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Chemical Thermodynamics — Notes

6 sections 6 min readBeta

1. System, surroundings and state functions

The system is the part of the universe under study; the surroundings exchange energy with it across a boundary.

  • Open: exchanges matter and energy (tea in an open cup).
  • Closed: exchanges energy only (a sealed tube in a water bath).
  • Isolated: exchanges neither (an ideal vacuum flask).

State functions (p,V,T,U,H,S,Gp, V, T, U, H, S, G) depend only on the present state, so their changes depend only on the initial and final states. Path functions (qq, ww) depend on the route.

Extensive properties depend on the amount (m,V,U,H,S,Gm, V, U, H, S, G); intensive ones do not (T,pT, p, density, molar quantities). The ratio of two extensive properties is intensive.

Types of thermodynamic system
SystemMatterEnergy
OpenExchangedExchanged
ClosedNot exchangedExchanged
IsolatedNot exchangedNot exchanged

Key terms

State function

A property whose value depends only on the state of the system, not on the path by which it was reached.

Isolated system

A system that exchanges neither matter nor energy with its surroundings.

Intensive property

A property that does not depend on the amount of substance, such as temperature or density.

2. Work, heat and the first law

IUPAC sign convention: viewed from the system, energy entering is positive. Work done on the system: w>0w > 0. Work done by the system (expansion): w<0w < 0. Heat absorbed: q>0q > 0.

Pressure–volume work:

  • Free expansion into vacuum: w=0w = 0.
  • Constant external pressure: w=−pexΔVw = -p_\text{ex}\Delta V (1 L bar=100 J1\ \text{L bar} = 100\ \text{J}).
  • Isothermal reversible, ideal gas: w=−2.303 nRTlog⁡(V2/V1)w = -2.303\,nRT\log(V_2/V_1); this is the maximum work obtainable.

First law: ΔU=q+w\Delta U = q + w. For an isolated system ΔU=0\Delta U = 0; for an adiabatic change ΔU=w\Delta U = w; for an isothermal change of an ideal gas ΔU=0\Delta U = 0, so q=−wq = -w.

Reversible work is the maximum work the gas can do.

Formulas

  • First law
    ΔU=q+w\Delta U = q + w
  • Irreversible expansion work
    w=−pex ΔVw = -p_\text{ex}\,\Delta V
  • Isothermal reversible work
    w=−2.303 nRTlog⁡V2V1w = -2.303\,nRT\log\dfrac{V_2}{V_1}

Key terms

Internal energy (U)

Total energy of a system: kinetic, potential and bonding energy of its particles. A state function.

Reversible process

A process carried out infinitely slowly so that the system is always in equilibrium with its surroundings.

Adiabatic process

A process in which no heat is exchanged, q = 0.

3. Enthalpy and heat capacity

Enthalpy H=U+pVH = U + pV is a state function. At constant pressure qp=ΔHq_p = \Delta H; at constant volume qV=ΔUq_V = \Delta U.

For reactions involving gases: ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT, where Δng\Delta n_g is moles of gaseous products minus moles of gaseous reactants. If Δng=0\Delta n_g = 0, ΔH=ΔU\Delta H = \Delta U.

Heat capacity C=q/ΔTC = q/\Delta T (extensive); molar heat capacity per mole, specific heat per gram. ΔU=nCVΔT\Delta U = nC_V\Delta T, ΔH=nCpΔT\Delta H = nC_p\Delta T, and for an ideal gas Cp−CV=RC_p - C_V = R.

Exothermic: ΔH<0\Delta H < 0 (products lower in enthalpy). Endothermic: ΔH>0\Delta H > 0.

Formulas

  • Enthalpy and internal energy
    ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT
  • Molar heat capacities of an ideal gas
    Cp−CV=RC_p - C_V = R

Key terms

Enthalpy (H)

H = U + pV; the heat absorbed at constant pressure equals its change.

Heat capacity

Heat needed to raise the temperature of a system by one kelvin.

4. Calorimetry and standard enthalpy changes

Bomb calorimeter (constant volume) measures ΔU\Delta U; q=CcalΔTq = C_\text{cal}\Delta T; then convert with ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT. A constant-pressure (coffee-cup) calorimeter measures ΔH\Delta H: q=mcΔTq = mc\Delta T.

