Chapter 5 · Thermodynamics

5.3 Thermochemistry: Enthalpies of Reaction and Hess's Law

Thermochemistry puts numbers on the heat of chemical change: standard enthalpies of reaction, of phase changes and of formation, Hess's law for combining them, enthalpies of combustion, atomisation and bonds, and the Born–Haber cycle that reveals the lattice enthalpy of an ionic crystal.

Nobody has ever measured the enthalpy change for turning graphite into diamond directly — the change is far too slow at ordinary pressures. Nor can anyone measure the heat released when gaseous sodium ions and chloride ions crash together to form a crystal of salt. Yet both values are known to a fraction of a kilojoule. They come from reactions that can be measured — burning graphite and diamond in a bomb calorimeter, forming salt from sodium and chlorine — combined by a single principle: because enthalpy is a state function, the enthalpy change between two states is the same whatever route connects them.

This section builds the toolkit that rests on that principle: thermochemical equations, standard enthalpies of reaction, of phase changes and of formation, Hess's law, enthalpies of combustion and atomisation, bond enthalpies, the Born–Haber cycle for lattice enthalpy, and the enthalpies of solution and dilution. With a short table of formation enthalpies you will be able to predict the heat of thousands of reactions without doing a single experiment.

Standard Enthalpies of Reaction and Phase Changes

The enthalpy change of a reaction depends on the conditions — temperature, pressure, the physical states of the substances. To compare reactions fairly, chemists quote values for a set of agreed standard conditions.

📘 Definition 5.11 — Standard state and standard enthalpy of reaction
The standard state of a substance at a given temperature is its pure form at a pressure of $1\ \mathrm{bar}$ (for a solute, a concentration of $1\ \mathrm{mol\,L^{-1}}$); the temperature is usually $298\ \mathrm{K}$ and is stated. The standard enthalpy of reaction $\Delta_{r}H^{\circ}$ is the enthalpy change when the reaction occurs as written, in molar amounts given by the coefficients, with all reactants and products in their standard states. Its unit is $\mathrm{kJ\,mol^{-1}}$, "per mole of reaction as written". A thermochemical equation is a balanced equation with state symbols together with its $\Delta_{r}H^{\circ}$, for example $$\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)};\quad \Delta_{r}H^{\circ}=-890.3\ \mathrm{kJ\,mol^{-1}}$$
🧭 Rules for thermochemical equations
  1. The coefficients are numbers of moles, so fractions such as $\tfrac12\mathrm{O_{2}(g)}$ are allowed.
  2. State symbols are essential: $\mathrm{H_{2}O(l)}$ and $\mathrm{H_{2}O(g)}$ give different $\Delta_{r}H^{\circ}$ values.
  3. Multiplying the equation by a factor multiplies $\Delta_{r}H^{\circ}$ by the same factor.
  4. Reversing the equation changes the sign of $\Delta_{r}H^{\circ}$ (the law of Lavoisier and Laplace).

Physical changes have enthalpies too. Melting ice, boiling water and the direct conversion of a solid to a vapour all need heat, to overcome the attractions between the particles.

📘 Definition 5.12 — Enthalpies of phase transformation
  • Standard enthalpy of fusion $\Delta_{fus}H^{\circ}$: the enthalpy change when one mole of a solid melts at its melting point at $1\ \mathrm{bar}$. Water: $\mathrm{H_{2}O(s)\rightarrow H_{2}O(l)}$, $+6.01\ \mathrm{kJ\,mol^{-1}}$ at $273\ \mathrm{K}$.
  • Standard enthalpy of vaporisation $\Delta_{vap}H^{\circ}$: one mole of liquid vaporises at its boiling point. Water: $+40.66\ \mathrm{kJ\,mol^{-1}}$ at $373\ \mathrm{K}$.
  • Standard enthalpy of sublimation $\Delta_{sub}H^{\circ}$: one mole of solid turns directly into vapour (iodine, solid $\mathrm{CO_{2}}$, naphthalene). At the same temperature, $\Delta_{sub}H^{\circ}=\Delta_{fus}H^{\circ}+\Delta_{vap}H^{\circ}$.
All three are positive (endothermic); the reverse processes — freezing, condensation, deposition — have the same values with a negative sign. $\Delta_{vap}H$ is much larger than $\Delta_{fus}H$ because vaporisation separates the molecules almost completely, while melting only loosens them.
✏️ Example 1 — Working with thermochemical equations
$\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)}$, $\Delta_{r}H^{\circ}=-571.6\ \mathrm{kJ\,mol^{-1}}$. (a) Find $\Delta_{r}H^{\circ}$ for $\mathrm{H_{2}O(l)\rightarrow H_{2}(g)+\tfrac12O_{2}(g)}$. (b) Find $\Delta_{r}H^{\circ}$ for $\mathrm{4H_{2}(g)+2O_{2}(g)\rightarrow 4H_{2}O(l)}$. (c) How much heat is released at constant pressure when $10.0\ \mathrm{g}$ of hydrogen burns to liquid water? ($M(\mathrm{H_{2}})=2.016\ \mathrm{g\,mol^{-1}}$)
(a) Reverse and halve: $\Delta_{r}H^{\circ}=-\tfrac12(-571.6)=+285.8\ \mathrm{kJ\,mol^{-1}}$.
(b) Double: $2(-571.6)=-1143.2\ \mathrm{kJ\,mol^{-1}}$.
(c) $n(\mathrm{H_{2}})=10.0/2.016=4.960\ \mathrm{mol}$. $-571.6\ \mathrm{kJ}$ goes with $2\ \mathrm{mol}$ of $\mathrm{H_{2}}$, i.e. $285.8\ \mathrm{kJ}$ per mole; heat released $=4.960\times285.8=1.42\times10^{3}\ \mathrm{kJ}$.
✏️ Example 2 — From ice to steam
How much heat is needed to turn $36.04\ \mathrm{g}$ of ice at $0\ ^{\circ}\mathrm{C}$ into steam at $100\ ^{\circ}\mathrm{C}$ at $1\ \mathrm{bar}$? Use $\Delta_{fus}H=6.01\ \mathrm{kJ\,mol^{-1}}$, $\Delta_{vap}H=40.66\ \mathrm{kJ\,mol^{-1}}$ and $C_{p,m}(\mathrm{H_{2}O,l})=75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$. Which step needs the most heat?
Amount $n=36.04/18.02=2.000\ \mathrm{mol}$.
Three steps Melt: $2.000\times6.01=12.02\ \mathrm{kJ}$. Warm the liquid by $100\ \mathrm{K}$: $2.000\times75.3\times100=15.06\ \mathrm{kJ}$. Boil: $2.000\times40.66=81.32\ \mathrm{kJ}$.
Total $12.02+15.06+81.32=108.4\ \mathrm{kJ}$. Enthalpy is a state function, so the steps simply add.
Most heat Vaporisation — three-quarters of the total. This is why a steam burn is so much worse than a burn from water at the same temperature: condensing steam releases $40.66\ \mathrm{kJ\,mol^{-1}}$ on the skin.

