4. Thermodynamics
🎯 Learning outcomes
After completing this lesson, you should be able to:
- Define thermodynamics and explain its basic terms: system, surroundings, boundary and universe.
- Classify systems as open, closed or isolated, and give real-life examples of each.
- Define state functions and distinguish them from path functions.
- Distinguish between extensive and intensive properties with examples.
- Define internal energy and explain its characteristics.
- Explain the energy change in chemical reactions.
- State and explain enthalpy and enthalpy changes in various processes.
- State and explain the first law of thermodynamics and derive ΔH = ΔE + PΔV.
- Prove that ΔE = qv and ΔH = qp.
- Define and calculate enthalpy of reaction, formation, combustion and neutralisation.
- Draw and interpret energy profile diagrams for exothermic and endothermic reactions.
- State the laws of thermochemistry.
- State Hess’s law and apply it to calculate unknown enthalpy changes.
- Estimate enthalpy changes using bond enthalpies.
- Define entropy, state and explain the second law of thermodynamics, and predict spontaneity.
- Define Gibbs free energy and use ΔG = ΔH − TΔS to predict spontaneity and calculate ΔG for a reaction.
- Relate ΔG° to the equilibrium constant K.
1Basic terms and types of system
Thermodynamics is the branch of science that studies energy transformations — particularly between heat and work — and the direction of spontaneous change in physical and chemical processes.
- How much heat is released when fuel burns?
- Why does sugar dissolve in water but not in oil?
- Can a reaction occur on its own, or does it need energy input?
- What is the maximum efficiency of a heat engine?
The specific part of the universe chosen for study — e.g. the contents of a beaker, a reaction mixture, or a gas in a cylinder. Systems may be homogeneous or heterogeneous.
Everything outside the system that can interact with it. It can exchange energy (and sometimes matter) with the system.
The real or imaginary surface separating the system from its surroundings.
Dissolving sugar in water in a beaker: system = sugar + water; surroundings = the beaker, the lab and everything else; boundary = the walls and surface of the beaker.
Open, closed and isolated systems
| Type | Matter exchange | Energy exchange | Example |
|---|---|---|---|
| Open | Yes | Yes | Hot tea in an open cup |
| Closed | No | Yes | Gas in a sealed, flexible balloon |
| Isolated | No | No | Coffee in a sealed thermos flask |
A well-stoppered thermos flask containing ice cubes is the classic isolated system: thick walls prevent heat exchange and the stopper prevents matter escaping.
2State functions and properties of a system
A property whose value depends only on the current state of the system (pressure, temperature, volume, composition), not on the path taken to reach it.
Examples: T, P, V, n, internal energy E, enthalpy H, entropy S, Gibbs free energy G.
Depends on the specific route between two states. Examples: heat (q) and work (w).
Climbing from Kathmandu (1,400 m) to Nagarkot (2,100 m): the change in altitude (700 m) is a state function — the same whichever road you take. The distance walked is a path function — it depends on your route.
Extensive vs intensive properties
Depend on the amount of matter; double the substance and they double.
V, mass, E, H, S, G.
Independent of the amount of matter.
T, P, density, boiling point, refractive index, molar volume.
Intensive = Independent of amount (the letter “i” in both words). Density is intensive — a drop of water and a bucket of water have the same density.
3Heat, work and internal energy
Heat
Energy transferred between system and surroundings because of a temperature difference.
| Situation | Sign of q |
|---|---|
| Heat absorbed by the system | q = positive (+) |
| Heat released by the system | q = negative (−) |
Work
Mechanical work in chemistry is usually pressure–volume (P–V) work: a gas expanding or being compressed against an external pressure.
| Situation | Sign of w |
|---|---|
| Work done on the system (compression) | w = positive (+) |
| Work done by the system (expansion) | w = negative (−) |
Unit conversion: 1 calorie = 4.184 joule.
Internal energy (E or U)
The total energy stored within a system: kinetic energy (translational, rotational, vibrational) plus potential energy (intermolecular attractions, bond energies).
- Extensive property — depends on the amount of substance.
- State function — only the change ΔE = E2 − E1 matters.
- Not directly measurable — we measure changes, not absolute values.
- Excludes nuclear energy, gravitational energy, and the kinetic energy of the whole system moving through space.
