Engineering Thermodynamics for Fuel Cells

8/17/2019 Engineering Thermodynamics for Fuel Cells

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THERMODYNAMIC EVALUATION OF GREEN ENERGY

TECHNOLOGIES

COURSE READER

May 2016


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Contents

1 Carbon Dioxide Separation and Entropy 1

1.1 The Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 The Thermodynamic Problem . . . . . . . . . . . . . . . . . . 3

1.1.2 The Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.2 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.3 Internal Energy and the First Law . . . . . . . . . . . . . . . 12

1.2.4 Heating an Ob ject . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2.5 Reversible and Irreversible Processes . . . . . . . . . . . . . . 151.2.6 Defining the Second Law . . . . . . . . . . . . . . . . . . . . . 17

1.3 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.4 Minimum Work for Separating Carbon Dioxide . . . . . . . . . . . . 20

1.4.1 Minimum Work for Gas Separation . . . . . . . . . . . . . . . 25

2 Fuel Cells and Chemical Equilibria 27

2.1 Introduction to Fuel Cells . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Thermodynamic Problem . . . . . . . . . . . . . . . . . . . . 29

2.2 Chemical Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2.1 The Chemical Potential . . . . . . . . . . . . . . . . . . . . . 32

2.3 The Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Chemical Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.5 Chemical Equilibrium in a Fuel Cell . . . . . . . . . . . . . . . . . . . 39

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2.6 Voltage of a Fuel Cell . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.6.1 A Note on Units . . . . . . . . . . . . . . . . . . . . . . . . . 432.6.2 Dependence of Voltage on Partial Pressure . . . . . . . . . . . 44

2.6.3 Voltages with Different Fuels . . . . . . . . . . . . . . . . . . . 45

2.7 Law of Mass Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.8 Fuel Cell vs. Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . 48

3 Solar Thermal, Phase Transitions, Heat Engines 50

3.1 The Three Phases of Water . . . . . . . . . . . . . . . . . . . . . . . 52

3.2 Phase Transition in Water . . . . . . . . . . . . . . . . . . . . . . . . 55

3.3 The Clausius-Clapeyron Relation . . . . . . . . . . . . . . . . . . . . 58

3.4 Evaporation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.5 The Carnot Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.5.1 Isothermal Expansion of Gas . . . . . . . . . . . . . . . . . . . 63

3.5.2 The Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.6 The Carnot Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.6.1 Stacking a Carnot Heat Engine and a Heat Pump . . . . . . . 69

3.6.2 Case Studies of a Carnot Engine and a Heat Pump . . . . . . 71

3.6.3 Carnot Engine as the Most Efficient Heat Engine . . . . . . . 733.6.4 η for a Heat Pump . . . . . . . . . . . . . . . . . . . . . . . . 73

3.7 Efficiency of a Solar Thermal Power Plant . . . . . . . . . . . . . . . 74

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List of Tables

2.1 Chemical species and reactions in a fuel cell . . . . . . . . . . . . . . 40

3.1 Table describing the four parts of the Carnot cycle given in Fig. 3.10. 67

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List of Figures

1.1 Correlation between fossil fuel combustion and atmospheric concentra-

tion of CO2. Adapted and used with permission from M. L. Machala,

Carbon-neutral Synthetic Fuels , large.stanford.edu/PHYS240. . . . . . 1

1.2 Correlation between atmospheric concentration of CO2, CH4 and T.

Courtesy of the Intergovernmental Panel on Climate Change, 2001;

http://www.ipcc.ch, N. Oreskes, Science 306, 1686, 2004; D. A. Stain-

forth et al, Nature 433, 403, 2005. . . . . . . . . . . . . . . . . . . . . 2

1.3 Minimum work required to separate one mole of CO2 at difference con-

centrations of CO2. From Wilcox, J. Carbon Capture , 2012, Springer

Publishing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 The initial state of mixed gases and the final state with CO2 separation

and other gases. From Wilcox, J. Carbon Capture , 2012, Springer

Publishing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5 Representation of differential volume dV . . . . . . . . . . . . . . . . 6

1.6 Schematic of the difference between heat and work . . . . . . . . . . 11

1.7 Representation of how the internal energy of a closed system may change. 12

1.8 Depending on the arrangement of the molecules, the entropy–or disorder–

can differ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.9 A partitioned box containing carbon dioxide and vacuum. . . . . . . . 15

1.10 A gas lattice with J sites and N molecules. . . . . . . . . . . . . . . . 19

1.11 The processes of mixing and demixing gases. . . . . . . . . . . . . . . 20

1.12 A barrier divides α moles of C O2 with 1− α moles of other gas. Both

gases have the same pressure. . . . . . . . . . . . . . . . . . . . . . . 22

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1.13 The entropy of mixing as a function of the C O2 fraction. . . . . . . . 24

1.14 The specific entropy of mixing as a function of the C O2 fraction. . . . 241.15 The partial molar entropy as a function of the C O2 fraction. . . . . . 25

1.16 The minimum work required to separate a mol of CO2 as a function

of the C O2 fraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.1 Hydrogen and oxygen gas separated by a membrane that is permeable

to protons but not electrons . . . . . . . . . . . . . . . . . . . . . . . 28

2.2 A fuel cell with a completed circuit, in the form of a light bulb. The

net reaction is 2H 2 + O2 ⇔ 2H 2O. . . . . . . . . . . . . . . . . . . . . 29

2.3 Hydrogen and oxygen gas at the same temperature and pressure, sep-arated by a barrier. All walls, including the internal purple one, are

impermeable to heat, work and matter. . . . . . . . . . . . . . . . . . 30

2.4 Using the slope and intercept, we can map the monotonically-decreasing

function U to a value of G for every value of U and V . It is important

for U to be monotonically decreasing such that there is only one value

of V for every value of U . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.5 An illustration of a system in a heat and work reservoir. . . . . . . . 36

2.6 The voltage of a fuel cell as a function of the pressure. The interceptis given by µ0rxn and can be changed by modifying the chemistry of the

reaction. Increasing the reactant concentration increases the voltage,

while increasing the product concentration decreases it. . . . . . . . . 45

2.7 The efficiency of a fuel cell as compared to the Carnot efficiency of a

heat engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.1 Schematic of a solar thermal power plant. This module will focus on

the thermodynamics of phase transitions and heat engines. . . . . . . 51

3.2 The specific Gibbs free energy ( Ḡ) as a function of the temperature

(T ). The stable phase of water is the one with the lowest Ḡ at a certain

T . Here, we neglect the temperature dependence of H̄ and S̄ . . . . . 53

3.3 Schematic of Gibbs free energy before and after boiling. . . . . . . . . 55

3.4 Gibbs free energy of water vapor and liquid water during boiling. . . 57

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3.5 Sketch of the Clausius-Clapeyron relation given by equation 3.13. The

boiling temperature rises with higher pressure. . . . . . . . . . . . . . 603.6 Schematic of boiling from vacuum (top) and evaporation from a dry

environment (bottom) with liquid water at room temperature. In both

cases, the liquid water evaporates/boils into water vapor (left) initially,

but the rate of evaporation (blue arrows) and condensation (red ar-

rows) eventually equilibrate after waiting a long time, leaving a partial

pressure of water equal to the water vapor pressure. . . . . . . . . . . 62

3.7 A piston in a heat reservoir isothermally expanding. Here the wall of

the piston is impermeable to matter but allows heat transfer from the

reservoir. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.8 Pressure, volume, temperature, and entropy during isothermal expan-

sion of a gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.9 A piston that isothermally expands and compresses between two pres-

sures. No net work is being done in a cycle. . . . . . . . . . . . . . . 66

3.10 The T vs S diagram of the Carnot engine. . . . . . . . . . . . . . . . 67

3.11 The Carnot efficiency, defined in equation 3.15, as a function of the

hot reservoir temperature. . . . . . . . . . . . . . . . . . . . . . . . . 68

3.12 A heat pump and a heat engine stacked. The work from the heat

engine is being used to power the heat pump. Here, ∆QH,1 and ∆QC,2

are positive because heat flows out, whereas ∆QC,1 and ∆QH,2 are

negative as heat flows in. . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.13 The maximum efficiency of a solar thermal power plant at different

temperatures and light concentration (C). This expression accounts

for both efficiency of sunlight to heat, after account for re-radiation,

and the efficiency of the heat to work through a Carnot engine. . . . 75

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CHAPTER 1. CARBON DIOXIDE SEPARATION AND ENTROPY 2

Industrialized and developing nations emit tens of millions of tons of carbon diox-

ide (CO2) into the atmosphere every year. Natural processes–such as CO2 uptakeby biomass through photosynthesis and absorption by oceans–cannot recapture and

sequester CO2 at the rate humans are producing it, particularly through the combus-

tion of fossil fuels. This leads to an increase in CO2 concentration in the atmosphere

shown in Fig. 1.1.

