2.625 - Electrochemical Systems Fall 2013

Lecture 2 - Thermodynamics Overview Dr.Yang Shao-Horn

Reading: Chapter 1 & 2 of Newman, Chapter 1 & 2 of Bard & Faulkner, Chapters 9 & 10 of Physical Chemistry

I. Lecture Topics:

A. Review 1st and 2nd laws of thermodynamics

• First law: dU = δQ − δW

• Second law: δIS ≥ 0

B. Chemical Equilibrium

ν II iaprod i• ΔGr = ΔG◦ + nRT ln Ka = 0; where Ka = Ir νiareact i ◦• µi = µ + RT ln ai; µiνin = 0 i

C. Electrochemical Equilibrium

• ΔGr = ΔG◦ + nRT ln Ka + nzrFEr = 0 r

• Nernst Equation: Er = E◦ + RT

ln Kar zrF

II. Review 1st and 2nd laws of thermodynamics

A. 1st Law of Thermodynamics

For a closed system of constant mass,

i. internal energy is constant: δIU = 0

ii. external energy changes through work and heat: δEU = δQ − δW , where δQ is heat trans­ferred into the system, and δW is work done by the system. Note: in most (fluid) systems, δW is PdV; for electrochemical systems, we will expand this definition to include electro­chemical work.

iii. the total energy change of the system is the sum the internal and external energy changes: dU = δIU + δEU = δQ − δW

δIU and δEU are intensive properties whereas δQ and δW are extensive properties

A. Intensive property: A physical property which does not depend on the size (volume, mass, number, etc.) of the system

B. Extensive property: A physical property which does depend on the size (volume, mass, number, etc.) of the system

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13

δIU

δEU

System

Environment

Figure 1: The canonical thermodynamic system, with internal energy change δIU and external energy change δEU .

B. 2nd Law of Thermodynamics

i. change in internal entropy is equal to generated entropy, which is never less than zero:

δIS = δgS ≥ 0

ii. external entropy change is the change in heat scaled by the temperature of the system: δQ δES = T

iii. the total entropy change is, similarly, the sum of the internal and external entropy changes: δQ dS = δIS + δES = δIS + T

Multiplying through by T, we have: T dS = TδIS + δQ And since δIS is always greater than or equal to zero, T dS ≥ δQ. These expressions can be mapped into state space. The change in entropy while U and V are held constant, or H and P are held constant, are shown below.

C. Combining the 1st and 2nd Laws, δQ = dU + P dV (where δW is given as P dV )

Subbing in T dS − δIS for δQ and rearranging, we have

−TδIS = dU + P dV − T dS ≤ 0 (1)

For reversible processes, δIS = 0, so T dS = dU + P dV . This is Gibbs’ fundamental equation. Further, a closed system approaches equilibrium by maximizing entropy, S.

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13

States

SAt constant U, V

equilibrium

Figure 2: Equilibrium is achieved when entropy is maximized for a system where U and V are constant.

D. Alternatively, we can define equilibrium by using Gibbs free energy.

i. First, we define enthalpy, H: H = U + PV

ii. Then define Gibbs free energy, G: G = H − TS

iii. ⇒ G = U + PV − TS, and dG = dU + d(PV ) − d(TS)

By considering the combined first and second law equation, we can now write:

−δIS = dG − V dP + SdT ≤ 0.

Thus, a system approaches equilibrium by minimizing G such that dG ≤ 0 :

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13

States

G At constant P, T

equilibrium

Figure 3: With the proper choice of variables held constant, the Helmholtz and Gibbs functions show similar profiles. Equilibrium is achieved when Helmholtz or Gibbs functions are minimized.

Note: for a reversible system, dG = V dP − SdT

III. Gibbs free energy of multicomponent systems

The general form of the Gibbs free energy for a multicomponent system or phase is G(T, P, n1, n2, ....ni), where ni is the number of moles of species i in the system or phase.

∂G ∂G ∂G dG(P, T, ni) = dT + dP + dni (2)

∂T ∂P ∂niP,n T,n T,P,nji

We can define the partial differentials in the above equation as follows: ∂G • −S = ∂T P,n

∂G • V = ∂P T,n g r ∂G • µi = ∂ni T,P,nj =ni

Then we can write the differential from of the Gibbs free energy for a multicomponent system as:

dG(P, T, ni) = −SdT + V dP + µidni (3)

i

A. Comparing this to the expression for dG above, we see that the internal entropy is expressed here as the sum of the chemical potentials: TδIS = i µidni

This entropy is generated when a new component i is introduced into the system, with a chemical g r ∂G potential, µi = ∂ni T,P,nj =ni

Looking back, recall that at constant T and P, −TδI S = dG = i µidni ≤ 0

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13 B. Thermodynamic quantities are typically tabulated on a molar basis, meaning that the quantity is

provided in units per mole of reactant/product. Examples are:

• g : molar Gibbs free energy

• s : molar entropy

• h : molar enthalpy

C. We define mole fraction: Xi = ini ni

G = nigi; where gi = Xigi; and n = ni

D. We define the partial molar Gibbs free energy with respect to composition as: g r g r ∂g ∂G gi = = = µi∂Xi ∂niT,P,Xj =Xi T,P,nj =ni

E. For an ideal gas, we define the partial pressure as:

Pi = XiP

F. We can define a more convenient form of the chemical potential, where we write the chemical potential of a species i in terms of its chemical potential in the reference state, µ◦ and its activity, i ai: µi = µ◦ + RT ln aii

Defining the chemical potential of a species in terms of its activity is slightly more convenient because the activity can be described by simple models, which will be discussed next.

