Thermodynamicscontent.njctl.org/courses/science/chemistry/thermodynamics-2/... · energy from one form to another don't just happen naturally. For example, gold does not rust in the

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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 ThermodynamicsWithin 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.

.Initialstate

Finalstate

E of system decreases

Inte

rnal

ene

rgy,

E

E < E 0

#E < 0 (-)

E

E0

Energy lost tosurroundings

Initialstate

Finalstate

E of system increases

Inte

rnal

ene

rgy,

E

Energy gained fromsurroundings

E > E0

#E > 0 (+)

E

E0

http://www.njctl.org

http://www.njctl.org

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 object; 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

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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.

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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 occur if a system is left on its own.

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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

Slide 14 (Answer) / 93

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 [This object is a pull tab]

Ans

wer

B

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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 occur at an observable rate.

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


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 occur at an observable rate.

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

[This object is a pull tab]

Ans

wer

D

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Reversible ProcessesIn 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-#TT

System

Exothermic

Surroundings

Heat -q

T+#T T

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Irreversible Processes

Irreversible processes cannot be undone by exactly reversing the change to the system.

Thermodynamically favorable processes are irreversible.

workGasVacuum

Movable partitionPiston

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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


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[This object is a pull tab]

Ans

wer

A

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Entropy

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

S = qT

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 of 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

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EntropyFor 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

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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

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4 The thermodynamic quantity that expresses the degree of disorder in a system is ______.

A enthalpyB internal energyC bond energyD entropyE heat flow


4 The thermodynamic quantity that expresses the degree of disorder in a system is ______.

A enthalpyB internal energyC bond energyD entropyE heat flow


Ans

wer

D

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5 For an isothermal (constant temperature) process, #S = __________. A q

B qrev / T

C qrev

D Tqrev

E q + w


5 For an isothermal (constant temperature) process, #S = __________. A q

B qrev / T

C qrev

D Tqrev

E q + w[This object is a pull tab]

Ans

wer

B

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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


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[This object is a pull tab]

Ans

wer

C

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7 The entropy of the universe is __________. A constant

B continually decreasing

C continually increasing

D zero

E the same as the energy, E


7 The entropy of the universe is __________. A constant

B continually decreasing

C continually increasing

D zero

E the same as the energy, E [This object is a pull tab]

Ans

wer

C

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Entropy on the Molecular Scale

Ludwig Boltzmann described the concept of entropy on the molecular level by using statistical analysis

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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

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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.


**

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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.


**

Slide 32 / 93


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 or tails is extremely small.

Macrostate

# of microstates Probability

heads tails

100 0 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 0 100 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 on 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 odds 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.


**

Slide 34 / 93

Entropy on the Molecular ScaleDegrees 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.

**

Slide 37 / 93

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**

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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.


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. [This object is a pull

tab]A

nsw

er

D

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


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[This object is a pull tab]

Ans

wer

A

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


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 [This object is a pull tab]

Ans

wer

E

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Third Law of ThermodynamicsThe entropy of a pure crystalline substance at

absolute zero is 0.

Solid crystalordered arrangement Liquid

less orderedmore freedom

Can entropy ever equal 0?

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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

SubstanceH2(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-K130.6191.5205.0188.8192.5237.6269.2

69.9126.8172.8

29.151.464.7

27.23142.3

72.3

Slide 45 / 93

Standard Entropies

Larger and more complex molecules have greater entropies.

CC

C

H

H

H

H

HH

H

H

Propane, C 3H8

#S = 270.3 J/mol K

C

H H

H

H

Methane, CH4

#S = 186.3 J/mol K

Slide 46 / 93

Predicting Entropy ChangesTo 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?

AThe 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 overall 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.


11 Which of the following statements is false?

AThe 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 overall 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.