Standard enthalpy changes (per mole, 1 bar, usually 298 K):

  • Formation ΔfH∘\Delta_fH^\circ: one mole of compound from elements in reference states; zero for an element.
  • Combustion ΔcH∘\Delta_cH^\circ: one mole burnt completely in oxygen; always negative.
  • Atomisation ΔaH∘\Delta_aH^\circ: one mole of gaseous atoms from the element (or compound).
  • Sublimation, fusion, vaporisation (phase transitions): positive.
  • Ionisation enthalpy: removing one electron from a gaseous atom (Na: +496 kJ mol⁻¹).
  • Solution: dissolving one mole in a large excess of solvent (NaOH −44.5, NH₄NO₃ +25.7 kJ mol⁻¹).
  • Dilution: enthalpy change on adding solvent to a solution.
  • Neutralisation of a strong acid and strong base: about −57 kJ per mole of water.
Examples of standard enthalpy changes
Enthalpy changeExamplekJ mol⁻¹
Formation of H₂O(l)H₂ + ½O₂−285.8
Combustion of graphiteC + O₂−393.5
Fusion of iceH₂O(s) → H₂O(l)+6.01
Vaporisation of waterH₂O(l) → H₂O(g)+40.7
Ionisation of Na(g)Na → Na⁺ + e⁻+496

Formulas

  • Heat from a calorimeter
    q=Ccal ΔTq = C_\text{cal}\,\Delta T

Key terms

Standard state

Pure substance at 1 bar pressure; 298 K is the usual reference temperature.

Enthalpy of combustion

Enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.

5. Hess’s law and bond enthalpies

Hess’s law: the enthalpy change of a reaction is the same whether it occurs in one step or in several, because HH is a state function. Reversing an equation changes the sign of ΔH\Delta H; multiplying scales it; equations and enthalpies can be added.

ΔrH∘=∑ΔfH∘(products)−∑ΔfH∘(reactants)\Delta_rH^\circ = \sum \Delta_fH^\circ(\text{products}) - \sum \Delta_fH^\circ(\text{reactants})

Bond enthalpy is the energy to break one mole of a bond in the gas phase (average value for many compounds). For gaseous reactions, ΔrH≈∑(bonds broken)−∑(bonds formed)\Delta_rH \approx \sum(\text{bonds broken}) - \sum(\text{bonds formed}). Typical values (kJ mol⁻¹): H–H 436, C–H 414, C–C 347, C=C 611, O=O 498, O–H 463, C=O (CO₂) 799, Cl–Cl 242, H–Cl 431.

Enthalpy changes in kJ mol⁻¹ at 298 K.

Formulas

  • Enthalpy of reaction from formation data
    ΔrH∘=∑ΔfH∘(P)−∑ΔfH∘(R)\Delta_rH^\circ = \sum\Delta_fH^\circ(\text{P}) - \sum\Delta_fH^\circ(\text{R})

Key terms

Hess’s law

The total enthalpy change of a reaction is independent of the route between the same initial and final states.

Bond enthalpy

Energy needed to break one mole of a particular bond in gaseous molecules; an average over compounds.

6. Entropy, Gibbs energy and spontaneity

Entropy SS is a measure of disorder, a state function. For a reversible process ΔS=qrev/T\Delta S = q_\text{rev}/T. Entropy rises on melting, boiling, dissolving and when the number of gas moles increases. Second law: ΔStotal=ΔSsys+ΔSsurr>0\Delta S_\text{total} = \Delta S_\text{sys} + \Delta S_\text{surr} > 0 for a spontaneous process.

Gibbs energy: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S (constant TT, pp). ΔG<0\Delta G < 0: spontaneous; =0= 0: equilibrium; >0> 0: non-spontaneous.

ΔHΔSSpontaneous?
−+At all temperatures
+−At no temperature
−−At low temperatures
++At high temperatures

Equilibrium: ΔG∘=−2.303 RTlog⁡K\Delta G^\circ = -2.303\,RT\log K. If ΔG∘<0\Delta G^\circ < 0, K>1K > 1.

ΔG = ΔH − TΔS is a straight line of slope −ΔS.

Formulas

  • Gibbs equation
    ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
  • Gibbs energy and equilibrium constant
    ΔG∘=−2.303 RTlog⁡K\Delta G^\circ = -2.303\,RT\log K

Key terms

Entropy (S)

A state function measuring the disorder of a system; for a reversible change, ΔS = q_rev/T.

Gibbs energy (G)

G = H − TS; its decrease at constant T and p is the criterion of spontaneity.

Exam tips

Common mistakes

  • Writing w = +pΔV for an expansion (the work done by the gas is negative).
  • Forgetting that 1 L bar = 100 J, or mixing J and kJ in ΔG = ΔH − TΔS.
  • Counting solids and liquids in Δn_g (only gases count).
  • Taking formed minus broken for bond enthalpies (it is broken minus formed).
  • Not multiplying ΔfH° by the stoichiometric coefficients; forgetting that ΔfH° of an element is zero.
  • Quoting spontaneity from ΔH alone.

How marks are usually awarded

  • 1-mark items: sign convention, state function or the correct relation.
  • Numericals (2–3 marks): formula (½–1), substitution with units (1), answer with sign and unit (½–1).
  • Reasoning (2–3 marks): the relation used and the argument in terms of ΔH, ΔS and T.