Thermodynamics & Energetics

flowchart TD
    A[Thermodynamics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Thermodynamics is like a bank account for energy, you can’t create or destroy it, only transfer or transform it.

A thermodynamic system is the part of the universe under study. The surroundings are everything else. The universe is the system plus the surroundings.

• Open system: exchanges both matter and energy with surroundings.
• Closed system: exchanges energy but not matter.
• Isolated system: exchanges neither matter nor energy.

Enthalpy is a thermodynamic state function defined as:

Where $U$ is internal energy, $p$ is pressure, and $V$ is volume. For constant-pressure processes (the usual condition in chemistry):

The enthalpy change $\Delta H$ is the heat exchanged at constant pressure. It cannot be measured absolutely; only changes in enthalpy are measurable.

The internal energy of a system is the total kinetic and potential energy of all particles within the system. At constant volume:

The relationship between $\Delta H$ and $\Delta U$:

For reactions involving only solids and liquids, $\Delta(pV) \approx 0$ So $\Delta H \approx \Delta U$. For reactions involving gases:

Where $\Delta n_g$ is the change in moles of gas.

Standard enthalpy changes are measured under the following conditions:

• Pressure: $100\,\mathrm{kPa}$ ($1\,\mathrm{bar}$ Approximately $1\,\mathrm{atm}$).
• Temperature: $298\,\mathrm{K}$ ($25^\circ\mathrm{C}$), unless otherwise stated.
• Concentration: $1\,\mathrm{mol/dm}^3$ for solutions.
• All substances in their standard states (most stable form at the specified conditions).

The standard symbol is $\Delta H^\circ$ with the superscript circle.

• Exothermic: $\Delta H \lt 0$; heat is released to the surroundings.
• Endothermic: $\Delta H \gt 0$; heat is absorbed from the surroundings.

Hess’s Law states that the enthalpy change for a chemical reaction is the same regardless of the route by which the reaction occurs, provided the initial and final conditions are the same.

This is a direct consequence of enthalpy being a state function — its value depends only on the current state of the system, not on the path taken to reach it.

Worked Example 1. Calculate $\Delta H_f^\circ$ of $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)$.

Given:

• $\Delta H_c^\circ(\mathrm{C}) = -394\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{H}_2) = -286\,\mathrm{kJ/mol}$
• $\Delta H_c^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -1367\,\mathrm{kJ/mol}$

Formation: $2\mathrm{C}(s) + 3\mathrm{H}_2(g) + \tfrac{1}{2}\mathrm{O}_2(g) \to \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)$

Combustion route: $2\mathrm{C}(s) + 3\mathrm{H}_2(g) + 3\frac{1}{2}\mathrm{O}_2(g) \to 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$

By Hess’s Law:

The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

• $\Delta H_f^\circ$ of an element in its standard state is zero by definition.
• Example: $\Delta H_f^\circ(\mathrm{H}_2\mathrm{O}(l)) = -286\,\mathrm{kJ/mol}$.

The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.

• Combustion reactions are always exothermic ($\Delta H_c^\circ \lt 0$).
• The products must be in their standard states ($\mathrm{CO}_2(g)$, $\mathrm{H}_2\mathrm{O}(l)$ Etc.).

The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

For a diatomic molecule $\mathrm{X}_2$:

Example: $\Delta H_\mathrm{at}^\circ(\mathrm{Cl}_2) = \tfrac{1}{2}(242) = 121\,\mathrm{kJ/mol}$.

For a solid element (e.g. Na), atomisation includes sublimation and bond breaking:

The bond dissociation enthalpy is the enthalpy change required to break one mole of a specific bond in a specific molecule in the gaseous phase.

For a diatomic molecule, this is unambiguous. For polyatomic molecules, each bond-breaking step may have a different enthalpy. Mean bond enthalpies are averages across a range of compounds.

Note: $\mathrm{C-H}$ bond energy in $\mathrm{CH}_4$ is not the same as in $\mathrm{CH}_3\mathrm{Cl}$. Mean bond enthalpies introduce systematic error in thermochemical calculations.

The lattice enthalpy $\Delta H_\mathrm{lat}^\circ$ is the enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions under standard conditions. This is always exothermic ($\Delta H_\mathrm{lat}^\circ \lt 0$).

