Worked examples — Hess's law — enthalpy is a state function; enthalpy cycles

This page is a complete drill. We built the machinery in the parent note; here we hit every kind of problem the topic can throw at you, so no exam surprise is ever new.

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The scenario matrix

Cell	Case class	What's tricky	Hit by
A	Reverse + add (no scaling)	one step flips sign	Ex 1
B	Scale + reverse together	multiply and flip $\Delta H$	Ex 2
C	Formation data (${\Delta }_f{H}^{\circ }$)	products − reactants, elements = 0	Ex 3
D	Combustion data (${\Delta }_c{H}^{\circ }$)	reactants − products (sign role swaps!)	Ex 4
E	Hidden phase change	$H_{2}O(l)$ vs $(g)$ costs vaporization	Ex 5
F	Zero / degenerate input	element or identity reaction, $\Delta H=0$	Ex 6
G	Multi-step Born–Haber-style cycle	5 legs, one unknown	Ex 7
H	Real-world word problem	translate words → equations first	Ex 8
I	Exam twist: given "per gram" / limiting reagent	unit + amount conversion	Ex 9

Every cell A–I gets its own worked example below.

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Ex 1 — Cell A: Reverse and add (no scaling)

Forecast: guess — will the answer be positive or negative? (Both given steps are positive; but we must reverse one, so hold that guess.)

1. Orient (ii). Our target has $N{O}_2$ as a product, but (ii) has it as a reactant. Reverse (ii): $2NO+O_2\to 2N{O}_2$, $\Delta H=-114\text{kJ}$. Why this step? Rule 1 — flip the arrow, flip the sign, so $N{O}_2$ lands on the correct side.
2. Add (i) + reversed (ii). $N_2+O_2\to 2NO(+180)$ $2NO+O_2\to 2N{O}_2(-114)$ Why this step? $NO$ appears as a product in (i) and reactant in reversed-(ii) — same species, same state (both gas), so it cancels (telescoping).
3. Collect. $N_2+2O_2\to 2N{O}_2$, $\Delta H=180-114=+66\text{kJ}$.

$\boxed{\Delta H=+66\text{kJ}}$

Verify: the $O_2$ tally on the left is $1+1=2$, matching the target's $2O_2$. $NO$ ($+2$ then $-2$) cancels cleanly. Units kJ throughout. Positive answer — nitrogen oxides are endothermic to form, chemically sensible.

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Ex 2 — Cell B: Scale AND reverse together

Forecast: the target is exactly twice step (i). Guess the answer before reading on.

1. Compare stoichiometry. Target has $N_2$ (coefficient 1) and $2N{H}_3$; step (i) has $\frac{1}{2}{N}_2$ and $1N{H}_3$. Target $=2\times$ (i). Why this step? You must scale so species and coefficients match the target before adding.
2. Scale $\Delta H$ by 2. Rule 2 — enthalpy is extensive: $\Delta H=2\times (-46)=-92\text{kJ}$. Why this step? Double the amount of stuff reacting ⟹ double the heat released.

$\boxed{\Delta H=-92\text{kJ}}$

Verify: $2\times (\frac{1}{2}{N}_2)=N_2$ ✓, $2\times (\frac{3}{2}{H}_2)=3H_2$ ✓, $2\times N{H}_3=2N{H}_3$ ✓. Exothermic (matches the real Haber process, $\approx -92\text{kJ}$).

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Ex 3 — Cell C: From formation enthalpies

Forecast: combustion of a smelly gas — expect a large negative number.

1. Write the master formula. ${\Delta }_r{H}^{\circ }=\sum {\nu }_p{\Delta }_f{H}_{\text{prod}}^{\circ }-\sum {\nu }_r{\Delta }_f{H}_{\text{react}}^{\circ }$ Why this step? This is the "reactants → elements → products" Hess cycle collapsed into one line (see Standard states and conventions).
2. Products side: $2(-296.8)+2(-285.8)=-593.6-571.6=-1165.2$. Why coefficient 2? Extensive property × stoichiometric coefficient.
3. Reactants side: $2(-20.6)+3(0)=-41.2$. ($O_2$ is an element ⟹ ${\Delta }_f{H}^{\circ }=0$.) Why zero for $O_2$? Forming an element from itself is no change.
4. Subtract: ${\Delta }_r{H}^{\circ }=-1165.2-(-41.2)=-1124.0\text{kJ}$.

