured heat is reduced by whatever non-expansion work was extracted. This distinction matters directly in electrochemistry, where $\Delta G$ , not $\Delta H$ , connects to the maximum non-expansion work.

Why is ΔH negative for exothermic reactions?

By convention, thermodynamic quantities are defined from the system's perspective. When a reaction releases heat to the surroundings at constant pressure, the system's enthalpy decreases, so $q_p = \Delta H < 0$ . This matches the everyday sense that "exothermic" reactions feel hot to the surroundings because they are dumping energy outward, which corresponds to the system's own enthalpy content dropping. Endothermic reactions absorb heat from the surroundings, giving $\Delta H > 0$ .

How does bomb calorimetry relate to enthalpy if it measures constant volume, not constant pressure?

A bomb calorimeter holds the reaction at essentially constant volume, so no expansion work occurs and the heat measured is $q_v = \Delta U$ directly, not $\Delta H$ . To report $\Delta H$ from bomb calorimetry data, chemists apply the correction $\Delta H = \Delta U + \Delta n_{gas}RT$ , using the change in moles of gas in the balanced reaction. This is exactly the calculation performed in Example 2 above, just run in reverse to go from $\Delta U$ back to $\Delta H$ .