Enthalpy of Formation and Hess's Law

📘 Definition 5.13 — Standard enthalpy of formation
The standard enthalpy of formation $\Delta_{f}H^{\circ}$ of a compound is the enthalpy change when one mole of the compound is formed from its elements, each in its most stable form (its reference state) at $1\ \mathrm{bar}$ and the stated temperature. For example $$\mathrm{H_{2}(g)+\tfrac12O_{2}(g)\rightarrow H_{2}O(l)};\quad\Delta_{f}H^{\circ}=-285.8\ \mathrm{kJ\,mol^{-1}}$$ By definition, $\Delta_{f}H^{\circ}=0$ for an element in its reference state: $\mathrm{O_{2}(g)}$, $\mathrm{N_{2}(g)}$, $\mathrm{C(graphite)}$, $\mathrm{Br_{2}(l)}$, $\mathrm{Na(s)}$. Other forms of an element have non-zero values: $\mathrm{C(diamond)}$ $+1.9$, $\mathrm{O_{3}(g)}$ $+142.7\ \mathrm{kJ\,mol^{-1}}$.
Substance$\mathrm{CO_{2}(g)}$$\mathrm{H_{2}O(l)}$$\mathrm{H_{2}O(g)}$$\mathrm{CO(g)}$$\mathrm{CH_{4}(g)}$$\mathrm{C_{2}H_{6}(g)}$$\mathrm{C_{2}H_{4}(g)}$$\mathrm{NH_{3}(g)}$$\mathrm{NO(g)}$$\mathrm{C_{2}H_{5}OH(l)}$$\mathrm{CaCO_{3}(s)}$$\mathrm{CaO(s)}$
$\Delta_{f}H^{\circ}$ / $\mathrm{kJ\,mol^{-1}}$−393.5−285.8−241.8−110.5−74.8−84.7+52.3−46.1+90.3−277.7−1206.9−635.1

Standard enthalpies of formation at $298\ \mathrm{K}$ (rounded literature values). Use them throughout this section.

📐 Rule 5.5 — Enthalpy of reaction from enthalpies of formation
For a reaction $\;a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}$ with all substances in their standard states, $$\Delta_{r}H^{\circ}=\sum_{\text{products}}\nu\,\Delta_{f}H^{\circ}-\sum_{\text{reactants}}\nu\,\Delta_{f}H^{\circ}$$ where $\nu$ are the coefficients: $\Delta_{r}H^{\circ}=[c\,\Delta_{f}H^{\circ}(\mathrm{C})+d\,\Delta_{f}H^{\circ}(\mathrm{D})]-[a\,\Delta_{f}H^{\circ}(\mathrm{A})+b\,\Delta_{f}H^{\circ}(\mathrm{B})]$. Imagine taking the reactants apart into their elements (reverse formation) and building the products from the same elements (formation): the elements cancel.
📐 Law 5.6 — Hess's law of constant heat summation
If a reaction takes place in several steps, its standard enthalpy of reaction is the sum of the standard enthalpies of the steps, at the same temperature: $$\Delta_{r}H^{\circ}=\Delta_{r}H^{\circ}_{1}+\Delta_{r}H^{\circ}_{2}+\Delta_{r}H^{\circ}_{3}+\cdots$$ The steps may be real or imaginary. Hess's law (Germain Hess, 1840) follows from the fact that $H$ is a state function: the change between the same initial and final states cannot depend on the route. Thermochemical equations can therefore be added, subtracted, reversed and multiplied like algebraic equations, with their $\Delta_{r}H^{\circ}$ values treated the same way.
✏️ Example 3 — Using enthalpies of formation
Use the table to find $\Delta_{r}H^{\circ}$ at $298\ \mathrm{K}$ for (a) $\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)}$; (b) $\mathrm{CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)}$. (c) Find the heat released per gram of methane burned ($M=16.04\ \mathrm{g\,mol^{-1}}$).
(a) $\Delta_{r}H^{\circ}=[(-393.5)+2(-285.8)]-[(-74.8)+2(0)]=-965.1+74.8=-890.3\ \mathrm{kJ\,mol^{-1}}$ ($\mathrm{O_{2}}$ is an element: $0$).
(b) $\Delta_{r}H^{\circ}=[(-635.1)+(-393.5)]-(-1206.9)=+178.3\ \mathrm{kJ\,mol^{-1}}$ — endothermic, which is why limestone must be strongly heated in a lime kiln.
(c) $890.3/16.04=55.5\ \mathrm{kJ\,g^{-1}}$.
Check (b) is the same value used in Section 5.2 to compare $\Delta H$ and $\Delta U$ for this reaction ✓.
✏️ Example 4 — Hess's law: enthalpies that cannot be measured directly
(a) Burning carbon in a limited supply of oxygen always gives a mixture of CO and $\mathrm{CO_{2}}$, so $\Delta_{f}H^{\circ}(\mathrm{CO})$ cannot be measured directly. Find it from (1) $\mathrm{C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)}$, $-393.5\ \mathrm{kJ\,mol^{-1}}$ and (2) $\mathrm{CO(g)+\tfrac12O_{2}(g)\rightarrow CO_{2}(g)}$, $-283.0\ \mathrm{kJ\,mol^{-1}}$. (b) Find $\Delta_{f}H^{\circ}(\mathrm{CH_{4},g})$ from the enthalpy of combustion of methane, $-890.3\ \mathrm{kJ\,mol^{-1}}$, and those of graphite ($-393.5$) and hydrogen ($-285.8\ \mathrm{kJ\,mol^{-1}}$, giving liquid water).
(a) plan Target: $\mathrm{C(graphite)+\tfrac12O_{2}(g)\rightarrow CO(g)}$. Keep (1); reverse (2) to put CO on the right.
(a) add (1) − (2): $\mathrm{C+O_{2}+CO_{2}\rightarrow CO_{2}+CO+\tfrac12O_{2}}$, which simplifies to the target. $\Delta_{f}H^{\circ}=-393.5-(-283.0)=-110.5\ \mathrm{kJ\,mol^{-1}}$.
(b) plan Target: $\mathrm{C(graphite)+2H_{2}(g)\rightarrow CH_{4}(g)}$. Use: $\mathrm{C+O_{2}\rightarrow CO_{2}}$ ($-393.5$); $2\times[\mathrm{H_{2}+\tfrac12O_{2}\rightarrow H_{2}O(l)}]$ ($2\times-285.8$); reverse of the combustion of methane ($+890.3$).
(b) add $\Delta_{f}H^{\circ}=-393.5-571.6+890.3=-74.8\ \mathrm{kJ\,mol^{-1}}$; the $\mathrm{CO_{2}}$, $\mathrm{H_{2}O}$ and $\mathrm{O_{2}}$ all cancel.
Check Both answers agree with the table, as they must: this is how many table entries were obtained.
Interactive 1 — Hess's-Law Energy-Cycle Builder

