Thermodynamics Slide 3 / 93 Slide 4 / 93 First Law of - - PDF document

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Thermodynamics

Slide 3 / 93 Chemical Thermodynamics

The Golden Gate Bridge is painted regularly to slow down the inevitable rusting of the iron on the bridge. In this unit you will study how heat and temperature relate to work and energy and apply principles of thermodyamics to predict when chemical reactions will occur.

Slide 4 / 93 First Law of Thermodynamics

Recall a system is a portion of the universe that has been chosen for studying the changes that take place within it in response to varying conditions. A system can be relatively simple, like a glass of water, or it can be complex, like a planet, or the entire Universe can be considered a system.

Slide 5 / 93 First Law of Thermodynamics

Within every system exits a property called energy. In physics we learned about kinetic and potential energy. This year, we extend that by adding another way to change the energy of a system; by the flow of Heat (q). When two objects of different temperature are in contact, heat flow results in an increase of the energy of the cooler object and an identical decrease of the energy of the hotter object.

A B

heat flow T = 20 C T = 10 C

#E = W #E = w + q

Slide 6 / 93 The First Law of Thermodynamics

#E = w + q

The First Law of Thermodynamics tells us that energy cannot be created or destroyed. In other words the total energy of the universe is a constant. The same is true of any closed system. The First Law allows any process in which the total energy is conserved, including those where energy changes forms. .

Initial state Final state E of system decreases Internal energy, E E < E 0

#E < 0 (-)

E E0 Energy lost to surroundings Initial state Final state E of system increases Internal energy, E Energy gained from surroundings E > E0

#E > 0 (+)

E E0

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Slide 7 / 93 First Law of Thermodynamics

Most of the processes in the natural world that involve transfer of energy from one form to another don't just happen naturally. For example, gold does not rust in the same way iron does. 4Au(s) + 3O2(g) --> 2Au2O3(s) doesn't happen 4Fe(s) + 3O2(g) --> 2Fe2O3(s) does happen As reserves of fossil fuels run low, people say we have an energy

• crisis. But if the First law of Thermodynamics is true, energy cannot

be created or destroyed, so we're not actually running out of energy. What do people really mean?

Slide 8 / 93 The First Law of Thermodynamics

The First Law of Thermodynamics applies to any closed system. If our system is a cup, resting on a ledge at a certain height, we know the cup has potential energy and if it falls that energy is transfered to kinetic energy and thermal energy. In this process as our system is definited, total energy remains

• conserved. If the initial and final energy of the system are equal

to each other, why can't the process happen in reverse? Why don't we ever see a broken cup reassemble and return back to its initial position on the ledge?

E0 + W = Ef

Slide 9 / 93

The Second Law is a statement about which processes occur and which do not. There are many ways to state the second law: Heat can flow spontaneously from a hot object to a cold

• bject; but not from a cold object to a hot object.

It is impossible to build a perpetual motion machine. The universe always gets more disordered with time. Your bedroom will get increasingly messy unless you keep cleaning it up.

The Second Law of Thermodynamics Slide 10 / 93 2nd Law: Order to Disorder

Natural processes tend to move toward a state of greater disorder. Stir sugar into coffee and you get coffee that is uniformly

• sweet. No amount of stirring will get the sugar back out.

When a tornado hits a building, there is major damage. You never see a tornado pass through a pile of rubble and leave a building behind. You never walk past a lake on a summer day and see a puff of steam rise up, leaving a frozen lake behind. The First Law of Thermodynamics maintains that the above scenarios are possible. The Second Law maintains that they won't naturally occur.

Slide 11 / 93 2nd Law: Order to Disorder

The Second Law tell us which processes are naturally favorable - that is they can occur without more energy being put in than is released. Favorable doesn't mean fast, it just means that it will naturally

• ccur if a system is left on its own.

