Equilibrium and Stability

Phase Separation in Ethanol Blended Gasoline 1. Three-component system: Ethanol, water, and gasoline 2. Up to three phases depending on concentration X etOH, X H2O, X gas 3. Phase separation can be triggered by drop in temperature 4. Different levels of engine failure depending on phase fed

An Arbitrary Thermodynamic System n components m phases Surroundings System Closed System: dn = 0 1. What happens if the system is at equilibrium? 2. What happens if the system is not at equilibrium? 3. Why I need to know what is the equilibrium state of a system?

Moving Toward Equilibrium State n components m phases Assumption 1 T and P are uniform throughout the system System is in thermal and mechanical equilibrium = 0, and = 0 Assumption 2 System is in thermal and mechanical equilibrium with surroundings Heat transfer and/or expansion work with/on surroundings occurs reversibly (why?) 1. Are changes occurring in the system reversible or irreversible? Notice: Changes will occur in the system, because it IS NOT at chemical/phase equilibrium

Moving Toward Equilibrium State n components m phases Consequence 1 1. When does the inequality applies? Notice: Since U, S, and V (and T and P) are state functions. Consequence 3 is true for ANY closed-system of uniform T and P dS surr = dQ surr /T surr = -dQ/T From 2 nd law  dS universe ≥ 0 dS surr + dS ≥ 0 universe Consequence 2 dQ ≤ TdS From 1 st law  dQ = dU + PdV Consequence 3 dU + PdV – Tds ≤ 0

dU + PdV –TdS ≤ 0 Minimum Energy Maximum Entropy

Criterion for equilibrium (dS) U,V ≥ 0 Rigid and Isentropic Isolated Isothermal and IsobaricRigid and Isothermal (dU) S,V ≤ 0 (dG) T,P ≤ 0 (dA) T,V ≤ 0

Criterion for equilibrium Isothermal and Isobaric (dG) T,P ≤ 0 dU + PdV – TdS ≤ 0 (dU + d(PV) – d(TS) ≤ 0) T,P d(kx) = kdx d(x + y) = dx + dy (d(U + PV – TS) ≤ 0) T,P G = U + PV - TS (d(G) ≤ 0) T,P 1. What state functions are more easily controlled in a chemical process? Processes occur spontaneously in the direction that G decreases (at constant T and P) At equilibrium, dG = 0 (at constant T and P)

Analogy with a mechanical system Equilibrium Position Potential Energy z U = mg.( z ) x U = mg.( x 2 ) 0.0 Energy Derivative dU = mg.( x ) dx At equilibrium dU = 0 Gibbs Free Energy of Mixing Equilibrium Position At equilibrium dG = 0 Gibbs Free Energy G = G( x A )

 G mix = G –  x i G i 1. What is the difference between system I and system II? A  G mix A > α (  G mix ) α + β(  G mix ) β  G mix A < α (  G mix ) α + β(  G mix ) β System I System II

To see video showing temperature-induced phase separation in E10, click here Clear Liquid (one phase) Clear Liquid (phase I) Turbid Liquid (phase II) SYSTEM: Ethanol-Gasoline-Water In cold weather (winter) storage tank in car can be colder than storage tank in gas station Shape of ΔG mix changes with temperature

Analytical approach  Stability in terms of G E Not only does (ΔG mix ) T,P have to be negative, but also:  (d 2 ΔG mix /dx 1 2 > 0) T,P Since T is constant, we can divide both sides by RT  (d 2 (ΔG mix /RT)/dx 1 2 > 0) T,P For a binary system  ΔG mix /RT = x 1 lnx 1 + x 2 lnx 2 + G E /RT

Stability criteria in terms of G E Constant T and P

Analytical Approach  In terms of  i Alternative criteria, at constant T and P, valid for each of the components: See derivation of this criterion posted in the web sitederivation