Choose a target reaction. Below are three measured thermochemical equations. Give each one a multiplier — a negative multiplier reverses it — and the lab adds them up, cancels species that appear on both sides, and draws the steps as an enthalpy-level diagram starting from the reactants. When the summed equation matches the target, the cycle closes and $\Delta_{r}H^{\circ}$ is the sum of the steps. Press ▶ Play to watch the lab find the right combination.

target—
your sum—
Σ ΔH of steps—
cycle—

Enthalpies of Combustion, Atomisation and Bonds

📘 Definition 5.14 — Standard enthalpy of combustion
The standard enthalpy of combustion $\Delta_{c}H^{\circ}$ is the enthalpy change when one mole of a substance burns completely in oxygen, all reactants and products in their standard states (carbon → $\mathrm{CO_{2}(g)}$, hydrogen → $\mathrm{H_{2}O(l)}$). Combustion is always exothermic. Examples: $\mathrm{CH_{4}}$ $-890.3$, $\mathrm{C_{2}H_{5}OH(l)}$ $-1366.7$, $\mathrm{C_{6}H_{12}O_{6}(s)}$ $-2808\ \mathrm{kJ\,mol^{-1}}$. Divided by the molar mass, $\Delta_{c}H^{\circ}$ gives the fuel value (calorific value) in $\mathrm{kJ\,g^{-1}}$.
📘 Definition 5.15 — Enthalpy of atomisation
The enthalpy of atomisation $\Delta_{a}H^{\circ}$ is the enthalpy change when one mole of a substance is broken completely into separate atoms in the gas phase: $$\mathrm{H_{2}(g)\rightarrow 2H(g)},\ 436\ \mathrm{kJ\,mol^{-1}};\qquad \mathrm{CH_{4}(g)\rightarrow C(g)+4H(g)},\ 1665\ \mathrm{kJ\,mol^{-1}};\qquad \mathrm{Na(s)\rightarrow Na(g)},\ 108.4\ \mathrm{kJ\,mol^{-1}}$$ For a solid metal it equals the enthalpy of sublimation; for a diatomic molecule it equals the bond dissociation enthalpy.

Section 4.1 introduced the bond enthalpy, the energy needed to break one mole of a given bond in the gas phase. For a diatomic molecule there is only one bond and one value. In a polyatomic molecule such as methane the four C–H bonds are broken one after another, and each step needs a different amount of energy, because the fragment left behind changes ($\mathrm{CH_{4}}$, $\mathrm{CH_{3}}$, $\mathrm{CH_{2}}$, $\mathrm{CH}$). Only their sum, the atomisation enthalpy $1665\ \mathrm{kJ\,mol^{-1}}$, is fixed.

📘 Definition 5.16 — Bond dissociation enthalpy and mean bond enthalpy
The bond dissociation enthalpy is the enthalpy change for breaking one mole of a particular bond in a particular gaseous molecule (for diatomic molecules, the bond enthalpy). The mean bond enthalpy of a bond such as C–H is the average value over the bonds of that type in one molecule (methane: $1665/4=416\ \mathrm{kJ\,mol^{-1}}$) or, as tabulated, over many different compounds (C–H: $413\ \mathrm{kJ\,mol^{-1}}$).
📐 Rule 5.7 — Estimating ΔrH° from bond enthalpies
For a reaction in which all reactants and products are gases, $$\Delta_{r}H^{\circ}\approx\sum\text{bond enthalpies}_{\text{reactants}}-\sum\text{bond enthalpies}_{\text{products}}$$ (energy in to break the old bonds, minus energy out when the new bonds form). Because mean bond enthalpies are averages, the result is an estimate, usually within about $10\%$ or a few tens of kJ; values from enthalpies of formation are more reliable when available.

Mean bond enthalpies used here, in $\mathrm{kJ\,mol^{-1}}$ (as in Section 4.1): H–H 436, Cl–Cl 243, C–H 413, C–C 348, C=C 614, O=O 498, O–H 464; and H–Cl 431, C–Cl 328, N≡N 946, N–H 391.

✏️ Example 5 — Bond enthalpies
(a) Estimate $\Delta_{r}H^{\circ}$ for $\mathrm{H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)}$. (b) Estimate $\Delta_{r}H^{\circ}$ for the hydrogenation of ethene, $\mathrm{C_{2}H_{4}(g)+H_{2}(g)\rightarrow C_{2}H_{6}(g)}$, and compare with the value from the table of enthalpies of formation. (c) Use the atomisation enthalpy of methane to find the mean C–H bond enthalpy in methane, and explain why it differs from the tabulated $413\ \mathrm{kJ\,mol^{-1}}$.
(a) Broken: $436+243=679$. Formed: $2\times431=862$. $\Delta_{r}H^{\circ}\approx679-862=-183\ \mathrm{kJ\,mol^{-1}}$ (measured $2\times(-92.3)=-184.6$).
(b) bonds Broken: C=C $614$ + H–H $436=1050$. Formed: C–C $348$ + two new C–H $2\times413=826$, total $1174$ (the four C–H bonds of ethene survive and cancel). $\Delta_{r}H^{\circ}\approx1050-1174=-124\ \mathrm{kJ\,mol^{-1}}$.
(b) table $\Delta_{r}H^{\circ}=(-84.7)-(+52.3+0)=-137.0\ \mathrm{kJ\,mol^{-1}}$. The bond estimate is about $10\%$ smaller in magnitude — the usual accuracy of mean bond enthalpies.
(c) $1665/4=416\ \mathrm{kJ\,mol^{-1}}$. The tabulated value is an average over C–H bonds in many molecules, whose environments differ slightly from methane's.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — resonance energy from a Hess cycle

An extension, not a CBSE requirement. It uses Hess's law and bond enthalpies, which JEE Main, NEET and JEE Advanced all list; no syllabus names resonance energy itself.