Slide 12 / 93 Thermodynamically Favorable

Once the valve is opened, the gas in vessel B will effuse into vessel A and vice versa, but once the the gases are mixed, they will not spontaneously unmix. The mixing of these gases is favorable because there is much higher probability of the gases being mixed than unmixed. A thermodynamically favorable process is not reversible.

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FOR EXAMPLE favorable at T > 0 C favorable at T < 0 C

Favorable Processes

Processes that are favorable at one temperature may be not favorable at other temperatures.

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1 A reaction that is thermodynamically favorable _____. A is very rapid B will proceed without a net increase in energy C is also spontaneous in the reverse direction D has an equilibrium position that lies far to the left E is very slow

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2 Which of the following statements is true? A Processes that are favorable in one direction are not favorable in the opposite direction. B Processes are favorable because they

• ccur at an observable rate.

C Favorability can depend on the temperature. D A and C are true

Slide 16 / 93 Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. System Endothermic Surroundings Heat +q

T-#T T

System Exothermic Surroundings Heat

• q

T+#T T

Slide 17 / 93 Irreversible Processes

Irreversible processes cannot be undone by exactly reversing the change to the system. Thermodynamically favorable processes are irreversible. work Gas Vacuum Movable partition Piston

Slide 18 / 93

3 A reversible process is one that __________. A can be reversed with no net change in either system or surroundings B is thermodynamically favorable C is thermodynamically unfavorable D must be carried out at low temperature E must be carried out at high temperature

SLIDE 4

Slide 19 / 93 Entropy

Entropy (S) is a term coined by Rudolph Clausius in the 19th century. S = q

T

Entropy refers to the ratio of heat to the temperature at which the heat is delivered:

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Entropy can be thought of as a measure of the randomness of a system, or as a measure of the number

• f ways of arranging particles.

It is related to the various modes of motion in molecules. Like total energy, E, and enthalpy, H, entropy is a state

• function. As a result, we are interested in measuring the

change in entropy #S, as opposed to the absolute entropy, S

#S = Sfinal - Sinitial

Entropy Slide 21 / 93 Entropy

For a process occurring at constant temperature, the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:

#S = qrev T

Isothermal process Slide 22 / 93 Second Law of Thermodynamics

The entropy of the universe increases for thermodynamically favorable processes and The entropy of the universe does not change for reversible processes.

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In other words: For reversible processes:

∆S= #Ssystem + #Ssurroundings = 0

For irreversible processes:

#S= #Ssystem + #Ssurroundings > 0

This means that the entropy of the universe constantly increases.

Second Law of Thermodynamics Slide 24 / 93

4 The thermodynamic quantity that expresses the degree of disorder in a system is ______. A enthalpy B internal energy C bond energy D entropy E heat flow

SLIDE 5

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5 For an isothermal (constant temperature) process, #S = __________. A q B qrev / T C qrev D Tqrev E q + w

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6 Which one of the following is always positive when a thermodynamically favorable process occurs? A #Ssystem B #Ssurroundings C #Suniverse D #Huniverse E #Hsurroundings

Slide 27 / 93

7 The entropy of the universe is __________. A constant B continually decreasing C continually increasing D zero E the same as the energy, E

Slide 28 / 93 Entropy on the Molecular Scale

Ludwig Boltzmann described the concept of entropy

• n the molecular

level by using statistical analysis

Slide 29 / 93 Statistical Interpretation of Entropy and the Second Law

A macrostate of a system is specified by giving its macroscopic properties – temperature, pressure, and so on. A microstate of a system describes the position and velocity of every particle. For every macrostate, there are one or more microstates.

**

T = 16 C P = 1 atm

Slide 30 / 93

A simple example: tossing four coins. The macrostates describe how many heads and tails there are; the microstates list the different ways of achieving that macrostate.

Statistical Interpretation of Entropy and the Second Law **

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Assume each microstate is equally probable; the probability of each macrostate then depends on how many microstates are in it. The number of microstates quickly becomes very large if we have even 100 coins instead of four.

