Intuition What this page is forThe parent note ΔH°f gave you the definition and the master formula. Here we stress-test that formula against every situation an exam (or reality) can throw at you: exothermic and endothermic, positive and negative formation values, zeros and non-obvious "elements", back-solving for an unknown, fractional coefficients, allotropes, and a real-world word problem. If a case exists, it lives in the matrix below and gets its own worked example.
We lean on one tool the whole way — the master formula built in the parent note:
Here is every case class this topic can present. Each row names a trap or twist; the last column tells you which worked example nails it.
#
Case class
What makes it tricky
Example
A
Exothermic combustion, all values known
pure plug-in, sign should be negative
Ex 1
B
Endothermic reaction, positive answer
sign flips; must trust the arithmetic
Ex 2
C
An "element" that is not zero (allotrope)
C ( diamond ) = 0 Ex 3
D
Back-solve for an unknown Δ H f ∘
algebra, isolate the mystery term
Ex 4
E
Fractional coefficients (formation of 1 mol)
must produce exactly 1 mol
Ex 5
F
Zero / degenerate input (Δ H f ∘ = 0
knowing what silently drops out
Ex 6
G
Real-world word problem (heat for a mass, not a mole)
convert grams → moles, scale Δ H
Ex 7
H
Exam twist: combine with Standard enthalpy of combustion
use Δ H c ∘ Δ H f ∘
Ex 8
A shared data table (all in kJ/mol, at 298.15 K, 1 bar) — used across the examples:
Species
Δ H f ∘ Species
Δ H f ∘
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 ( diamond ) + 1.9 C ( graphite ) 0
O 2 ( g ) , N 2 ( g ) , H 2 ( g ) 0 N H 3 ( g ) − 46.1
C 2 H 2 ( g ) + 226.7 C 3 H 8 ( g ) − 103.8
Worked example Burn methaneC H 4 ( g ) + 2 O 2 ( g ) → C O 2 ( g ) + 2 H 2 O ( ℓ ) Δ H r x n ∘
Forecast: combustion releases heat, so guess a large negative number (hundreds of kJ). Write your guess before reading on.
Sum the products, each × its coefficient.
Σ prod = 1 ( − 393.5 ) + 2 ( − 285.8 ) = − 965.1 Why this step? The master formula weights every Δ H f ∘ 2 × Sum the reactants. O 2 0 Σ react = 1 ( − 74.8 ) + 2 ( 0 ) = − 74.8 Why this step? We must "refund" the cost of what we tore apart.Products minus reactants.
Δ H r x n ∘ = − 965.1 − ( − 74.8 ) = − 890.3 kJ Why this step? P − R is the formula; subtracting a negative adds the refund back.
Verify: sign is negative → exothermic ✓ (matches "combustion gets warm"). Units: kJ per mole of C H 4
Worked example Decompose water vapour into its elements... partly2 H 2 O ( g ) → 2 H 2 ( g ) + O 2 ( g ) Δ H r x n ∘
Forecast: we are tearing water apart — that is the reverse of forming it, which was exothermic. Reversing an exothermic step should cost energy → guess positive .
Products: all elements in standard states → 2 ( 0 ) + 1 ( 0 ) = 0 Why this step? By definition Δ H f ∘ H 2 O 2 Reactants: 2 H 2 O ( g ) gas value − 241.8 Σ react = 2 ( − 241.8 ) = − 483.6 Why this step? The state matters — vapour, not liquid, so we pick − 241.8 − 285.8 P − R:
Δ H r x n ∘ = 0 − ( − 483.6 ) = + 483.6 kJ
Verify: positive ✓ = endothermic, exactly the "reverse of formation" logic. Cross-check: forming 2 H 2 O ( g ) − 483.6 + 483.6
Worked example The diamond trapC ( diamond ) → C ( graphite ) Δ H r x n ∘
Forecast: both sides are "just carbon", so a lazy reader says 0 standard state (graphite) is zero. Guess: negative (diamond sits higher, sliding to graphite releases energy).
Product is graphite = standard state → Δ H f ∘ = 0 Why this step? Graphite is the most stable form of carbon at 1 bar.Reactant is diamond, a non-standard allotrope → Δ H f ∘ = + 1.9 Why this step? Diamond is not the reference form, so it carries a real, nonzero price.P − R:
Δ H r x n ∘ = 0 − ( + 1.9 ) = − 1.9 kJ
Verify: negative ✓ — diamond → graphite is (extremely slowly) downhill, matching the Enthalpy H and ΔH "height above sea level" picture. See the energy-ladder figure below.
Δ H f ∘ [ C 3 H 8 ] C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( ℓ ) , Δ H r x n ∘ = − 2220.0 kJ Δ H f ∘ [ C 3 H 8 ]
Forecast: propane is a stable fuel, so its formation value should be a modest negative (tens of kJ). Guess before solving.
Write the master formula with the unknown x
− 2220.0 = [ 3 ( − 393.5 ) + 4 ( − 285.8 ) ] − [ x + 5 ( 0 ) ] Why this step? The one equation contains one unknown — algebra will free it.Compute the product sum.