Bond enthalpies predict $\Delta_{f}H^{\circ}$ for a gaseous molecule: atomise the elements, then form the bonds. For benzene drawn as one Kekulé structure (3 C–C, 3 C=C, 6 C–H), with $\Delta_{a}H(\mathrm{C,graphite})=716.7$ and $\tfrac12\Delta_{a}H(\mathrm{H_{2}})=218.0\ \mathrm{kJ\,mol^{-1}}$ per H atom: $\Delta_{f}H^{\circ}_{\text{est}}=6(716.7)+6(218.0)-[3(348)+3(614)+6(413)]=5608.2-5364=+244\ \mathrm{kJ\,mol^{-1}}$. The measured value for $\mathrm{C_{6}H_{6}(g)}$ is $+82.9\ \mathrm{kJ\,mol^{-1}}$. Real benzene is about $161\ \mathrm{kJ\,mol^{-1}}$ more stable than any single Kekulé structure: this is an estimate of its resonance energy (Section 4.1; values of 150–160 are commonly quoted, depending on the bond data used).

Lattice Enthalpy, Solution and Dilution

The lattice enthalpy of an ionic solid (Section 4.1) is the enthalpy change for separating one mole of the solid completely into its gaseous ions, for example $\mathrm{NaCl(s)\rightarrow Na^{+}(g)+Cl^{-}(g)}$. It cannot be measured directly, but Hess's law gives it from quantities that can.

📘 Definition 5.17 — The Born–Haber cycle
A Born–Haber cycle is a Hess's-law cycle that connects the formation of an ionic solid from its elements with the steps that make gaseous ions and then pack them into the crystal. For sodium chloride (values in $\mathrm{kJ\,mol^{-1}}$):
  1. Sublimation of sodium: $\mathrm{Na(s)\rightarrow Na(g)}$, $\Delta_{sub}H^{\circ}=+108.4$.
  2. Atomisation of chlorine: $\mathrm{\tfrac12Cl_{2}(g)\rightarrow Cl(g)}$, $\tfrac12\Delta_{bond}H^{\circ}=+121.5$.
  3. Ionisation of sodium: $\mathrm{Na(g)\rightarrow Na^{+}(g)+e^{-}}$, $\Delta_{i}H^{\circ}=+495.8$.
  4. Electron gain by chlorine: $\mathrm{Cl(g)+e^{-}\rightarrow Cl^{-}(g)}$, $\Delta_{eg}H^{\circ}=-348.7$.
  5. Formation of the lattice: $\mathrm{Na^{+}(g)+Cl^{-}(g)\rightarrow NaCl(s)}$, $-\Delta_{lattice}H^{\circ}$.
The steps add up to the formation reaction $\mathrm{Na(s)+\tfrac12Cl_{2}(g)\rightarrow NaCl(s)}$, $\Delta_{f}H^{\circ}=-411.2$, so $$\Delta_{lattice}H^{\circ}=-\Delta_{f}H^{\circ}+\Delta_{sub}H^{\circ}+\tfrac12\Delta_{bond}H^{\circ}+\Delta_{i}H^{\circ}+\Delta_{eg}H^{\circ}$$

When an ionic solid dissolves, its lattice must be broken (endothermic) and the separated ions are surrounded by water molecules — they are hydrated (strongly exothermic). The balance between the two decides whether the solution warms up or cools down.

📘 Definition 5.18 — Enthalpies of solution, hydration and dilution
  • The enthalpy of solution $\Delta_{sol}H^{\circ}$ is the enthalpy change when one mole of a substance dissolves in a specified amount of solvent; at infinite dilution the ions no longer interact with each other.
  • The enthalpy of hydration $\Delta_{hyd}H^{\circ}$ is the enthalpy change when one mole of gaseous ions dissolves in water at infinite dilution, e.g. $\mathrm{Na^{+}(g)+Cl^{-}(g)\rightarrow Na^{+}(aq)+Cl^{-}(aq)}$ (always negative).
  • The enthalpy of dilution is the enthalpy change when a solution containing one mole of solute is diluted by adding more solvent. It depends on the initial and final concentrations and becomes zero once the solution is very dilute.
📐 Rule 5.8 — Enthalpy of solution from a Hess cycle
$$\Delta_{sol}H^{\circ}=\Delta_{lattice}H^{\circ}+\Delta_{hyd}H^{\circ}$$ The salt dissolves exothermically when hydration releases more than the lattice costs ($\mathrm{NaOH}$, $\mathrm{CaCl_{2}}$ — "hot packs"), and endothermically when it releases less ($\mathrm{NH_{4}NO_{3}}$, $\mathrm{KCl}$ — "cold packs"). Salts with a very large lattice enthalpy and only moderate hydration enthalpy are poorly soluble.
✏️ Example 6 — Born–Haber cycle and enthalpy of solution for NaCl
(a) Use the data of Definition 5.17 to calculate the lattice enthalpy of sodium chloride. (b) Draw the cycle as an enthalpy-level diagram and say which step makes the formation of NaCl favourable. (c) The enthalpy of solution of NaCl at infinite dilution is $+3.9\ \mathrm{kJ\,mol^{-1}}$. Find the enthalpy of hydration of $\mathrm{Na^{+}(g)+Cl^{-}(g)}$.
(a) $\Delta_{lattice}H^{\circ}=411.2+108.4+121.5+495.8+(-348.7)=+788.2\ \mathrm{kJ\,mol^{-1}}\approx788\ \mathrm{kJ\,mol^{-1}}$, the value quoted in Section 4.1.
(b) Starting from $\mathrm{Na(s)+\tfrac12Cl_{2}(g)}$ at zero, the levels climb by $108.4$, $121.5$ and $495.8$ to $+725.7$, fall by $348.7$ to $+377.0$ ($\mathrm{Na^{+}(g)+Cl^{-}(g)}$), then fall by $788.2$ to $-411.2$. Making the gaseous ions costs $377\ \mathrm{kJ}$ overall; the huge release of lattice enthalpy is what makes NaCl stable (Interactive 2).
(c) $\Delta_{hyd}H^{\circ}=\Delta_{sol}H^{\circ}-\Delta_{lattice}H^{\circ}=3.9-788.2=-784.3\ \mathrm{kJ\,mol^{-1}}$. The two large terms almost cancel, which is why dissolving salt hardly changes the temperature of water.
Interactive 2 — Born–Haber Cycle Ladder

Choose a salt and step through the cycle with the slider, or press ▶ Play. Each arrow is one step: up for endothermic, down for exothermic, with lengths to scale. The lab adds the steps as it goes; the last arrow, from the gaseous ions down to the solid, is the lattice enthalpy (with its sign reversed). Compare NaCl with KCl: which steps change, and why is the lattice enthalpy of KCl smaller?

