Enthalpy and Enthalpy of Reaction

Chemists rarely trap reactions in sealed, rigid boxes. Beakers, flasks, and the human body all react while open to the atmosphere. So pressure — not volume — stays fixed. This creates a bookkeeping problem. Some of the energy released by a reaction quietly does work pushing back the atmosphere. That portion never shows up as heat. Enthalpy, $H$ , is the state function engineered to solve exactly this problem. Building on the First Law of Thermodynamics, this article defines enthalpy. It shows why $\Delta H$ equals heat flow at constant pressure. It then applies that idea to reactions. You will learn to read thermochemical equations. You will classify reactions as exothermic or endothermic. You will also compute $\Delta H$ under simple, realistic conditions. The tools built here — sign conventions, state-function reasoning, and the mole-based scaling of $\Delta H$ — are the direct foundation for Calorimetry, Hess's Law, and Enthalpies of Formation.

Key takeaways

Enthalpy is defined as $H = U + PV$ ; it is a state function with units of energy (kJ or J).
At constant pressure, the heat exchanged equals $\Delta H$ , i.e. $q_p = \Delta H$ — this is the whole point of inventing $H$ .
$\Delta H < 0$ means exothermic (heat released to surroundings); $\Delta H > 0$ means endothermic (heat absorbed).
Thermochemical equations pair a balanced reaction with a numeric $\Delta H$ and must specify physical states, since $\Delta H$ depends on them.
$\Delta H$ scales with the amount of substance reacted — double the moles, double the enthalpy change.
Key idea: $\Delta H$ is just the constant-pressure heat flow of a reaction, repackaged as a state function so it can be added, scaled, and tabulated like a number.

The core insight

Enthalpy exists so that "heat released by a reaction" becomes a property of the reaction itself — independent of path — rather than a process-dependent quantity you must recompute every time.

The physical picture

Imagine a reaction happening in an open beaker on a lab bench. The system (reacting chemicals) can expand or contract against the constant pressure of the atmosphere, $P_{\text{ext}}$ . If gas is produced, the system pushes the atmosphere back and does work on it; if gas is consumed, the atmosphere pushes in and does work on the system. The First Law of Thermodynamics states

\Delta U = q + w

where $w = -P_{\text{ext}}\Delta V$ for a constant-pressure process with only expansion work. Substituting,

\Delta U = q_p - P\Delta V

so the heat at constant pressure is

q_p = \Delta U + P\Delta V

This combination, $\Delta U + P\Delta V$ , appears so often for constant-pressure processes that chemists gave it a name: the change in enthalpy, $\Delta H$ . Defining $H \equiv U + PV$ makes $\Delta H = \Delta U + \Delta(PV)$ , which at constant $P$ reduces to $\Delta U + P\Delta V$ — precisely $q_p$ . The payoff is enormous: $H$ is built from $U$ , $P$ , and $V$ , all state functions, so $H$ itself is a state function. Its change depends only on initial and final states, never on the path taken, the container shape, or the reaction rate. That is what allows enthalpies to be tabulated, scaled by stoichiometric coefficients, and added together — the entire machinery behind Hess's Law rests on this property.

Figure

Energy diagram comparing exothermic and endothermic reactions

The diagram shows enthalpy plotted vertically: exothermic reactions drop from a higher reactant level to a lower product level, releasing the difference as heat to the surroundings; endothermic reactions climb, absorbing heat from the surroundings to make up the difference.

The core relations

H \equiv U + PV

where: $H$ is enthalpy (J or kJ), $U$ is internal energy, $P$ is pressure, $V$ is volume.

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

where: $\Delta H$ is the enthalpy change, $\Delta U$ is the internal-energy change, $P\Delta V$ is the pressure–volume work term at constant pressure.

q_p = \Delta H

where: $q_p$ is heat exchanged with the surroundings at constant pressure — this equality is the reason enthalpy is useful.

\Delta H_{\text{reaction}} = \sum n_p \Delta H_p^\circ(\text{products}) - \sum n_r \Delta H_r^\circ(\text{reactants})

(a preview relation used fully under Enthalpies of Formation) where: $n_p, n_r$ are stoichiometric coefficients.

