Chemical Thermodynamics

1. System, Surroundings & Types of Systems

Thermodynamics deals with energy changes in chemical reactions and processes, and enables us to study these changes quantitatively and to make useful predictions. Thermodynamics is not concerned about how or at what rate these energy transformations are carried out — it is based only on the initial and final states of a system undergoing change. Laws of thermodynamics apply only when a system is in equilibrium, or moves from one equilibrium state to another.

A system is that part of the universe in which observations are made, and the remaining universe constitutes the surroundings. The system and the surroundings together constitute the universe. The wall that separates the system from the surroundings is called the boundary; it may be real or imaginary, and controls the flow of matter and energy in or out of the system.

• Open system: exchange of both matter and energy with the surroundings is possible (e.g. reactants in an open beaker).
• Closed system: no exchange of matter, but exchange of energy is possible (e.g. reactants in a closed vessel made of a conducting material).
• Isolated system: no exchange of either matter or energy with the surroundings (e.g. reactants kept in a thermos flask).

The state of a system: the system must be described in order to make any useful calculations, by specifying quantitatively its pressure ($p$), volume ($V$), temperature ($T$), and composition. These are called state variables or state functions, because their values depend only on the state of the system, and not on how that state was reached. Note: the state of the surroundings can never be completely specified, and fortunately it is not necessary to do so.

2. Internal Energy, Work, Heat & the First Law

The sum total of all forms of energy (chemical, electrical, mechanical, or any other) possessed by a system is called its internal energy, $U$. Internal energy is a state function — J. P. Joule (1840–50) showed experimentally that a given amount of work done on a system, no matter how it was done (irrespective of path), produced the same change in temperature and hence the same change in internal energy.

There are two ways the internal energy of a system can be changed:

(a) By doing work: if no heat is exchanged between the system and surroundings (an adiabatic process, $q = 0$), the change in internal energy equals the adiabatic work done: $\Delta U = U_2 - U_1 = \text{w}_{ad}$. By IUPAC convention, w is positive when work is done on the system (internal energy increases), and negative when work is done by the system.

(b) By transfer of heat: without expenditure of work, energy can also flow as heat, $q$, through thermally conducting walls, from higher to lower temperature. By IUPAC convention, $q$ is positive when heat is transferred from the surroundings to the system.

The general case — First Law of Thermodynamics: when a change of state is brought about by both heat transfer and work,

$$\Delta U = q + \text{w} \tag{6.1}$$

This is the mathematical statement of the first law of thermodynamics, which states that the energy of an isolated system is constant. It is the law of conservation of energy: energy can neither be created nor destroyed, although it can be transformed from one form to another. While $q$ and w individually depend on the path taken, their sum $\Delta U$ depends only on the initial and final states.

Work done in expansion/compression against external pressure: for a gas expanding against a constant external pressure $p_{ex}$ (irreversible, single step),

$$\text{w} = -p_{ex}(V_f - V_i) = -p_{ex}\Delta V \tag{6.2}$$

The negative sign ensures the correct convention: during expansion ($V_f > V_i$), work is done by the system, so w is negative.

For a reversible process (where $p_{ex}$ is always only infinitesimally different from the internal pressure, so the process occurs through a continuous series of equilibrium states):

$$\text{w}_{rev} = -\int_{V_i}^{V_f} p_{in}\,dV = -2.303\,nRT\log\frac{V_f}{V_i} \tag{6.5 (isothermal, ideal gas)}$$

Isothermal free expansion of an ideal gas ($p_{ex} = 0$, into vacuum): w $= 0$. Also, Joule showed experimentally that for an ideal gas, $q = 0$ too; therefore $\Delta U = 0$. No work is done during free expansion of an ideal gas, whether the process is reversible or irreversible.

3. Enthalpy: ΔH and ΔU

Most reactions are carried out not at constant volume, but in open vessels, i.e. under constant (atmospheric) pressure. It is useful to define a new state function suitable under these conditions. Starting from $\Delta U = q_p - p\Delta V$ at constant pressure, we define enthalpy, $H$:

$$H = U + pV \tag{6.7}$$

so that $q_p = \Delta H = \Delta U + p\Delta V$ (finite changes at constant pressure). Since $H$ depends only on $U$, $p$, $V$ — all state functions — $\Delta H$ is independent of path, i.e. it is itself a state function, even though $q$ is not.

$\Delta H$ is negative for exothermic reactions (heat evolved) and positive for endothermic reactions (heat absorbed).

For a reaction involving ideal gases, using $pV = nRT$ for both reactants and products at constant $T$ and $p$:

$$\Delta H = \Delta U + \Delta n_g RT \tag{6.10}$$

where $\Delta n_g$ = (moles of gaseous products) − (moles of gaseous reactants). This equation is the key link between the two state functions, and is used constantly to convert between calorimetrically measured $\Delta U$ (bomb calorimeter, constant volume) and $\Delta H$ (constant-pressure processes).