5 / 5
this step—
level reached—
lattice enthalpy—
ΔfH° (check)—

Exercises

I
Thermochemical equations and phase changes. Use $M(\mathrm{H_{2}O})=18.02$, $M(\mathrm{NH_{3}})=17.03\ \mathrm{g\,mol^{-1}}$.
  • 1 What is meant by the standard state of a substance? What does the symbol $^{\circ}$ in $\Delta_{r}H^{\circ}$ tell you?
  • 2 Write the thermochemical equation for the standard enthalpy of formation of liquid water ($-285.8\ \mathrm{kJ\,mol^{-1}}$).
  • 3 $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$, $\Delta_{r}H^{\circ}=-92.2\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta_{r}H^{\circ}$ for $\mathrm{NH_{3}(g)\rightarrow\tfrac12N_{2}(g)+\tfrac32H_{2}(g)}$ and for $\mathrm{2N_{2}(g)+6H_{2}(g)\rightarrow 4NH_{3}(g)}$.
  • 4 Using part 3, find the heat released when $3.40\ \mathrm{g}$ of ammonia is formed from its elements.
  • 5 $\Delta_{fus}H(\mathrm{H_{2}O})=6.01\ \mathrm{kJ\,mol^{-1}}$. How much heat is needed to melt $45.0\ \mathrm{g}$ of ice at $0\ ^{\circ}\mathrm{C}$?
  • 6 At $273\ \mathrm{K}$, $\Delta_{fus}H=6.01$ and $\Delta_{vap}H=45.07\ \mathrm{kJ\,mol^{-1}}$ for water. Find $\Delta_{sub}H$ at $273\ \mathrm{K}$ and explain why the two can be added.
  • 7 Why is the enthalpy of vaporisation of a substance always much larger than its enthalpy of fusion?
  • 8 How much heat is needed to turn $18.0\ \mathrm{g}$ of water at $25\ ^{\circ}\mathrm{C}$ into steam at $100\ ^{\circ}\mathrm{C}$? ($C_{p,m}=75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $\Delta_{vap}H=40.66\ \mathrm{kJ\,mol^{-1}}$)
1 The pure substance in its stable physical form at $1\ \mathrm{bar}$ (solutes at $1\ \mathrm{mol\,L^{-1}}$) and a stated temperature. The $^{\circ}$ means every reactant and product is in its standard state.
2 $\mathrm{H_{2}(g)+\tfrac12O_{2}(g)\rightarrow H_{2}O(l)}$; $\Delta_{f}H^{\circ}=-285.8\ \mathrm{kJ\,mol^{-1}}$.
3 Reverse and halve: $+46.1\ \mathrm{kJ\,mol^{-1}}$. Double: $-184.4\ \mathrm{kJ\,mol^{-1}}$.
4 $n=3.40/17.03=0.1996\ \mathrm{mol}$; heat released $=0.1996\times46.1=9.20\ \mathrm{kJ}$.
5 $n=45.0/18.02=2.497\ \mathrm{mol}$; $q=2.497\times6.01=15.0\ \mathrm{kJ}$.
6 $6.01+45.07=51.08\ \mathrm{kJ\,mol^{-1}}$. Solid → liquid → vapour and solid → vapour join the same two states, and $H$ is a state function (Hess's law).
7 Melting only loosens the particles (most attractions remain in the liquid); vaporisation separates them almost completely and also pushes back the atmosphere.
8 $n=18.0/18.02=0.999\ \mathrm{mol}$. Warming: $75.3\times75=5.65\ \mathrm{kJ\,mol^{-1}}$; boiling: $40.66$; total $46.31\ \mathrm{kJ\,mol^{-1}}\times0.999=46.3\ \mathrm{kJ}$.
II
Enthalpy of formation and Hess's law. Use the table of $\Delta_{f}H^{\circ}$ values.
  • 1 Which of these have $\Delta_{f}H^{\circ}=0$ at $298\ \mathrm{K}$: $\mathrm{O_{2}(g)}$, $\mathrm{O_{3}(g)}$, $\mathrm{C(diamond)}$, $\mathrm{C(graphite)}$, $\mathrm{Br_{2}(l)}$, $\mathrm{Br_{2}(g)}$, $\mathrm{H_{2}O(l)}$?
  • 2 Find $\Delta_{r}H^{\circ}$ for $\mathrm{C_{2}H_{4}(g)+3O_{2}(g)\rightarrow 2CO_{2}(g)+2H_{2}O(l)}$.
  • 3 Find $\Delta_{r}H^{\circ}$ for the first step of nitric acid manufacture, $\mathrm{4NH_{3}(g)+5O_{2}(g)\rightarrow 4NO(g)+6H_{2}O(g)}$.
  • 4 Find the standard enthalpy of combustion of liquid ethanol to $\mathrm{CO_{2}(g)}$ and $\mathrm{H_{2}O(l)}$.
  • 5 $\mathrm{C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)}$, $-393.5$; $\mathrm{C(diamond)+O_{2}(g)\rightarrow CO_{2}(g)}$, $-395.4\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta_{r}H^{\circ}$ for graphite → diamond. Which form has the lower enthalpy?
  • 6 $\Delta_{c}H^{\circ}$ of ethyne, $\mathrm{C_{2}H_{2}(g)}$ (to $\mathrm{H_{2}O(l)}$), is $-1299.5\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta_{f}H^{\circ}(\mathrm{C_{2}H_{2},g})$.
  • 7 State Hess's law. Which property of enthalpy makes it true?
  • 8 Find $\Delta_{r}H^{\circ}$ for $\mathrm{C_{2}H_{4}(g)+H_{2}(g)\rightarrow C_{2}H_{6}(g)}$ from the table.
1 $\mathrm{O_{2}(g)}$, $\mathrm{C(graphite)}$, $\mathrm{Br_{2}(l)}$ — elements in their reference states. ($\mathrm{O_{3}}$, diamond and bromine vapour are not the most stable forms; water is a compound.)
2 $[2(-393.5)+2(-285.8)]-[52.3+0]=-1358.6-52.3=-1410.9\ \mathrm{kJ\,mol^{-1}}$.
3 $[4(90.3)+6(-241.8)]-[4(-46.1)+0]=361.2-1450.8+184.4=-905.2\ \mathrm{kJ\,mol^{-1}}$.
4 $\mathrm{C_{2}H_{5}OH(l)+3O_{2}(g)\rightarrow 2CO_{2}(g)+3H_{2}O(l)}$: $[2(-393.5)+3(-285.8)]-(-277.7)=-1644.4+277.7=-1366.7\ \mathrm{kJ\,mol^{-1}}$.
5 First equation minus the second: $-393.5-(-395.4)=+1.9\ \mathrm{kJ\,mol^{-1}}$. Graphite has the lower enthalpy (it is the reference state).
6 $\mathrm{C_{2}H_{2}+\tfrac52O_{2}\rightarrow 2CO_{2}+H_{2}O(l)}$: $-1299.5=[2(-393.5)+(-285.8)]-\Delta_{f}H^{\circ}$, so $\Delta_{f}H^{\circ}=-1072.8+1299.5=+226.7\ \mathrm{kJ\,mol^{-1}}$.
7 The enthalpy change of a reaction is the same whether it occurs in one step or several; it follows because $H$ is a state function (a consequence of the first law).
8 $(-84.7)-(52.3+0)=-137.0\ \mathrm{kJ\,mol^{-1}}$.
III
Combustion, atomisation and bond enthalpies (use the bond values listed above).
  • 1 Define the standard enthalpy of combustion. Why is it always negative?
