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

This page is your self-test. Read each problem, try it on paper first, then open the collapsible solution. Problems climb from simple recognition to full synthesis. Every number here is machine-checked.

Before we start, one reminder of the tools we lean on (all built in the parent note):

The picture below is the map for the whole page — it shows where $q_{P}$ and $q_{V}$ live and how enthalpy is drawn as an energy "hill" from reactants to products.

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

Related tools you may want open in another tab: Internal Energy U and First Law, PV Work and Expansion, Hess's Law, Standard Enthalpy of Formation, Calorimetry, Bond Enthalpies, Entropy and Gibs Free Energy.

Level 1 — Recognition

Can you read the sign and label the process?

Exercise 1.1 — Reading a sign

A reaction has $Δ H = - 241.8 kJ/mol$ . Is it exothermic or endothermic, and does the surrounding water warm up or cool down?

Recall Solution 1.1

$Δ H < 0$ means the system's enthalpy went down — the "missing" energy left as heat. Look at the left panel of the figure above: an exothermic reaction is the hill that goes downhill, products sitting lower than reactants.

Sign negative $\Rightarrow$ exothermic (heat exits, "EXo = EXit").
Heat flows out of the system into the surroundings, so the water warms up.

Answer: Exothermic; surrounding water warms up.

Exercise 1.2 — Which quantity did the calorimeter measure?

A student runs a reaction in an open coffee-cup calorimeter (constant atmospheric pressure) and reads off $q = - 57.3 kJ$ . Is this number $Δ U$ or $Δ H$ ?

Recall Solution 1.2

The key is the condition, not the apparatus name. Recall the twin equations from the opening panel: $q_{P} = Δ H$ (constant pressure) and $q_{V} = Δ U$ (constant volume).

Open cup $\Rightarrow$ constant pressure $\Rightarrow$ the heat you read is $q_{P}$ , so it equals $Δ H$ .
A bomb calorimeter is rigid: its volume cannot change, so the heat you read is $q_{V}$ , which equals $Δ U$ instead.

Answer: It is $Δ H = - 57.3 kJ$ .

Level 2 — Application

Plug into a formula and scale correctly.

Exercise 2.1 — Scaling by moles

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O (l) Δ H^{\circ} = - 2220 kJ/mol$ How much heat is released when $22 g$ of propane ( $M = 44 g/mol$ ) burns at constant pressure?

Recall Solution 2.1

Why does $Δ H$ scale linearly with amount? Enthalpy is a state function (it depends only on start and end states, not the path). Burning $2$ mol of propane is just "the $1$ -mol reaction done twice," and a state function's change simply adds up over identical steps — so twice the propane gives twice the $Δ H$ , half the propane gives half. That is why $Δ H^{\circ}$ is quoted "per mole of reaction as written": multiply it by however many moles you actually run. Step 1 — moles. $n = \frac{22 g}{44 g/mol} = 0.50 mol$ Step 2 — scale. $q_{P} = n Δ H^{\circ} = 0.50 \times (- 2220) = - 1110 kJ$ The minus sign says the system loses the energy; the amount released is its magnitude.

Answer: $1110 kJ$ released ( $q_{P} = - 1110 kJ$ ).

Exercise 2.2 — $Δ (P V)$ from a gas count

$2 CO (g) + O_{2} (g) \to 2 CO_{2} (g)$ at $T = 298 K$ . Compute $Δ (P V) = Δ n_{g a s} R T$ in kJ.

Recall Solution 2.2

Step 1 — count gas moles. Only gases count. $Δ n_{g a s} = products 2 - reactants (2 + 1) = - 1 mol$ Step 2 — multiply. $Δ (P V) = (- 1) (8.314) (298) = - 2477.6 J = - 2.48 kJ$ Negative because the gas volume shrank — the surroundings did compression work on the system.

Answer: $Δ (P V) = - 2.48 kJ$ .

Level 3 — Analysis

Convert between $Δ H$ and $Δ U$ ; decide when the correction matters.

Exercise 3.1 — $Δ U$ from $Δ H$

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) Δ H^{\circ} = - 571.6 kJ/mol (T = 298 K)$ Find $Δ U^{\circ}$ .

