Visual walkthrough — Enthalpy H = U + PV; ΔH for reactions at constant P

This page rebuilds the parent result Enthalpy H = U + PV from the ground up, in pictures. We will not use a single symbol until you can see what it means. By the end you will understand, at a gut level, why chemists invented the letter $H$ at all.

Prerequisites we lean on (each linked so you can top up): Internal Energy U and First Law, PV Work and Expansion, and later Calorimetry.

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Step 1 — Draw the system and its energy pot

WHAT. Look at Figure s01. The box is our system. Inside it lives one number, drawn as a filled tank: the internal energy, written $U$. Think of $U$ as all the energy the system already owns — the jiggling of molecules, the pull between them, and the energy locked inside chemical bonds.

WHY. Before we can talk about changes, we need one honest quantity that says "how much energy is in the box right now." That is $U$. It is a state function: it depends only on the current state, not on how we got there.

PICTURE. The red tank level is $U$. Nothing is flowing yet — this is the starting snapshot.

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Step 2 — Two doors let energy in and out

WHAT. A reaction happens. Energy can now cross the box wall through exactly two doors (Figure s02):

• the heat door, $q$ — energy flowing because of a temperature difference;
• the work door, $w$ — energy flowing because something is pushed.

The bookkeeping rule that ties them together is the First Law of thermodynamics:

$\underset{\text{change in the tank}}{\underbrace{\Delta U}}=\underset{\text{heat door}}{\underbrace{q}}+\underset{\text{work door}}{\underbrace{w}}$

WHY this equation and not another? Because energy is never created or destroyed. Whatever the tank level changes by ($\Delta U$, read "delta U" = final minus initial) must equal exactly what came in through the two doors. No leaks. This is the whole content of Internal Energy U and First Law.

PICTURE. Two arrows point into the box. When an arrow points in, that term is positive (energy added); out, negative (energy lost).

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Step 3 — What the work door actually looks like: a piston

WHAT. In most chemistry, the "work" is the gas pushing a wall outward — a piston, or the atmosphere itself acting like a lid. Figure s03 shows a gas pushing a piston up by a height, sweeping out a new volume.

The pressure outside, $P_{\text{ext}}$, presses down on the piston of area $A$ with a force $P_{\text{ext}}A$. When the gas lifts the piston a distance $d$, the volume increases by $\Delta V=Ad$. The work the system does against that outside push is force $\times$ distance:

$w=-\underset{\text{outside pressure}}{\underbrace{P_{\text{ext}}}}\underset{\text{volume swept out}}{\underbrace{\Delta V}}$

WHY the minus sign? Trace it in the figure. When the gas expands, $\Delta V>0$, but the system is spending energy to shove the atmosphere out of the way — so its own energy must go down. A quantity that lowers the tank must be negative. Hence $w=-P_{\text{ext}}\Delta V$. This is the heart of PV Work and Expansion.

PICTURE. The red arrow is the gas pushing up; the shaded slab is $\Delta V$, the volume it just conquered.

Recall Check the sign with a squeeze

If instead we compress the gas (push the piston in), what is the sign of $\Delta V$, and of $w$? $\Delta V<0$, so $w=-P_{\text{ext}}\Delta V>0$: work is done on the system, its energy rises. ✓

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Step 4 — Freeze the pressure: the "constant P" world

WHAT. Now we make one physical assumption that matches real lab life: an open beaker under the sky. The atmosphere always presses with the same pressure, so:

$P_{\text{ext}}=P_{\text{system}}=P=\text{constant}$

Figure s04 shows this: the lid can slide freely, so the inside pressure is always dragged to equal the fixed outside pressure. The piston moves, but $P$ never changes.

WHY do this? Because a constant $P$ is exactly what lets us do the magic factoring in the next step. Under a fixed sky, $P$ can be pulled out in front of any volume change like a constant multiplier.

PICTURE. The pressure gauge reads the same before and after; only the piston height (volume) differs.

Now substitute $w=-P\Delta V$ into the First Law and call the heat $q_P$ (the subscript $P$ shouts "measured at constant pressure"):

$\Delta U=q_P+(-P\Delta V)⟹\Delta U=q_P-P\Delta V$

Rearrange to isolate the heat — the thing our thermometer actually reads:

$\boxed{q_P=\Delta U+P\Delta V}$

• $q_P$ ::: heat exchanged, and only at constant $P$
• $\Delta U$ ::: change in internal energy
• $P\Delta V$ ::: the expansion-work correction

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Step 5 — The factoring trick that births $H$

WHAT. Look hard at $q_P=\Delta U+P\Delta V$. Expand each "delta" into (final − initial):

$q_P=(U_{\text{f}}-U_{\text{i}})+P(V_{\text{f}}-V_{\text{i}})$

Because $P$ is constant we can slide it inside each bracket and regroup by time-slice — bundle everything "final" together, everything "initial" together (Figure s05):