  • 2 Find the fuel values (in $\mathrm{kJ\,g^{-1}}$) of hydrogen ($\Delta_{c}H^{\circ}=-285.8$, $M=2.016$), methane ($-890.3$, $16.04$) and ethanol ($-1366.7$, $46.07$). Which is the best fuel per gram?
  • 3 How much heat does $1.00\ \mathrm{kg}$ of methane release on complete combustion?
  • 4 Define the enthalpy of atomisation. Why is it equal to the bond enthalpy for $\mathrm{H_{2}}$ but not for $\mathrm{CH_{4}}$?
  • 5 Distinguish the bond dissociation enthalpy of a C–H bond in methane from the mean C–H bond enthalpy.
  • 6 Estimate $\Delta_{r}H^{\circ}$ for $\mathrm{CH_{4}(g)+Cl_{2}(g)\rightarrow CH_{3}Cl(g)+HCl(g)}$.
  • 7 Estimate $\Delta_{r}H^{\circ}$ for $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)}$ from bond enthalpies and compare it with the value from $\Delta_{f}H^{\circ}(\mathrm{H_{2}O,g})$.
  • 8 Why would a bond-enthalpy estimate for $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)}$ be badly wrong?
1 The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. The products ($\mathrm{CO_{2}}$, $\mathrm{H_{2}O}$) contain very strong bonds, so more energy is released in forming them than is used in breaking the bonds of the fuel and $\mathrm{O_{2}}$.
2 $\mathrm{H_{2}}$: $285.8/2.016=141.8$; $\mathrm{CH_{4}}$: $890.3/16.04=55.5$; ethanol: $1366.7/46.07=29.7\ \mathrm{kJ\,g^{-1}}$. Hydrogen, by far.
3 $1000/16.04=62.34\ \mathrm{mol}$; $62.34\times890.3=5.55\times10^{4}\ \mathrm{kJ}$ ($55.5\ \mathrm{MJ}$).
4 The enthalpy change to break one mole of a substance into gaseous atoms. $\mathrm{H_{2}}$ has a single bond, so atomisation = breaking that bond ($436\ \mathrm{kJ\,mol^{-1}}$). $\mathrm{CH_{4}}$ has four C–H bonds, broken one at a time with different energies; the atomisation enthalpy is their sum ($1665$).
5 The bond dissociation enthalpy is for removing one particular H atom from $\mathrm{CH_{4}}$ to give $\mathrm{CH_{3}}$; the mean bond enthalpy is the average over all four steps ($416$ in methane) or over many compounds ($413$).
6 Broken: C–H $413$ + Cl–Cl $243=656$. Formed: C–Cl $328$ + H–Cl $431=759$. $\Delta_{r}H^{\circ}\approx656-759=-103\ \mathrm{kJ\,mol^{-1}}$.
7 Broken: $2(436)+498=1370$. Formed: $4(464)=1856$. $\Delta_{r}H^{\circ}\approx-486\ \mathrm{kJ\,mol^{-1}}$; from the table $2(-241.8)=-483.6\ \mathrm{kJ\,mol^{-1}}$ — excellent agreement.
8 Bond enthalpies refer to gases. Forming liquid water also releases the enthalpy of condensation ($2\times44\ \mathrm{kJ}$ at $298\ \mathrm{K}$), which bonds within molecules do not account for.
IV
Lattice enthalpy, solution and dilution.
  • 1 Calculate the lattice enthalpy of KCl from: $\Delta_{sub}H(\mathrm{K})=89.2$, $\tfrac12\Delta_{bond}H(\mathrm{Cl_{2}})=121.5$, $\Delta_{i}H(\mathrm{K})=418.8$, $\Delta_{eg}H(\mathrm{Cl})=-348.7$, $\Delta_{f}H^{\circ}(\mathrm{KCl,s})=-436.7\ \mathrm{kJ\,mol^{-1}}$.
  • 2 Why can a lattice enthalpy not be measured directly?
  • 3 For KCl, $\Delta_{lattice}H^{\circ}=718$ and $\Delta_{hyd}H^{\circ}=-701\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta_{sol}H^{\circ}$ and say whether the solution warms or cools as KCl dissolves.
  • 4 Dissolving one mole of $\mathrm{HCl(g)}$ in $10\ \mathrm{mol}$ of water releases $69.0\ \mathrm{kJ}$; dissolving it in a very large amount of water releases $74.8\ \mathrm{kJ}$. Find the enthalpy of dilution of the first solution to infinite dilution, and explain why it is not zero.
  • 5 The lattice enthalpy of NaCl is $788\ \mathrm{kJ\,mol^{-1}}$. Use the other data of Definition 5.17 to find $\Delta_{eg}H(\mathrm{Cl})$ — the way electron gain enthalpies were first obtained.
  • 6 Applied. A hot pack contains $40.0\ \mathrm{g}$ of $\mathrm{CaCl_{2}}$ ($M=110.98\ \mathrm{g\,mol^{-1}}$, $\Delta_{sol}H=-82.8\ \mathrm{kJ\,mol^{-1}}$) and $200\ \mathrm{g}$ of water. Estimate the temperature rise, treating the $240\ \mathrm{g}$ of solution as having $c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$ and ignoring heat losses.
1 $\Delta_{lattice}H^{\circ}=436.7+89.2+121.5+418.8-348.7=+717.5\ \mathrm{kJ\,mol^{-1}}$. Smaller than NaCl's $788$ because $\mathrm{K^{+}}$ is larger than $\mathrm{Na^{+}}$, so the ions are further apart.
2 Gaseous ions cannot be collected and made to form a crystal in a calorimeter; the process is imaginary, so its enthalpy must come from a Hess cycle.
3 $\Delta_{sol}H^{\circ}=718+(-701)=+17\ \mathrm{kJ\,mol^{-1}}$: endothermic, the solution cools.
4 $\Delta H=-74.8-(-69.0)=-5.8\ \mathrm{kJ\,mol^{-1}}$. In the concentrated solution the ions are not yet fully hydrated and still attract each other; adding water completes their hydration and releases more heat.
5 $788=411.2+108.4+121.5+495.8+\Delta_{eg}H$ → $\Delta_{eg}H=788-1136.9=-349\ \mathrm{kJ\,mol^{-1}}$.
6 $n=40.0/110.98=0.3604\ \mathrm{mol}$; heat released $=0.3604\times82.8=29.8\ \mathrm{kJ}$; $\Delta T=\dfrac{29\,840}{240\times4.18}=29.7\ \mathrm{K}$ — the pack warms by about $30\ \mathrm{K}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The standard enthalpy of formation is zero for (A) $\mathrm{O_{3}(g)}$ (B) $\mathrm{C(diamond)}$ (C) $\mathrm{Br_{2}(l)}$ (D) $\mathrm{H_{2}O(l)}$