The reverse process (separating one mole of solid into gaseous ions) is the lattice dissociation enthalpy and is always endothermic.

The Born-Haber cycle for an ionic compound $\mathrm{M}_a\mathrm{X}_b$ constructs a thermochemical cycle:

Rearranging gives the lattice enthalpy.

Worked Example. Calculate the lattice enthalpy of $\mathrm{MgO}$ given:

• $\Delta H_f^\circ(\mathrm{MgO}) = -602\,\mathrm{kJ/mol}$
• $\Delta H_\mathrm{at}^\circ(\mathrm{Mg}) = +148\,\mathrm{kJ/mol}$
• $\Delta H_\mathrm{at}^\circ(\mathrm{O}) = +248\,\mathrm{kJ/mol}$ (per atom; for $\tfrac{1}{2}\mathrm{O}_2$)
• $\mathrm{IE}_1(\mathrm{Mg}) = +738\,\mathrm{kJ/mol}$
• $\mathrm{IE}_2(\mathrm{Mg}) = +1451\,\mathrm{kJ/mol}$
• $\mathrm{EA}_1(\mathrm{O}) = -141\,\mathrm{kJ/mol}$
• $\mathrm{EA}_2(\mathrm{O}) = +798\,\mathrm{kJ/mol}$

The theoretical lattice enthalpy is calculated from the electrostatic model:

Where:

• $N_A$ is Avogadro’s constant
• $M$ is the Madelung constant (depends on the lattice geometry)
• $z^+$, $z^-$ are the ion charges
• $e$ is the elementary charge
• $\varepsilon_0$ is the permittivity of free space
• $r_0$ is the sum of the ionic radii
• $n$ is the Born exponent (related to the electron configuration of the ions)

This equation predicts purely ionic bonding. Differences between theoretical and experimental (Born-Haber) lattice enthalpies indicate the degree of covalent character in the ionic bond.

A small, highly charged cation (high charge density) can polarise a large, deformable anion, drawing electron density towards itself and introducing covalent character.

Fajans’ Rules predict increasing covalent character when:

1. The cation is small and/or highly charged (e.g. $\mathrm{Al}^{3+}$).
2. The anion is large and/or highly charged (e.g. $\mathrm{I}^-$).
3. The cation has an electronic configuration resembling a noble gas with no inert pair (e.g. $\mathrm{Li}^+$ vs $\mathrm{Tl}^+$).

Example: $\mathrm{AgI}$ has significant covalent character because $\mathrm{Ag}^+$ is a relatively soft cation and $\mathrm{I}^-$ is a large, polarisable anion. The experimental lattice enthalpy is less exothermic than the theoretical value.

Entropy is a measure of the number of ways energy can be distributed among the particles of a system — a measure of disorder or randomness.

The second law of thermodynamics states that the total entropy of the universe increases in any spontaneous process:

Standard entropies are always positive (absolute values, not relative like enthalpy).

Worked Example. Calculate $\Delta S^\circ$ for:

Given: S^\circ(\mathrm{CaCO}_3) = 92.9\,\mathrm{J\,mol^{-1}\,K^{-1}}$$S^\circ(\mathrm{CaO}) = 38.1\,\mathrm{J\,mol^{-1}\,K^{-1}}$$S^\circ(\mathrm{CO}_2) = 213.7\,\mathrm{J\,mol^{-1}\,K^{-1}}.

The entropy increases because a gas is produced from a solid (more microstates).

For an exothermic reaction ($\Delta H \lt 0$), $\Delta S_\mathrm{surroundings} \gt 0$ (the surroundings gain heat, increasing their disorder).

The Gibbs free energy combines enthalpy and entropy into a single criterion for spontaneity:

This is valid at constant temperature and pressure (standard laboratory conditions).

The relationship between $\Delta G^\circ$ and the equilibrium constant:

Where $R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $T$ is in Kelvin.

When $\Delta G^\circ = 0$: $K = 1$ (equal amounts of reactants and products at equilibrium).

When $\Delta G^\circ \lt 0$: $K \gt 1$ (products favoured).

When $\Delta G^\circ \gt 0$: $K \lt 1$ (reactants favoured).

