Thermodynamics

Every chemical change moves energy about. Burning a cylinder of LPG in a Kathmandu kitchen releases heat; dissolving ammonium nitrate in water makes the beaker go cold. Thermodynamics is the book-keeping that tells us how much energy moves, in which direction, and — most usefully — whether a reaction will happen at all without being pushed. Three quantities do all the work: enthalpy H, entropy S and Gibbs free energy G.

4.1 Introduction; energy in chemical reactions

Thermodynamics is the study of energy changes accompanying physical and chemical processes. It deals only with bulk, measurable properties — pressure, volume, temperature, heat — and it says nothing about how fast a change occurs (that is kinetics, Unit 3) or about individual molecules.

The part of the universe under study is the system; everything else is the surroundings. The imaginary boundary between them decides what can cross it.

Type of system	Exchanges matter?	Exchanges energy?	Example
Open	yes	yes	tea cooling in an open cup
Closed	no	yes	reaction in a sealed glass bulb
Isolated	no	no	hot water in a perfect thermos flask

Any property that depends only on the present state of the system, and not on the route taken to reach it, is a state function: p, V, T, U, H, S, G. Heat q and work w are path functions — their values depend on how the change was carried out.

Process	Held constant	Condition
Isothermal	temperature	\Delta T = 0
Isobaric	pressure	\Delta p = 0
Isochoric	volume	\Delta V = 0, so w = 0
Adiabatic	no heat exchange	q = 0
Cyclic	system returns to start	\Delta U = 0, \Delta H = 0

Where the energy comes from. A reaction is bond-breaking followed by bond-making. Breaking a bond always costs energy; making a bond always releases it. The net enthalpy change is the difference:

For H₂ + Cl₂ → 2HCl, the bonds broken are one H–H (436 kJ mol⁻¹) and one Cl–Cl (242 kJ mol⁻¹); two H–Cl bonds are formed (431 kJ mol⁻¹ each):

The negative sign means the products hold less energy than the reactants, so 184 kJ escapes into the surroundings for every mole of H₂ consumed. The measured value is −184.6 kJ mol⁻¹.

4.2 Internal energy

Absolute values of U cannot be measured, but changes can:

If the volume is kept constant no mechanical work is possible, so all the heat supplied goes into raising the internal energy:

This is why \Delta U of combustion is measured in a bomb calorimeter, a sealed steel vessel of fixed volume.

4.3 First law of thermodynamics

Everything depends on getting the signs right. The modern (IUPAC) convention is written from the system's point of view:

Quantity	Positive when	Negative when
q	heat is absorbed by the system	heat is released by the system
w	work is done on the system (compression)	work is done by the system (expansion)

The only work a chemist normally meets is pressure–volume work. If a gas expands by \Delta V against a constant external pressure p_{ext}, the system pushes the surroundings back, so it loses energy:

4.4 Enthalpy and enthalpy changes (endothermic and exothermic)

Most reactions in a laboratory are run in open vessels, at the constant pressure of the atmosphere, not at constant volume. The state function suited to constant pressure is enthalpy:

For reactions involving gases, p\Delta V = \Delta n_{g}RT where \Delta n_{g} is the change in the number of moles of gas, so

A reaction is exothermic if it gives heat out (\Delta H negative, products below reactants on an enthalpy diagram) and endothermic if it takes heat in (\Delta H positive, products above reactants).

The sign is part of the answer

"ΔH = 890 kJ for the combustion of methane" is wrong and loses the mark. Heat is released, so \Delta H = -890\ \text{kJ mol}^{-1}. Equally, an exothermic reaction has a negative \Delta H but makes the thermometer reading rise — the two signs are opposite because the thermometer is in the surroundings.

4.5 Enthalpy of reaction, solution, formation, combustion

The standard state of a substance is its most stable form at 1 bar pressure and the stated temperature (usually 298 K). A standard enthalpy change carries the superscript circle, \Delta H^{\circ}.

