Chemical Thermodynamics and Chemical Equilibrium

6.1Basic ideas

• System and surroundings: an open system exchanges matter and energy, a closed system energy only, an isolated system neither.
• Thermodynamic equilibrium needs three things at once: thermal (uniform temperature), mechanical (no unbalanced pressure) and chemical (no net change of composition). If any one is missing, heat flows, the boundary moves or a reaction runs.
• Extensive properties grow with the amount of matter: mass, V, U, H, S, G, A, heat capacity C_p. Intensive properties do not: T, p, density, concentration, molar quantities (H_m, C_p,m), chemical potential, surface tension, viscosity, dielectric constant, refractive index, electrode potential, emf. The ratio of two extensive properties is intensive.

State functions and path functions

• A state function (U, H, S, G, A, p, V, T) depends only on the present state. Three equivalent tests: its change depends only on the initial and final states; its integral round any cycle is zero (∮dX = 0); its differential is exact.
• Heat q and work w are path functions: their small amounts đq, đw are inexact differentials. Their sum q + w = ΔU is a state function. So ΔU is the same for a reversible and an irreversible path between the same two states, while q and w differ.
• For any cycle that returns to the starting state, ΔU = ΔH = ΔS = 0, however many steps it has.

Partial derivatives used in thermodynamics

• Cyclic (triple product) rule: for three variables linked by one relation the product is −1, not +1. Check with pV = RT: (R/V)(p/R)(−RT/p²) = −1.
• Chain rule with the same variable held constant throughout has no minus sign (second relation above).
• Response functions: expansion coefficient α = (1/V)(∂V/∂T)_p; isothermal compressibility κ_T = −(1/V)(∂V/∂p)_T (the minus sign makes it positive). For an ideal gas α = 1/T and κ_T = 1/p.

6.2First law, work and heat

• Sign convention (IUPAC): q and w are positive when energy enters the system. Work done on the gas (compression) is positive; work done by the gas (expansion) is negative. A balloon that swells as dry ice sublimes does work on the atmosphere: w < 0 for the system.
• Constant volume: w = 0, so ΔU = q_V. If dw = 0 and dq < 0 for an ideal gas, dU = nC_VdT < 0 and the temperature falls. A reaction in a closed rigid vessel does no work.
• Constant pressure: ΔH = q_p, with H = U + pV. Heat released in an open dish is −ΔH.
• For an ideal gas, U and H depend on temperature only; C_p,m − C_V,m = R; γ = C_p/C_V = 5/3 (monatomic), 7/5 (diatomic, room temperature).

Δn_g counts only gaseous moles (products − reactants). H₂F₂(g) → H₂ + F₂: Δn_g = +1, ΔH = ΔU + RT. For burning a solid such as benzoic acid, C₇H₆O₂(s) + 15/2 O₂ → 7CO₂ + 3H₂O(l), Δn_g = −½. For glucose, C₆H₁₂O₆(s) + 6O₂ → 6CO₂ + 6H₂O(l), Δn_g = 0, so Δ_cH = Δ_cU.

Work in different processes (ideal gas, n mol)

• Reversible means the external pressure differs from the gas pressure only infinitesimally at every step, so đw = −pdV with p the gas pressure. A slow expansion in a constant-temperature bath is reversible isothermal. The reversible path gives the maximum work in expansion (and the minimum work needed in compression).
• ∫pdV = ∫nRdT only at constant pressure; in general pdV + Vdp = nRdT. Work cannot be found from the end states alone.
• Reversible adiabatic relations: pV^γ = const, TV^γ−1 = const, T p^(1−γ)/γ = const. Hence T_f = T_i(p_f/p_i)^R/C_p,m and w = −nC_V,mT_i[1 − (p_f/p_i)^R/C_p,m].
• On a log p–log V plot the isotherm has slope −1 and the reversible adiabat slope −γ. Since γ > 1 the adiabat is steeper, and for the same expansion the adiabatic work is smaller than the isothermal work.
• For a van der Waals gas, attractions make the reversible isothermal work slightly smaller than for an ideal gas at ordinary temperatures (below the Boyle temperature).

6.3Joule–Thomson effect

A gas is forced slowly through a porous plug (throttle) from a high constant pressure p₁ to a low constant pressure p₂ in an insulated tube. Then q = 0 and the net work on the gas is p₁V₁ − p₂V₂, so U₂ − U₁ = p₁V₁ − p₂V₂, i.e. U₂ + p₂V₂ = U₁ + p₁V₁. The expansion is isenthalpic (H constant). It is irreversible, so the entropy rises.

• The middle form follows from the cyclic rule. For a perfect gas H depends on T only, so (∂H/∂p)_T = 0, μ_JT = 0 and there is no temperature change on throttling.
• Since the pressure falls (Δp < 0), μ_JT > 0 means cooling and μ_JT < 0 means heating. ΔT ≈ μ_JTΔp.
• Attractions (a) cause cooling: the gas does work against them. Repulsions (b) cause heating. Cooling when 2a/RT > b, heating when 2a/RT < b.
• Inversion temperature T_i = 2a/Rb (twice the Boyle temperature a/Rb): below it the gas cools on expansion, above it warms. H₂ (T_i ≈ 200 K) and He (≈ 40 K) therefore warm on throttling at room temperature; they must be pre-cooled before they can be liquefied this way. N₂, O₂, CO₂ cool.
• At the inversion temperature μ_JT = 0, so (∂H/∂p)_T = (∂U/∂V)_T(∂V/∂p)_T + (∂(pV)/∂p)_T = 0: the second term is positive and equal in size to the first (which is negative).