Figure 1.2: Correlation between atmospheric concentration of CO2, CH4 and T. Cour-tesy of the Intergovernmental Panel on Climate Change, 2001; http://www.ipcc.ch,N. Oreskes, Science 306, 1686, 2004; D. A. Stainforth et al, Nature 433, 403, 2005.

It has been shown that atmospheric greenhouse gas concentration track average global

temperature. Figure 1.2 shows the correlation over 0.4 million years between concen-

trations of methane (CH4) and CO2 in relation to temperature, T . However, atmo-

spheric CO2 levels never surpassed ∼325 parts per million by volume (ppmv) during

these hundreds of millennia, until the modern day. In 2015, CO2 concentrations levelsexceed 400 ppmv.

As anthropogenic climate change and its adverse effects have been linked to a green

house gas emissions (largely CO2), many have proposed capturing and sequestering

CO2 from the atmosphere as well as from industrialized processes (e.g. from flue


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gas) before it escapes into the atmosphere. But which process is more reasonable

both economically and from an energetic perspective? Figure 1.3 shows the minimum

Figure 1.3: Minimum work required to separate one mole of CO2 at difference con-centrations of CO2. From Wilcox, J. Carbon Capture , 2012, Springer Publishing.

amount of energy it takes to extract 1 mol of CO2 as a function of CO2 concentration in

the original gas. Notice the stark difference between the energy needed to remove onemole of CO2 at atmospheric concentrations versus industrial flue gas concentrations.

During this chapter you we learn how to construct this graph and will be able to

evaluate the thermodynamic viability of one capture process versus another.

1.1.1 The Thermodynamic Problem

As we will show in this module, principles of thermodynamics sets the absolute mini-

mum amount of energy required to capture and separate CO2. Let’s begin by precisely

defining the problem. Initially, CO2 is mixed with other gases. In the atmosphere,

this is mostly N2 and O2, with CO2 concentration ∼0.04 vol%. At the smokestack of

a coal combustion power plant, CO2 is mixed with H2O and N2, and has a concen-

tration typically in the range of a few tens of percent. The process of capture and

separation will separate CO2 into a pure stream from the other gases.


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Figure 1.4: The initial state of mixed gases and the final state with CO2 separation

and other gases. From Wilcox, J. Carbon Capture , 2012, Springer Publishing

Let’s more precisely define the problem using initial and final state. Because thermo-

dynamics does not describe the time-dependent behavior of a system, we will describe

the composition of the gas mixtures rather than the flow rates. Specifically, we can

use a closed volume of gas with appropriate internal partitioning to describe the ini-

tial and final states of the system (Fig. 1.4).

The initial state is a single volume of gas consisting of CO2 mixed with “other gases”

(which we are not specifying). The final state is the same volume but divided into

two partitions with a wall that does not allow the gas to go back and forth: the first

chamber contains pure CO2, while the second chamber contains the “other gases”.

Our goal is to calculate the minimum amount of energy needed to carry out the

separation process per mole of CO2. This represents the absolute best case scenario.

As we will show later, a process requiring energy less than this minimum value is

thermodynamically impossible and is a perpetual motion machine.

1.1.2 The Ideal Gas Law

Before discussing such topics as the First and Second Law of Thermodynamics in the

context of CO2 capture, we must first define an important relation between variables

that we can control in a gaseous system–namely pressure, volume, number of gas


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molecules, and temperature. An equation relating these variables is known as the

Ideal Gas Law. Recall that pressure (P) is defined as force applied per unit area:

P = Force

Area (1.1)

and that momentum is defined as:

Momentum = Force× Time (1.2)

Combining Equations 1.1 and 1.2 gives another relation for pressure

P = ForceArea

= MomentumTime×Area

(1.3)

Consider a gas molecule moving in a box of volume V. This molecule bounces around

inside the box hitting the walls periodically. As the molecule bounces it imparts

outward momentum to the walls of the box. If we add trillions of gas molecules to

the box, the momentum imparted the walls increases per equation 1.3. This outward

momentum is what we experience as pressure. But how does this relate to tempera-

ture and number of gas molecules?

First, we make three assumptions. The molecules:

1. collide with the wall elastically, meaning no energy is lost

2. are very small

3. do not interact with one another

These assumptions define an ideal gas.

The rate at which molecules collide with the wall depends on the volume of the box

and the velocity of the gas molecules. For a given period of time, only a fraction of

the gas molecules in the box will strike the wall, transfer momentum, and contribute


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Since we assumed that the collision of a gas molecule with the wall is elastic, the

kinetic energy of the molecule is conserved and the initial velocity of the moleculevx will rebound with −vx. By conservation of momentum and per Equation 1.2 the

momentum imparted to the wall is 2mvx.

Substituting this result and Eq. 1.6 into Eq. 1.4, we obtain:

P =

i

1

2

2mvx,i

×

vx,iV

=i

mv2x,iV

(1.7)

summed over the number of molecules. The 1

2 accounts for the probability of half of the molecules striking one wall, dA and the other half striking the opposite wall,

which does not exist in reality.

Now we consider the internal energy (U ) of the gases in the box defined as the

translational kinetic energy of the total number of gas molecules, N .

U = 1

2 imv2i

= 12

i

m(v2x,i + v2y,i + v

2z,i) (1.8)

If we sum over all of the molecules in the box, N , and consider the average kinetic

energy of all of the gas molecules, < KE >trans= 1

2m < v >2, then we can rewrite

Equation 1.8 as

U = N < KE >trans (1.9)

Notice similarities between Eqns. 1.7 and 1.8. Let’s begin by manipulating Equation

1.7–which included the momentum of the gas molecules striking the wall. Because weonly consider the x-direction, we multiply the righthand side of Eq. 1.7 by a factor


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of 13

–assuming the molecules move in all directions randomly.

P =i

miv2x,i

V =

1

3

2

2

i

miv2i

V

= 2

3

1

2

i

miv2i

V

Substituting Eqns. 1.8 and 1.9 yields

P = i

miv2x,i

V =

2

3N < KE >trans

V

= 2

3

U

V

(1.10)

Let’s do a sanity check to confirm that U V

has the same unit as pressure

U

V =

Energy

Volume

= Force× Length

Volume

= Force

Area

(1.11)

Rearranging the equation for pressure gives

P V = 2

3U (1.12)

It’s helpful to rescale U to a convenient value with which we are familiar. Let’s define

“temperature” as being linearly proportional to < KE >trans:

U ∝ T (1.13)

Moreover, we select a prefactor for T to scale with the boiling point and freezing

point of water, such that

T H 2Oboiling − T H 2Ofreezing = 100 (1.14)


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and results in

N < KE >trans= U = 32

NkBT (1.15)

This is where the Boltzmann constant (kB) originated. If we substitute Eq. 1.15 into

1.12 we obtain

P V = N kBT (1.16)

The Boltzmann factor is related to another variable to which you may be familiar–theuniversal gas constant, R, by a factor of Avagadro’s number, N A, or one mole.

R = N AkB (1.17)

Substituting Eq. 1.17 into 1.16 gives

P V = N

N ART = nRT (1.18)

where n is the number of moles of gas molecules. Thus, the Ideal Gas Law can bewritten as

P V = nRT (1.19)

We have shown how pressure, volume, number of molecules, and temperature are

related for an ideal gas and that pressure exerted on a container is due to the thermal

motion of gas molecules.

1.2 First Law of ThermodynamicsThere are two ways that energy can be transferred into and out of a thermodynamic

system: work and heat.


