Chapter 5: Free Energy and Chemical Thermodynamicsodessa.phy.sdsmt.edu/~bai/teaching_2014Fall_done/... · Chapter 5: Free Energy and Chemical Thermodynamics Applications of the laws

Outline Free Energy as Available Work Free Energy as Force toward Equilibrium

Chapter 5: Free Energy and ChemicalThermodynamics

Applications of the laws of thermodynamicsPart 1: Chemical reactions

X Bai

SDSMT, Physics

Fall 2014

X Bai Chapter 5: Free Energy and Chemical Thermodynamics


1 Free Energy as Available WorkBasic toolsElectrolysis, Fuel Cells, BatteriesThermodynamic Identities revisit: In chemical reactions

2 Free Energy as Force toward EquilibriumNon-isolated system



Basic tools

What are the typical thermal conditions for chemical reactions andother transformations of matter?

Electrolysis: H2O → H2 + 12O2

Energy: may vary (reactions may release or absorb heat),chemical reactions happen in non-isolated system

Volume: often not fixed, new substances can be created

Number of particles: often change in reactions

Pressure: often fixed, equal to the environment

Temperature: equal to the environment, approximately fixed

We are looking for Tools for fixed T and P !!!



Basic tools

To describe the energy budget:

1. Enthalpy :

H ≡ U + PV (1)

H: The total energy to create the system AND let it stay

in the environment with pressure P - Need the work.

2. Helmholtz Free Energy :

F ≡ U − TS (2)

U: The total energy to create the system;

TS = Q: The heat from the environment, with T fixed.

F : energy needed in the format of work if the

system is created from nothing.

3. Gibbs Free Energy :

G ≡ U − TS + PV = F + PV = H − TS (3)

PV is the work done under the constant pressure P.



H, G, Q=TS, W=PV: An example

Nothing ⇒ Rabbit

Left: in an environment with pressure P (without heat exchange with theenvironment)

W = PV : push out a volume for the rabbitH = Urabbit + PV : Required entire energy, the Enthalpy.

Right: in chemical reactions that can have heat exchange with the environment.Q = TS : can flow in as heat from the environment.G = H − TS : Energy the magician needs to provide.



Basic tools - cnt.

U,H,F , and G are called the Thermodynamic potentials.We are interested in the correlations among the changes of themunder conditions for chemical reactions. For example,

1, At fix temperature:

δF = δU − TδS = Q + W − TδS (4)

2. At constant pressure:

δG = δU − TδS + PδV = Q + W − TδS + PδV (5)

Because H = U + PV , in terms of U, we have:

δG = δH − TδS (6)

In-Class Exercise Ch5-01:Consider the production of ammonia from the reaction N2 + 3H2 → 2NH3.Assume the reaction was at 298 K and 1 bar . (1) Find out the values of δHand S the table on page 404-405 in the textbook; (2) Calculate δG for thisreaction, (3) Crosscheck the value of δG with that in the table.



Electrolysis

Let’s analyze a reaction:

Reactions: (i) Burning hydrogen into water:H2O ← 1

2O2 + H2

(ii) Reverse: Electrolysis of water:H2O → 1

2O2 + H2

Energy budget in electrolysis:From Table on p. 405 (burning process):(1) Total energy required:δH/mole = +286 kJ(2) Work to make the volume for H2,O2:PδV ≈ −3.7 kJ (see example page)(3) The Total energy remaining in the system:δH/mole − PδV = +286 kJ − 3.7 kJ = 282.3 kJ

I I

V

- +

H2 O2



Electrolysis

Energy budget (cnt.):(4) Heat for the change of entropy:δS = SH2 + SO2 − SH2O in J/(K ·mole)δS = 131 + 1

2205− 70 = 163.5 J/K - Positive, needs heat!

The heat Q = TδS = 298 K · 163 J/K = 49 kJ, from environment.(5) The energy required from the electricity:Ee = 286 kJ − 49 kJ = 237 kJ

What is this Ee? Ee = δH − TδSAccording to Eq. (6), δG = δH − TδS . So, Ee is the Gibbs Free Energy G !To summarize:



Fuel Cells: Reverse reaction

(1) 49 kJ: waste heat. (Cause the change in entropy - reduced S !)(2) Efficiency: 1− 49

286= 82.8%

Please see if you can get these numbers yourself.



Battery

The reaction: Pb + PbO2 + 4H+ + 2SO2−4 → 2PbSO4 + 2H2O

1 δG = 2∗(−237.13)+2∗(−813.0)−0−(−217.33)−4∗0−2∗(−744.53) =−393.87 kJ/mol : Total power produced for this reaction at the standardpressure, temperature and concentration of the solution.