Statistical Interpretation of Entropy and the Second Law ** Slide 32 / 93 Statistical Interpretation of Entropy and the Second Law

This table lists some of the possible outcomes (macrostates) for 100 coin tosses , how many microstates they have, and the relative probability that each macrostate will occur. Note that the probability of getting fewer than 20 heads

• r tails is extremely small.

Macrostate # of microstates Probability heads tails 100

1

8.0x10-31

99 1 100

8.0x10-29

90 10

1.7x1013 1.0x10-17

80 20

5.4x1020 4.0x10-10

60 40

1.4x1028 0.01

55 45

6.1x1028 0.05

45 55

6.1x1028 0.05

20 80

5.4x1020 4.1x10-10

10 90

1.7x1013 1.0x10-17

1 99 100

8.0x10-29 100 1 8.0x10-31

probabilities of various macrostates for 100 coin tosses

** Slide 33 / 93

The second law does not forbid certain processes; all microstates can occur with equal probability. However, some processes are extremely unlikely – a lake freezing

• n a hot summer day, broken mug re-assembling itself; all the air in

a room moving into a single corner. Tossing a coin 100 times led to some macrostates being so unlikely that they will probably not ever occur...Now think of the

• dds with even one mole of matter (6 x 1023 particles)...some

macrostates become so rare as to be effectively impossible in the lifetime of the universe.

Statistical Interpretation of Entropy and the Second Law ** Slide 34 / 93 Entropy on the Molecular Scale

Degrees of Freedom: Some molecules exhibit several types of motion Translational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about s bonds.

** Slide 35 / 93

Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy can then be defined as: S = k lnW As the microstates increases, entropy (S) increases where k is the Boltzmann constant, 1.38 # 10# 23 J/K.

Entropy on the Molecular Scale **

Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the molecules.

Slide 36 / 93 Entropy on the Molecular Scale

The change in entropy for a process, then, is

#S = k lnWfinal # k lnWinitial

Wfinal Winitial

#S = k ln

Entropy increases with the number of possible microstates.

**

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The number of microstates and, therefore, the entropy tends to increase with increases in: · Temperature · Volume · The number of independently moving molecules · The number of degrees of freedom of each molecule

Entropy on the Molecular Scale ** Slide 38 / 93

8 Which of the following has the largest entropy? A He @10 C B He @15 C C He @25 C D He @500 C E They are all the same. Entropy is independent of temperature.

Slide 39 / 93 Entropy and Physical States

Entropy increases with the freedom of motion of molecules. S(g) > S(l) > S(s)

** Slide 40 / 93

9 Which of the following phase changes would result in an increase in entropy? A L --> g B g --> s C L --> s D A and B E B and C

Slide 41 / 93 Entropy

When substances are dissolved in one another, such as when a solid is dissolved in a solvent, entropy increases.

_ 2+ _ _ _ _ _ _ 2+ _ _ 2+ _ 2+ 2+

Slide 42 / 93

10 In which of the following instances would entropy increase A Gases are formed from liquids and solids B Liquids or solutions are formed from solids C The number of gas molecules increases D The number of moles increases E All of the above

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Slide 43 / 93 Third Law of Thermodynamics

The entropy of a pure crystalline substance at absolute zero is 0. Solid crystal

• rdered

arrangement Liquid less ordered more freedom

Can entropy ever equal 0?

Slide 44 / 93 Standard Entropies

Entropy values tend to increase with increasing molar mass as the molecules have more degrees of freedom. Entropy values are small compared to enthalpy. Therefore entropy values are expressed in J/mol-K. Although entropy values are always positive, the change in entropy , #S, can be either positive or negative since Substance

H2(g) N2(g) O2(g) H2O(g) NH3(g) CH3OH(g) C6H6(g) H2O(l) CH3OH(l) C6H6(l) Li(s) Na(s) K(s) Fe(s) FeCl3(s) NaCl(s)

S0, J/mol-K

130.6 191.5 205.0 188.8 192.5 237.6 269.2 69.9 126.8 172.8 29.1 51.4 64.7 27.23 142.3 72.3

Slide 45 / 93 Standard Entropies

Larger and more complex molecules have greater entropies.