Idea Only an element in its most stable form at $1\ \mathrm{bar}$ and $298\ \mathrm{K}$ has $\Delta_{f}H^{\circ}=0$. Bromine is a liquid at $298\ \mathrm{K}$; ozone and diamond are not the reference forms; water is a compound.
Answer (C).

E2. JEE Main Numerical answer. The standard enthalpies of combustion of $\mathrm{C(graphite)}$, $\mathrm{H_{2}(g)}$ and propane $\mathrm{C_{3}H_{8}(g)}$ are $-393.5$, $-285.8$ and $-2220.0\ \mathrm{kJ\,mol^{-1}}$ (water as liquid). The magnitude of the standard enthalpy of formation of propane, in $\mathrm{kJ\,mol^{-1}}$, to the nearest integer is ____.

Target $\mathrm{3C(graphite)+4H_{2}(g)\rightarrow C_{3}H_{8}(g)}$ = $3\times$(C combustion) + $4\times$(H₂ combustion) − (propane combustion).
Hess $\Delta_{f}H^{\circ}=3(-393.5)+4(-285.8)-(-2220.0)=-1180.5-1143.2+2220.0=-103.7\ \mathrm{kJ\,mol^{-1}}$.
Answer 104.

E3. Board · 3 marks Draw a labelled Born–Haber cycle for the formation of sodium chloride and use it to calculate the lattice enthalpy of NaCl, given (in $\mathrm{kJ\,mol^{-1}}$): $\Delta_{sub}H(\mathrm{Na})=108.4$, $\Delta_{bond}H(\mathrm{Cl_{2}})=243$, $\Delta_{i}H(\mathrm{Na})=495.8$, $\Delta_{eg}H(\mathrm{Cl})=-348.7$, $\Delta_{f}H^{\circ}(\mathrm{NaCl})=-411.2$.

1 mark Cycle: $\mathrm{Na(s)\rightarrow Na(g)\rightarrow Na^{+}(g)}$; $\mathrm{\tfrac12Cl_{2}(g)\rightarrow Cl(g)\rightarrow Cl^{-}(g)}$; $\mathrm{Na^{+}(g)+Cl^{-}(g)\rightarrow NaCl(s)}$; direct route $\mathrm{Na(s)+\tfrac12Cl_{2}(g)\rightarrow NaCl(s)}$.
1 mark Hess: $\Delta_{f}H^{\circ}=\Delta_{sub}H+\tfrac12\Delta_{bond}H+\Delta_{i}H+\Delta_{eg}H-\Delta_{lattice}H$ (note half of $243$, because only $\tfrac12$ mol $\mathrm{Cl_{2}}$ is used).
1 mark $\Delta_{lattice}H=108.4+121.5+495.8-348.7+411.2=+788.2\ \mathrm{kJ\,mol^{-1}}$.

E4. JEE Main Using the bond enthalpies N≡N $946$, H–H $436$ and N–H $391\ \mathrm{kJ\,mol^{-1}}$, $\Delta_{r}H$ for $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$ is approximately (A) $-92\ \mathrm{kJ}$ (B) $+92\ \mathrm{kJ}$ (C) $-46\ \mathrm{kJ}$ (D) $-1162\ \mathrm{kJ}$

Bonds Broken: $946+3(436)=2254$. Formed: $6(391)=2346$ (two $\mathrm{NH_{3}}$, three N–H each).
Sum $\Delta_{r}H\approx2254-2346=-92\ \mathrm{kJ}$ per mole of reaction as written. (C) is per mole of $\mathrm{NH_{3}}$; (D) forgets to subtract the bonds broken.
Answer (A).

E5. JEE Advanced One or more options correct. (A) The enthalpy of atomisation of $\mathrm{H_{2}(g)}$ equals its bond dissociation enthalpy. (B) The standard enthalpy of formation of diamond is zero. (C) At the same temperature, $\Delta_{sub}H=\Delta_{fus}H+\Delta_{vap}H$. (D) The enthalpy of neutralisation of every acid by NaOH is $-57.1\ \mathrm{kJ\,mol^{-1}}$.

