Chemical Thermodynamics and Phase Equilibria

Excess Gibbs energy models and activity coefficients

The molar excess Gibbs energy g^E is the difference between the molar Gibbs energy of a real liquid solution and that of an ideal solution at the same temperature, pressure and composition, so g^E = g - g^{id}. Activity coefficients follow from a g^E model through \ln\gamma_i = \partial(nG^E/RT)/\partial n_i, so one model for g^E/RT fixes every \ln\gamma_i. The two-parameter Redlich–Kister form is the standard example. For g^E/RT = x_1x_2[A + B(x_1 - x_2)], one obtains \ln\gamma_1 = (A+3B)x_2^2 - 4Bx_2^3 and \ln\gamma_2 = (A-3B)x_1^2 + 4Bx_1^3, and at infinite dilution \ln\gamma_1^\infty = A - B. For a binary liquid at constant T and P, the Gibbs–Duhem equation is x_1\,d\ln\gamma_1 + x_2\,d\ln\gamma_2 = 0, so the composition slope of one \ln\gamma fixes the other: \dfrac{d\ln\gamma_2}{dx_1} = -\dfrac{x_1}{x_2}\dfrac{d\ln\gamma_1}{dx_1}. For example, if d\ln\gamma_1/dx_1 = 0.8 at x_1 = 0.3, then d\ln\gamma_2/dx_1 = -(0.3/0.7)(0.8) = -0.343.

Maxwell relations for thermodynamic derivatives

Maxwell relations are equalities among mixed second derivatives of the thermodynamic potentials U, H, A and G. For example, dA = -S\,dT - P\,dV gives (\partial S/\partial V)_T = (\partial P/\partial T)_V. They allow entropy changes, which cannot be measured directly, to be written in terms of measurable P–V–T quantities. The cyclic rule (\partial x/\partial y)_z(\partial y/\partial z)_x(\partial z/\partial x)_y = -1 is often needed to convert (\partial P/\partial T)_V into \beta/\kappa_T, where \beta is the thermal expansion coefficient and \kappa_T the isothermal compressibility. As a check, an ideal gas has \beta = 1/T and \kappa_T = 1/P, so \beta/\kappa_T = P/T, which agrees with (\partial P/\partial T)_V = P/T from P = RT/V. A zero thermal expansion coefficient therefore makes the isothermal volume dependence of entropy vanish, as happens for water at its density maximum near 4\,^\circC. The same habit of checking what each group contains applies to dimensionless numbers. The Bond number \mathrm{Bo} = \Delta\rho\,gL^2/\sigma contains surface tension, and the Jakob number \mathrm{Ja} = c_p\Delta T/h_{fg} contains latent heat. The Prandtl number \mathrm{Pr} = \mu c_p/k contains no liquid–vapour density difference, and the Biot number \mathrm{Bi} = hL/k contains no gravitational term.

Carnot engine efficiency and reversibility

A Carnot engine is a reversible heat engine operating between a hot reservoir at T_H and a cold reservoir at T_L, and its efficiency is \eta = 1 - T_L/T_H, with temperatures in kelvin. The efficiency depends on the reservoir temperatures, so a larger temperature ratio gives a higher efficiency. The Carnot theorem states that no engine operating between the same two reservoirs can exceed this value, and that the Carnot engine itself is reversible. For a reversible cycle, Q_L/T_L = Q_H/T_H, so the entropy change of the engine over one cycle is zero and the total entropy of the engine plus reservoirs does not increase. Irreversible engines between the same reservoirs have lower efficiency and generate entropy. Example: reservoirs at 500 K and 300 K give \eta_C = 0.40, and a real engine operating at half this efficiency produces 2 kJ of work from 10 kJ of heat input. When an efficiency is stated as a fraction of the Carnot efficiency between the extreme reservoirs, the overall power is W = \eta\,\dot Q_H, where the efficiency is defined with respect to the heat taken from the source. An intermediate reservoir that receives the rejected heat of one stage and drives another does not change this product.

Gibbs phase rule degrees of freedom

The Gibbs phase rule gives the number of degrees of freedom for a non-reactive system as F = C - P + 2, where C is the number of components and P is the number of phases in equilibrium. The 2 accounts for temperature and pressure, and F counts the intensive variables that must be fixed to define the state. For a reactive system, the number of independent reactions R is subtracted as well, giving F = C - R - P + 2. For a ternary mixture in vapour–liquid–liquid equilibrium, C = 3 and P = 3, so F = 2 intensive variables must be specified. A pure substance at its triple point has C = 1 and P = 3 (solid, liquid and vapour), so F = 0 and the point is invariant; no intensive variable can be changed without losing a phase.

Heat of reaction from heats of combustion

The heat of combustion of a substance is the enthalpy change when it is completely burned to its final oxidised products, usually CO_2 and liquid water. Because every species in a reaction is burned to the same final products, the heat of reaction is found from the heats of combustion by \Delta H_r = \sum_{\text{reactants}} \nu_i\,\Delta H_c,i - \sum_{\text{products}} \nu_j\,\Delta H_c,j. A species that is already fully oxidised, such as water, has a heat of combustion of zero, so it does not contribute. Example: for ethylene plus hydrogen forming ethane, with approximate heats of combustion of -1411 kJ/mol for ethylene, -286 kJ/mol for hydrogen and -1560 kJ/mol for ethane, \Delta H_r = (-1411) + (-286) - (-1560) = -137 kJ/mol.