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1.2.1 Work

Work is the transfer of energy through directed motion. In our convention, we

define positive work as work done by the system on the surrounding (e.g. a piston–

the system–expanding against atmosphere–the surrounding), and negative work is

done by the surrounding on the system (a piston being compressed). Figure 1.6

shows a schematic of work. Recall from mechanics that work W done is given by

∆W = F ∆x (1.20)

where F is the force and x the displacement. Here the operator ∆ refers to the

difference between the initial and final state. In differential form, this is written as

dW = F dx

= P Adx

= P dV

(1.21)

Here, P is the pressure (which is defined as F A

, or force per unit area), A the area, and

V the volume. This expression states that the differential work equals the pressure

multiplied by the differential change in volume.

For example, consider doing work on a piston that is held at constant temperature

(e.g. by contacting it to a thermal reservoir), Fig. 1.6. From the Ideal Gas Law, we

know that P V = nRT = constant. Substituting this relation for P (V ), we can solve

for the work done by the system when the piston expands from V 1 to V 2

∆W =

dW

= V 2

V 1

nRT V

dV

= nRT ln

V 2V 1

(1.22)

This is known as isothermal expansion.


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1.2.2 Heat

Figure 1.6: Schematic of the difference between heat and work

Heating, on the other hand, is the transfer of energy through random motion. An

example is the random motion of gas molecules. The sign for heat is positive for heat

added into the system.

The difference between heat and work is that for work, the atoms are all traveling

in the same direction. For heat, on the other hand, the molecules are traveling

in random directions. Both heat and work added to a system involve a transfer of

kinetic energy, but they are in different directions. Figure 1.6 shows a schematic of the


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difference between heat and work. For work, gas molecules can all move in a specific

direction–the piston moving upward–and energy in the form of work is transferred forthis movement. When heat is added to the system, the random velocities of the gas

molecules increases.

1.2.3 Internal Energy and the First Law

Change in the internal energy (dU) = Heat In (dQ) - Work Done (dW)

dU = dQ − dW (1.23)

Figure 1.7: Representation of how the internal energy of a closed system may change.

In this class we will use the following definitions to classify systems: open, closed,

and isolated. An open system allows the passage of heat, work and mass into and out

of the system. A closed system is impermeable to mass but still allows the transferof heat and work. An isolated system allows nothing to pass in and out. Consider

the isolated system shown in Fig. 1.7 where there is no energy flow into or out of the

system. There is a closed system contained within this larger, isolated system. The


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first law of thermodynamics states that

dU = (dQ1 − dQ2)− (dW 1 − dW 2) (1.24)

In other words, the total energy of an isolated system is conserved. The net heat

and work coming into and out of the inner box must be zero. Since the universe is a

isolated system, the total energy of the universe is conserved.

1.2.4 Heating an Object

To heat up a gas is to increase the random motion of the gas molecules. There aretwo components to heating:

1. Increasing randomness or disorder of the gas molecules

2. Increasing the average velocity (< KE >) of the gas molecules

Let’s consider four indistinguishable molecules sitting in a 4x4 “lattice”, shown in Fig.

1.8, with each lattice position identical to the other (i.e. no edge or corner effects).

Figure 1.8: Depending on the arrangement of the molecules, the entropy–or disorder–can differ.

• (A): the four molecules are clustered in groups of 4

• (B): they are clustered in groups of 2

• (C): there is no cluster (only groups of 1)


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For simplicity, assume that the clusters cannot rotate.

The measure of disorder is related to the number of unique ways to arrange the

molecules in the lattice. The greater the number of unique arrangements, the greater

the disorder. We count the number of possible arrangements in Fig. 1.8.

• (A): 9 Ways

• (B): 12 × 12 = 144 ways

• (C) 16 × 16 × 16 × 16 = 65536 ways

Note that we are overcounting here by allowing molecules to occupy the same lattice

position (because they are identical). We are ignoring the few cases in (B) where the

configuration is in clusters of 4, as well as for (C).

Now, suppose that the average kinetic energy < KE > for (A), (B), and (C) are all

equal, meaning the molecules have the same average velocity. Which of the following

“transformations” requires a greater amount of heat?

• (A) → (B), or

• (A) → (C)

Recall that heating is the increase of random motion. Since the average speed of

the three systems are the same, the change in the randomness also determines the

extent of heating. Thus, (A) → (C) requires a greater amount of heating.

From this, we make the following definition:

Heating = < KE > × Change in Disorder

In differential form, heating is represented by dQ, average kinetic energy is repre-

sented by T (at constant temperature) and changes in disorder are represented by a

change in entropy (dS ). Here, S is the entropy which is a measure of disorder. We


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will give a more quantitative definition later.

In the following equation we substitute variables in for < KE >, disorder and heating.

dQ = T dS (1.25)

1.2.5 Reversible and Irreversible Processes

Figure 1.9: A partitioned box containing carbon dioxide and vacuum.

As we will show through an example, Eq. 1.25 only holds under very specific con-

ditions. Consider a partitioned box (Fig. 1.9) with walls impermeable to energy a

matter, where the left partition contains a gas (e.g. CO2) and the right partitioncontains no gas (i.e. vacuum). Now, let’s break the wall. The gas from the left

partition expands to fill the entire box. This process is known as the free expansion

of gas and results in the increase of disorder in the system–the number of potential

gas lattice sites has increased. Therefore, the entropy of the system increases and

∆S = S final − S init > 0. In contrast to isothermal compression or expansion (see

1.2.1), no work is done as we assume that breaking the wall requires negligible work,

like popping a balloon. So we have:

∆W = 0 (1.26)

Again, ∆ refers to the difference between the final and initial states. Moreover,


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because the box is an isolated system, the internal energy remains constant:

∆U = 0 (1.27)

Combining these two equalities with the First Law of Thermodynamics, we obtain:

∆Q = ∆U + ∆W = 0 (1.28)

This result contradicts with Eq. 1.25 and our earlier statement that the entropy of

the system increases when the wall is broken. It turns out Eq. 1.25 only applies to

a reversible process, that is, a process that can be reverted back to the initial statewithout doing additional work. In this free expansion of gas thought problem, it is not

a reversible process because no work is needed to expand the gas, yet work is required

to compress the gas back to the initial state. There are a few more equivalent ways

to define an irreversible process:

• Occurs in one direction

• Occurs spontaneously without work being done on the system

A reversible process, at any given point during the process, must not be far from

equilibrium (where equilibrium reflects a state of balance, with no net flow of matter

or energy). In other words, a reversible process occurs sufficiently slowly such that

the system is essentially in equilibrium at all time. This is also known as a quasistatic

process.

Note that a reversible process must be quasistatic, but not all quasistatic processes are

reversible. As an example, consider the system in Fig. 1.9. Now as a thought prob-

lem, add a special valve on the partition wall which lets one molecule flow through at

a time. The system is near equilibrium (i.e. static) at all times, but the gas expands

spontaneously and irreversibly as work must be done to remove the gas molecules.

Returning to Eq. 1.25, it only applies to reversible processes. Irreversible processes,


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such as the free expansion of gas, cause the entropy to increase more than the heat

added to the system. Therefore, we modify Eq. 1.25:

dQ ≤ T dS (1.29)

with dQrev = T dS (1.30)

where the subscript “rev” indicates a reversible process.

1.2.6 Defining the Second Law

There are three ways to state the Second Law of Thermodynamics

• An isolated system always tends to a configuration that maximizes disorder

(entropy)

• The disorder of an isolated system (such as the universe) increases due to oc-

currence of spontaneous and irreversible processes

• dQ/T ≤ dS , where dQrev/T = dS is the reversible limit

Whereas the First Law specifies the conservation of energy, the Second Law specifies

the direction of the process. For example, the Second Law says that gas expands when

a barrier to vacuum is broken. The First Law does not state the order of events, only

that the internal energy of the system is the same whether or not it is confined by

the barrier.

How do reversibility and the Second Law connect to carbon dioxide sepa-

ration?