3 ( − 393.5 ) + 4 ( − 285.8 ) = − 1180.5 − 1143.2 = − 2323.7 Isolate x
− 2220.0 = − 2323.7 − x ⇒ x = − 2323.7 + 2220.0 = − 103.7 kJ/mol Why this step? Add x − 2220
Verify: − 103.7 − 103.8 Standard enthalpy of reaction ΔH°rxn ).
Worked example Write and evaluate the formation reaction of ammoniaGive the formation equation for N H 3 ( g ) Δ H ∘ Δ H f ∘ [ N H 3 ]
Forecast: a formation reaction makes exactly 1 mole of the compound from elements, so expect fractions on N 2 H 2 − 46.1
Balance for 1 mol N H 3 One N H 3 2 1 N 2 ( g ) + 2 3 H 2 ( g ) → N H 3 ( g ) Why this step? Definition demands 1 mol product; half an N 2 H 2 Apply the master formula. Reactants are elements → 0 Δ H ∘ = 1 ( − 46.1 ) − [ 2 1 ( 0 ) + 2 3 ( 0 ) ] = − 46.1 kJ Why this step? When products are a single compound from pure elements, Δ H r x n ∘ is Δ H f ∘
Verify: result = − 46.1 Δ H f ∘ [ N H 3 ] reactants are legal and required here ✓.
Worked example A reaction where almost everything is zeroFormation of ozone from oxygen:
2 3 O 2 ( g ) → O 3 ( g ) , Δ H f ∘ [ O 3 ( g )] = + 142.7 kJ/mol Δ H r x n ∘
Forecast: the reactant O 2 be the ozone value, positive .
Reactant sum: 2 3 O 2 2 3 ( 0 ) = 0 Why this step? O 2 Product sum: 1 × O 3 = + 142.7 P − R:
Δ H r x n ∘ = 142.7 − 0 = + 142.7 kJ
Verify: equals the given Δ H f ∘ [ O 3 ] O 2 O 3 not a standard state.
Worked example Heating your morning teaYour stove burns methane. How much heat is released by burning 8.0 g of C H 4 − 890.3
Forecast: 8 g of C H 4 half a mole , giving roughly half of 890, i.e. ~ − 445
Convert grams to moles. Molar mass C H 4 = 12.0 + 4 ( 1.0 ) = 16.0 n = 16.0 g/mol 8.0 g = 0.50 mol Why this step? Δ H ∘ mole of reaction; a mass must first become moles.Scale the per-mole enthalpy.
q = n × Δ H ∘ = 0.50 × ( − 890.3 ) = − 445.15 kJ Why this step? Enthalpy is extensive — double the substance, double the heat.
Verify: − 445.15
Δ H f ∘ combustion of acetylene is
C 2 H 2 ( g ) + 2 5 O 2 ( g ) → 2 C O 2 ( g ) + H 2 O ( ℓ ) , Δ H c ∘ = − 1299.6 kJ/mol Δ H f ∘ [ C O 2 ] = − 393.5 Δ H f ∘ [ H 2 O ( ℓ )] = − 285.8 Δ H f ∘ [ C 2 H 2 ]
Forecast: acetylene stores lots of energy (used in welding torches), so unlike most fuels its Δ H f ∘ positive — guess ~ +200 kJ.
Master formula, unknown y = Δ H f ∘ [ C 2 H 2 ]
− 1299.6 = [ 2 ( − 393.5 ) + 1 ( − 285.8 ) ] − [ y + 2 5 ( 0 ) ] Why this step? Combustion enthalpy is just a reaction enthalpy; feed it into the same formula.Product sum.
2 ( − 393.5 ) + ( − 285.8 ) = − 787.0 − 285.8 = − 1072.8 Isolate y
− 1299.6 = − 1072.8 − y ⇒ y = − 1072.8 + 1299.6 = + 226.8 kJ/mol Why this step? Rearranged exactly as in Ex 4; one unknown, one equation.
Verify: + 226.8 + 226.7 positive as forecast — acetylene is a genuinely high-energy molecule ✓. This is the Standard enthalpy of combustion "back-out" trick in action.
Recall Which cell am I in? (self-quiz)A combustion with all data given ::: Case A — pure plug-in, expect negative.
The answer came out positive ::: Case B — endothermic, or a formation of a high-energy species.
The problem mentions diamond, ozone, or white phosphorus ::: Case C/F — allotrope, the "element" is NOT zero.
One Δ H f ∘ Δ H c ∘ Δ H r x n ∘
Mnemonic The universal move"Products Pay, Reactants Refunded, Elements Free — then check the sign against exo/endo."
Parent: ΔH°f — definition and master-formula derivation.Hess's Law — why the two-step-through-elements path is valid.Standard enthalpy of reaction ΔH°rxn — the direct output here.Standard enthalpy of combustion — Case H uses it to back out Δ H f ∘ Bond enthalpies — a rougher alternative estimate.State functions vs path functions — the foundation for path independence.Enthalpy H and ΔH — the "only changes are measurable" idea behind the zero reference. 