Standard enthalpy of formation ΔH°f: Exercises — Chemistry deep dive

Throughout, the master formula we lean on is $Δ H_{r x n}^{\circ} = \sum_{products} ν Δ H_{f}^{\circ} - \sum_{reactants} ν Δ H_{f}^{\circ}$ where $ν$ (the Greek letter "nu") is the stoichiometric coefficient — the number written in front of a formula in a balanced equation, telling you how many moles take part. This formula is justified by Hess's Law, because enthalpy is a state function.

A shared data table (all values in $kJ/mol$ , at $298.15 K$ , $1 bar$ ) that the exercises draw from:

Level 1 — Recognition

L1.1

Which of these has $Δ H_{f}^{\circ} = 0$ by definition? (a) $O_{3} (g)$ (b) $O_{2} (g)$ (c) $C (diamond)$ (d) $H_{2} O (ℓ)$

Recall Solution

What we check: is the substance an element in its most stable form at $1 bar$ , $298.15 K$ ? Only that gets a zero.

(a) $O_{3}$ — ozone is an element but not the most stable form of oxygen → not zero.
(b) $O_{2}$ — oxygen's standard state → $0$ . ✓
(c) diamond — carbon, but graphite is the stable form → not zero ( $+ 1.9 kJ/mol$ ).
(d) water — a compound → not zero.

Answer: (b).

L1.2

Write the balanced formation reaction for $N H_{3} (g)$ , using $N_{2} (g)$ and $H_{2} (g)$ .

Recall Solution

Rule: a formation reaction makes exactly 1 mole of the compound from elements in their standard states. Nitrogen's standard state is $N_{2} (g)$ , hydrogen's is $H_{2} (g)$ .

To make 1 mol $N H_{3}$ we need 1 N atom and 3 H atoms. One $N_{2}$ has 2 N atoms, so we take half of it; one $H_{2}$ has 2 H atoms, so we take $3/2$ of it: $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⟶ N H_{3} (g)$ Fractional coefficients on the elements are required here, because the product coefficient is pinned to exactly 1.

Level 2 — Application

L2.1

Combustion of ethanol: compute $Δ H^{\circ}$ for $C_{2} H_{5} O H (ℓ) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 3 H_{2} O (ℓ)$

Recall Solution

Master formula, products − reactants, each weighted by $ν$ : $Δ H^{\circ} = [2Δ H_{f}^{\circ} (C O_{2}) + 3Δ H_{f}^{\circ} (H_{2} O (ℓ))] - [Δ H_{f}^{\circ} (C_{2} H_{5} O H) + 3 = 0 Δ H_{f}^{\circ} (O_{2})]$ Substitute: $= [2 (- 393.5) + 3 (- 285.8)] - [(- 277.7) + 0]$ $= (- 787.0 - 857.4) - (- 277.7) = - 1644.4 + 277.7 = - 1366.7 kJ$ Negative → exothermic, exactly what we expect for burning fuel. ✓

L2.2

Formation of water vapour vs liquid. Given $Δ H_{f}^{\circ} [H_{2} O (g)] = - 241.8$ and $Δ H_{f}^{\circ} [H_{2} O (ℓ)] = - 285.8 kJ/mol$ , find the enthalpy of condensation: $H_{2} O (g) ⟶ H_{2} O (ℓ)$

Recall Solution

Treat the two waters as "product" and "reactant" in the master formula: $Δ H^{\circ} = Δ H_{f}^{\circ} [H_{2} O (ℓ)] - Δ H_{f}^{\circ} [H_{2} O (g)] = (- 285.8) - (- 241.8) = - 44.0 kJ$ What it means: gas → liquid releases $44.0 kJ$ per mole. Condensation gives off heat (that's why steam scalds worse than boiling water). ✓

Figure — Standard enthalpy of formation ΔH°f

Level 3 — Analysis

L3.1

Back-solve for an unknown. For $4 N H_{3} (g) + 5 O_{2} (g) ⟶ 4 N O (g) + 6 H_{2} O (g), Δ H^{\circ} = - 905.4 kJ$ verify this is consistent with the table (treat $Δ H_{f}^{\circ} [N O]$ as unknown $x$ and solve for it).