For a closed system, the First Law tells us that the internal energy is fixed, thus

dU = dQ − dW = 0

dQ = dW


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The Second Law states that the disorder of a closed system stays constant or increases

but never decreases.dQ ≤ T dS (1.31)

Combining the two, we obtain the final expression for a closed system, where the

differential work is less than the temperature multiplied by the differential entropy.

dW ≤ T dS (1.32)

In other words, to decrease the entropy of the system, such as separating a gas mixture

containing carbon dioxide, oxygen, and nitrogen to just carbon dioxide and oxygenand nitrogen, work must be done on the system (dW


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Figure 1.10: A gas lattice with J sites and N molecules.

where the exclamation mark ! represents the factorial. If the molecules are indistin-

guishable, then we overcounted by N !. Accounting for this, Eq. 1.34 becomes:

Ω = J !

(J − N )!N ! (1.35)

Thus, Ω is the number of ways to configure indistinguishable gas molecules in the lat-

tice. For gas, the size of each “equivalent cell” is very small, which can be obtained

from quantum mechanics.

Boltzmann postulated that the entropy is related to the number of configurations of the molecules in the lattice.

S = kB ln Ω (1.36)

Here, kB is the Boltzmann constant, which is 1.38×10−23JK −1. For indistinguishable

molecules, it is given by Eq. 1.35. Combining Eqns. 1.35 and 1.36 yields the following

expression for entropy

S = kB ln J !

(J − N )!N ! (1.37)Using the Stirling approximation, ln N ! ≈ N ln N − N , which holds for large values


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of N

S = kB (J ln J − J − (J − N )ln(J − N ) + J − N − N ln N + N )

= kB (J ln J − (J − N )ln(J − N )−N ln N )

= kB

J ln

J

J − N −N ln

N

J − N

(1.38)

If J N , the first term in the expression approaches kBJln(1) = 0. This approxi-

mation is valid when the number of gas molecules is much less than the number of

lattice sites (i.e. low density), which is typically true for a gas except under very high

pressures. Thus, in this limit:

S = −N kB ln N

J (1.39)

This expression describes the entropy of a gas in a lattice.

1.4 Minimum Work for Separating Carbon Diox-

ide

As described in the problem statement, carbon dioxide separation can be depicted as

follows:

Figure 1.11: The processes of mixing and demixing gases.

We want to calculate the minimum work needed to carry out the demixing (separa-

tion) process. For an isolated system, the First Law states that dU = dQ − dW = 0.

The Second Law states that dQ ≤ T dS . Combining this with the First Law yields


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dW ≤ T dS . We call the lower bound the minimum work, W min.

During gas demixing, the system goes from the final state to the initial state shown

in Fig. 1.11, whereas during mixing, the system goes from the initial state to the final

state. Because demixing and mixing involves two identical end states, the associated

entropy changes are precisely opposite of one another:

∆S demix = −∆S mix (1.40)

Conceptually, it is easier to think about mixing, so we write:

∆W min = −T dS mix (1.41)

The minimum work is the work required to carry out the process reversibly. To

evaluate the minimum work, we define the system depicted in 1.11 with the following

parameters:

• α = volume fraction of CO2

• ntot = total moles of gas

• V tot = total volume of gas

Conceptually, what happens to the gas when they mix, as shown in Fig. 1.11?

1. The number of moles of the each gas component does not change, assuming no

exchange of matter with the outside world and no chemical reactions.

2. While the number of moles does not change, the volume occupied by each gas

changes.

Recall from Eq. 1.39, S = −NkB ln N J

, where N is the number of molecules and J is

the number of gas lattice sites. Remember that we can convert N kB = nR, where n

is the number of moles, N is the total number of molecules, and R is the gas constant.


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Thus, we see that the extent of volume expansion for each gas depends on α, the

volume fraction of CO2. Let us look at the entropy of mixing:

∆S mix = S final− S init (1.44)

where ∆ refers to the difference before and after breaking the wall. We consider the

entropy of each component:

S init = S CO2init + S

otherinit

= −αntotR ln αntotαV tot

+ constant+ −(1− α)ntotR ln (1 − α)ntot(1− α)V tot

+ constant= −ntotR ln

ntotV tot

+ constant

where ntot is the total number of moles of gas molecules. After mixing by breaking

the wall

S final = S CO2final + S

otherfinal

= −αntotR ln αntot

V tot+ constant+ −(1− α)ntotR ln

(1 − α)ntotV tot

+ constant= (−αntotR ln α) + (−(1− α)ntotR ln(1− α)) +

−ntotR ln

ntotV tot

+ constant

It can be shown that the constant term is equal for both S init and S final. Thus, we

can write the entropy of mixing as

∆S mix = S final − S init = −ntotR (α ln α + (1 − α) ln(1− α)) (1.45)

Does this make sense? Let’s consider the boundary cases. When α = 0 or α = 1,

∆S mix = 0. In other words, if there is no wall to begin with, there is no change in

the entropy of the system.

The first term represents the entropy gained by expanding CO 2. The second term

represents the entropy gained by the “other” gas. To understand the behavior of the


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Therefore, we also consider ∆S mix = ∆S mixnCO2

. This is known as the specific entropy,

plotted in Fig. 1.14.

∆S mix = ∆S mix

nCO2= −

1

αR (α ln α + (1 − α)ln(1− α)) (1.46)

Figure 1.15: The partial molar entropy as a function of the C O2 fraction.

As the fraction of CO2 increases, the magnitude of the specific entropy of mixing (and

demixing) tends to zero. Finally, sometimes it’s useful to calculate the incremental

change in ∆S mix per incremental addition of CO2. This is known as the partial molar

entropy and is given by the derivative of ∆S mix with respect to nCO2, or ∂ ∆S mix∂nCO2

, and

is plotted in Fig. 1.15.

1.4.1 Minimum Work for Gas Separation

To wrap up, we now have enough information to determine the minimum work re-

quired to separate carbon dioxide as a function of the number of moles of gas ntot

and the fraction of carbon dioxide α.

The minimum work to separate a mole of gas into CO2 and other gases is given

by:

∆W min = T ∆S demix = −T ∆S mix

= ntotRT (α ln α + (1 − α)ln(1− α))(1.47)


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The minimum work for a single mole of carbon dioxide is given by

∆W min = 1α

RT (α ln α + (1 − α) ln(1− α)) (1.48)

Here, the unit is J mol−1 CO2. This equation shows that it takes more work to cap-

ture a mole of CO2 from a source with low CO2 concentrations than from a source

with a high CO2 concentration. Thus, capturing a mole of CO2 from flue gas would

require less work than capturing a mole of CO2 from the atmosphere.

Sometimes, it’s also convenient to express α in terms of partial pressures instead of vol-

ume fractions, where pCO2 = αptot and pother = (1−α)P tot. Note that pCO2 + pother =

P tot, where P tot is the total pressure.

The minimum work for a mol of CO2 is given by

∆W min = RT

ln

pCO2P tot

+ pother pCO2

ln pother

P tot

(1.49)

This is plotted in Fig. 1.16.

Figure 1.16: The minimum work required to separate a mol of CO2 as a function of the CO2 fraction.


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Chapter 2

Fuel Cells and Chemical Equilibria

2.1 Introduction to Fuel Cells

Modern society requires a steady supply of electricity. There are many ways to

generate electricity, including hydroelectric dams, wind turbines, and solar cells. The

dominant source of electricity arises from burning of fossil fuels like coal or natural

gas, and then using a heat engine to convert the heat of combustion to electricity.

As we will learn in module 3, such heat engines are often fairly inefficient (around 30

% heat to electrical), and cannot be scaled down: for example, you cannot create a

small and efficient natural gas heat engine to power your computer.

An alternative way to convert fuel to electricity is a fuel cell. Here, fuel such as

hydrogen or methane (natural gas) reacts with oxygen to form water and/or carbon

dioxide, and this process directly generates electricity. This process can be much more

energy efficient than combustion, often nearing 70 %. The increased efficiency not

only reduces fuel consumption but also minimizes the release of harmful greenhouse

gases. Additionally, fuel cells are scalable, and can be used to power a device as small

as an integrated circuit to as large as a city.

To understand fuel cells, consider what happens if we mix two reacting gases,

such as hydrogen and oxygen, in a chamber. The fuel will combust, releasing heat.