C C C H H H H H H H H

Propane, C 3H8 #S = 270.3 J/mol K

C H H H H

Methane, CH

4

#S = 186.3 J/mol K

Slide 46 / 93 Predicting Entropy Changes

To determine whether entropy increases or decreases in a reaction: FIRST, look at how the phases of the reactants and products are changing: Increase in S (l) to (g) vaporization (s) to (g) sublimation (s) to (l) melting Decrease in S (g) to (l) condensation (g) to (s) deposition (l) to (s) freezing

Slide 47 / 93 Predicting Entropy Changes *

Increase in S 2SO3(g) --> 2S(s,rhombic) + 3O2 Decrease in S 2H2(g) + O2(g) --> 2H2O Negligible change in S H2(g) + Cl2(g) --> 2HCl(g) Lastly, check to see whether the number of moles increases or decreases. Second, look at whether the number of gas moles is increasing or decreasing:

Slide 48 / 93

11 Which of the following statements is false? A The change in entropy in a system depends on the initial and final states of the system and the path taken from one state to the other. B Any irreversible process results in an

• verall increase in entropy.

C The total entropy of the universe increases in any spontaneous process. D Entropy increases with the number of microstates of the system.

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12 Of the following, the entropy of __________ is the largest. A HCl (l) B HCl (s) C HCl (g) D HBr (g) E HI (g)

Slide 50 / 93

13 Of the following, the entropy of gaseous __________ is the largest at 25oC and 1 atm. A H2 B C2H6 C C2H2 D CH4 E C2H4

Slide 51 / 93

14 The entropy of a pure crystalline substance at 0oC is zero. True False

Slide 52 / 93

15 Which one of the following processes produces a decrease in the entropy of the system? A boiling water to form steam B dissolution of solid KCl in water C mixing of two gases into one container D freezing water to form ice E melting ice to form water

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16 #S is positive for __________. A 2H2(g) + O2(g) --> 2H2O(g) B 2NO2(g) --> N2O4(g) C CO2(g) --> CO2(s) D BaF2(s) --> Ba2+(aq) + 2F-(aq) E 2Hg(l) + O2(g) --> 2HgO(s)

Slide 54 / 93

17 For which of the following processes would #S be negative? A H2O(l) --> H2O (g) B CaCO3(s) --> CaO(s) + CO2(g) C Ca2+(aq) + CO32-(aq) --> CaCO3(s) D 2NH3(g) --> N2(g) + 3H2(g) E A and C

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Slide 55 / 93 Calculating Entropy Changes

The entropy change, #S# # for a reaction can be estimated in the same way we estimated enthalpy change, #H: S# = # # nS#(products)) - # # mS#(reactants)) where n and m are the coefficients in the balanced chemical equation.

Slide 56 / 93

18 The value of #So for the catalytic hydrogenation of acetylene to ethane is _____ J/K∙ mol. A +18.6 B +550.8 C +112.0 D

• 112.0

C2H2(g) + H2(g) # C2H4(g) E

• 18.6

Slide 57 / 93

19 The value of #So for the oxidation of carbon to carbon dioxide, is ______ J/K. A +424.3 B +205.0 C

• 205.0

D

• 2.9

C(s, graphite) + O2(g) # CO2(g) E +2.9

Slide 58 / 93 Entropy Changes in Surroundings

Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: At constant pressure, qsys is simply #Ho for the system, so

#Ssurr = - qsys T #Ssurr =

• #Hsys

T

Slide 59 / 93 Entropy Change in the Universe

The universe is composed of the system and the surroundings. Therefore,

#Suniverse = #Ssystem + #Ssurroundings

For thermodynamically favorable processes... #Suniverse > 0

Slide 60 / 93

• T#Suniverse = #Hsystem - T#Ssystem

#Suniverse = #Ssystem + #Ssurroundings #Suniverse = #Ssystem +

• #Hsys

T

Multiply both sides by (-T)

• T#Suniverse = #G

#G = #Hsystem - T#Ssystem #Ssurr =

• #Hsys

T

Gibbs Free Energy

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#G = #H - T#S

This is called the Gibbs equation.

