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 XetOH, XH2O, Xgas

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 in equilibrium?

2. What happens if the system is not in 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

dSsurr = dQsurr/Tsurr = -dQ/T

From 2nd law dSuniverse ≥ 0 dSsurr + dS ≥ 0

universe

Consequence 2

dQ ≤ TdS

From 1st 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 Isobaric Rigid and Isothermal

(dU)S,V ≤ 0

(dG)T,P ≤ 0 (dF)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 systemEquilibrium

Position

Potential Energy

z

U = mg .( z )

x

U = mg .( x2)

0.0

Energy DerivativedU = mg .( x ) dx

At equilibriumdU = 0

Gibbs Free Energy of Mixing

Equilibrium Position

At equilibriumdG = 0

Gibbs Free EnergyG = G( xA)

DGmix = G – SxiGi

1. What is the difference between system I and system II?

Mixing requires: G < SxiGi

A

DGmixA > α (DGmix) α + β(DGmix) β

DGmixA < α (DGmix) α + β(DGmix) 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 ΔGmix changes with temperature

Analytical approach Stability in terms of GE

Not only does (ΔGmix)T,P have to be negative, but also:

(d2ΔGmix/dx12 > 0)T,P (why?)

Since T is constant, we can divide both sides by RT

(d2(ΔGmix/RT)/dx12 > 0)T,P

For a binary system ΔGmix/RT = x1lnx1 + x2lnx2 + GE/RT

Constant T and P

Analytical Approach In terms of gi

0ln1)/(

lnln

1)/(

1

1

221

2

2211

2121

2

dx

d

xdx

RTGd

xxRT

G

xxdx

RTGd

E

E

E

Alternative criteria, at constant T and P, valid for each of the components: 0

0ˆ

1ln

1

1

1

1

11

1

dx

d

dx

fd

xdx

d

See derivation of this criterion posted in the web site

Implications for VLE equilibrium

d(ln u)/dv = (1/u) (du/dv)

Calculus note:

1. What is the equivalent criterium for component 2?

Reminder:

dx1 = -dx2

Implications for VLE equilibrium

Since T is constant, we can divide both sides by RT

Constant T and P

Let us demonstrate that for an ideal gas:

What is the connection to variations in compositions?

dy1/dx1 > 0 dy1 > 0; dx1 > 0 or dy1 < 0; dx1 < 0

For low pressure VLE, an ideal gas phase, let us demonstrate that:

1

1

21

11

1

)(1

dx

dy

yy

xy

dx

dP

P

What does the above equation implies for the sign of dP/dx1?

Implications for VLE equilibrium

Gibbs-Duhem Equation

What does the above equation implies at the azeotropic point?

Implications for VLE equilibrium

11

1

1 /

/

dxdy

dxdP

dy

dP du/dv = (du/dw)/(dv/dw)

Calculus note:

What does the above equation implies for the sign of dP/dy1?

What does the above equation implies at the azeotropic point?

Summary• In an isolated system entropy is maximized at equilibrium

• In an T, P controlled system Gibbs free energy is minimized at equilibrium

• Two components will be mixed into a single phase if ΔGmix < 0, and DGmix

A < α (DGmix) α + β(DGmix) β

For VLE equilibrium

sign same thehave )(,,

,0

1111

1

1

xydy

dP

dx

dP

dx

dy