$\boxed{{\Delta }_r{H}^{\circ }=-1124.0\text{kJ}}$

Verify: Atom balance: S ($2=2$), H ($4=4$), O (left $6$, right $4+2=6$) ✓. Sign negative ✓ (combustion releases heat).

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Ex 4 — Cell D: From combustion enthalpies (the sign role SWAPS)

Forecast: hydrogenation is mildly exothermic — expect a small negative number.

1. Realise combustion flips the roles. With formation data we do (products − reactants). With combustion data it's the opposite: (reactants − products). Why this step? ${\Delta }_c{H}^{\circ }$ measures burning a species to its combustion products. A reactant's combustion is a forward leg of the cycle; a product's combustion must be reversed — so products get the minus sign flipped relative to the formation formula. ${\Delta }_r{H}^{\circ }=\sum {\nu }_r{\Delta }_c{H}_{\text{react}}^{\circ }-\sum {\nu }_p{\Delta }_c{H}_{\text{prod}}^{\circ }$
2. Reactants side: ${\Delta }_c(C_2{H}_4)+{\Delta }_c(H_2)=-1411+(-286)=-1697$. Why this step? Both reactants get burned in the cycle's first leg.
3. Products side: ${\Delta }_c(C_2{H}_6)=-1560$.
4. Subtract: ${\Delta }_r{H}^{\circ }=-1697-(-1560)=-137\text{kJ}$.

$\boxed{{\Delta }_r{H}^{\circ }=-137\text{kJ}}$

Verify: Small exothermic value (real hydrogenation of ethene $\approx -137\text{kJ}$) ✓. See Enthalpy of combustion for why the sign convention reverses versus formation data.

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Ex 5 — Cell E: A hidden phase change

Forecast: making the product gas costs extra energy, so the reaction releases less heat — the number should become less negative.

1. Chain the two processes. $C{H}_4+2O_2\to C{O}_2+2H_{2}O(l)\to C{O}_2+2H_{2}O(g)$. Why this step? Reaching gaseous water = the liquid-water reaction, then vaporize both moles. Look at the figure: the pale-mint arrow (combustion) drops far down; the coral arrow (vaporization) climbs back up.
2. Vaporize 2 moles. $2\times (+44.0)=+88.0\text{kJ}$. Why coefficient 2? Two moles of water are formed; extensive scaling.
3. Add legs (telescoping cycle). $\Delta H=-890.3+88.0=-802.3\text{kJ}$.

$\boxed{\Delta H=-802.3\text{kJ}}$

Verify: Less negative than $-890.3$ ✓ (steam carries away hidden heat). This is exactly the lower vs higher heating value distinction of fuels. Phases kept explicit throughout.

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Ex 6 — Cell F: Zero / degenerate inputs

Forecast: all three are traps that should give exactly zero.

1. (a) $O_2$ is an element in its standard state. ${\Delta }_f{H}^{\circ }=0$. Why this step? Forming a thing from itself — identical initial and final state ⟹ $\Delta H=H_{\text{final}}-H_{\text{initial}}=0$.
2. (b) Same species, same phase on both sides. $\Delta H=0$. Why this step? A null cycle; endpoints coincide.
3. (c) Bromine's standard state is $B{r}_2(l)$ (not gas!). So this is again "element from itself." ${\Delta }_f{H}^{\circ }(B{r}_2(l))=0$, and this identity gives $0$. Why the warning? Students often think ${\Delta }_f{H}^{\circ }(B{r}_2)$ is nonzero because bromine is often drawn as a gas — but the standard state fixes it as liquid. See Standard states and conventions.