Entropy Change in the Universe

#G refers to the change in Gibbs free energy This is the energy available to do work.

Slide 62 / 93

When #Suniverse is positive, #G is negative. If #Suniverse is positive this means the reaction is irreversible. Irreversible reactions/processes are spontaneous. When #G is negative, a process is thermodynamically favorable.

• T∆Suniverse = #G

Gibbs Free Energy # Slide 63 / 93

N2 + 3H2 <--> 2NH3 s p

• n

t a n e

• u

s spontaneous

F r e e e n e r g y

Equilibrium Course of reaction

Gibbs Free Energy

If #G is negative , the reaction is favorable. If #G is 0, the system is at equilibrium. If #G is positive, the reaction is unfavorable. However, the reverse reaction would be favorable.

HH N N N N N N N N N N N N HH HH H H HH HH HH HH HH N HH N HH N HH N HH N HH H H H H N N N N H H N N H H H H N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH N HH pure N2 +H2

Equilibrium mixture ΔG=0 pure NH3

# Slide 64 / 93

20 Which of the following MUST be true for a reaction to be thermodynamically favorable?

A #G is positive B #G is negative C #Suniverse is positive D #Ssystem is positive E B and C

Slide 65 / 93 Standard Free Energy Changes

Analogous to standard enthalpies of formation are standard free energies of formation, Gf#. where n and m are the stoichiometric coefficients As it was for ΔHf0 (but not ΔSf0), the ΔGf0 for elements in their standard state is zero #G# = # # n#G#

f(products) - #(m#G# f (reactants))

# Slide 66 / 93

21 The standard Gibbs free energy of formation of __________ is zero. (I) Al (s) (II) Br

2 (l) (III) Hg (l)

A I only B II only C III only D II and III E I, II, and III

#

SLIDE 12

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22 The value of ΔGo at 25oC for the decomposition of gaseous sulfur trioxide to solid elemental sulfur and gaseous oxygen, as shown below, is __________ kJ/ mol.

A +740.8

2SO3(g) à 2S(s, rhombic) + 3O2(g)

B

• 370.4

C +370.4 D

• 740.8

E +185.2

*

Sulfur S (s, rhombic) 31.88 SO2(g)

• 269.0 -300.4

248.5 SO3(g)

• 395.2 -370.4

256.2 Oxygen O2(g) 205.1 #H0f #G0f

S (kJ/mol) (J/k-mol) (kJ/mol)

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23 The standard Gibbs free energy of formation of __________ is zero. (I) H2O(l) (II) Na(s) (III) H2(g)

A I only B II only C III only D II and III E I, II, and III

# Slide 69 / 93

24 The value of ΔGo at 25oC for the formation of phosphorous trichloride from its constituent elements, as shown below, is ____ kJ/mol.

A

• 539.2

P2(g) + 3Cl2(g) à 2PCl3(g)

B +539.2 C

• 642.9

D +642.9 E

• 373.3

*

Phosphorous P2 (g) 144.3 103.7 218.1 PCl3(g)

• 288.1 -269.6

311.7 POCl3(g)

• 542.2 -502.5

325 Chlorine Cl2 (g) 222% Cl-(aq)

• 167.2 -131.2

56.5 #H0f #G0f

S (kJ/mol) (J/k-mol) (kJ/mol)

Slide 70 / 93 Free Energy Changes

At temperatures other than 25 oC,

#G#

= #H# - T #S# How does #G# change with temperature?

Slide 71 / 93 Free Energy and Temperature

There are two parts to the free energy equation:

#H# is the enthalpy term

T#S# is the entropy term Note that the temperature dependence of free energy comes from the entropy term.