Van der Waals equation at critical point

The van der Waals equation P = RT/(v-b) - a/v^2 has critical constants v_c = 3b, T_c = 8a/(27Rb) and P_c = a/(27b^2). These give the universal critical compressibility factor Z_c = P_cv_c/(RT_c) = 3/8. In reduced variables the equation becomes P_r = 8T_r/(3v_r - 1) - 3/v_r^2. Because Z = Pv/RT and each variable is scaled by its critical value, the compressibility factor is Z = Z_c\,P_r v_r/T_r. Example: at T_r = 1 and v_r = 3, P_r = 8/8 - 3/9 = 2/3, so Z = (3/8)(2/3)(3)/1 = 0.75.

Minimum work of separation for ideal mixtures

When a mixture is separated into pure components at the same temperature and pressure, the minimum reversible work equals the negative of the Gibbs energy of mixing, W_{min} = -\Delta G_{mix}. For an ideal gas or ideal solution, \Delta G_{mix} = nRT\sum_i x_i\ln x_i, so W_{min} = -nRT\sum_i x_i\ln x_i. This quantity is positive because each \ln x_i is negative, and the most dilute component contributes the largest term in magnitude. Example: one mole of an equimolar binary ideal gas mixture at 298 K requires W_{min} = RT\ln 2 \approx 1.7 kJ.

Single-component VLE thermodynamic identities

In a single-component system at vapour–liquid equilibrium, the two phases share the same temperature, pressure and chemical potential. The derivatives (\partial A/\partial V)_{T,N} = -P and (\partial A/\partial N)_{T,V} = \mu therefore hold in each phase and take the same value in both phases at equilibrium. The molar Gibbs energy (A + PV)/N equals \mu, so it is also the same in the two phases. In contrast, (A + TS)/N = U/N is the molar internal energy, which differs between vapour and liquid by an amount related to the latent heat, so it cannot be equal across the two phases.

Ideal solution mixing Gibbs energy

Mixing components to form an ideal solution produces no enthalpy change, so the Gibbs energy of mixing is governed by entropy. The relations are \Delta H_{mix} = 0, \Delta S_{mix} = -R\sum_i n_i\ln x_i and \Delta G_{mix} = RT\sum_i n_i\ln x_i. Because every mole fraction is less than one, the entropy of mixing is always positive and depends only on composition, not on the chemical identity of the species. Pure liquids mixed at the same temperature and pressure therefore have a total entropy change equal to this mixing term. Example: 2 mol of A and 3 mol of B give x_A = 0.4 and x_B = 0.6, so \Delta S = -R(2\ln 0.4 + 3\ln 0.6) \approx 28 J/K.

Ideal gas two-step process work and heat

For an ideal gas, internal energy and enthalpy depend only on temperature, so for any path \Delta U = nC_V\,\Delta T. The first law gives the heat gained as Q = \Delta U + W, where W is the work done by the gas. An isochoric step does no P\,dV work, whereas an isobaric step does W = P\,\Delta V at the constant pressure of that step. Since C_P - C_V = nR for an ideal gas, the heat of an isobaric step equals nC_P\,\Delta T. In a two-step path the work therefore appears only in the isobaric step. Example: 1 mol of a monatomic ideal gas heated at constant volume from 300 K to 400 K gains Q = \Delta U = 1.5R(100) \approx 1247 J.

Clausius-Clapeyron vapor pressure

The Clausius–Clapeyron equation relates the slope of the saturation curve to the latent heat of vaporisation: d\ln P^{sat}/dT = \Delta H_{vap}/(RT^2). The integrated form assumes that the vapour behaves ideally, that the liquid molar volume is negligible compared with the vapour volume, and that \Delta H_{vap} is constant over the temperature range. It then gives \ln(P_2/P_1) = -(\Delta H_{vap}/R)(1/T_2 - 1/T_1), so a plot of \ln P^{sat} against 1/T is a straight line with slope -\Delta H_{vap}/R. Example: with \Delta H_{vap} = 30 kJ/mol, a vapour pressure of 0.5 bar at 330 K rises to about 0.93 bar at 350 K.

Temperature dependence of equilibrium constant

The van't Hoff equation describes how the equilibrium constant changes with temperature: d\ln K/dT = \Delta H^\circ_r/(RT^2). For a reaction whose heat of reaction is constant over the temperature range, integration gives \ln K = -\Delta H_r/(RT) + \text{const}, so \ln K is linear in 1/T with slope -\Delta H_r/R. For an endothermic reaction, \Delta H_r > 0, so K increases with temperature. For an exothermic reaction, \Delta H_r < 0, so K decreases with temperature. This agrees with Le Chatelier's principle, since heating favours the endothermic side. Example: an exothermic reaction with K = 10 at 400 K has a smaller K at 450 K.