$\boxed{\Delta H=0\text{ in all three cases}}$

Verify: In every case $H_{\text{final}}=H_{\text{initial}}$, so the difference is $0$ by definition of a state function. Degenerate-input check passes.

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Ex 7 — Cell G: Multi-step cycle with one unknown (Born–Haber flavour)

Forecast: lattice formation (gaseous ions → solid) is strongly exothermic — expect a large negative number.

1. State the cycle equation. Going from elements to $NaCl(s)$ by the long route (1→2→3→4→ then lattice) must equal the direct route (5): ${\Delta }_f{H}^{\circ }=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_L$ Why this step? Both routes share the same start (elements) and end ($NaCl(s)$); Hess ⟹ equal totals. Follow the arrows in the figure around the box.
2. Insert numbers. $-411=108+121+496+(-349)+\Delta H_L$
3. Sum the knowns: $108+121+496-349=376$. Why this step? Collapse all known legs into one number.
4. Solve: $\Delta H_L=-411-376=-787\text{kJ/mol}$.

$\boxed{\Delta H_L=-787\text{kJ/mol}}$

Verify: Large negative (real $NaCl$ lattice enthalpy $\approx -787\text{kJ/mol}$) ✓. See Born–Haber cycle and Bond enthalpies for each leg's meaning.

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Ex 8 — Cell H: Real-world word problem

Forecast: they need about a quarter of the heat one mole gives, so guess ~$11\text{g}$.

1. Translate words → number. "Releases $2220\text{kJ}$" means ${\Delta }_{c}H=-2220\text{kJ/mol}$. Heat available per mole $=2220\text{kJ}$. Why this step? Word problems must first become clean quantities with signs.
2. Moles needed. $n=\frac{555\text{kJ}}{2220\text{kJ/mol}}=0.25\text{mol}$. Why this step? Enthalpy is extensive — heat scales linearly with moles, so divide.
3. Convert to grams. $m=n\times M=0.25\times 44=11\text{g}$. Why this step? The question asks mass, not moles.

$\boxed{m=11\text{g of propane}}$

Verify: $0.25\text{mol}\times 2220\text{kJ/mol}=555\text{kJ}$ ✓ (matches requirement). Matches the forecast of ~11 g.

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Ex 9 — Cell I: Exam twist — "per gram" heating value

Forecast: hydrogen is famously the best fuel by mass — guess it wins big.

1. Hydrogen per gram. $\frac{286\text{kJ}}{2\text{g}}=143\text{kJ/g}$. Why this step? Divide molar heat by molar mass to convert kJ/mol → kJ/g.
2. Methane per gram. $\frac{890\text{kJ}}{16\text{g}}=55.625\text{kJ/g}$. Why this step? Same conversion, fair comparison.
3. Compare. $143>55.625$; ratio $=143/55.625\approx 2.57$. Why this step? The question asks "by how much," so form the ratio.

$\boxed{{\text{H}}_2\text{ gives }143\text{kJ/g}{\text{ vs CH}}_4\text{’s }55.6\text{kJ/g}(\approx 2.6\times )}$

Verify: Units kJ/g consistent; hydrogen's low molar mass makes it dominate per-mass (matches its use as rocket fuel). Ratio $143/55.625=2.571$ ✓.

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Recall

Recall When do you use (products − reactants) vs (reactants − products)?

Formation data: products − reactants. Combustion data: reactants − products. (Combustion is contrary.)

Recall Why did the steam version of methane combustion give a

less negative $\Delta H$? Because vaporizing the product water costs $+44\text{kJ/mol}$; that energy is subtracted from the heat released, so $-890.3+88=-802.3\text{kJ}$.

Recall In the Born–Haber cycle, how do you isolate one unknown leg?

Set (long path total) = (direct path total) via Hess, plug all known legs, and solve the single remaining unknown algebraically.

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Connections

• Parent: Hess's law
• Enthalpy of formation
• Enthalpy of combustion
• Born–Haber cycle
• Bond enthalpies
• Standard states and conventions
• First law of thermodynamics
• State functions vs path functions

Which formula, which data?