Enthalpy H = U + PV; ΔH for reactions at constant P: Worked examples

This page is a drill. The parent note Enthalpy H = U + PV built the idea $Δ H = q_{P}$ . Here we hit every kind of question it can throw at you. First we map the terrain, then we walk one worked example per cell of that map.

Before anything, one reminder in plain words. Enthalpy $H$ is a number attached to a chemical system that behaves like "heat content" when pressure is held constant. The symbol $Δ$ (Greek "delta") just means "final minus initial" — the change in a quantity. So $Δ H = H_{after} - H_{before}$ . A negative $Δ H$ means the system ended with less stored energy, and that lost energy walked out as heat.

The scenario matrix

Every enthalpy problem lives in one of these boxes. Each column is a way the question can vary; each example below is tagged with the box it fills.

#	Case class	What varies	Example
A	Sign: exothermic ( $Δ H < 0$ )	heat leaves system	Ex 1
B	Sign: endothermic ( $Δ H > 0$ )	heat enters system	Ex 2
C	Scaling by moles	given per-mole $Δ H$ , asked for a real mass	Ex 1, Ex 3
D	$Δ H \leftrightarrow Δ U$ conversion, $Δ n_{g a s} < 0$	gas moles shrink	Ex 4
E	$Δ H \leftrightarrow Δ U$ , $Δ n_{g a s} > 0$	gas moles grow	Ex 5
F	Degenerate: $Δ n_{g a s} = 0$	no gas change → $Δ H = Δ U$	Ex 6
G	Zero / limiting input	$Δ H = 0$ , or $T \to 0$	Ex 7
H	Real-world word problem	translate a story into numbers	Ex 8
I	Exam twist: sign trap	surroundings vs system perspective	Ex 9

We will also lean on the geometry of an enthalpy diagram (a picture of energy levels) to make the signs visible rather than memorised.

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

Look at the figure: enthalpy runs up the vertical axis (higher = more stored energy). Reactants sit on one shelf, products on another. If the product shelf is lower (blue arrow points down), energy fell out as heat → exothermic. If it is higher (red arrow up), energy had to be pumped in → endothermic. Every sign question is just "which shelf is lower?"

The worked examples

Cell A + C — exothermic, scale by moles

Worked example Example 1: Burning methane

$CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l) Δ H^{\circ} = - 890.3 kJ/mol$ Constant pressure, $2 5^{\circ} C$ . How much heat is released when 8.0 g of methane burns?

Forecast: the number $- 890.3$ is per mole. We have less than a mole (8 g, and a mole is 16 g), so expect less than 890 kJ. Guess before reading on.

Moles of methane. $n = \frac{8.0 g}{16.0 g/mol} = 0.50 mol$ Why this step? $Δ H^{\circ}$ is quoted per mole of reaction. To use it we need "how many moles of reaction happened", and that is set by the limiting reactant, here methane.
Scale the enthalpy. $q_{P} = n \times Δ H^{\circ} = 0.50 \times (- 890.3) = - 445.15 kJ$ Why this step? At constant $P$ , $Δ H = q_{P}$ (parent result). Enthalpy is extensive — twice the stuff, twice the heat — so we just multiply.
Interpret the sign. $Δ H < 0$ → this is the blue downward arrow in the figure. Energy left the system as heat.

Verify: Half a mole should give half of $- 890.3 = - 445.15$ . ✔ Units: $mol \times kJ/mol = kJ$ . ✔ Sign negative = exothermic = flame is hot. ✔

Answer: 445.15 kJ released.

Cell B — endothermic

Cell C — awkward scaling (fractional coefficient)

Worked example Example 3: Half-reaction bookkeeping

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) Δ H^{\circ} = - 571.6 kJ/mol (as written)$ What is $Δ H^{\circ}$ for forming one mole of liquid water, $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ ?

Forecast: the given equation makes 2 waters. Halving the equation should halve the heat → about $- 286$ kJ.