4. Extensive/Intensive Properties & Heat Capacity

An extensive property depends on the quantity or size of matter present in the system (e.g. mass, volume, internal energy, enthalpy, heat capacity). An intensive property does not depend on the amount of matter present (e.g. temperature, density, pressure). A molar property, $\chi_m = \chi/n$, is the value of an extensive property for 1 mole of substance, and is itself intensive.

Heat capacity: the increase of temperature is proportional to the heat transferred, $q = C\Delta T$, where $C$ is the heat capacity. The molar heat capacity, $C_m = C/n$, is the heat needed to raise the temperature of one mole of a substance by one degree.

At constant volume, $q_V = C_V\Delta T = \Delta U$; at constant pressure, $q_p = C_p\Delta T = \Delta H$. For one mole of an ideal gas, using $\Delta H = \Delta U + \Delta(RT) = \Delta U + R\Delta T$:

$$C_p\Delta T = C_V\Delta T + R\Delta T \quad\Rightarrow\quad C_p - C_V = R \tag{6.13}$$

5. Measurement of ΔU and ΔH: Calorimetry

Calorimetry measures energy changes associated with chemical or physical processes experimentally.

• $\Delta U$ measurement — bomb calorimeter: a combustible substance is burnt in pure dioxygen inside a sealed steel vessel (the "bomb") immersed in a water bath. Since the vessel's volume is fixed, no work is done ($\Delta V = 0$), so the heat released is measured directly as $q_V = \Delta U$.
• $\Delta H$ measurement: carried out in a calorimeter open to the atmosphere (constant pressure), so the heat measured is directly $q_p = \Delta H$.

6. Enthalpies of Reactions & Hess's Law

The enthalpy change accompanying a reaction is called the reaction enthalpy:

$$\Delta_r H = \sum_i a_i H_{products} - \sum_i b_i H_{reactants} \tag{6.14}$$

The standard enthalpy of reaction ($\Delta_r H^\ominus$) is the enthalpy change when all participating substances are in their standard states (pure form, 1 bar, usually 298 K).

Named enthalpy changes you must recognise:

• Standard enthalpy of fusion / vaporisation / sublimation ($\Delta_{fus}H^\ominus$, $\Delta_{vap}H^\ominus$, $\Delta_{sub}H^\ominus$): enthalpy change for melting / vaporising / directly subliming one mole of a substance at constant temperature under standard pressure.
• Standard enthalpy of formation ($\Delta_f H^\ominus$): enthalpy change when one mole of a compound is formed from its elements in their most stable states of aggregation (reference states). $\Delta_f H^\ominus$ of every element in its reference state is, by convention, zero.
• Standard enthalpy of combustion ($\Delta_c H^\ominus$): enthalpy change per mole when a substance undergoes complete combustion, with all reactants and products in standard states.
• Bond enthalpy ($\Delta_{bond}H^\ominus$): for diatomic molecules, the bond dissociation enthalpy equals the enthalpy of atomization (e.g. H₂(g) → 2H(g)). For polyatomic molecules, since successive bond-breaking steps require different energies, a mean bond enthalpy (average over all equivalent bonds) is used instead.
• Lattice enthalpy ($\Delta_{lattice}H^\ominus$): enthalpy change when one mole of an ionic compound dissociates into its ions in the gaseous state, e.g. Na⁺Cl⁻(s) → Na⁺(g) + Cl⁻(g). Since it cannot be measured directly, it is obtained indirectly via a Born–Haber cycle, which applies Hess's law to a closed loop of sublimation, ionization, dissociation, electron gain, and lattice steps.
• Enthalpy of solution ($\Delta_{sol}H^\ominus$): enthalpy change when one mole of a substance dissolves in a specified amount of solvent. $\Delta_{sol}H^\ominus = \Delta_{lattice}H^\ominus + \Delta_{hyd}H^\ominus$.

Hess's Law of Constant Heat Summation: since enthalpy is a state function, if a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided, at the same temperature.

Hess's law lets us calculate enthalpy changes that cannot be measured directly (e.g. C(graphite) + ½O₂(g) → CO(g), which always produces some CO₂ alongside CO) by combining other, measurable reactions algebraically — reversing an equation reverses the sign of its $\Delta H$, and multiplying an equation by a factor multiplies its $\Delta H$ by the same factor.

7. Spontaneity: Is ΔH Alone Enough?

A spontaneous process is one that has the potential to proceed without the assistance of external agency (it may be slow or fast) and, once started, cannot reverse its direction on its own — a spontaneous process is an irreversible process and may only be reversed by some external agency.