#G = #H# -T#S#

When you use the Gibbs equation above, be sure to have both enthalpy and entropy terms in the same units. Generally, it is easiest to divide #S by 1000 in order for it to be in kJ/mol-K.

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25 The value of ΔGo at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide, as shown below, is ______ kJ/mol. The ΔHo for this reaction is -269.9 kJ/mol, and ΔSo is +11.6 J/K.

S(s, rhombic) + O2(g) à SO2(g) A

• 300.4

B +300.4 C

• 4,597

D +4,597 E

• 274.2

#G = #H -T#S #G = # # n#G#

f(products) - #(m#G# f (reactants))

SLIDE 13

Slide 73 / 93 #G and Thermodynamic Favorability

As we have seen, the enthalpy, entropy, and the temperature influence both the sign and magnitude of

#G for a given process.

Let's examine four different processes and determine the qualitative impact of these factors on #G.

Slide 74 / 93

Case 1: Freezing of water @ 1 atm: H2O(s) --> H2O(l)

#H = - (energy is released as bonds form) #S = - (disorder decreases as the crystal forms) #G = #H -T#S

At what temperatures will this reaction be thermodynamically favorable?

#G = (-) + (-T(-))

= (-) + (T(+)) To be thermodynamically favorable, #G must be negative, so a relatively small T is required.

#G and Thermodynamic Favorability Slide 75 / 93

Case 2: Combustion of gasoline(octane) @ 1 atm: 2C8H18(g) + 25O2(g) --> 16 CO2(g) + 18H2O(g)

#H = - (energy is released as new bonds form) #S = + (few moles are converted to many)

At what temperatures will this reaction be thermodynamically favorable? G = (-) + (-T(+)) = (-) + (T(-)) This reaction is favorable at all temperatures as both the enthalpy and entropy terms are negative.

#G and Thermodynamic Favorability

#G = #H -T#S

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#G = #H -T#S

Case 3: Photosynthesis 6CO2(g) + 6H2O(g) --> 6O2(g) + C6H12O6(s)

#H = + (energy is absorbed as to break existing bonds) #S = - (many moles are converted to few)

At what temperatures will this reaction be thermodynamically favorable?

#G = (+) + (-T(-))

= (+) + (T(+)) This reaction is unfavorable at all temperatures as both the enthalpy and entropy terms are positive. Plants need energy from the sun for this reaction to occur.

move for answer #G and Thermodynamic Favorability Slide 77 / 93

#G = #H -T#S

Case 4: Dissolving of ammonium chloride in water NH4Cl(s) --> NH4+(aq) + Cl-(aq)

#H = + (energy is absorbed as to break existing bonds) #S = + (increase in disorder as ions mix with water)

At what temperatures will this reaction be thermodynamically favorable?

#G = (+) + (-T(+))

= (+) + (T(-)) This reaction is favorable only at relatively high temperatures.

move for answer

#G and Thermodynamic Favorability Slide 78 / 93

Fill in the following summary chart

#G and Thermodynamic Favorability

#G = #H -T#S #H #S

+ + +

T #G

+

• all

high low high low all

SLIDE 14

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26 For a reaction to be thermodynamically favorable under standard conditions at all temperatures, the signs of ΔHo and ΔSo must be _____ and _____, respectively.

A +, + B +, - C

• , +

D

• , -

E +, 0

#G = #H -T#S

Slide 80 / 93

27 For a reaction to be unfavorable under standard conditions at all temperatures, the signs of ΔHo and ΔSo must be _____ and _____, respectively.

A +, + B +, - C

• , +

D

• , -

E +, 0

#G = #H -T#S

Slide 81 / 93

28 A reaction that is unfavorable at low temperature can become favorable at high temperature if ΔH is ____ and ΔS is ____.