Ans

wer

A

Slide 49 / 93

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)


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)


Ans

wer

E

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


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


Ans

wer

B

Slide 51 / 93

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

TrueFalse


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

TrueFalse


Ans

wer

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 steamB dissolution of solid KCl in water

C mixing of two gases into one container

D freezing water to form iceE melting ice to form water


15 Which one of the following processes produces a decrease in the entropy of the system? A boiling water to form steamB dissolution of solid KCl in water

C mixing of two gases into one container

D freezing water to form iceE melting ice to form water


Ans

wer

D

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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)


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)[This object is a pull tab]

Ans

wer

D

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


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 [This object is a pull tab]

Ans

wer

C

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 +18.6B +550.8 C +112.0 D -112.0

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

E -18.6


18 The value of #So for the catalytic hydrogenation of acetylene to ethane is _____ J/K∙ mol.

A +18.6B +550.8 C +112.0 D -112.0

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

E -18.6


Ans

wer

D

Slide 57 / 93

19 The value of #So for the oxidation of carbon to carbon dioxide, is ______ J/K.

A +424.3 B +205.0C -205.0 D -2.9

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

E +2.9


19 The value of #So for the oxidation of carbon to carbon dioxide, is ______ J/K.

A +424.3 B +205.0C -205.0 D -2.9

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

E +2.9


Ans

wer

E

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

Slide 61 / 93

#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

spontaneous

spontaneousFr

ee e

nerg

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.

HHNN

NNNN

NNNNNN

HHHH HHHH HH

HH HHHH

NHHNHHNHH

NHHNHH

HH HH

NNNN

HH NNHHHH

NHH

NHHNHHNHH

NHH

NHHNHH

NHHNHH

NHHNHH

NHHNHH

NHH

NHH

NHHpure N2 +H2 Equilibrium mixture

ΔG=0pure 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


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 [This object is a pull tab]

Ans

wer

E

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 ΔSf

0), 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) Br2 (l) (III) Hg (l)

A I only B II onlyC III onlyD II and IIIE I, II, and III

#



(I) Al (s) (II) Br2 (l) (III) Hg (l)


#


Ans

wer

E

Slide 67 / 93

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 -370.4C +370.4D -740.8E +185.2

*

Sulfur

S (s, rhombic) 0 0 31.88

SO2(g) -269.0 -300.4 248.5

SO3(g) -395.2 -370.4 256.2

Oxygen

O2(g) 0 0 205.1

#H0f #G0

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


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 -370.4C +370.4D -740.8E +185.2

*

Sulfur

S (s, rhombic) 0 0 31.88

SO2(g) -269.0 -300.4 248.5

SO3(g) -395.2 -370.4 256.2

Oxygen

O2(g) 0 0 205.1

#H0f #G0



Ans

wer

D

Slide 68 / 93


(I) H2O(l) (II) Na(s) (III) H2(g)


#



(I) H2O(l) (II) Na(s) (III) H2(g)


#


Ans

wer

A

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 +539.2C -642.9D +642.9E -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) 0 0 222%

Cl-(aq) -167.2 -131.2 56.5

#H0f #G0



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 +539.2C -642.9D +642.9E -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) 0 0 222%

Cl-(aq) -167.2 -131.2 56.5

#H0f #G0



Ans

wer

C

Slide 70 / 93

Free Energy Changes

At temperatures other than 25 oC,

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

How does #G# change with temperature?

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Free Energy and TemperatureThere 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))


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))[This object is a pull tab]

Ans

wer

E

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#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.

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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.


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 = #H -T#S

Slide 76 / 93

#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)


#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


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)


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

= (+) + (T(-))

This reaction is favorable only at relatively high temperatures.

move for answer


Slide 78 / 93

Fill in the following summary chart


#G = #H -T#S

#H #S

+

+ +

T #G

+

-

--

-

allhighlow

highlow

all

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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


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


Ans

wer

C

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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


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


Ans

wer

B

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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


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


Ans

wer

A

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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


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


Ans

wer

B

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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


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


Ans

wer

B

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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 - 0

#G = #H -T#S


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 - 0

#G = #H -T#S


Ans

wer

B

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

or

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.


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.


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

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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


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


Ans

wer

E

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 of 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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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

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