$q_P=\underset{\text{final combo}}{\underbrace{(U_{\text{f}}+P{V}_{\text{f}})}}-\underset{\text{initial combo}}{\underbrace{(U_{\text{i}}+P{V}_{\text{i}})}}$

WHY regroup like this? Because the same recipe — "take $U$, add $PV$" — appears twice, once at the end and once at the start. Whenever one recipe shows up in two states and we subtract, it is begging to be given its own name. Define:

$H\equiv U+PV$

• $H$ ::: enthalpy, the new combined quantity
• $U$ ::: internal energy (the tank)
• $PV$ ::: the "space rent" — energy tied up occupying volume $V$ under pressure $P$

PICTURE. Two stacked bars: the lower block is $U$, the upper block is $PV$; their combined height is $H$. Two such towers stand side by side (initial and final), and $q_P$ is simply the difference in total height.

With the name in hand:

$\boxed{q_P=H_{\text{f}}-H_{\text{i}}=\Delta H}$

At constant pressure, the heat you measure IS the change in enthalpy. That is the whole point — the messy work term was quietly swallowed into $H$.

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Step 6 — Edge case: no gas moves ($\Delta V=0$)

WHAT. Suppose the reaction makes no change in volume — think of a reaction among only solids and liquids, where $V_{\text{f}}\approx V_{\text{i}}$ (Figure s06). Then $\Delta V=0$, the shaded work-slab collapses to nothing.

WHY it matters. Put $\Delta V=0$ into $q_P=\Delta U+P\Delta V$:

$q_P=\Delta U+P(0)=\Delta U⟹\Delta H=\Delta U$

So in condensed-phase-only reactions, enthalpy and internal energy changes coincide. The $PV$ tower on top has equal height before and after, so it cancels.

PICTURE. The piston does not move; the two towers differ only in their lower ($U$) blocks.

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Step 7 — Edge case: gases change — the $\Delta n_{\text{gas}}RT$ bridge

WHAT. When the number of moles of gas changes, the volume swings a lot and the two energies genuinely diverge. For ideal gases, $PV=nRT$, so at constant $T$:

$\Delta (PV)=\Delta n_{\text{gas}}RT$

• $\Delta n_{\text{gas}}$ ::: moles of gas products − moles of gas reactants
• $R$ ::: gas constant, $8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$
• $T$ ::: absolute temperature in kelvin

Therefore the exact bridge between the two energy changes is:

$\boxed{\Delta H=\Delta U+\Delta n_{\text{gas}}RT}$

WHY. This is where a bomb calorimeter (constant volume, measures $\Delta U$) and a coffee-cup calorimeter (constant pressure, measures $\Delta H$) part ways — see Calorimetry.

Worked check — ammonia synthesis (the parent's Example 3):

${\text{N}}_2(g)+3{\text{H}}_2(g)\to 2{\text{NH}}_3(g),\Delta H^{\circ }=-92.4\text{kJ/mol}$

$\Delta n_{\text{gas}}=2-(1+3)=-2$ $\Delta (PV)=(-2)(8.314)(298)=-4955\text{J}=-4.96\text{kJ}$ $\Delta U=\Delta H-\Delta (PV)=-92.4-(-4.96)=-87.4\text{kJ/mol}$

PICTURE. Figure s07 shows four gas molecules collapsing into two: the gas count drops, the piston is sucked inward, so the $PV$ tower shrinks and $\Delta H$ ends up more negative than $\Delta U$.

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The one-picture summary

Figure s08 compresses the entire journey: the First Law splits energy into heat and work; the work is $-P\Delta V$; locking $P$ constant lets us fold $PV$ into $U$ to build $H$; and the heat you read off a thermometer at constant pressure is exactly $\Delta H$.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a box holding a reacting gas, with a tank inside it labelled "energy it already has" — that's $U$. Energy can only get in or out through two doors: a heat door and a push door. That's the First Law: whatever the tank level changes by equals what came through the two doors.

The push door is a piston. When the gas expands under the open sky, it has to shove the atmosphere aside, and that costs energy — so the push term carries a minus sign, $-P\Delta V$.

Now here's the trick. If we keep the pressure fixed (open beaker), we can rearrange the bookkeeping so the heat equals "energy tank plus space-rent" at the end minus the same thing at the start. That combo, $U+PV$, shows up twice — so we give it a name: enthalpy, $H$. The reward is huge: the heat your thermometer reads at constant pressure is just $\Delta H$. No need to separately measure the pushing work — it's already baked in.

Finally the special cases. If nothing changes volume (solids and liquids), the push does nothing and $\Delta H=\Delta U$. If gases appear or vanish, the piston lurches and the two differ by exactly $\Delta n_{\text{gas}}RT$. That single correction is the bridge between a sealed bomb calorimeter and an open coffee cup.

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See also: Hess's Law · Standard Enthalpy of Formation · Bond Enthalpies · Entropy and Gibs Free Energy