Recall Solution 3.1

Step 1 — rearrange. From $Δ H = Δ U + Δ (P V)$ and the just-derived $Δ (P V) = Δ n_{g a s} R T$ : $Δ U = Δ H - Δ n_{g a s} R T .$ Step 2 — count gas. Water is a liquid here, so it contributes no gas. $Δ n_{g a s} = 0 - (2 + 1) = - 3 mol$ Step 3 — correction. $Δ n_{g a s} R T = (- 3) (8.314) (298) = - 7432.7 J = - 7.43 kJ$ Step 4 — solve. $Δ U^{\circ} = - 571.6 - (- 7.43) = - 564.2 kJ/mol$ $Δ U$ is less negative: part of the enthalpy drop was compression work as $3$ mol of gas vanished, not a loss of internal energy.

Answer: $Δ U^{\circ} \approx - 564.2 kJ/mol$ .

Exercise 3.2 — The reverse trip

A gas-forming decomposition has $Δ U^{\circ} = + 120.0 kJ/mol$ with $Δ n_{g a s} = + 2$ at $T = 500 K$ . Find $Δ H^{\circ}$ .

Recall Solution 3.2

Step 1 — forward formula. $Δ H = Δ U + Δ n_{g a s} R T$ . Step 2 — correction. $Δ n_{g a s} R T = (+ 2) (8.314) (500) = 8314 J = 8.314 kJ$ Step 3 — add. $Δ H^{\circ} = 120.0 + 8.314 = 128.3 kJ/mol$ Here $Δ H > Δ U$ : as new gas appears the system must also push the atmosphere back (expansion work), so it "costs" extra enthalpy.

Answer: $Δ H^{\circ} \approx 128.3 kJ/mol$ .

Exercise 3.3 — The edge case $Δ n_{g a s} = 0$

For $H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)$ at any temperature, decide without any arithmetic how $Δ H$ and $Δ U$ compare.

Recall Solution 3.3

Step 1 — count gas moles. $Δ n_{g a s} = HCl 2 - H_{2} + Cl_{2} (1 + 1) = 0.$ Step 2 — read the correction. If $Δ n_{g a s} = 0$ then $Δ (P V) = Δ n_{g a s} R T = 0$ , no matter what $T$ is. The piston in the figure does not move — same number of gas molecules before and after. Step 3 — conclude. $Δ H = Δ U + 0 \Rightarrow Δ H = Δ U .$ Answer: They are exactly equal — recognise this the instant gas moles balance, and skip the calculation. (The same holds for reactions among only solids/liquids, since those never enter $Δ n_{g a s}$ .)

Level 4 — Synthesis

Combine reactions (Hess) and formation data, then correct to $Δ U$ .

Exercise 4.1 — Hess's Law assembly

Given $C (s) + O_{2} (g) \to CO_{2} (g) Δ H_{1} = - 393.5 kJ/mol$ $CO (g) + \frac{1}{2} O_{2} (g) \to CO_{2} (g) Δ H_{2} = - 283.0 kJ/mol$ Find $Δ H$ for $C (s) + \frac{1}{2} O_{2} (g) \to CO (g)$ .

The figure below draws these three reactions as an energy staircase: because enthalpy is a state function, the two routes from $C + O_{2}$ down to $CO_{2}$ must arrive at the same floor, which pins down the missing step.

Recall Solution 4.1

Idea (Hess's Law): enthalpy is a state function, so we may add reactions like algebra. Step 1 — target. We want $C + \frac{1}{2} O_{2} \to CO$ . Step 2 — arrange. Take reaction 1 forward and reaction 2 reversed (flip $\Rightarrow$ flip sign): $C (s) + O_{2} \to CO_{2} (- 393.5)$ $CO_{2} \to CO + \frac{1}{2} O_{2} (+ 283.0)$ Step 3 — add. $CO_{2}$ cancels, one $\frac{1}{2} O_{2}$ cancels: $Δ H = - 393.5 + 283.0 = - 110.5 kJ/mol$

Answer: $Δ H = - 110.5 kJ/mol$ .

Exercise 4.2 — From formation enthalpies to $Δ U$

For $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l)$ use (Standard Enthalpy of Formation): $Δ H_{f}^{\circ} [CH_{4} (g)] = - 74.8$ , $Δ H_{f}^{\circ} [CO_{2} (g)] = - 393.5$ , $Δ H_{f}^{\circ} [H_{2} O (l)] = - 285.8$ kJ/mol, $Δ H_{f}^{\circ} [O_{2}] = 0$ . Find $Δ H_{r x n}^{\circ}$ , then $Δ U^{\circ}$ at $298 K$ .