G. Activity models

• Ideal (Raoultian) Solution: ai = Xi

An ideal solution assumes that the activity of species i is equal to the mole fraction of species i.

• Ideal Gas: ai = Pi

For an ideal gas, the activity of species i is equal to the partial pressure of species i. This is an extension of the ideal solution model.

i. Figure ?? demonstrates how activities can be calculated from the partial pressures of gases in a mixture.

ii. The enthalpy of mixing can be found by calculating the enthalpy before and after mixing: hinit = Xihi(T, Pi)i

¯hfinal = i Xihi(T, Pi) = i Xihi

Δhmix = hfinal − hinit = 0

iii. The entropy of mixing can be found in a similar way: sinit = i Xisi = i Xi s◦

i − R ln PPi

◦

◦sfinal = si = i Xi s − R ln Pi i Xi i P ◦

Δsmix = sfinal − sinit = − i XiR ln Xi

Note that since ΔSmix is greater than zero, mixing is irreversible. This should line up with your intuitive understanding of gas behavior.

iv. Next, the Gibbs free energy of a mixture: ◦ginit = Xigi = Xi g + RT ln Pi

i i i P ◦

◦gfinal = i Xigi = i Xi gi + RT ln P P

◦ i

Δgmix = gfinal − ginit = XiRT ln Xi ≤ 0i When Δgmix ≤ 0, mixing will occur spontaneously and work will be generated.

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6= 6=6

∑∑ ∑∑ ∑ ( )∑ ∑ ( )∑

∑ ∑ ( )∑ ∑ ( )∑

Lecture 2 Thermodynamics Overview 2.625, Fall ’13

System

Before Mixing: P1 = P2 = P3 = Pi = P

After Mixing: P = p1 + p2 + p3 + ∑pi

P1

n1

P2

n2

P3

n3

P4

n4

p1, p2, p3, ... pi

n1, n2, n3, ... ni

Figure 4: Initially, a box contains several species of a gas, each having the same pressure Pi, the same as the overall box pressure P. After the partitions are removed, the gases mix and P is the sum of each gas’ partial pressure Pi.

IV. Chemical Equilibrium

A. Condition for chemical equilibrium

µidni = 0 , or ΔGr = 0 at constant T and P.

Where ΔGr = µiνin, and νi is the stoichiometric coefficient of species i.

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13

States

∑µidni At constant T, P

equilibrium

Figure 5: Equilibrium is achieved at constant T and P when the sum of chemical potentials of components in the system is minimized.

B. Example 1i. Consider the formation of water: H2(g) + O2(g) → H2O2

= 0.5, and νH2O = 1 ii. νH2(g) = 1, νO2(g)

C. Deriving the condition for chemical equilibrium: ◦ i + RT ln ai)i. ΔGr = µiνin = n νi (µ

◦ii. ΔGr νi + nRT νi ln ai = n µi ◦ ◦iii. Let ΔG νi = n µir

II νiaprod i

I ◦iv. ΔGr = ΔG + nRT lnr ν

iareact i II

νi prod ia

I νiareact i

v. Remember that Ka =

vi. At equilibrium: ΔGr = ΔG◦ r + nRT ln Ka = 0

ΔG◦ = −nRT ln Karvii. Thus, at we can define the chemical equilibrium condition:

D. Two interesting relations

i. The natural logarithm of the equilibrium constant, Ka, can be written as follows: h r h r −ΔG◦ −ΔH◦

r rKa = exp = A expnRT nRT

◦ΔHrln Ka = lnA − nRT

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13

1/T

ln Ka

(high T) (low T)

H2 + 0.5O2 → H2O

CO2 → CO + 0.5O2

1 1Figure 6: Example plot of ln Ka vs. ln Ka varies approximately linearly with . .T T

ii. Assuming that ΔH◦ r

range of T), the following relationship can be developed: is not a function of temperature (this is a good assumption for a wide

r

rrSince ΔG◦(T, P ◦) = ΔH◦ − T ΔS◦

r

∂ΔG◦

= −ΔS◦ , assuming that ΔH◦

∂T

r

r

,

= f(T )

1Example: H2(g) + O2(g) → H2O2

T (K) Δh◦ r (kJ/mol) Δg◦

r (kJ/mol) E◦ r

300 -286 -237 1.23 500 -242.4 -219 1.13 1000 -245 -192 0.99

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Lecture 2 Thermodynamics Overview 2.625, Fall ’13

T

ΔGr˚

0

-ΔSr˚

Figure 7: Visual representation of how the change in Gibbs free energy with temperature is approximately equal to the change in entropy (i.e. the slope of ΔG◦ as a function of temperature is equal to −ΔS◦). r r

E. Electrochemical Equilibrium

i. Relative to a chemical system, equilibrium for an electrochemical system has an additional work term.

ii. δW = Edq for charged particles with a charge of dq

iii. Condition for electrochemical equilibrium (at constant T and P): µiνi + zrFEr = 0

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2.625 / 10.625J Electrochemical Energy Conversion and Storage: Fundamentals, Materials, and ApplicationsFall 2013

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