Name	Definition	Example
Enthalpy of reaction \Delta H_{r}	Heat change when the molar amounts in the balanced equation react completely	N₂ + 3H₂ → 2NH₃, \Delta H^{\circ} = -92.2 kJ
Enthalpy of formation \Delta H_{f}	Heat change when one mole of a compound is formed from its elements in their standard states	C(s) + 2H₂(g) → CH₄(g), \Delta H_{f}^{\circ} = -74.8 kJ mol⁻¹
Enthalpy of combustion \Delta H_{c}	Heat change when one mole of a substance burns completely in excess oxygen	C₂H₅OH(l) + 3O₂ → 2CO₂ + 3H₂O(l), \Delta H_{c}^{\circ} = -1367 kJ mol⁻¹
Enthalpy of solution \Delta H_{sol}	Heat change when one mole of a solute dissolves in so much solvent that further dilution causes no further heat change	NH₄Cl(s) + water → NH₄⁺(aq) + Cl⁻(aq), \Delta H_{sol} = +15.1 kJ mol⁻¹
Enthalpy of neutralisation	Heat change when one gram-equivalent of acid is neutralised by a base in dilute solution	H⁺(aq) + OH⁻(aq) → H₂O(l), \Delta H = -57.1 kJ

Two consequences worth remembering. First, \Delta H_{f}^{\circ} of any element in its standard state is zero — O₂(g), C(graphite), Br₂(l) all have \Delta H_{f}^{\circ} = 0. Second, once a table of formation enthalpies exists, any reaction enthalpy follows from

Enthalpy of solution is itself the sum of two opposing terms: the lattice enthalpy (energy needed to pull the ionic lattice apart, always positive) and the hydration enthalpy of the separated ions (always negative). NH₄Cl dissolves endothermically because its lattice enthalpy slightly outweighs hydration — which is why an instant cold pack works.

4.6 Laws of thermochemistry: Laplace law and Hess's law

For example, if C(s) + O₂(g) → CO₂(g) has \Delta H = -393.5 kJ mol⁻¹, then CO₂(g) → C(s) + O₂(g) has \Delta H = +393.5 kJ mol⁻¹. The law is just a restatement of the first law: if it were not true you could run the forward and backward reactions in a loop and manufacture energy from nothing.

Hess's law follows because H is a state function: \Delta H depends only on the endpoints, never the route. So around any closed cycle the enthalpy changes sum to zero, and an unmeasurable \Delta H can be found from measurable ones.

The Born–Haber cycle

The most important application of Hess's law in inorganic chemistry is the Born–Haber cycle, which finds the lattice enthalpy of an ionic solid — the enthalpy change when one mole of the crystal is formed from its gaseous ions. Lattice enthalpy cannot be measured directly; every other step in the cycle can.

4.7 Entropy and spontaneity

A spontaneous process is one that occurs on its own, without continuous outside help: ice melting in a glass at room temperature, iron rusting, a gas filling a vacuum. Exothermicity alone does not explain spontaneity — ice melts spontaneously above 0 °C even though melting is endothermic. A second driving force is at work: the tendency towards disorder.

Entropy increases when disorder increases:

• Phase changes: S_{\text{solid}} < S_{\text{liquid}} \ll S_{\text{gas}}.
• Number of gas molecules: CaCO₃(s) → CaO(s) + CO₂(g) creates a gas, so \Delta S > 0. N₂ + 3H₂ → 2NH₃ destroys two moles of gas, so \Delta S < 0.
• Dissolving an ionic solid and mixing are entropy-increasing.
• Heating anything raises its entropy.

For a phase change at its transition temperature the process is reversible, so \Delta S = \Delta H / T. For the melting of ice, \Delta S_{fus} = 6010 / 273 = 22.0 J K⁻¹ mol⁻¹; for the boiling of water, \Delta S_{vap} = 40700 / 373 = 109.1 J K⁻¹ mol⁻¹. Boiling disorders far more, as expected.

The third law of thermodynamics completes the picture: the entropy of a perfectly crystalline substance at absolute zero is zero. That gives entropy an absolute scale, which is why tables list S^{\circ} itself, not just \Delta S^{\circ}.

4.8 Second law of thermodynamics

The first law says energy is conserved; it does not say which direction a change will go. The second law supplies the direction.

Writing the universe as system plus surroundings,

Note that \Delta S_{system} alone may be negative — water freezing in a Mustang winter becomes more ordered — provided the surroundings gain more entropy than the system loses.