Recall Solution 4.2

Step 1 — products minus reactants. $Δ H_{r x n}^{\circ} = [(- 393.5) + 2 (- 285.8)] - [(- 74.8) + 2 (0)]$ $= (- 965.1) - (- 74.8) = - 890.3 kJ/mol$ Step 2 — gas count. Water is liquid. $Δ n_{g a s} = 1 - (1 + 2) = - 2 mol$ Step 3 — correction. $Δ n_{g a s} R T = (- 2) (8.314) (298) = - 4955.1 J = - 4.96 kJ$ Step 4 — $Δ U$ . $Δ U^{\circ} = Δ H^{\circ} - Δ n_{g a s} R T = - 890.3 - (- 4.96) = - 885.3 kJ/mol$

Answer: $Δ H^{\circ} = - 890.3 kJ/mol$ , $Δ U^{\circ} \approx - 885.3 kJ/mol$ .

Level 5 — Mastery

Multi-step reasoning where $Δ H$ meets spontaneity and calorimetry.

Exercise 5.1 — Crossover temperature

For $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$ : $Δ H^{\circ} = + 178.0 kJ/mol$ and $Δ S^{\circ} = + 160.0 J K^{- 1} mol^{- 1}$ . Above what temperature does it become spontaneous ( $Δ G < 0$ )?

Recall Solution 5.1

Idea (Entropy and Gibs Free Energy): $Δ G = Δ H - T Δ S$ . Spontaneity turns on when $Δ G = 0$ . Step 1 — set $Δ G = 0$ and solve for $T$ . (Match units: $178.0 kJ = 178000 J$ .) $T = \frac{Δ H}{Δ S} = \frac{178000}{160.0} = 1112.5 K$ Step 2 — interpret. Both $Δ H$ and $Δ S$ are positive, so heat helps only when $T$ is large enough that $T Δ S$ beats $Δ H$ . That happens above $\approx 1112 K$ ( $\approx 839^{\circ} C$ ).

Answer: Spontaneous above $\approx 1113 K$ .

Exercise 5.2 — Bomb $\to$ coffee-cup conversion

Benzoic acid burned in a bomb calorimeter (constant $V$ ) gives $Δ U = - 3226.7 kJ/mol$ : $C_{7} H_{6} O_{2} (s) + \frac{15}{2} O_{2} (g) \to 7 CO_{2} (g) + 3 H_{2} O (l)$ Convert to the constant-pressure $Δ H$ at $298 K$ .

Recall Solution 5.2

The bomb reads $q_{V} = Δ U$ ; the coffee cup would read $q_{P} = Δ H$ . To bridge them we add the $Δ (P V)$ term. Step 1 — gas count. Solid acid and liquid water contribute nothing. $Δ n_{g a s} = 7 - \frac{15}{2} = - 0.5 mol$ Step 2 — correction. $Δ n_{g a s} R T = (- 0.5) (8.314) (298) = - 1238.8 J = - 1.24 kJ$ Step 3 — $Δ H$ . $Δ H = Δ U + Δ n_{g a s} R T = - 3226.7 + (- 1.24) = - 3227.9 kJ/mol$ A tiny correction here — but for reactions with large $∣Δ n_{g a s} ∣$ it can be tens of kJ, so never skip it in (Calorimetry).

Answer: $Δ H \approx - 3227.9 kJ/mol$ .

Exercise 5.3 — Cross-check with bond enthalpies

Estimate $Δ H$ for $H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)$ using bond enthalpies (Bond Enthalpies): $B (H-H) = 436$ , $B (Cl-Cl) = 242$ , $B (H-Cl) = 431$ kJ/mol.

Recall Solution 5.3

Idea: $Δ H \approx (bonds broken) - (bonds formed)$ . Breaking costs energy ( $+$ ), forming releases it ( $-$ ). Step 1 — bonds broken (reactants). One H–H and one Cl–Cl: $436 + 242 = 678 kJ$ Step 2 — bonds formed (products). Two H–Cl: $2 \times 431 = 862 kJ$ Step 3 — subtract. $Δ H \approx 678 - 862 = - 184 kJ/mol$ Exothermic — stronger bonds formed than broken. (Note this is the same reaction as Exercise 3.3, where $Δ n_{g a s} = 0$ , so this $Δ H$ equals $Δ U$ .)

Answer: $Δ H \approx - 184 kJ/mol$ .

Recall One-line self-quiz

Constant P measures which quantity? ::: $q_{P} = Δ H$ (heat at constant pressure). Constant V measures which quantity? ::: $q_{V} = Δ U$ (heat at constant volume). The bridge between them? ::: $Δ H = Δ U + Δ n_{g a s} R T$ (gases only). When is $Δ H = Δ U$ with no arithmetic? ::: When $Δ n_{g a s} = 0$ (gas moles balance).