However, this thermal energy cannot be converted to work unless it is combined with

a heat engine. Next, consider what happens if we separate the hydrogen and oxygen

27


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CHAPTER 2. FUEL CELLS AND CHEMICAL EQUILIBRIA 28

gas into separate chambers. The wall dividing the two is replaced by a membrane

that can only conduct protons, but not electrons (Fig. 2.1).

Figure 2.1: Hydrogen and oxygen gas separated by a membrane that is permeable toprotons but not electrons

In Fig. 2.1, a small amount of oxygen gas molecules will dissociate into oxygen ions

and travel across the membrane to react with hydrogen. However, once that happens,

an electrostatic force will develop between the two chambers, which will attract the

protons back into the hydrogen chamber, where they recombine with the electrons to

form hydrogen. As a result, no net reaction will take place and no work is generated.

To create an operational fuel cell, the electron must also be able to conduct from

the hydrogen to the oxygen chamber, enabling a complete, charge-balanced reaction.

Thus, creating an electrical pathway from the hydrogen to the oxygen chamber is

necessary to operate a fuel cell, as shown in Fig. 2.2

In the fuel cell, the oxygen gas in the anode receives electrons through an external

circuit and dissociates into oxygen ions through the half reaction O2 + 4e− ⇔ 2O2−.

The oxygen ions conduct through the membrane, and migrate to the anode and react

with hydrogen to form water through the half reaction 2O2− + 2H 2 ⇔ 2H 2O + 4e−.

Because the hydrogen wants to react with the oxygen to form water (in this module,you will learn why), this driving force gives rise to a voltage between the cathode and

the anode. Electrical work is generated when current flows.


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Figure 2.2: A fuel cell with a completed circuit, in the form of a light bulb. The netreaction is 2H 2 + O2 ⇔ 2H 2O.

2.1.1 Thermodynamic Problem

In this module, we aim to understand the maximum electrical work that can be

derived from a fuel cell. Electrical work equals voltage multiplied by the amount

of charge passed. The charge passed is proportional to the amount of gas reacted:

for example, reacting one oxygen molecule results in a transfer of 4 electrons. The

current is defined as the charge passed per unit of time, so increasing the current

results in an increased fuel consumption.

The voltage, on the other hand, depends strongly on the thermodynamics of the

system. In particular, the voltage under zero current condition, also known as the

open-circuit voltage, is governed purely by thermodynamic conditions. Thermody-

namic variables like temperature, pressure, and the composition of the gas all directly

affect the voltage of a fuel cell. To understand how much electrical work can be pro-duced by a fuel cell, we have to be able to calculate the voltage of a fuel cell, and

understand what factors affect the voltage.


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2.2 Chemical Work

Figure 2.3: Hydrogen and oxygen gas at the same temperature and pressure, sepa-rated by a barrier. All walls, including the internal purple one, are impermeable toheat, work and matter.

In module 1, we learned that mechanical work is defined as P dV . However, this

is not the only form of work possible. Let us consider this thought experiment,

shown schematically in Fig. 2.3: An isolated enclosure of gas is separated by a

barrier impermeable to matter (purple). One chamber contains oxygen, and the other

contains hydrogen. Both sides are at the same temperature and pressure. What

happens when the barrier is broken? Intuitively, we expect the following chemical

reaction to take place

2H 2 + O2 → 2H 2O (2.1)

Specifically

1. The number of hydrogen and oxygen atoms remain unchanged, consistent with

the enclosure being impermeable to matter.

2. 2 mol of hydrogen and 1 mol of oxygen react to form 2 mol of water. In other

word, the chemical bonds rearrange while conserving the number of constituent

atoms, but not the number of molecules.

3. Because the reaction is exothermic, we expect the temperature to increase.


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Since the enclosure is isolated, no heat flows in and no external mechanical work

is being done on the system. Thus, dQ = dW = 0, and the First Law gives dU = 0.Since the internal energy is constant, temperature should not change. This contradicts

point 3. However, since the First Law (dU = dQ−dW ) must be obeyed, we conclude

that our existing definition of work (dW = P dV ) is incomplete. In other words, we

are missing contributions to dW .

To be more precise, we are missing chemical work , which is the transfer of potential

energy contained in the chemical bonds of the molecules to kinetic energy of a system,

just as mechanical work (P dV ) can be converted to internal energy that increases the

temperature, like by a compressing piston. Here, we define chemical work as

Chemical Work = hdn − Tsdn (2.2)

where h is the molar enthalpy, or the bond energy, T the temperature, s the molar

entropy, and dn is the change in the number of moles. The first term in equation

2.2 relates to the change in the bond energy of the molecules (enthalpy). The second

term relates to the change in the degree of randomness in the molecules (entropy).

Equation 2.2 considers a system with a single chemical species, such as water. For

a system composed of many species, such as hydrogen, oxygen, and water, we cansum the chemical work for the ith component of the system:

Chemical Work =i

(hidni − T sidni) (2.3)

By incorporating chemical work as well as mechanical work, we can write an

expanded first law of thermodynamics

dU = dQ − dW (2.4)

dU rev = T dS − P dV +i

(hi − T si)dni (2.5)

Equation 2.5 summarizes this section. In the expression for the internal energy, the

first term relates to the heat added to or removed from the system. The second term


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relates to the mechanical work done on or by the system. The third term is the

chemical energy transferred into the system via the rearrangement of bonds and/orthe exchange of mass.

2.2.1 The Chemical Potential

Now that we have defined chemical work, we introduce a new variable, the chemical

potential of species i, denoted by the variable µi:

µi = hi − T si (2.6)

Like T and P , the chemical potential is independent of the quantity of the material.

Substituting equation 2.6 into equation 2.5 yields

dU rev = T dS − P dV +i

µidni (2.7)

2.3 The Gibbs Free Energy

The internal energy is given by equation 2.7. As written, the expression has three

independent variables: (S,V,n). Consider the special case that S and V are held

constant in a system, such that dS = 0 and dV = 0. Based on equation 2.7, this

indicates that the chemical work done on the system equals the change in the internal

energy. We can write the expression below:

dU =i

µidni|S,V (2.8)

The subscripted S and V after | indicate that those variables are held constant. This

equation can also be rearranged to provide an explicit definition of µi:


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S,V,ni are the independent variables that we control, whereas U, T , P are the depen-

dent variables.STEP 2

We consider the special case where dS = 0 and dni = 0. The Taylor series

expansion for U (equation 2.7) gives

dU =

∂U

∂V

S,ni

dV = −P dV (2.12)

Fig. 2.4 plots U as an arbitrary function of V . The local derivative of the curve is

equal to −P ; since the pressure must be positive, U must monotonically decrease as

a function of V . Suppose we pick an arbitrary point (V 0, U 0). We draw a tangent

line at the point (V 0, U 0) and label the y-intercept of this line as G0, which equals

U 0 − P V 0. More generally, we can map every value of U to a value of G via the

equation U = G − P V , where G is the intercept and −P is slope of the dashed line

in Fig. 2.4.

Figure 2.4: Using the slope and intercept, we can map the monotonically-decreasingfunction U to a value of G for every value of U and V . It is important for U to bemonotonically decreasing such that there is only one value of V for every value of U .

Combining U = G − P V , equation 2.12, and the product rule yields:


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dG = dU + d(P V )

dG = −P dV + P dV + V dP

dG = V dP

Now, we have created a thermodynamic function G whose independent variable

is P rather than V . In this expression, if the pressure is kept constant, there is no

change to G.

STEP 3

Applying the same transformation we did in STEP 2 to V to the variable S , we

can convert U (S,V,ni) to G(T , P , ni). First, the full expression for G is given:

Grev = U + P V − T S (2.13)

It follows that:

dGrev

= dU + d(P V )− d(T S )

= T dS − P dV +i

µidni + P dV + V dP − SdT + T dS

= V dP − SdT +i

µidni

(2.14)

This completes the transformation from U (S,V,ni) to G(T , P , ni). We have created

a thermodynamic function that is dependent on T and P rather than S and V . Now,

by holding T and P constant, as is commonly done during chemical reactions, G is

only dependent on dni, the change in the amount of species in the systems:


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dGrev =i

µidni|T,P (2.15)

µi =

∂G

∂ni

T,P,nj=i

(2.16)

The variable G is called the Gibbs free energy, and is the greatest amount of work

that can be extracted from a system at constant T and P . Moreover, equation 2.16

tells us that the chemical potential of species i is the partial molar Gibbs free energy

at constant T , P , and n j=i.