4.9 Gibbs free energy and prediction of spontaneity

Testing \Delta S_{total} means measuring the surroundings, which is awkward. J. W. Gibbs removed the need by defining a function of the system only:

\Delta H	\Delta S	\Delta G = \Delta H - T\Delta S	Spontaneous?
−	+	always negative	Yes, at all temperatures
−	−	negative at low T	Only below T = \Delta H/\Delta S
+	+	negative at high T	Only above T = \Delta H/\Delta S
+	−	always positive	Never

\Delta G also has a mechanical meaning: -\Delta G is the maximum useful (non-expansion) work obtainable from the change. At equilibrium \Delta G = 0 and no further work can be extracted.

4.10 Relationship between \Delta G and equilibrium constant

For a reaction not at standard concentrations, the free energy change is

where Q is the reaction quotient — the same expression as the equilibrium constant but with the actual, not equilibrium, concentrations. As the reaction proceeds, Q changes and \Delta G rises towards zero.

\Delta G^{\circ}	K	Position of equilibrium
negative	K > 1	products favoured
zero	K = 1	neither favoured
positive	K < 1	reactants favoured

Chapter summary

• A system exchanges matter and/or energy with its surroundings. U, H, S, G, p, V, T are state functions; q and w are not.
• First law: \Delta U = q + w with w = -p_{ext}\Delta V. At constant volume q_{v} = \Delta U; at constant pressure q_{p} = \Delta H.
• H = U + pV, so \Delta H = \Delta U + \Delta n_{g}RT for gas reactions. \Delta H < 0 exothermic, \Delta H > 0 endothermic.
• \Delta H_{r}^{\circ} = \sum\Delta H_{f}^{\circ}(\text{products}) - \sum\Delta H_{f}^{\circ}(\text{reactants}), and \Delta H_{f}^{\circ} of an element in its standard state is zero.
• Lavoisier–Laplace: reversing a reaction reverses the sign of \Delta H. Hess's law: \Delta H is route-independent — the basis of the Born–Haber cycle, which gives lattice enthalpy (−787 kJ mol⁻¹ for NaCl).
• Entropy S measures disorder, \Delta S = q_{rev}/T in J K⁻¹ mol⁻¹; S rises solid → liquid → gas.
• Second law: \Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0 for every spontaneous change.
• Gibbs: \Delta G = \Delta H - T\Delta S; spontaneous if \Delta G < 0, at equilibrium if \Delta G = 0; crossover at T = \Delta H/\Delta S.
• \Delta G = \Delta G^{\circ} + RT\ln Q and \Delta G^{\circ} = -RT\ln K = -2.303\,RT\log K.

Practice questions

Group A — Multiple choice (1 mark each)

1. Which of the following is not a state function? [1]ainternal energybenthalpycworkdentropy
2. For the isothermal expansion of an ideal gas, \Delta U is [1]apositivebnegativeczerodequal to w
3. For N₂(g) + 3H₂(g) → 2NH₃(g), the value of \Delta H - \Delta U is [1]a+2RTb-2RTc+RTdzero
4. A reaction is spontaneous at all temperatures when [1]a\Delta H < 0, \Delta S < 0b\Delta H > 0, \Delta S > 0c\Delta H < 0, \Delta S > 0d\Delta H > 0, \Delta S < 0
5. If \Delta G^{\circ} = 0 for a reaction at temperature T, then K equals [1]a0b1cinfinityde
6. The entropy of a perfectly crystalline solid at 0 K is [1]azerobpositivecnegativedinfinite

Answers to Group A

1. (c) — work depends on the path taken, not just on the endpoints.

2. (c) — U of an ideal gas depends only on T, and T is constant.

3. (b) — \Delta n_{g} = 2 - 4 = -2, and \Delta H - \Delta U = \Delta n_{g}RT.

4. (c) — then -T\Delta S is negative too, so \Delta G is negative at every T.

5. (b) — \Delta G^{\circ} = -RT\ln K, so \ln K = 0 and K = 1.

6. (a) — third law of thermodynamics: perfect order means zero disorder.