Divide the whole equation by 2. $Δ H^{\circ} = \frac{- 571.6}{2} = - 285.8 kJ/mol$ Why this step? $Δ H$ is tied to the equation as written. Scale the coefficients by a factor and $Δ H$ scales by the same factor — even fractional ones like $\frac{1}{2}$ .

Verify: Two of these half-equations add back to the original: $2 \times (- 285.8) = - 571.6$ . ✔ (This is a mini Hess's Law check.) This value, $- 285.8 kJ/mol$ , is exactly the Standard Enthalpy of Formation of liquid water.

Answer: $Δ H^{\circ} = - 285.8 kJ/mol$ .

Cell D — $Δ H \to Δ U$ with gas moles shrinking

Worked example Example 4: Ammonia synthesis (

$Δ n_{g a s} < 0$ ) $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g) Δ H^{\circ} = - 92.4 kJ/mol$ Find $Δ U^{\circ}$ at $298$ K.

Forecast: gas moles fall from 4 to 2, so the surroundings push in on the shrinking system (compression does work on it). Compression adds energy the internal store keeps, so $Δ U$ should be less negative than $Δ H$ .

Link the two. From $H = U + P V$ (parent definition): $Δ H = Δ U + Δ (P V)$ Why? $H$ is $U + P V$ ; take the change of both sides.
Turn $Δ (P V)$ into gas counting. For ideal gases at fixed $T$ , $P V = n R T$ , so $Δ (P V) = Δ n_{g a s} R T$ . Why only gases? Solids/liquids have negligible volume, so only the gas count moves $P V$ appreciably. (This uses Internal Energy U and First Law and PV Work and Expansion.)
Count gas moles. $Δ n_{g a s} = 2 - (1 + 3) = - 2$ .
Compute the correction. $Δ (P V) = (- 2) (8.314) (298) = - 4955.1 J = - 4.955 kJ$
Solve. $Δ U^{\circ} = Δ H^{\circ} - Δ (P V) = - 92.4 - (- 4.955) = - 87.4 kJ/mol$

Verify: $Δ U$ ( $- 87.4$ ) is indeed less negative than $Δ H$ ( $- 92.4$ ), matching the forecast. ✔ Units: $mol \cdot \frac{J}{mol\cdotK} \cdot K = J$ . ✔

Answer: $Δ U^{\circ} = - 87.4 kJ/mol$ .

Cell E — $Δ H \to Δ U$ with gas moles growing

Cell F — degenerate case, $Δ n_{g a s} = 0$

Cell G — zero / limiting input

Worked example Example 7: The

$Δ H = 0$ boundary and the $T \to 0$ limit (a) A gas expands into a vacuum, then is recompressed to the identical starting state. What is $Δ H$ for the round trip? (b) For the reaction in Ex 5, what does the $Δ H \to Δ U$ correction approach as $T \to 0$ K?

Forecast: (a) same start and end state → state function → change is zero. (b) the correction is $Δ n_{g a s} R T$ , and anything $\times T$ with $T \to 0$ vanishes.

(a) Use the state-function property. $H$ depends only on the current state, not the path. Start state = end state ⇒ $Δ H = H_{final} - H_{initial} = 0$ Why this step? This tests the degenerate zero case: a closed loop always gives $Δ H = 0$ for any state function.
(b) Take the limit. $lim_{T \to 0} Δ n_{g a s} R T = (1) (8.314) (0) = 0 J$ Why this step? At absolute zero, the atmosphere has no thermal "push" ( $P V = n R T \to 0$ ), so $Δ H$ and $Δ U$ merge. This is the limiting behaviour row of the matrix.

Verify: (a) A cyclic process returns every state variable to its start, so $Δ$ of any of them is 0. ✔ (b) Plugging $T = 0$ gives exactly $0$ , and for any small $T$ the gap grows linearly, $\approx 8.31 T$ J. ✔

Answers: (a) $Δ H = 0$ . (b) correction $\to 0$ J.