Figure 2.5: An illustration of a system in a heat and work reservoir.


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To illustrate this, consider a system in contact with an infinite heat and mechanical

work reservoirs, schematically shown in Fig. 2.5. We label the system with thesubscript sys and the reservoir with the subscript res. Thus, both T and P of the

system and the reservoir remain constant. If the heat and work is transferred and done

quasi-statically, then the maximum work that can withdrawn from the combination

of the system and the reservoir is given by:

Total Work Withdrawn = −dU sys − T dS res + P dV res

The first term relate to the change in the internal energy of the system. However,

the change in internal energy due to heat transfer (second term) or work done on thereservoir (third term) cannot be withdrawn.

Since dV res = −dV sys, dS res = −dS sys, and T and P are constant, we can write

Total Work Withdrawn = −dU sys + T dS sys − P dV sys

= −d(U sys − T S sys + P V sys)

= −dGsys

Thus, dGsys is the maximum work extracted from a system at constant P and T

when it is in contact with external reservoirs. In a fuel cell, P and T can be controlled

and kept constant, so dG yields the maximum work.

2.4 Chemical Equilibrium

The Gibbs free energy allows us to define equilibrium for a system at constant T and

P . These three conditions must be satisfied by a system when it is in equilibrium.

1. There is no net flow of energy or matter within the system

2. There is no net exchange of energy or matter with the outside world

3. There is no unbalanced driving forces or potentials


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Note the term “net” in “no net flow” and “no net exchange.” Equilibrium does not

mean there is no flow and no exchange of matter or energy, only that the exchangeis balanced in the forward and reverse directions. More formally, G is minimized

against perturbations in the composition of the system (n1, n2...) at constant T and

P when the system is at equilibrium:

∂G

∂ni

T,P,nj=i

= µi = 0 for all i (2.17)

In other words, an infinitesimally small modification in the composition of the system

yields no change in the Gibbs free energy. This is analogous to how the derivative of

the gravitational potential energy of a ball in a well is 0.

We can also apply this formalism to chemical reactions. Consider the following

reaction at constant T and P :

aA + bB ⇔ cC + dD (2.18)

where the lowercase letters are the stoichiometric coefficients. An example is 2H 2 +

O2 ⇔ 2H 2O, where the stoichiometric coefficients are 2, 1, and 2. Now, consider

what happens to dG when the reaction in equation 2.18 occurs, where dG is equal to

the change in free energy associated with each species, or dGi:

dG = dGA + dGB + dGC + dGD

= µAdnA + µBdnB + µC dnC + µDdnD(2.19)

As written, dG has four degrees of freedom, namely the change in the number of mol of

each species (dnA, dnB, dnC , dnD). In actuality, however, the dni are not independent

from one another. The reason is that the reaction must proceed while obeying thestoichiometry given in equation 2.18 in order to conserve mass. For example, for

every a mol of species A consumed, b mol of species B must also be consumed, while

c mol of species C and d mol of species D must be produced.

Thus, for the reaction shown in equation 2.18, we can substitute dnA = −adx,


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dnB = −bdx, dnC = cdx, and dnD = ddx, where dx is the extent of the reaction.

Combining the above substitutions for dnA, dnB, dnC , dnD with equation 2.19 yields

dG = − aµAdx− bµBdx + cµC dx + dµDdx

= (−aµA − bµB + cµC + dµD)dx

⇒

∂G

∂x

T,P

= −aµA − bµB + cµc + dµD

(2.20)

Now, there is only one degree of freedom, x, or the extent of reaction. At equilibrium,

the Gibbs free energy is minimized with regard to the change in the composition,at a constant temperature and pressure. The more general statement of chemical

equilibrium is given below.

∂G

∂x

T,P

=i

viµi = 0 (2.21)

where, vi is the stoichiometric coefficient of species i. It is negative for reactants and

positive for products.

2.5 Chemical Equilibrium in a Fuel Cell

We want to determine the voltage of the fuel cell (shown schematically in Fig. 2.2) at

open circuit, when the current is infinitesimally small. The reason is that any flowing

current would cause a change in voltage due to resistances in the cell (on the zeroth

order, you can think of this as Ohm’s Law, ∆V = I R). The first step is to consider

the components of the fuel cell where equilibrium is attained at open circuit, so that

we can apply equation 2.21 to these components/species. To attain open circuit, we

would remove the lightbulb and replace it with an infinitely resistive open circuit:

Table 2.1 lists the different species and reactions in a fuel cell. There, only the first,

fourth, and fifth reactions can reach equilibrium because all the participating species

can access each other. The other reactions cannot equilibrate because the species are

physically blocked from each other. The electrons are blocked by the broken circuit,


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Species Location Can Equilibrate ReactionO2− Membrane Yes O2−A ⇔ O

2−C

e− Anode/Cathode NoH 2O Anode No

H 2/e−/O2−/H 20 Anode Yes 2H 2 + 2O

2−A ⇔ 2H 2O + 4e

−A

O2/O2−/e− Cathode Yes O2 + 4e

−C ⇔ 2O

2−C

H 2/H 2O/O2 Anode/Cathode No 2H 2 + O2 ⇔ 2H 2O

Table 2.1: Chemical species and reactions in a fuel cell

and the gases are blocked by the membrane. To calculate the open-circuit voltage of

the fuel cell, we recognize that the following reactions must reach equilibrium at open

circuit:

1. 2H 2 + 2O2−A ⇔ 2H 2O + 4e

−A

2. O2 + 4e−C ⇔ 2O

2−C

3. O2−A ⇔ O2−C

The subscripts denote whether the reaction is happening at the cathode (C ) or the

anode (A). Note that, strictly speaking, number 3 is not a chemical reaction, but a

transport process. Applying the equilibrium condition (equation 2.21), we obtain

1. 2µH 2 + 2µO2−A= 2µH 2O + 4µe−

A

2. µO2 + 4µe−C = 2µO2−C

3. µO2−C = µO2−A

Combining these equations and rearranging, we obtain the chemical potential differ-

ence between electrons in the cathode and the anode as a function of the chemical

potential of the gas species.

µe−C − µe−

A=

1

4 (2µH 2O − µO2 − 2µH 2) (2.22)

The right hand side of equation 2.22 gives the difference in the chemical potential

between the product (H 2O) and reactants (H 2 and O2), multiplied by the appropriate


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stoichiometric coefficients. The left hand side gives the difference in the electron

chemical potential between the anode and the cathode. Equation 2.22 tells us that thedifference in the chemical potential of electrons in the anode and cathode is precisely

balanced by the difference in chemical potential of the reactants and products. The

factor of 4 results from 4 electrons transferred for every molecule of oxygen consumed.

Note for advanced readers: we left out the electrostatic potential to simplify this

problem.

2.6 Voltage of a Fuel Cell

At the end of the last section, equation 2.22 shows that the difference in the chemical

potential of the reactants and products create a difference in the chemical potential

of electrons between the anode and the cathode. We can use a voltmeter to measure

the open-circuit voltage, which is related to the difference in the electron chemical

potential between the two electrodes:

V = −1

F

µe−C

− µe−A

where F is Faraday’s constant, which equals 96,500 Coloumbs of charge per mol of electron, and µ is given in units of J/mol. The negative sign results from electrons

being negatively charged. Thus, a higher electric potential results in a lower energy.

Combining the definition of the voltage with equation 2.22, we arrive at the following

equation for the open-circuit voltage of a fuel cell

V = −1

F

µe−

C − µe−

A

= −

1

4F (2µH 2O − µO2 − 2µH 2) (2.23)

Here, the voltage (also known as the Nernst potential) is a direct measure of the

change in chemical potential of the species (H 2, O2, H 2O) upon reaction. In other

words, the fuel cell develops a precise voltage difference between the anode and cath-

ode to balance out the chemical potential difference between the gas in the two elec-

trodes (H 2 and O2/H 2O), so that equilibrium is achieved locally in the anode and

the cathode and electrolyte as in Table. 2.1.