Worked Example. For the thermal decomposition of $\mathrm{CaCO}_3$:

\Delta H^\circ = +178\,\mathrm{kJ/mol}$$\Delta S^\circ = +160\,\mathrm{J\,mol^{-1}\,K^{-1}}.

At what temperature does the reaction become spontaneous?

Above 1113\,\mathrm{K}$$\Delta G^\circ \lt 0 and the decomposition is spontaneous.

This shows how $\Delta G/T$ varies with temperature. In the simplified form used at A-Level, it justifies the linear relationship:

A plot of $\Delta G$ vs $T$ is linear with gradient $= -\Delta S$ and y-intercept $= \Delta H$.

The relationship between $\Delta G^\circ$ and the equilibrium constant $K$ is one of the most important equations in physical chemistry:

This equation allows prediction of the equilibrium position from thermodynamic data:

• When $\Delta G^\circ \lt 0$: $K \gt 1$ Products are favoured at equilibrium.
• When $\Delta G^\circ = 0$: $K = 1$ Reactants and products are present in equal amounts.
• When $\Delta G^\circ \gt 0$: $K \lt 1$ Reactants are favoured at equilibrium.

Worked Example. Calculate $K$ at $298\,\mathrm{K}$ for the reaction $\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)$ given $\Delta G^\circ = +4.72\,\mathrm{kJ/mol}$.

$K = 0.149$ So reactants ($\mathrm{N}_2\mathrm{O}_4$) are favoured at equilibrium.

Using $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ with the equilibrium expression:

Combining the two equations:

This is the van ‘t Hoff equation (linear form), which is identical in structure to the Arrhenius equation.

Coffee-cup calorimetry: A polystyrene cup minimises heat exchange with the surroundings. The heat absorbed by the solution equals the heat released by the reaction:

Where $m$ is the mass of solution (g), $c$ is the specific heat capacity ($4.18\,\mathrm{J\,g^{-1}\,K^{-1}}$ for water), and $\Delta T$ is the temperature change.

Sources of error in calorimetry:

1. Heat loss to the surroundings (exothermic reactions underestimate $\Delta H$; endothermic reactions overestimate it).
2. The calorimeter itself absorbs heat ($q_\mathrm{calorimeter} = C_\mathrm{cal}\Delta T$). For accurate work, include this.
3. Incomplete reaction (if the reaction does not go to completion).
4. Non-standard conditions (reactions are not at $298\,\mathrm{K}$ or $100\,\mathrm{kPa}$).

Mean bond enthalpies are averages across different molecules. They introduce systematic error because the actual bond enthalpy in a specific molecule may differ from the mean.

Example of systematic error: The C—O bond enthalpy in $\mathrm{CH}_3\mathrm{OH}$ differs from the mean C—O bond enthalpy (which averages C—O bonds in alcohols, ethers, esters, carboxylic acids). Using mean values for a specific molecule gives only an estimate of $\Delta H$.

Rule: Calculations using mean bond enthalpies give estimated values. Calculations using standard enthalpies of formation give accurate values (provided the data are reliable). Always use formation enthalpies in preference to bond enthalpies when both are available.

The enthalpy of neutralisation is the enthalpy change when one mole of water is formed from the reaction of an acid and a base:

This value is approximately constant for all strong acid-strong base reactions because the net ionic equation is always the same. Weak acid-strong base or strong acid-weak base neutralisations have less exothermic values because some energy is consumed in dissociating the weak acid or weak base.

Calculate $\Delta H_f^\circ$ of $\mathrm{CH}_4(g)$ given:

• $\mathrm{C}(s) + \mathrm{O}_2(g) \to \mathrm{CO}_2(g)$; $\Delta H_1 = -394\,\mathrm{kJ/mol}$
• $\mathrm{H}_2(g) + \tfrac{1}{2}\mathrm{O}_2(g) \to \mathrm{H}_2\mathrm{O}(l)$; $\Delta H_2 = -286\,\mathrm{kJ/mol}$
• $\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \to \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)$; $\Delta H_3 = -890\,\mathrm{kJ/mol}$

Formation: $\mathrm{C}(s) + 2\mathrm{H}_2(g) \to \mathrm{CH}_4(g)$

By Hess’s Law, route the formation through combustion:

1. Sign errors in Hess’s Law cycles. When constructing a cycle, ensure all arrows point in the correct direction. If a step is reversed, change the sign of the enthalpy.