4The first law of thermodynamics
“Energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy of the universe (system + surroundings) remains constant.”
Also: it is impossible to build a perpetual motion machine of the first kind — one that produces work without consuming energy.
Mathematical form
When a system absorbs heat q and has work w done on it:
If the only work is P–V work, w = −PΔV, so:
Deriving ΔH = ΔE + PΔV
- From the first law at constant pressure: ΔE = qp − PΔV
- So qp = ΔE + PΔV
- Enthalpy is defined as H = E + PV
- So ΔH = ΔE + Δ(PV) = ΔE + PΔV (at constant P)
- Therefore ΔH = qp
At constant volume ΔV = 0, so PΔV = 0. Then ΔE = q + w = q − PΔV = qv.
The enthalpy change equals the heat measured at constant pressure; the internal energy change equals the heat measured at constant volume. Most laboratory reactions run at atmospheric pressure, so we measure ΔH. For gases, PΔV = ΔngRT, so ΔH = ΔE + ΔngRT.
Different forms of process
| Process | Condition |
|---|---|
| Isobaric | Constant pressure |
| Isothermal | Constant temperature |
| Isochoric | Constant volume |
| Adiabatic | No heat exchange (q = 0) |
It does not tell us the direction of a process — it only accounts for conservation of energy. It cannot explain why heat flows spontaneously from hot to cold and never the reverse. The second law addresses this.
A system absorbs 10 kJ of heat and does 6 kJ of work on the surroundings. Find ΔE.
q = +10 kJ (heat absorbed). Work done by the system, so w = −6 kJ.
ΔE = q + w = +10 + (−6) = +4 kJ — the internal energy increases by 4 kJ.
5Enthalpy and energy profile diagrams
A state function equal to the sum of internal energy and the pressure–volume product.
Absolute values cannot be measured meaningfully, so we use the change in enthalpy, ΔH — the heat exchanged at constant pressure, in kJ mol−1. The standard state means 298 K (25 °C) and 1 atm (101 kPa).
Every reaction breaks some bonds (energy needed) and forms others (energy given out). These rarely balance, so energy is absorbed or released — usually as heat, but sometimes as sound, light or electricity.
Gives out heat; ΔH is negative. e.g. combustion, neutralisation.
Takes in heat; ΔH is positive. e.g. all thermal decomposition.
The minimum energy colliding particles must have to break bonds and start a reaction. It is always positive. The transition state (activated complex) is the unstable, high-energy species at the top of the energy “hill”.
| Feature | Exothermic | Endothermic |
|---|---|---|
| Heat flow | Released to surroundings | Absorbed from surroundings |
| Sign of ΔH | Negative (ΔH < 0) | Positive (ΔH > 0) |
| Energy of products | Lower than reactants | Higher than reactants |
| Surroundings | Temperature increases | Temperature decreases |
| Example | CH4 + 2O2 → CO2 + 2H2O | CaCO3 → CaO + CO2 |
6Standard enthalpy changes
The enthalpy change when the amounts of reactants shown in the stoichiometric equation react to give products under standard conditions, all in their standard states.
The enthalpy change when one mole of a compound is formed from its elements under standard conditions. ΔH°f of any element in its standard state is 0.
Formation means exactly 1 mole of compound from elements in standard states.
Correct: 2Na(s) + ½O2(g) → Na2O(s)
Incorrect: 4Na(s) + O2(g) → 2Na2O(s) (forms 2 mol)
The enthalpy change when one mole of a substance is burnt in excess oxygen under standard conditions. Always negative.
The enthalpy change when one mole of water is formed by reaction of an acid with an alkali under standard conditions.
Why is ΔH°neut constant for strong acid + strong base?
Both are fully ionised, so the net ionic equation is always the same, with the same bonds broken and formed:
Some energy is needed to fully ionise the weak acid/base. This extra energy must be supplied, so |ΔH°neut| is less than 57.1 kJ mol−1.
7Laws of thermochemistry and Hess’s law
The enthalpy change of decomposing a compound is equal in magnitude but opposite in sign to its enthalpy change of formation: ΔH(decomposition) = −ΔH(formation).
Hess applied the law of conservation of energy to enthalpy changes in 1840.