2 δH = 2∗(−285.83)+2∗(−920.0)−0−(−277.4)−4∗0−2∗(−909.27) =−315.72 kJ/mol : The solution system energy decrease.

3 Heat absorbed from the environment: Q = δH − δG = 78 kJ.

4 Entropy increase by the heat:δS = Q/T = 78 kJ/298 K = 260 J/(K ·mole).

5 The voltage created: We need to know the reaction cycle that shows thegeneration of electrons and ions in 3 steps:(i) in solution: 2SO2−

4 + 2H+ → 2HSO−4(ii) at − electrode: Pb + HSO−4 → PbSO4 + H+ + 2e−

(iii) at + electrode: PbO2 + HSO−4 + 3H+ + 2e− → PbSO4 + 2H2O



Battery

Answer:(1) For each reaction cycle, 2 electrons are produced and pushed around thecircuit(2) The electric work produced by one electron is W = total electric work/mol

number of electrons/mol

W = 394 kJ/mol

2×6.02×1023/mole= 3.27× 10−19 J = 2.04 eV

(3) The voltage V = 2.04 V . (*) NA = 6.02× 1023: Avogadro’s number.

In-Class Exercise Ch5-02:In a hydrogen fuel cell, the steps of the chemical reaction are:at negative electrode: H2 + 2OH− → 2H2O + 2e−

at positive electrode: 12O2 + H2O + 2e− → 2OH−

Calculate the voltage of the cell? What is the minimum voltage required forelectrolysis? Why?



The thermodynamic identities revisit: In chemical reactions

If we know the enthalpy (H) or free energy (F or G) of a substance under acertain set of conditions (T ,V ,N, S , ..., ex. before the reaction), how to obtaintheir value under some other conditions (T + dT ,V + dV ,N + dN, S + dS , ...,ex. after the reaction)? Or, is this possible at all?

Generally speaking, for a thermal system of a substance, the state of the systemcan be described by the equation of state, f (U,V ,P,N,H,G ,H,S ,T ) = 0.

We know that NOT all these variables, U,V ,P, ..., are independent.

If we know f (U,V ,P,N,H,G ,H,S ,T )Cndt=a = 0, what is the equation ofstate f ′(U,V ,P,N,H,G ,H, S ,T )Cndt=b = 0 ?



The thermodynamic identities (cnt.)

These relations resemble the Thermodynamic Identity:

dU = TdS − PdV + µdN (7)

µ is the chemical potential. We have learned (Chapter 3) that two systems indiffusive equilibrium will have the same chemical potential. The + sign beforeµ means particles tend to flow from the system with higher µ into the systemwith lower µ. µ ≡ −T

(∂S∂N

)U,V

Eq. (7) is for U. We can get similar formula for H, F , or G :

Enthalpy (H)

H = U + PV (8)

dH = dU + PdV + VdP (9)

Using Eq. (7)

dH = TdS + VdP + µdN (10)



The thermodynamic identities (cnt.)

Helmholtz Free Energy (F),

F = U − TS (11)

dF = dU − TdS − SdT (12)

dF = (TdS − PdV + µdN)− TdS − SdT (13)

dF = −SdT − PdV + µdN (14)

Gibbs Free Energy (G).

G = U − TS + PV (15)

dG = dU − SdT − TdS + PdV + VdP (16)

dG = (TdS − PdV + µdN)− SdT − TdS + PdV + VdP (17)

dG = −SdT + VdP + µdN (18)



Why are they useful?

These formulae tell us how different quantities can be derived from interactionsunder certain ”environmental” conditions.

For example, from dF = −SdT − PdV + µdN, we can get

S = −(∂F

∂T

)V ,N

(19)

P = −(∂F

∂V

)T ,N

(20)

µ =

(∂F

∂N

)T ,V

(21)

From dG = −SdT + VdP + µdN, we can get

S = −(∂G

∂T

)P,N

(22)

V =

(∂G

∂P

)T ,N

(23)

µ =

(∂G

∂N

)T ,P

(24)



Two types of thermodynamic variables

Extensive quantities: V ,N, S ,U,H,F ,G ,mThey change when the amount of substance changes. They are additive forindependent, noninteracting subsystems. They are proportional to the amountof material in the system. (Like all types of energies!)

Intensive quantities: T ,P, µ, ρThey do NOT change when the amount of substance changes.

First introduced by Richard C. Tolman in 1917.



Three notes

1 Gibbs free energy and chemical potentialµ =

(∂G∂N

)T ,P

. We have G = Nµ (with T ,P fixed!!). This is OK.

Because G is an extensive quantity.

2 µ =(∂F∂N

)T ,V

. Can we have F = Nµ (with T ,V fixed!!). NO! Because

holding P and V (!) fixed may cause µ varies as the number of particleschanges - density changes when V is fixed.