Recall Solution

Set up master formula with $x = Δ H_{f}^{\circ} [N O]$ : $Δ H^{\circ} = [4 x + 6 (- 241.8)] - [4 (- 46.1) + 5 (0)]$ $- 905.4 = 4 x - 1450.8 + 184.4$ $- 905.4 = 4 x - 1266.4$ Why isolate $x$ : it's the only unknown, so ordinary algebra frees it. $4 x = - 905.4 + 1266.4 = 361.0 \Rightarrow x = + 90.25 kJ/mol$ This matches the table value $+ 90.3$ (rounding). Note $x > 0$ : making NO from the elements costs energy — it's endothermic, which is why NO forms only in the hot flame. ✓

L3.2

Reverse and scale. The thermite reaction is $2 A l (s) + F e_{2} O_{3} (s) ⟶ A l_{2} O_{3} (s) + 2 F e (s)$ Find $Δ H^{\circ}$ .

Recall Solution

$A l (s)$ and $F e (s)$ are elements in standard states → their $Δ H_{f}^{\circ} = 0$ . $Δ H^{\circ} = [Δ H_{f}^{\circ} (A l_{2} O_{3}) + 2 \cdot 0] - [2 \cdot 0 + Δ H_{f}^{\circ} (F e_{2} O_{3})]$ $= (- 1675.7) - (- 824.2) = - 851.5 kJ$ Massively exothermic — this is the reaction that melts iron to weld railway tracks. ✓

Level 4 — Synthesis

L4.1

Combine combustion data to get a formation enthalpy (Hess-style). You are told $C (s) + O_{2} (g) \to C O_{2} (g), Δ H_{1}^{\circ} = - 393.5 kJ$ $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ), Δ H_{2}^{\circ} = - 285.8 kJ$ $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (ℓ), Δ H_{3}^{\circ} = - 890.3 kJ$ Find $Δ H_{f}^{\circ} [C H_{4}]$ , i.e. $Δ H^{\circ}$ for $C (s) + 2 H_{2} (g) \to C H_{4} (g)$ .

Recall Solution

Strategy: algebraically combine the three equations so everything cancels except the target. We want $C H_{4}$ as a product, so we must reverse equation 3 (flip its sign). We keep eqn 1 as is (gives us $C$ consumed) and take 2× eqn 2 (we need $2 H_{2}$ consumed).

Eqn 1: $C + O_{2} \to C O_{2}$ , $Δ H = - 393.5$
$2 \times$ Eqn 2: $2 H_{2} + O_{2} \to 2 H_{2} O$ , $Δ H = 2 (- 285.8) = - 571.6$
Reverse Eqn 3: $C O_{2} + 2 H_{2} O \to C H_{4} + 2 O_{2}$ , $Δ H = + 890.3$

Add them. Check the cancellations: $C O_{2}$ appears as product (eqn 1) and reactant (rev-3) → gone. $2 H_{2} O$ produced (2×eqn 2) and consumed (rev-3) → gone. $O_{2}$ : $+ 1 + 1 - 2 = 0$ → gone. Net left: $C + 2 H_{2} \to C H_{4}$ . ✓

$Δ H_{f}^{\circ} [C H_{4}] = - 393.5 - 571.6 + 890.3 = - 74.8 kJ/mol$ Matches the table. This is exactly Hess's Law doing the heavy lifting. ✓

L4.2

Fuel comparison per gram. Using $Δ H_{f}^{\circ} [C_{3} H_{8} (g)] = - 104.7$ , compute the combustion enthalpy of propane $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (ℓ)$ then find the energy released per gram (molar mass of $C_{3} H_{8} = 44.1 g/mol$ ).

Recall Solution

Step 1 — reaction enthalpy (master formula): $Δ H^{\circ} = [3 (- 393.5) + 4 (- 285.8)] - [(- 104.7) + 0]$ $= (- 1180.5 - 1143.2) + 104.7 = - 2323.7 + 104.7 = - 2219.0 kJ$ Step 2 — per gram: divide the per-mole release by the molar mass. $\frac{- 2219.0 kJ}{44.1 g} = - 50.3 kJ/g$ The negative sign says energy leaves the system; the magnitude $50.3 kJ/g$ is why propane is a great portable fuel. ✓

Level 5 — Mastery

L5.1

Everything at once — a designed trap. Ozone decomposes: $2 O_{3} (g) ⟶ 3 O_{2} (g), Δ H^{\circ} = - 285.4 kJ$ Find $Δ H_{f}^{\circ} [O_{3} (g)]$ .