Derivation AWhy

q_p = \Delta H

Goal. Show that at constant pressure, with only pressure–volume work, the heat exchanged equals the enthalpy change — no more, no less.

Step 1. Start from the First Law for a closed system doing only expansion work:

\Delta U = q + w

Step 2. At constant external pressure $P$ , the work done on the system during a volume change $\Delta V$ is

w = -P\Delta V

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

Step 3. Substitute Step 2 into Step 1 and solve for the heat at constant pressure, $q_p$ :

q_p = \Delta U - w = \Delta U + P\Delta V

Step 4. Define enthalpy as $H = U + PV$ . At constant pressure the differential form gives

\Delta H = \Delta U + P\Delta V + V\Delta P

but $\Delta P = 0$ at constant pressure, so the last term vanishes:

\Delta H = \Delta U + P\Delta V

Step 5. Compare Step 3 and Step 4 — both equal $\Delta U + P\Delta V$ , so

q_p = \Delta H

This holds whenever pressure is constant and no non-expansion work is done, which describes essentially every reaction run in an open flask or beaker.

Limiting case. If a reaction produces no net change in moles of gas, $\Delta V \approx 0$ , so $P\Delta V \approx 0$ and $\Delta H \approx \Delta U$ : enthalpy and internal-energy changes converge whenever gas volume does not change.

Worked examples

Example 1 — Combustion of propane, heat vs. enthalpy. The reaction is

\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)

with $\Delta H = -2220\ \text{kJ}$ per mole of propane burned, measured at constant pressure ( $1.00\ \text{atm}$ , $298\ \text{K}$ ). Step 1: identify moles of gas before and after. Reactant gas moles: $1 + 5 = 6$ . Product gas moles: $3$ (water is liquid, not counted). So

\Delta n_{\text{gas}} = 3 - 6 = -3

Step 2: compute the $P\Delta V$ work term using the ideal gas law, $P\Delta V = \Delta n_{\text{gas}}RT$ :

P\Delta V = (-3)(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K})

P\Delta V = -7432\ \text{J} = -7.43\ \text{kJ}

Step 3: relate $\Delta H$ and $\Delta U$ using $\Delta H = \Delta U + P\Delta V$ :

\Delta U = \Delta H - P\Delta V = -2220\ \text{kJ} - (-7.43\ \text{kJ})

\Delta U = -2212.6\ \text{kJ}

Step 4: relate to heat. Since the reaction runs at constant pressure, $q_p = \Delta H = -2220\ \text{kJ}$ per mole of propane. Note $\Delta U$ and $\Delta H$ differ by only about $0.3\%$ here — the gas-mole change is modest relative to the huge combustion energy, but the two quantities are not identical, confirming they measure genuinely different things.

Example 2 — Scaling and sign for an endothermic decomposition. Calcium carbonate decomposes as

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \qquad \Delta H = +178\ \text{kJ}

per mole of $\text{CaCO}_3$ . A lab technician decomposes $2.50\ \text{mol}$ of $\text{CaCO}_3$ completely. Step 1: scale $\Delta H$ by moles reacted, since enthalpy is extensive:

\Delta H_{\text{total}} = 2.50\ \text{mol} \times \left(\frac{+178\ \text{kJ}}{1\ \text{mol}}\right)

\Delta H_{\text{total}} = +445\ \text{kJ}

Step 2: interpret the sign. Because $\Delta H > 0$ , the process is endothermic: $445\ \text{kJ}$ must flow into the system from the surroundings (e.g., a furnace) to drive the decomposition, and $q_p = +445\ \text{kJ}$ . Step 3: sanity check with reverse reaction. If instead $1.20\ \text{mol}$ of $\text{CaO}$ reacted with $\text{CO}_2$ to re-form $\text{CaCO}_3$ (the reverse reaction), the enthalpy change would flip sign and scale with the new mole amount:

\Delta H_{\text{reverse}} = -1.20\ \text{mol} \times \left(\frac{178\ \text{kJ}}{1\ \text{mol}}\right) = -213.6\ \text{kJ}

confirming that reversing a reaction negates $\Delta H$ , while changing the amount scales it proportionally — both direct consequences of $H$ being a state function.