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Now that we have obtained the relationship between the fuel cell voltage and the

gas chemical potentials, we relate the chemical potential and the voltage to the partialpressure of each gas specie. Recall from equation 2.6 that the chemical potential is

related to the partial molar enthalpy (h) and the partial molar entropy (s) of the

species:

µi = hi − T si

Recall from module 1 that the partial molar entropy of a gas is related to the concen-

tration, or the partial pressure ( pi). In module 1, we learned that the entropy (capital

S ) of a gas is given by:

S i = −niR ln pi + constant (2.24)

where R is the gas constant and ni is the number of mol of a gas. It can be shown

that the partial molar entropy,∂S i∂ni

, is given by

si = −R ln pi (2.25)

Strictly speaking, the argument of a logarithm should be unitless, so we modify theabove expression:

si = s0i − R ln

pi pref

(2.26)

Here, pref is the standard pressure such that si = s0i when pi = pref . In the homework

and exams, we typically choose 1 atm or 1 bar as pref . Combining equation 2.26 with

equation 2.6, we can write the chemical potential of species i as:

µi = hi − T s0i + RT ln pi pref (2.27)

Here, s0i not only accounts for the configurational entropy at standard state, but also

all other sources of entropy, such as the number of ways a molecule can rotate. Now,

we combine the hi and T s0i terms into a single term that we refer to as µ

0i . This


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allows us to lump all terms that are independent of the partial pressure pi:

µi = µ0i + RT ln

pi pref

(2.28)

This equation divides the chemical potential into two contributions. The first term,

µ0i , is the standard chemical potential of species i. This is the chemical potential at a

defined temperature T and partial pressure pref . When the partial pressure of species

i changes, µi changes in accordance with the second term in equation 2.28.

By combining this expression for the chemical potential (equation 2.28) with the

expression for the voltage of the fuel cell (equation 2.23), we can write the partial-

pressure-dependent open-circuit voltage:

V = 1

4F

µ0O2 + 2µ

0H 2− 2µ0H 2O + RT ln

( pH 2/pref )

2 pO2/pref ( pH 2O/pref )

2

(2.29)

The first three terms give the standard chemical potential of the reaction, whereas

the last term describes the configurational entropy of the gas molecules. Equation 2.29

states that the voltage of a fuel cell is not only dependent on the types of reactants

and products, but also on the concentration of each species. We combine the firstthree terms into the standard reaction potential ∆µ0rxn = −(µ

0O2

+ 2µ0H 2 − 2µ0H 2O

),

where ∆ indicate the difference between the initial and final state of the reaction. The

standard reaction potential is the change in the Gibbs free energy of the system per

mol of reactants reacted when the reactants and products are have a partial pressure

equal to pref .

2.6.1 A Note on Units

You may also see the voltage written in terms of a single molecule, rather than a mol

of molecules as in equation 2.29. With the proper conversion factors, the following

equation is identical to the one in equation 2.29:

V = 1

4e

µ0O2 + 2µ

0H 2− 2µ0H 2O + kBT ln

( pH 2/pref )

2 pO2/pref ( pH 2O/pref )

2


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Here, µi is the chemical potential of a single molecule of a species, given in units of

electron-volts (eV), rather than the chemical potential of a mol of the species, whichwould have units of J/mol. An electron-volt is the potential energy of an energy

across a voltage of 1 V. e in this equation is the charge of a single electron, which

equals 1.6×10−19 Coulombs. All the constants here differ from ones in equation 2.29

by Avogadro’s number, N A. For example, F = N A × e, R = N A × kB, and the µi

in equation 2.29 (which has units of J/mol) equals Avogadro’s number multiplied by

the µi in this equation (which has units of eV ).

For this course, we will teach in terms of F , R, and µi in units of J/mol, as

in equation 2.29. These units are more typical of chemistry. More physics-oriented

courses will use e, kB, µi in terms of eV. You are welcome to use those units if you

feel more comfortable. However, make sure you do not mix up the units in a single

equation. If you mix up units, you may get unphysical quantities on the order of

10−20 V or 1020 V. If you do find yourself getting those values, check that you are

using consistent units.

2.6.2 Dependence of Voltage on Partial Pressure

To analyze the fuel cell voltage further, we sketch voltage as a function of the partialpressure of each species in Fig. 2.6. Here, one component’s partial pressure is varied

at a time. When P i = P ref , the fuel cell voltage is given only by the standard reaction

potential. The voltage increases when the partial pressure of oxygen or hydrogen is

increased. In contrast, the voltage decreases when the partial pressure of water is

increased. To rationalize the origins, we revisit the net reaction of hydrogen and

oxygen to form water (2H 2 + O2 ⇔ 2H 2O).

When this reaction equilibrates, there will be significantly more water than hydro-

gen and oxygen. In other words, we say that the equilibrium of this reaction lies far tothe right (i.e., favoring the product). We can use this knowledge of equilibrium to ex-

plain why the open-circuit voltage of a fuel cell modulates with the partial pressures.

Let us imagine a fuel cell with a lot of water, and very little hydrogen and oxygen.

In such a fuel cell, there will be almost no driving force toward creating more water.


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Figure 2.6: The voltage of a fuel cell as a function of the pressure. The intercept isgiven by µ0rxn and can be changed by modifying the chemistry of the reaction. In-creasing the reactant concentration increases the voltage, while increasing the productconcentration decreases it.

Accordingly, this fuel cell generates a small voltage because there is no driving force

for the hydrogen and oxygen to react and form water. Alternatively, if the partial

pressures of hydrogen and oxygen is very high, then there will be a significant driving

force for creating water, giving a high fuel cell voltage, which is consistent with thevoltage plot based on equation 2.29.

2.6.3 Voltages with Different Fuels

The voltage of a fuel cell depends strongly on the type of fuel and oxidant. So far,

we have considered hydrogen and oxygen as reactants and water as the product. For

this reaction, we define the reaction potential ∆µ0rxn, and equals -475,000 J/mol of

O2. Thus, −∆µ0rxn

4F = 1.23V at 300 K.

What happens if we change the mobile ion from oxygen ion to proton? In such a

fuel cell, the reactions become:


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Cathode : O2 + 4H + + 4e− ⇔ 2H 2O

Anode : 2H 2 ⇔ 4H + + 4e−

Total : O2 + 2H 2 ⇔ 2H 2O

Because the total reaction and the number of electrons transferred are the same,

the open-circuit voltage is the same and equals the expression in 2.29. This illustrates

an important concept in thermodynamics: thermodynamic variables like open-circuit

voltage and Gibbs free energy is independent of the path taken. Here, regardless of

whether the proton or the oxygen ion is the mobile carrier, the change in the Gibbs

free energy per mol to react hydrogen and oxygen into water is always equal. Thus,

the open-circuit voltage of a fuel cell based on proton conduction equals one based

on oxygen ion conduction.

Finally, let us consider changing the fuel from hydrogen to methane. If oxygen

ions are mobile, this undergoes the following reactions:

Cathode : 2O2 + 8e− ⇔ 4O2−

Anode : CH 4 + 4O2− ⇔ CO2 + 2H 2O + 8e

−

Sum : 2O2 + CH 4 ⇔ 2H 2O + CO2

Here, ∆µ0rxn = -763,000 J/mol at 473 K, and −∆µ0rxn

8F = 1.10V . The voltage

differs from the hydrogen fuel cell because the change in bond energy upon reaction

of methane and oxygen into water and carbon dioxide is significantly different from

the change in bond energy from the reaction of hydrogen and oxygen. Thus, while

changing the mobile ion cannot change the open-circuit voltage, changing the type of

reactants can.


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2.7 Law of Mass Action

The concepts in this module can be used to derive the Law of Mass Action, which is

a very useful relationship. You can find versions of this law throughout many topics

in the physical, chemical, and biological sciences. It governs relationships as diverse

as the number of free electrons in a semiconductor, the yield of ammonia produced in

a chemical reaction plant, the kinetics of a digestion enzyme, and the rate in which

an infectious disease spreads in a population.