A +, + B

• , -

C +, - D

• , +

E +, 0

#G = #H -T#S

Slide 82 / 93

29 A reaction that is not favorable at one temperature can become favorable if the temperature is lowered. Therefore, ΔH is ____ and ΔS is ____.

A +, + B

• , -

C +, - D

• , +

E +, 0

#G = #H -T#S

Slide 83 / 93

30 The values for #H and #S values are as follows for a reaction. #H = 137 kJ/mol #S = 120 J/K mol This reaction is __________________________.

A favorable only at low temperatures. B favorable only at high temperature. C favorable at all temperatures. D Not enough information is provided.

#G = #H -T#S

Slide 84 / 93

31 When ammonium chloride dissolves in water the temperature of the solution is less than that of the original water sample. Thus, we know that ΔH is _____ and that ΔS is _________.

A

• -

B + + C

• +

D + - E

#G = #H -T#S

SLIDE 15

Slide 85 / 93

Occasionally, you are asked to calculate one of the following: 1) the specific temperature at which a reaction changes from being favorable to unfavorable

• r

2) the specific temperature at which a reaction changes from being unfavorable to favorable

Free Energy and Temperature Slide 86 / 93

At what specific temperature will a reaction change from being favorable to unfavorable? In this situation, if #G < 0 at low temperatures and #G > 0 at high temperatures, then we can conclude that #H < 0 and #S < 0. You can calculate the specific temperature at which #G changes sign by setting #G = 0.

Free Energy and Temperature

Remember to note that enthalpy values are usually in kJ/mol while entropy values are usually given in J/mol-K. Therefore, divide entropy values by 1000. Below this temperature, the reaction will be favorable, while above it, the process is unfavorable.

#G = #H -T#S 0 = #H -T#S T#S = #H #T = #H/#S

Slide 87 / 93

At what specific temperature will a reaction change from being unfavorable to favorable? In this situation, if #G > 0 at low temperatures and #G < 0 at high temperatures, then we can conclude that #H > 0 and #S > 0.

Free Energy and Temperature

Below this temperature, the reaction will be unfavorable, while above it, the process is favorable. Remember to note that enthalpy values are usually in kJ/mol while entropy values are usually given in J/mol-K. Therefore, divide entropy values by 1000.

#G = #H -T#S 0 = #H -T#S T#S = #H #T = #H/#S

Slide 88 / 93

32 For the below reaction, ΔHo = 131.3 kJ/mol and ΔSo = 133.6 J/mol at 298 K. At temperatures greater than _____°C this reaction is spontaneous under standard conditions

A 273 B 325 C 552 D 710 E 983 C(s) + H2O(g) --> CO(g) + H2(g)

#G = #H -T#S

Slide 89 / 93 Unavailability of Energy: Heat Death

Another consequence of the second law: In any natural process, some energy becomes unavailable to do useful work. If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the heat death of the universe.

# # Slide 90 / 93 Evolution and Growth: “Time’s Arrow”

Growth of an individual, and evolution of a species, are both processes of increasing order. Do they violate the second law

• f thermodynamics?

No! These are not isolated systems. Energy comes into them in the form of food, sunlight, and air, and energy also leaves them. The second law of thermodynamics is the one that defines the arrow of time – processes will occur that are not reversible, and movies that run backward will look silly.

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SLIDE 16

Slide 91 / 93 Thermal Pollution and Climate Change

Air pollution is also emitted by power plants, industries, and

• consumers. Some of this pollution results in a buildup of CO2 in

the atmosphere, contributing to global warming. This can be minimized through careful choices of fuels and processes. Thermal pollution, however, is a consequence of the second law, and is unavoidable; it can be reduced only by reducing the amount of energy we use.

Slide 92 / 93 Thermal Pollution and Alternative Power

The generation of electricity using solar energy does not involve a heat engine, but fossil-fuel plants and nuclear plants do. However, any use of energy will always result in some loss of "heat death" as no system is 100% effective due to the second law of thermodynamics.

Slide 93 / 93