Cell H — real-world word problem

Worked example Example 8: Camping stove

A butane stove burns propane, $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O (l)$ , with $Δ H^{\circ} = - 2220 kJ/mol$ . You need to heat 1.0 kg of water from $2 0^{\circ} C$ to $10 0^{\circ} C$ . The water's specific heat is $4.18 J g^{- 1} K^{- 1}$ . Assuming all heat reaches the water, how many grams of propane ( $M = 44 g/mol$ ) do you burn?

Forecast: heating a kilo of water 80 degrees is a few hundred kJ, and one mole of propane gives 2220 kJ, so expect well under a mole — a small handful of grams.

Heat the water needs (this is a Calorimetry step). $q = m c Δ T = 1000 \times 4.18 \times 80 = 334400 J = 334.4 kJ$ Why this step? The stove's job is defined by how much energy the water demands.
Moles of propane to supply it. Each mole releases $2220$ kJ: $n = \frac{334.4}{2220} = 0.15063 mol$ Why this step? Match released heat to required heat; divide by the per-mole yield.
Convert to mass. $m = n \times 44 = 0.15063 \times 44 = 6.628 g$ Why this step? Camping is measured in grams of fuel, not moles.

Verify: $6.628$ g is a small handful, matching the forecast. ✔ Chain of units: $J \to kJ \to mol \to g$ all cancel cleanly. ✔ Reverse check: $0.15063 \times 2220 = 334.4$ kJ back. ✔

Answer: about 6.63 g of propane.

Cell I — exam twist, the sign trap

Worked example Example 9: Whose perspective?

In a coffee-cup calorimeter, dissolving $5.00$ g of ammonium nitrate ( $M = 80.0 g/mol$ ) in water makes the water temperature drop by $4.20 K$ . The solution (100 g total, $c = 4.18 J g^{- 1} K^{- 1}$ ) is the surroundings. Find $Δ H^{\circ}$ of dissolution per mole.

Forecast: the water got colder, so the salt pulled heat in → endothermic → $Δ H$ should come out positive. The trap is to copy the water's $- 4.2$ K straight into $Δ H$ and get a wrong minus sign.

Heat gained by the surroundings (water). $q_{surr} = m c Δ T = 100 \times 4.18 \times (- 4.20) = - 1755.6 J$ Why this step? The water lost energy ( $Δ T < 0$ ), so $q_{surr}$ is negative.
Flip to the system. $q_{sys} = - q_{surr}$ : $q_{sys} = + 1755.6 J$ Why this step? Energy leaving the water entered the dissolving salt. This is the exact sign-flip in Internal Energy U and First Law. At constant $P$ , $q_{sys} = Δ H$ .
Per mole. Moles of salt $n = 5.00/80.0 = 0.0625 mol$ . $Δ H^{\circ} = \frac{+ 1755.6 J}{0.0625 mol} = + 28089.6 J/mol = + 28.09 kJ/mol$

Verify: Positive $Δ H$ ⇒ endothermic ⇒ water cools — exactly what the cold pack does. ✔ The naive un-flipped answer would have been $- 28$ kJ/mol, contradicting the cold beaker — that is the trap. ✔

Answer: $Δ H^{\circ} = + 28.1 kJ/mol$ (endothermic).

Pulling the matrix together

Recall Which cell was which?

State-function loop gives $Δ H = 0$ ::: Cell G (Example 7a) Gas moles grow, $Δ U < Δ H$ ::: Cell E (Example 5) Gas moles shrink, $Δ U$ less negative than $Δ H$ ::: Cell D (Example 4) Water cools, system endothermic, sign flips ::: Cell I (Example 9) $Δ n_{g a s} = 0$ so $Δ U = Δ H$ exactly ::: Cell F (Example 6)

For the theory these drills rest on, return to the parent enthalpy note; to see where these $Δ H$ values get combined, continue to Hess's Law and Bond Enthalpies, and for the "will it actually happen?" question, Entropy and Gibs Free Energy.