(A) True: $\mathrm{H_{2}}$ has one bond; breaking it gives the atoms.
(B) False: graphite is the reference state; diamond has $+1.9\ \mathrm{kJ\,mol^{-1}}$.
(C) True: Hess's law (two routes from solid to vapour).
(D) False: only strong acids; weak acids give less negative values because part of the heat is used to ionise them.
Answer (A), (C).

E6. Board · Case study Natural gas is mostly methane. Use $\Delta_{f}H^{\circ}$ ($\mathrm{kJ\,mol^{-1}}$): $\mathrm{CH_{4}(g)}$ $-74.8$, $\mathrm{CO_{2}(g)}$ $-393.5$, $\mathrm{H_{2}O(l)}$ $-285.8$, $\mathrm{H_{2}O(g)}$ $-241.8$. (i) Calculate $\Delta_{c}H^{\circ}$ of methane (water as liquid). (2 marks) (ii) Find its fuel value in $\mathrm{kJ\,g^{-1}}$ ($M=16.04\ \mathrm{g\,mol^{-1}}$). (1 mark) (iii) In a gas burner the water leaves as vapour. Calculate $\Delta_{r}H^{\circ}$ for that case and explain the difference. (1 mark)

(i) $\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)}$: $[-393.5+2(-285.8)]-(-74.8)=-890.3\ \mathrm{kJ\,mol^{-1}}$.
(ii) $890.3/16.04=55.5\ \mathrm{kJ\,g^{-1}}$.
(iii) $[-393.5+2(-241.8)]-(-74.8)=-802.3\ \mathrm{kJ\,mol^{-1}}$. Less heat is released because the $2\times44.0=88.0\ \mathrm{kJ}$ that would be given out when the steam condenses is carried away with the flue gases.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Hydrogen, ammonia and methanol are all proposed as carbon-free or low-carbon fuels for ships, power stations and trucks. Per kilogram, and per litre of fuel tank, how do they compare with methane and petrol — and what does thermochemistry say about the trade-offs?

Why it matters. Japan and South Korea are trialling ammonia co-firing in coal power stations; shipping companies in Europe and Asia are ordering ammonia- and methanol-fuelled vessels; India's National Green Hydrogen Mission aims at millions of tonnes of hydrogen a year. All of these choices start from enthalpies of combustion.

What to do.

  1. Write balanced combustion equations for $\mathrm{H_{2}(g)}$, $\mathrm{NH_{3}(g)}$ (to $\mathrm{N_{2}(g)}$ and $\mathrm{H_{2}O(g)}$), $\mathrm{CH_{3}OH(l)}$ and $\mathrm{CH_{4}(g)}$, all with water as vapour, as in an engine.
  2. Calculate each $\Delta_{c}H^{\circ}$ from the table plus $\Delta_{f}H^{\circ}(\mathrm{CH_{3}OH,l})=-238.7\ \mathrm{kJ\,mol^{-1}}$, then the fuel value in $\mathrm{MJ\,kg^{-1}}$.
  3. Look up the densities of liquid hydrogen ($\approx71\ \mathrm{kg\,m^{-3}}$), liquid ammonia ($\approx680$) and methanol ($\approx792$), and convert to $\mathrm{MJ\,L^{-1}}$. Which fuel needs the biggest tank?
  4. Which fuel releases no $\mathrm{CO_{2}}$ when burned? What else must be considered (how the fuel is made, $\mathrm{NO_{x}}$ formation, toxicity)?

Tools: Rule 5.5, the table of $\Delta_{f}H^{\circ}$, a spreadsheet, an energy-agency report on fuels for shipping.

Hand in: A table of $\Delta_{c}H^{\circ}$, $\mathrm{MJ\,kg^{-1}}$ and $\mathrm{MJ\,L^{-1}}$ for the four fuels, and a 200-word recommendation for one use (a ferry, a city bus or a power station).

Going further. Ammonia is made from hydrogen ($\mathrm{N_{2}+3H_{2}\rightarrow 2NH_{3}}$, $-92.2\ \mathrm{kJ}$). What fraction of the energy of the hydrogen is "lost" as heat when it is converted to ammonia and then burned?

Instructor note (water as vapour): $\mathrm{H_{2}}$ $-241.8\ \mathrm{kJ\,mol^{-1}}$ → $120\ \mathrm{MJ\,kg^{-1}}$, $\approx8.5\ \mathrm{MJ\,L^{-1}}$ (liquid); $\mathrm{NH_{3}}$: $\mathrm{4NH_{3}+3O_{2}\rightarrow 2N_{2}+6H_{2}O(g)}$, $\Delta=6(-241.8)-4(-46.1)=-1266.4\ \mathrm{kJ}$ → $-316.6\ \mathrm{kJ\,mol^{-1}}$ → $18.6\ \mathrm{MJ\,kg^{-1}}$, $\approx12.6\ \mathrm{MJ\,L^{-1}}$; $\mathrm{CH_{3}OH}$: $-393.5+2(-241.8)+238.7=-638.4\ \mathrm{kJ\,mol^{-1}}$ → $19.9\ \mathrm{MJ\,kg^{-1}}$, $\approx15.8\ \mathrm{MJ\,L^{-1}}$; $\mathrm{CH_{4}}$ $-802.3\ \mathrm{kJ\,mol^{-1}}$ → $50.0\ \mathrm{MJ\,kg^{-1}}$. Going further: $3\ \mathrm{mol}\ \mathrm{H_{2}}$ release $3(241.8)=725.4\ \mathrm{kJ}$; via $2\ \mathrm{NH_{3}}$ only $633.2\ \mathrm{kJ}$, so about $13\%$ is released earlier, in the ammonia plant.

🧪 Self-check quiz

0/5 correct

Q1. Hess's law holds because

Q2. Which equation has $\Delta_{r}H^{\circ}$ equal to $\Delta_{f}H^{\circ}$ of liquid water?

Q3. If $\mathrm{A\rightarrow B}$ has $\Delta H=-40\ \mathrm{kJ}$ and $\mathrm{B\rightarrow C}$ has $\Delta H=+15\ \mathrm{kJ}$, then $\mathrm{C\rightarrow A}$ has $\Delta H=$

Q4. The lattice enthalpy of an ionic solid is normally obtained from

Q5. At a certain temperature $\Delta_{fus}H=6.0$ and $\Delta_{vap}H=45.0\ \mathrm{kJ\,mol^{-1}}$ for a substance. Its $\Delta_{sub}H$ at that temperature is

← 5.2 Enthalpy, Heat Capacity and Calorimetry5.4 Entropy, Gibbs Energy and Spontaneity →