Group B — Short answer (5 marks each)

1. Define internal energy. State the first law of thermodynamics and derive the expression w = -p_{ext}\Delta V for the work done in an isobaric expansion. [5]
2. Define enthalpy of formation, enthalpy of combustion, enthalpy of solution and enthalpy of neutralisation, giving one thermochemical equation for each. [5]
3. State Hess's law. Given C(s) + O₂(g) → CO₂(g), \Delta H = -393.5 kJ mol⁻¹ and CO(g) + ½O₂(g) → CO₂(g), \Delta H = -283.0 kJ mol⁻¹, calculate the enthalpy of formation of carbon monoxide. [5]
4. For a certain reaction \Delta H = 30.56 kJ mol⁻¹ and \Delta S = 66.0 J K⁻¹ mol⁻¹. Calculate \Delta G at 300 K and find the temperature at which the reaction becomes spontaneous. [5]
5. Derive \Delta G = \Delta H - T\Delta S from the second law, and use it to discuss the four possible sign combinations of \Delta H and \Delta S. [5]
6. The equilibrium constant of a reaction at 298 K is 1.0\times10^{5}. Calculate \Delta G^{\circ} and comment on the position of equilibrium. [5]

Answers to Group B

3. Hess's law: the enthalpy change is the same by whatever route the reaction is carried out. The target is C(s) + ½O₂(g) → CO(g). Subtract the second equation from the first (i.e. reverse the second and add):

C(s) + O₂ → CO₂ ..... -393.5; CO₂ → CO + ½O₂ ..... +283.0

\Delta H_{f}^{\circ}(\text{CO}) = -393.5 + 283.0 = -110.5\ \text{kJ mol}^{-1}

4. With \Delta S = 0.0660 kJ K⁻¹ mol⁻¹,

\Delta G_{300} = 30.56 - (300)(0.0660) = 30.56 - 19.80 = +10.76\ \text{kJ mol}^{-1}

so it is non-spontaneous at 300 K. Setting \Delta G = 0, T = 30.56/0.0660 = 463\ \text{K}; above 463 K the reaction is spontaneous.

6. \Delta G^{\circ} = -2.303RT\log K = -(2.303)(8.314)(298)\log(10^{5}) = -(5706)(5) = -28530\ \text{J mol}^{-1} = -28.53\ \text{kJ mol}^{-1}. Since \Delta G^{\circ} is large and negative and K \gg 1, the equilibrium lies far to the product side.

Group C — Long answer (8 marks each)

1. (a) State Hess's law of constant heat summation and explain why it follows from enthalpy being a state function. [3](b) Construct a Born–Haber cycle for potassium chloride and calculate its lattice enthalpy from: \Delta H_{f}^{\circ}(\text{KCl}) = -437 kJ mol⁻¹, enthalpy of sublimation of K = +89 kJ mol⁻¹, ionisation energy of K = +419 kJ mol⁻¹, bond dissociation enthalpy of Cl₂ = +242 kJ mol⁻¹, electron affinity of Cl = −349 kJ mol⁻¹. Explain why the lattice enthalpy of NaCl is more negative than that of KCl. [5]
2. (a) State the second law of thermodynamics in two different ways and define entropy. Explain, using \Delta S_{total}, how water can freeze spontaneously in winter although freezing decreases the entropy of the water. [4](b) The enthalpy of vaporisation of water at 373 K is 40.7 kJ mol⁻¹. Calculate \Delta S_{vap}, and hence \Delta G for the vaporisation at 373 K and at 383 K. Interpret both answers. [4]

Answers to Group C

1.(b) The steps are: K(s) → K(g), +89; ½Cl₂(g) → Cl(g), \tfrac{1}{2}(242) = +121; K(g) → K⁺(g) + e⁻, +419; Cl(g) + e⁻ → Cl⁻(g), −349; K⁺(g) + Cl⁻(g) → KCl(s), U. By Hess's law

-437 = 89 + 121 + 419 - 349 + U = 280 + U \;\Longrightarrow\; U = -717\ \text{kJ mol}^{-1}

Lattice enthalpy varies as 1/(r_{+} + r_{-}). Na⁺ (102 pm) is smaller than K⁺ (138 pm), so the ions in NaCl sit closer together, the electrostatic attraction is stronger and the lattice enthalpy is more negative (−787 against −717 kJ mol⁻¹).

2.(b) Vaporisation at the normal boiling point is a reversible change, so

\Delta S_{vap} = \frac{\Delta H_{vap}}{T} = \frac{40700}{373} = 109.1\ \text{J K}^{-1}\text{mol}^{-1}

At 373 K: \Delta G = 40700 - (373)(109.1) = 0. Liquid and vapour are in equilibrium — which is what "boiling point" means.

At 383 K: \Delta G = 40700 - (383)(109.1) = 40700 - 41785 = -1085\ \text{J mol}^{-1} = -1.09\ \text{kJ mol}^{-1}, negative, so above 373 K vaporisation is spontaneous and the liquid boils away.