Limiting cases & common traps

Situation	What happens to $\Delta H$	Common trap
No change in gas moles ( $\Delta n_{\text{gas}} = 0$ )	$\Delta H \approx \Delta U$	Assuming they are always equal — true only when $\Delta n_{\text{gas}} = 0$
Reaction reversed	Sign of $\Delta H$ flips	Forgetting to flip the sign when writing the reverse reaction
Reaction scaled by factor $k$	$\Delta H$ scales by $k$	Treating $\Delta H$ as an intensive property (it is extensive, like mass)
Physical state changed (e.g., $\text{H}_2\text{O}(l)$ vs. $\text{H}_2\text{O}(g)$ )	$\Delta H$ changes by the enthalpy of vaporization	Omitting phase labels in a thermochemical equation, making $\Delta H$ ambiguous
Constant-volume process (bomb calorimeter)	$q_v = \Delta U$ , not $\Delta H$	Reporting bomb-calorimeter heat directly as $\Delta H$ without correction

The chemist behind it

The enthalpy concept crystallized from nineteenth-century work on heat and mechanical equivalence, but the term "enthalpy" itself and its modern symbol $H$ were popularized in the early twentieth century as thermodynamics matured into a rigorous discipline for engineers and chemists alike. The deeper conceptual leap — recognizing that constant-pressure heat could be captured by a new state function built from $U$ , $P$ , and $V$ — grew directly out of the work of Josiah Willard Gibbs, whose systematic treatment of thermodynamic potentials gave chemists the mathematical language to define quantities like $H$ and later the Gibbs free energy.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist at Yale who founded chemical thermodynamics

Gibbs published his monumental "On the Equilibrium of Heterogeneous Substances" in the 1870s, introducing the systematic use of thermodynamic potentials — including the combinations that underlie enthalpy and free energy — to predict whether and how far a chemical process proceeds. His work was so mathematically dense that it went largely unnoticed in America for years, yet it became the backbone of physical chemistry once European scientists like Ostwald and van der Waals championed and popularized it.

Ask the tutor: How does specific heat capacity let us measure $\Delta H$ experimentally using a calorimeter?

Is enthalpy the same thing as heat?

Not quite. Heat, $q$ , is energy in transit and depends on the path a process takes — it's not a state function in general. Enthalpy, $H$ , is a state function defined as $U + PV$ . What makes chemists treat them almost interchangeably is that, specifically at constant pressure with only expansion work, $q_p$ happens to equal $\Delta H$ exactly. Under any other conditions (constant volume, for instance), heat and enthalpy change are different numbers.

Why do we even need enthalpy if we already have internal energy from the First Law?

Because most chemistry happens at constant atmospheric pressure, not constant volume. At constant volume, $q_v = \Delta U$ directly — no correction needed. But constant-pressure processes involve expansion or compression work that eats into or adds to the internal-energy change. Enthalpy is engineered specifically to absorb that $P\Delta V$ term, so that the measured heat at constant pressure equals a clean state function, $\Delta H$ , that can be tabulated and reused.

Can ΔH be exactly zero for a real reaction?

Yes, in principle — if the enthalpies of products and reactants happen to be equal (a special coincidence), or in the limiting case of a reaction with no net change in bonding energy relevant to enthalpy under given conditions. It's rare for common reactions but not forbidden by any law; it simply means no net heat flows at constant pressure.

Why must physical states be specified in a thermochemical equation?

Because $H$ depends on the state of matter — solid, liquid, or gas — through differences in internal energy and molar volume. Converting a liquid product to a gaseous one, for example, requires additional energy equal to the enthalpy of vaporization, which shifts the overall $\Delta H$ . Omitting states makes the reported $\Delta H$ ambiguous and irreproducible, since two labs could report different numbers for what looks like "the same" reaction.

How does this connect to Hess's Law and enthalpies of formation?

Because $H$ is a state function, $\Delta H$ for any reaction depends only on initial and final states, never on the reaction's mechanism or intermediate steps. That path-independence is exactly what allows chemists to add and subtract thermochemical equations to build up unknown $\Delta H$ values from known ones — the essence of Hess's Law — and to define standard Enthalpies of Formation as reusable building blocks for any reaction's $\Delta H$ .