“The total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place, provided the initial and final states are the same.” — Germain Henri Hess, 1840
Enthalpy cycle
If a reaction can occur in one step or in several, the total ΔH is the same. For carbon forming CO2:
- Route 1 (direct): C(s) + O2(g) → CO2(g); ΔH1
- Route 2, step A: C(s) + ½O2(g) → CO(g); ΔHA
- Route 2, step B: CO(g) + ½O2(g) → CO2(g); ΔHB
Applications
- Enthalpy of formation of compounds that cannot be formed directly from elements (e.g. CO, C2H2)
- Enthalpy of reactions difficult or impossible to measure directly
- Enthalpy of combustion from enthalpy of formation data
- Lattice energy via the Born–Haber cycle
- Verifying experimental thermochemical data
Method
- Write the target equation.
- Use the given equations: reversing one changes the sign of ΔH; multiplying by a factor multiplies ΔH by the same factor.
- Add the equations so that intermediate species cancel to give the target.
- Add the ΔH values algebraically.
Find ΔH°f of CS2(l), given (i) C + O2 → CO2, ΔH = −395 kJ; (ii) S + O2 → SO2, ΔH = −295 kJ; (iii) CS2 + 3O2 → CO2 + 2SO2, ΔH = −1110 kJ. Target: C(s) + 2S(s) → CS2(l).
Keep (i): ΔH = −395 kJ
Multiply (ii) by 2: 2S + 2O2 → 2SO2, ΔH = −590 kJ
Reverse (iii): CO2 + 2SO2 → CS2 + 3O2, ΔH = +1110 kJ
Adding and cancelling CO2, 2SO2 and 3O2 gives C + 2S → CS2.
ΔH = −395 − 590 + 1110 = +125 kJ mol−1
8Bond enthalpy
The energy required to break one mole of a specific covalent bond in the gaseous phase, forming gaseous atoms. Breaking bonds requires energy (endothermic); forming bonds releases energy (exothermic).
Values are average bond enthalpies (the exact value varies with the molecular environment), so results are only approximate.
Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g). H–H = 435, Cl–Cl = 243, H–Cl = 430 kJ mol−1.
Bonds broken: 435 + 243 = +678 kJ
Bonds formed: 2 × 430 = 860 kJ released
ΔH = 678 − 860 = −182 kJ mol−1 (exothermic)
9Spontaneity and entropy
Occurs naturally with no continuous external input. Water flowing downhill; a hot object cooling; sugar dissolving in hot tea; iron rusting; a gas expanding into a vacuum.
Does not occur without continuous energy input. Water flowing uphill (needs a pump); decomposing water (needs electrolysis).
Spontaneous does NOT mean fast. Rusting is spontaneous but can take months or years. Spontaneity is about thermodynamic tendency, not rate.
Entropy (S)
The first law cannot predict direction, so we need entropy: a measure of the disorder or randomness of a system. Units: J K−1 mol−1. S is a state function: ΔS = S2 − S1.
Factors affecting entropy
- State: S(gas) > S(liquid) > S(solid) — gases have the most freedom of movement.
- Temperature: entropy rises with temperature (more disordered motion).
- Number of particles: more moles of products than reactants → higher entropy.
- Complexity: larger molecules have more ways to store energy.
- Mixing increases entropy.
Predicting the sign of ΔS
| Process | ΔS | Reason |
|---|---|---|
| Solid → liquid → gas | + | Disorder increases |
| Dissolving a solid in a liquid | + | Disorder increases |
| Gas → liquid → solid | − | Disorder decreases |
| More moles of gas formed | + | More disorder |
| Fewer moles of gas formed | − | Less disorder |
Latent heat of fusion of ice = 336 J g−1. Find the molar entropy of fusion at 0 °C.
ΔHfus = 336 × 18 = 6048 J mol−1; T = 273 K
ΔS = 6048 / 273 = 22.15 J K−1 mol−1
10The second law and Gibbs free energy
1 (Entropy): for any spontaneous process, the total entropy of the universe always increases.
2 (Heat): heat cannot spontaneously flow from a colder body to a hotter body.
3 (Perpetual machine): it is impossible to build a machine of the second kind — one that converts heat completely into work with no other change.
- Spontaneous: ΔSuniverse > 0
- Equilibrium: ΔSuniverse = 0
- Non-spontaneous: ΔSuniverse < 0 (the reverse is spontaneous)
A decrease in enthalpy alone is not a sufficient criterion: dissolving ammonium nitrate is endothermic (ΔH > 0) yet spontaneous because ΔS > 0. An increase in entropy alone is not always sufficient either. Gibbs free energy combines both.