2. Confusing $\Delta S^\circ$ with $S^\circ$. $\Delta S^\circ$ is the change in entropy; $S^\circ$ is the absolute entropy of a single species. $S^\circ$ is always positive; $\Delta S^\circ$ can be positive or negative.

3. Using $\Delta H$ instead of $\Delta H^\circ$ in Gibbs calculations. Standard values must be used for standard free energy calculations.

4. Forgetting units in entropy calculations. Entropy is in $\mathrm{J\,mol^{-1}\,K^{-1}}$ Not $\mathrm{kJ}$. Always convert $\Delta H$ to $\mathrm{J}$ (or $\Delta S$ to $\mathrm{kJ}$) before combining in $\Delta G = \Delta H - T\Delta S$.

5. Misidentifying the theoretical vs experimental lattice enthalpy in Fajans’ Rules discussions. The experimental (Born-Haber) value is less exothermic when covalent character is present; the theoretical (Born-Lande) value assumes perfect ionic bonding.

6. Assuming $\Delta G = 0$ means the reaction has stopped. $\Delta G = 0$ means the reaction is at equilibrium, not that it has stopped. Both forward and reverse reactions continue at equal rates.

7. Using the wrong formula for the surroundings entropy. $\Delta S_\mathrm{surroundings} = -\Delta H / T$. The negative sign is critical: an exothermic reaction ($\Delta H \lt 0$) increases the entropy of the surroundings.

Standard enthalpies of formation ($\Delta H_f^\circ$) are measured experimentally. Mean bond enthalpies are calculated as averages. When both are available, $\Delta H_f^\circ$ data give more accurate results.

When to use bond enthalpies: Use mean bond enthalpies when $\Delta H_f^\circ$ data are not available for all species in the reaction (e.g. For gaseous atoms or free radicals).

Calculate $\Delta H$ for the reaction:

Using mean bond enthalpies: \mathrm{N}\equiv\mathrm{N} = 945\,\mathrm{kJ/mol}$$\mathrm{H}-\mathrm{H} = 436\,\mathrm{kJ/mol}$$\mathrm{N}-\mathrm{H} = 391\,\mathrm{kJ/mol}.

The experimental value is $-92\,\mathrm{kJ/mol}$ Showing good agreement.

• Lattice dissociation enthalpy: Enthalpy change when one mole of an ionic lattice is separated into its gaseous ions (endothermic, always positive).
• Lattice formation enthalpy: Enthalpy change when one mole of an ionic lattice is formed from its gaseous ions (exothermic, always negative).

1. Ionic charge: Higher charge $\to$ stronger electrostatic attraction $\to$ more exothermic lattice enthalpy. $\mathrm{MgO}$ (\mathrm{Mg}^{2+}$$\mathrm{O}^{2-}) has a much more exothermic lattice enthalpy than $\mathrm{NaCl}$ (\mathrm{Na}^+$$\mathrm{Cl}^-).
2. Ionic radius: Smaller ions $\to$ shorter internuclear distance $\to$ stronger attraction $\to$ more exothermic lattice enthalpy. $\mathrm{LiF}$ has a more exothermic lattice enthalpy than $\mathrm{NaF}$ because $\mathrm{Li}^+$ is smaller than $\mathrm{Na}^+$.

When the cation is small and highly charged (e.g. $\mathrm{Al}^{3+}$) and the anion is large and highly charged (e.g. $\mathrm{I}^-$), the electron cloud of the anion is distorted (polarised) towards the cation. This introduces covalent character, making the experimental lattice enthalpy less exothermic than the theoretical (purely ionic) value.

Construct a Born-Haber cycle for $\mathrm{CaCl}_2(s)$ and calculate the lattice enthalpy.

Steps (values are illustrative):

Lattice enthalpy (formation):

Haber process: $\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g)$

• $\Delta H^\circ = -92\,\mathrm{kJ/mol}$ (exothermic)
• $\Delta S^\circ = -199\,\mathrm{J\,mol^{-1}\,K^{-1}}$ (4 moles gas $\to$ 2 moles gas)
• $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

At $298\,\mathrm{K}$: $\Delta G^\circ = -92 - 298(-0.199) = -92 + 59.3 = -32.7\,\mathrm{kJ/mol}$ (spontaneous, $K \approx 5.6 \times 10^5$).