In this module, we present the Law of Mass Action in terms of chemical equilib-

rium. Consider the general reaction aA + bB ⇔ cC + dD, and recall the conditions

for equilibrium at a constant T and P from equation 2.21, reproduced below:

−aµA − bµB + cµC + dµD = 0 (2.30)

Substituting equation 2.28 (µi = µ0i + RT ln

piP ref

) and rearranging, we obtain the

following equilibrium condition:

− −aµ0A − bµ0B + cµ0C + dµ0D = RT − ln peqA

pref

a

− ln peqB

pref

b

+ ln peqC

pref

c

+ ln peqD

pref

(2.31)

where the superscript eq represents the equilibrium state. The left hand side of this

equation is −∆µ0rxn, the standard reaction potential. Rearranging yields:

K = exp

−

∆µ0rxnRT

=

peqC

pref

c peqD

pref

d

peqA

pref

a peqB

pref

B (2.32)Here, K , is referred to as the equilibrium constant. Equation 2.32 is the Law of

Mass Action, which relates the reactant and product partial pressure to the standard

chemical potential of the reaction and T . Going back to equation 2.31, we see that

the equilibrium occurs when the disorder of the system (entropy) perfectly balances

the change in the bond energy (enthalpy).

A more general relationship for the Law of Mass Action is shown here: for a


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chemical reaction with the stoichiometric coefficients written as vi:

exp

−

i viµi

RT

=i

peqi pref

vi

(2.33)

where a negative vi indicates a reactant and a positive vi indicates a product, and Π

indicates the product of all the terms enclosed.

2.8 Fuel Cell vs. Heat Engine

At the beginning of the module, we stated that a fuel cell can be much more efficient

than a heat engine. Now, we will close this module by defining and calculating the

efficiency of a fuel cell and a heat engine.

Figure 2.7: The efficiency of a fuel cell as compared to the Carnot efficiency of a heatengine

We define the efficiency as the maximum possible electrical work harvested divided

by the change in the bond energy, or the reaction enthalpy (∆µrxn). It can be shown

that the reaction enthalpy is the amount of heat released if we directly burn the fuel.

At constant pressure, this is known as the thermal efficiency ηtherm. Using oxygen

and hydrogen as reactants in the fuel cell, Fig. 2.7 shows that the efficiency is high


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at low temperatures and drops with increased temperature. The reason that the fuel

cell efficiency drops is because the reaction 2H 2 + O2 ⇔ 2H 2O decreases in entropyupon the creation of water. Lowering the entropy means that heat is released, which is

discarded to the environment rather than contribute to the open-circuit voltage of the

fuel cell. At high temperatures, T ds plays a much larger contribution to the chemical

potential (equation 2.6), and therefore reduces ∆µ0rxn and the open-circuit voltage.

However, there are other fuels where the voltage does not fall with temperature, as

you will see in your homework.

Figure 2.7 paints a rather pessimistic picture: the thermal efficiency of a hydro-

gen/oxygen fuel cell is a a little more than 90% at room temperature, and drops to

below 80% at 1000K. only a little more than 80 % at room temperature. In other

words, the electrical energy generated by such a fuel cell is less than 80 % of heat

released from burning the fuel directly. Why, then, should we use a fuel cell? It turns

out that the heat generated by combusting hydrogen with oxygen cannot be converted

to electrical work at 100 % efficiency. As you will learn in module 3, conversion of

thermal energy (heat) into work (electrical, mechanical, etc) is limited by the Carnot

efficiency, which is plotted in red in 2.7. Comparing the fuel cell thermal efficiency

to the Carnot efficiency, we see that fuel cells have a clear efficiency advantage at

temperatures below 1,000 K.


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Chapter 3

Solar Thermal, Phase Transitions,

Heat Engines

In module 2, we described fuel cells as one way to generate electricity from fuel. While

fuel cells are very efficient at converting energy from chemical bonds into electric work,

they require a steady source of fuel to operate, such as hydrogen. Because it readily

reacts with atmospheric oxygen, hydrogen gas is very rare on this planet. As a result,

hydrogen is typically created through methane reformation to meet demand, but

carbon dioxide is released in the process. Alternatively, some fuel cells directly use

methane but also release carbon dioxide that contributes to global warming.

One clean and abundant source of energy is the sun. Forty minutes of solar

irradiation (energy of sunlight reaching the earth’s surface) contains as much energy

as all of humanity ueses in an entire year. To put it another way, we can meet the

energy demands of the world by converting just 10% of the solar energy that shines

on the state of Arizona into electricity.

One way to harness solar power is to use solar cells. In these devices, solar energy is

converted directly to electricity through a semiconductor like silicon. While solar cell

technology has improved greatly in the last decade, they have several disadvantages:

firstly, these devices can only convert some of the sunlight to electricity (typically

15-20%); the rest is not absorbed or released as heat. Secondly, silicon processing

is expensive and energy-intensive. Third, solar cells cannot produce any electricity

50


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CHAPTER 3. SOLAR THERMAL, PHASE TRANSITIONS, HEAT ENGINES 51

when the sun does not shine, such as at night, while there is always demand for

electricity. In the absence of energy storage, solar cells can only produce about 20% of daily electrical demand until another source is required to supplement its production

deficiencies.

Figure 3.1: Schematic of a solar thermal power plant. This module will focus on thethermodynamics of phase transitions and heat engines.

An alternative method to create electricity from the sun is using a solar thermal

plant, shown in Fig. 3.1. Here, sunlight is concentrated by mirrors and used to heat

water or another working fluid. The water boils and passes through a heat engine.

In the heat engine, the difference in the temperature between the water vapor (or

steam) and the liquid water is converted to mechanical work using a turbine. Thismechanical work can be converted to electricity in a generator using the principles of

electromagnetic induction. Solar thermal electricity generation overcomes the three

problems of solar cells: solar thermal plants use most of the energy from the sun

and can attain higher efficiencies, the mirrors in a solar thermal plant are generally

less expensive than silicon, and such plants can store energy in the form of heat to


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continue to produce energy in the absence of sunlight.

As we see in the schematic in Fig. 3.1, there are two key components to a solarthermal plant. First, the water needs to transform from the liquid phase to the

gas phase during heating. Second, the water vapor runs a heat engine to generate

mechanical work. Thus, this module is divided into two parts: phase transitions (i.e.,

boiling water) and heat engines.

To simplify, we will describe the two processes separately and use the Carnot

engine as a model of a heat engine. Combining the two processes in a single ther-

modynamic cycle, as in a Rankine Cycle used in many real solar thermal plants, is

beyond the scope of this class. You may wish to take a more advanced thermody-

namics course in mechanical engineering if you are more interested in heat engines.

3.1 The Three Phases of Water

Water is highly abundant on earth and is the most important material for sustaining

life. Over 70% of the earth’s surface is covered by water, and the human body is more

than 60% water. Due to its abundance and lack of toxicity, it is the working fluid of

choice in most heat engines.

Water exists in three phases: solid ice, liquid water, and gaseous water vapor.

Figure 3.2 sketches the specific Gibbs free energy of water as a function of temperature

for all three phases of water. Recall that Ḡ = H̄ − T S̄ , so the slope of Ḡ vs T gives

−S̄ while the y-intercept gives H̄ . For simplicity, we assume that H̄ and S̄ do not

depend on temperature.

First, we will discuss the specific enthalpy, or the y-intercept of Fig. 3.2. Here,

the specific enthalpy of a molecule such as liquid water is the bond energy relative to

a defined state. In module 2, we chose the elemental form of O and H as the reference

state; however, because there is only one molecular species here, we can choose the

enthalpy of water vapor at 0◦K as the reference. Negative enthalpy implies stronger

intermolecular bonds, such as the hydrogen bonding between liquid water molecules.

However, we note that the specific value of H̄ does not matter (we can choose any

arbitrary point as our reference); choosing a different reference would move all the


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Figure 3.2: The specific Gibbs free energy ( Ḡ) as a function of the temperature (T ).The stable phase of water is the one with the lowest Ḡ at a certain T . Here, weneglect the temperature dependence of H̄ and S̄ .

curves in figure 3.2 up or down. The difference in the H̄ between, however, must be

specifically defined.

Documents

Engineering Thermodynamics for Fuel Cells