3 What about a system with different types of particles?We need to replace µdN with

∑µidNi , with the µi given by

µ1 =

(∂G

∂N1

)T ,P,N2

(25)

µ2 =

(∂G

∂N2

)T ,P,N1

(26)

...

N = N1 + N2 + N3 + ... (27)



More about chemical potential µ

µ = −T(∂S∂N

)U,V

µ =(∂F∂N

)T ,V

µ =(∂G∂N

)T ,P

(1) All with respect to the change in N.(2) µ may or may not change with the number of particles, depending on thedensity or pressure for example. See P.164 for more discussion.

Let’s look at an simple example: Chemical potential for Ideal Gas

G = Nµ → µ = GN

Using (23) V =(∂G∂P

)T ,N

,(∂µ∂P

)T ,N

= 1N

(∂G∂P

)T ,N

= VN

Ideal Gas, PV = NkT , VN

= kTP

. So(∂µ∂P

)T ,N

= kTP.

Integration with respect to P, with T and N fixed gives

µ(T ,P)− µ(T ,P0) = kTln(P/P0). (28)



Entropy and Helmholtz Free Energy

Isolated system - entropy increases - equilibriumWhat governs a non-isolated system? The typical method: Starting from thetotal entropy of the entire system.

Stotal = S + Senvironment (29)

dStotal = dS + dSenvironment (30)

The thermodynamic identity

dS =1

TdU +

P

TdV − µ

TdN (31)

For the environment, we can assume Ve and Ne are fixed. So,

dStotal = dS +1

TedUe (32)

Since (i) Te = T , (ii) dU + dUe = 0, and F = U − TS , we have

dUe = −dU (33)

dStotal = dS − 1

TdU = − 1

T(dU − TdS) = − 1

TdF (34)



Entropy and Gibbs Free Energy

dStotal = dS − 1TdU = − 1

T(dU − TdS) = − 1

TdF .

Conclusion: The change in Entropy can be measured by the change ofHelmholtz Free Energy - when Ve ,Ne ,Te are fixed.

Not a surprise, because F ≡ U − TS



Entropy and Gibbs Free Energy - cnt.

If we also allow volume Ve change (keep Te ,Ne the same) (like in phasetransition! - next lecture):

dStotal = dS + dSenvironment (35)

The thermodynamic identity

dS =1

TdU +

P

TdV − µ

TdN (36)

Keep Ne fixed to give

dStotal = dS +1

TedUe +

Pe

TedVe (37)

dStotal = dS − 1

TdU − P

TdV = − 1

T(dU − TdS + PdV ) (38)

dStotal = − 1

TdG (G = U − TS + PV ) (39)

Conclusion: The change in Entropy can be measured by the change of Gibbs

Free Energy - when Ne ,Te are fixed.



Entropy and Gibbs Free Energy - cnt.

dStotal = − 1

TdG



Summaries: Potentials

Now, let’s summarize what we have learned:

Table: Quantities that govern the thermal processes

Process What governs the process

constant E and V Entropy Sconstant T and V Helmholtz Free Energy Fconstant T and P Gibbs Free Energy G

Table: Thermodynamic quantities

Extensive quantities Intensive quantities

Do change when the Do NOT change when theamount of matter changes amount of matter changes

V, N, S, U, H, F, G , mass T, P, ρ, µ (chemical potential)



Summaries: Thermodynamic quantities, Potentials

Of course, you can see this by their definitions:µ ≡ −T

(∂S∂N

)U,V

ρ ≡ ∂M∂V

F ≡ U − TSQuestion: Is heat capacity C extensive or intensive? What aboutspecific heat capacity c? [Hint: Look at the definition: C ≡ Q

δT ]

Potential variables identity

U(S ,V ,N) S ,V ,N dU = TdS − PdV + µdNH(S ,P,N) S ,P,N dH = TdS + VdP + µdNF (T ,V ,N) V ,T ,N dF = −SdT − PdV + µdNG (T ,P,N) P,T ,N dG = −SdT + VdP + µdN



The concept

In-class Exercise Ch5-03: By subtracting µN from U,H,F ,Gone can obtain for new thermodynamic potentials. The GrandFree Energy is defined as φ = F − µN = U − TS − µN.(1) Derive the thermodynamic identity for φ, and related formulasfor the partial derivatives of φ with respect to T ,V , µ.(2) Prove φ = −PV


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Chapter 5: Free Energy and Chemical Thermodynamicsodessa.phy.sdsmt.edu/~bai/teaching_2014Fall_done/... · Chapter 5: Free Energy and Chemical Thermodynamics Applications of the laws