At $700\,\mathrm{K}$: $\Delta G^\circ = -92 - 700(-0.199) = -92 + 139.3 = +47.3\,\mathrm{kJ/mol}$ (non-spontaneous).

The Haber process uses high temperature for kinetic reasons (faster rate), despite the thermodynamic penalty. High pressure shifts equilibrium towards products (fewer moles of gas).

Contact process: $2\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)$

• $\Delta H^\circ = -198\,\mathrm{kJ/mol}$ (exothermic)
• $\Delta S^\circ = -190\,\mathrm{J\,mol^{-1}\,K^{-1}}$
• Optimal temperature is a compromise: $400$—$450^\circ\mathrm{C}$ gives acceptable rate and reasonable equilibrium yield. $\mathrm{V}_2\mathrm{O}_5$ catalyst lowers the activation energy.

The relationship between $\Delta G^\circ$ and the equilibrium constant:

This is one of the most important equations in A-Level chemistry. It connects thermodynamics (energetics) with equilibrium (composition).

Worked Example: Calculate $K_c$ at $298\,\mathrm{K}$ for a reaction with $\Delta G^\circ = -5.40\,\mathrm{kJ/mol}$.

Since \Delta G^\circ < 0$$K > 1 Confirming the reaction is spontaneous and products are favoured at equilibrium.

Worked Example: Calculate the temperature at which $\Delta G = 0$ for the reaction $\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)$ Given $\Delta H^\circ = +57.2\,\mathrm{kJ/mol}$ and \Delta S^\circ = +175.8\,\mathrm{J\,K^{-1}\,\mathrm{mol}^{-1}.

At $\Delta G = 0$: $0 = \Delta H^\circ - T\Delta S^\circ$

Below $325\,\mathrm{K}$, $\Delta G < 0$ and the forward reaction is spontaneous. Above $325\,\mathrm{K}$, $\Delta G > 0$ and the reverse reaction is spontaneous. At $325\,\mathrm{K}$ The system is at equilibrium ($K = 1$).

Worked Example: Calculate $\Delta S^\circ$ for the reaction $\mathrm{CaCO}_3(s) \to \mathrm{CaO}(s) + \mathrm{CO}_2(g)$.

$S^\circ$ values: \mathrm{CaCO}_3(s) = 92.9\,\mathrm{J\,K^{-1}\,\mathrm{mol}^{-1}}$$\mathrm{CaO}(s) = 38.1\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}}$$\mathrm{CO}_2(g) = 213.7\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}}.

The entropy change is positive, as expected: a solid decomposes to give a gas (increased disorder).

Worked Example: Use Hess’s Law to calculate $\Delta H_f^\circ$ for $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l)$.

Given data:

• $\mathrm{C}(s) + \mathrm{O}_2(g) \to \mathrm{CO}_2(g)$, $\Delta H = -393.5\,\mathrm{kJ/mol}$
• $\mathrm{H}_2(g) + \frac{1}{2}\mathrm{O}_2(g) \to \mathrm{H}_2\mathrm{O}(l)$, $\Delta H = -285.8\,\mathrm{kJ/mol}$
• $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(l) + 3\mathrm{O}_2(g) \to 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$, $\Delta H = -1367\,\mathrm{kJ/mol}$

By Hess’s Law:

$\Delta H_f^\circ(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -277\,\mathrm{kJ/mol}$ (to 3 s.f.)

Worked Example: Is the reduction of $\mathrm{TiO}_2$ to $\mathrm{Ti}$ by carbon thermodynamically feasible at $1500\,\mathrm{K}$?

$\Delta H^\circ = +877\,\mathrm{kJ/mol}$, \Delta S^\circ = +193\,\mathrm{J\,K^{-1}\mathrm{mol}^{-1}

$\Delta G^\circ > 0$ So the reaction is not thermodynamically feasible at $1500\,\mathrm{K}$.

At what temperature does it become feasible?

The reaction becomes feasible above approximately $4544\,\mathrm{K}$ (extremely high temperature, impractical). In practice, the Kroll process (reduction with $\mathrm{Mg}$ or $\mathrm{Cl}_2$) is used.