Recall Solution

The trap it tests: $O_{2}$ is the element's standard state ( $Δ H_{f}^{\circ} = 0$ ), but $O_{3}$ is not — so $O_{3}$ carries an unknown $Δ H_{f}^{\circ} = x$ . $Δ H^{\circ} = [3 0 Δ H_{f}^{\circ} (O_{2})] - [2 x] = - 2 x$ $- 285.4 = - 2 x \Rightarrow x = + 142.7 kJ/mol$ Positive: ozone sits above ordinary oxygen on our enthalpy "altitude" — it stores energy relative to $O_{2}$ , which is why it's reactive and decomposes exothermically. ✓

L5.2

Consistency across two data sets. From combustion data, the standard enthalpy of combustion of ethanol is $Δ H_{c}^{\circ} [C_{2} H_{5} O H] = - 1366.7 kJ/mol$ (the L2.1 reaction). Given $Δ H_{f}^{\circ} [C O_{2}] = - 393.5$ and $Δ H_{f}^{\circ} [H_{2} O (ℓ)] = - 285.8$ , recover $Δ H_{f}^{\circ} [C_{2} H_{5} O H]$ — proving the two viewpoints agree.

Recall Solution

Combustion reaction: $C_{2} H_{5} O H (ℓ) + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O (ℓ)$ . Set $y = Δ H_{f}^{\circ} [C_{2} H_{5} O H]$ : $Δ H_{c}^{\circ} = [2 (- 393.5) + 3 (- 285.8)] - [y + 0]$ $- 1366.7 = (- 787.0 - 857.4) - y = - 1644.4 - y$ $y = - 1644.4 - (- 1366.7) = - 1644.4 + 1366.7 = - 277.7 kJ/mol$ Exactly the table value — the formation and combustion routes are two faces of the same Hess's Law coin. See Standard enthalpy of combustion for this back-calculation as a general technique. ✓

L5.3

Degenerate / limiting check. Without a calculator, state $Δ H^{\circ}$ for $N_{2} (g) + O_{2} (g) \to N_{2} (g) + O_{2} (g) (a "do-nothing" reaction)$ and for the pure phase change $C (graphite) \to C (graphite)$ . Explain using state-function logic.

Recall Solution

Both are $0$ . Start state = end state, and enthalpy is a state function (State functions vs path functions), so $Δ H$ depends only on endpoints. Identical endpoints ⇒ zero change. Equivalently, master formula: every species is an element in standard state ( $Δ H_{f}^{\circ} = 0$ ), and products − reactants $= 0 - 0 = 0$ . This is the sanity anchor for the whole framework: if a reaction changes nothing, it must cost nothing. ✓

Active recall

Recall Quick self-test

When is an element's $Δ H_{f}^{\circ}$ zero? ::: Only when it is in its most stable form (standard state) at 1 bar and the stated temperature. In L2.1 why does the $O_{2}$ term vanish? ::: $O_{2} (g)$ is oxygen's standard state, so $Δ H_{f}^{\circ} = 0$ . In L4.1 what did we do to equation 3? ::: Reversed it, flipping the sign of its $Δ H$ from $- 890.3$ to $+ 890.3$ . Why is $Δ H_{f}^{\circ} [O_{3}]$ positive? ::: Ozone stores energy above $O_{2}$ ; forming it from $O_{2}$ costs energy. What guarantees a do-nothing reaction has $Δ H = 0$ ? ::: Enthalpy is a state function — identical start/end states give zero change.

Connections

Standard enthalpy of formation ΔH°f — parent note with the definition and master-formula derivation.
Hess's Law — the engine behind L4's equation-combining.
Standard enthalpy of reaction ΔH°rxn — every answer here is one.
Standard enthalpy of combustion — L2.1, L4.2 and L5.2 are combustion problems.
State functions vs path functions — justifies L5.3.
Enthalpy H and ΔH — why only differences matter.

Substance	$Δ H_{f}^{\circ}$	Substance	$Δ H_{f}^{\circ}$
$C H_{4} (g)$	$- 74.8$	$C O_{2} (g)$	$- 393.5$
$H_{2} O (ℓ)$	$- 285.8$	$H_{2} O (g)$	$- 241.8$
$C_{2} H_{5} O H (ℓ)$	$- 277.7$	$C O (g)$	$- 110.5$
$N H_{3} (g)$	$- 46.1$	$N O (g)$	$+ 90.3$
$C_{3} H_{8} (g)$	$- 104.7$	$S O_{2} (g)$	$- 296.8$
$F e_{2} O_{3} (s)$	$- 824.2$	$A l_{2} O_{3} (s)$	$- 1675.7$
any element in standard state	$0$