Gibbs free energy (G)
Introduced by J. Willard Gibbs: a state function combining enthalpy and entropy as one criterion for spontaneity at constant T and P.
Derivation of ΔG = ΔH − TΔS
- For a spontaneous process, ΔSuniverse > 0.
- ΔSuniverse = ΔSsystem + ΔSsurroundings.
- At constant T and P: ΔSsurroundings = −ΔHsystem / T.
- So ΔSuniverse = ΔSsystem − ΔHsystem / T.
- Multiplying by −T: −TΔSuniverse = ΔHsystem − TΔSsystem.
- Since ΔG = ΔH − TΔS for the system, ΔG = −TΔSuniverse.
Effect of ΔH, ΔS and temperature
| ΔH | ΔS | ΔG = ΔH − TΔS | Result |
|---|---|---|---|
| − | + | Always negative | Always spontaneous (all T) |
| + | − | Always positive | Never spontaneous (all T) |
| − | − | Depends on T | Spontaneous at low T only |
| + | + | Depends on T | Spontaneous at high T only |
Is Ag2O(s) → 2Ag(s) + ½O2(g) spontaneous at 500 K? ΔH = +30.56 kJ mol−1; ΔS = +66 J K−1 mol−1.
ΔS = 0.066 kJ K−1 mol−1
ΔG = 30.56 − (500 × 0.066) = 30.56 − 33.0 = −2.44 kJ mol−1
ΔG < 0, so it is spontaneous at 500 K.
Crossover temperature: T = ΔH/ΔS = 30560 / 66 = 463 K. Above 463 K the reaction is spontaneous, and 500 K > 463 K ✓
11ΔG and the equilibrium constant
At equilibrium ΔG = 0 and the system’s Gibbs energy is at a minimum. The standard free energy change is related to K by:
| ΔG° | K | Interpretation |
|---|---|---|
| Negative (< 0) | K > 1 | Equilibrium favours products; forward reaction spontaneous |
| Zero (= 0) | K = 1 | Reactants and products in equal amounts at equilibrium |
| Positive (> 0) | K < 1 | Equilibrium favours reactants; reverse reaction spontaneous |
The more negative ΔG°, the larger K, and the more the reaction favours products.
Water-gas reaction C(s) + H2O(g) → H2(g) + CO(g): ΔH° = +131.28 kJ mol−1, ΔS° = +133.57 J K−1 mol−1. Find ΔG° and K at 1000 °C, and the equilibrium temperature.
T = 1273 K; ΔS = 0.13357 kJ K−1 mol−1
ΔG° = 131.28 − (1273 × 0.13357) = 131.28 − 170.04 = −38.76 kJ mol−1
−38760 = −(8.314)(1273) ln K, so ln K = 3.664 and K ≈ 39
Equilibrium (ΔG = 0): T = ΔH/ΔS = 131280 / 133.57 ≈ 983 K ≈ 710 °C
Exam tips
- Sign conventions: heat absorbed and work done on the system are positive; expansion work is −PΔV.
- Convert ΔS from J to kJ (or ΔH to J) before using ΔG = ΔH − TΔS, and always use kelvin.
- ΔH°f means 1 mole of compound from elements; elements in standard state have ΔH°f = 0.
- Reversing an equation flips the sign of ΔH; multiplying an equation multiplies ΔH.
- “Spontaneous” does not mean “fast”.
- Bond enthalpy calculations are approximate (average values).
Quick summary
| Concept | Key points |
|---|---|
| System | Portion of the universe under study; open / closed / isolated |
| State function | Depends only on current state (E, H, S, G) |
| First law | ΔE = q + w; energy is conserved |
| Enthalpy | H = E + PV; ΔH = qp |
| Exo / endo | ΔH < 0 heat released / ΔH > 0 heat absorbed |
| Hess’s law | ΔH independent of pathway |
| Entropy | Measure of disorder; S(g) > S(l) > S(s) |
| Second law | ΔSuniverse > 0 for spontaneous processes |
| Gibbs energy | ΔG = ΔH − TΔS; ΔG < 0 spontaneous |
| ΔG° and K | ΔG° = −RT ln K |