It is tempting to think that a decrease in enthalpy (exothermic reactions like combination of H₂ and O₂, or neutralisation) is the driving force for spontaneity, since this mirrors familiar mechanical analogies (a ball rolling downhill). However, this cannot be the whole story: many endothermic reactions are also spontaneous, for example the dissolution of many salts, or:

$$\text{C(graphite, s)} + 2\text{S(l)} \rightarrow \text{CS}_2(l); \quad \Delta_r H^\ominus = +128.5\text{ kJ mol}^{-1}$$

So while a decrease in enthalpy may be a contributory factor for spontaneity, it is not true for all cases — enthalpy alone cannot decide the direction of spontaneous change.

8. Entropy & the Second Law of Thermodynamics

What actually drives spontaneous change? Consider two gases diffusing into each other in an isolated container: initially each molecule's identity (gas A or gas B) is known with certainty; once mixed, the system has become less predictable, more disordered/chaotic. This suggests a postulate: in an isolated system, there is always a tendency for the system's energy to become more disordered or chaotic, and this could be a criterion for spontaneous change.

Entropy, $S$, is the thermodynamic state function that measures the degree of randomness or disorder in a system. Like $U$ and $H$, entropy is a state function, and $\Delta S$ is independent of path. For a reversible process at temperature $T$:

$$\Delta S = \frac{q_{rev}}{T} \tag{6.18}$$

Second Law of Thermodynamics: the total entropy change of the system and surroundings ($\Delta S_{total}$) for a spontaneous process is always greater than zero:

$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} > 0 \tag{6.19}$$

At equilibrium, $\Delta S_{total} = 0$. This is the crucial refinement over judging spontaneity from $\Delta S_{system}$ alone: a reaction with negative $\Delta S_{sys}$ can still be spontaneous overall, provided $\Delta S_{surr}$ (which depends on how exothermic the reaction is, $\Delta S_{surr} = -\Delta H_{sys}/T$ at constant $p$) is large enough and positive to make $\Delta S_{total}$ positive.

Third Law of Thermodynamics: the entropy of a perfectly ordered crystalline substance approaches zero as the temperature approaches absolute zero.

9. Gibbs Energy & Criteria for Spontaneity

Tracking $\Delta S_{total}$ requires computing changes in the surroundings as well as the system — inconvenient for chemical reactions, which are usually studied from the system's point of view alone. We define a new state function, the Gibbs energy (or Gibbs function), $G$:

$$G = H - TS \tag{6.20}$$

At constant temperature, the change in Gibbs energy of the system works out to:

$$\Delta G = \Delta H - T\Delta S \tag{6.21}$$

This equation, known as the Gibbs equation, is derived directly from the second law ($\Delta S_{total} > 0 \Rightarrow -(\Delta H_{sys} - T\Delta S_{sys}) > 0 \Rightarrow \Delta G < 0$), so it packages both enthalpy and entropy considerations into a single criterion, evaluated using only the system's own properties:

• If $\Delta G$ is negative ($< 0$), the process is spontaneous.
• If $\Delta G$ is positive ($> 0$), the process is non-spontaneous.
• At equilibrium, $\Delta G = 0$.

Since $\Delta G = \Delta H - T\Delta S$, temperature can decide whether a reaction proceeds: if $\Delta H$ and $\Delta S$ are both positive, the reaction is non-spontaneous at low $T$ but becomes spontaneous once $T$ is large enough to make $T\Delta S$ outweigh $\Delta H$; if both are negative, the reverse pattern holds.

10. Gibbs Energy Change and Equilibrium

At equilibrium, $\Delta_r G = 0$, and the Gibbs energy of a reaction in which reactants and products are in their standard states is related to the equilibrium constant $K$ by:

$$\Delta_r G^\ominus = -RT\ln K = -2.303\,RT\log K \tag{6.23}$$

Also, combining with the Gibbs equation:

$$\Delta_r G^\ominus = \Delta_r H^\ominus - T\Delta_r S^\ominus = -RT\ln K \tag{6.24}$$

This is one of the most important results in the whole of chemical thermodynamics, because it connects a purely thermodynamic quantity ($\Delta_r G^\ominus$, calculable from tabulated $\Delta_f H^\ominus$ and standard entropies) to the equilibrium constant $K$, which describes the actual extent to which a reaction proceeds:

• Strongly exothermic reactions ($\Delta_r H^\ominus$ large and negative) tend to have large, negative $\Delta_r G^\ominus$, and hence $K \gg 1$ (reaction goes nearly to completion).
• Strongly endothermic reactions ($\Delta_r H^\ominus$ large and positive) tend to have large, positive $\Delta_r G^\ominus$, and hence $K \ll 1$ (reaction barely proceeds).
• $\Delta_r S^\ominus$ also matters: even an exothermic reaction with a sufficiently unfavourable (negative) entropy change can have a smaller $K$ than expected from enthalpy alone, and vice versa.