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Chemical Thermodynamics

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Chemical Thermodynamics

The Golden Gate Bridge is painted regularly to slow down the inevitable rusting of the iron on the bridge. This unit will help us understand how we determine whether

• r not a certain reaction will occur.

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First Law of Thermodynamics

The First Law tells us that energy cannot be created nor destroyed; the total energy of the universe is a constant. The First Law allows any process in which the total energy is conserved, including those where energy changes forms.

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First Law of Thermodynamics

However, we observe processes in the natural world that just don't happen "naturally". For example, gold does not rust in the same way that iron does. The First Law of thermodynamics doesn't help us here. 4Au(s) + 3O2(g) --> 2Au2O3(s) doesn't happen naturally 4Fe(s) + 3O2(g) --> 2Fe2O3(s) does happen naturally

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The First Law of Thermodynamics

For instance, the First Law would allow a broken cup shown below to reassemble itself, but it never will. The absence of processes like this shows that the conservation

• f energy is not the whole story. If it were, movies that were run

backwards would look perfectly normal to us!

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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

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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 says all these could happen, the Second Law says they won't.

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Spontaneous Processes and the Second Law

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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Thermodynamic Favorability

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.

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Favorable Processes

Processes that are favorable in

• ne direction are not favorable in

the reverse direction.

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FOR EXAMPLE

Favorable Processes

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

favorable at T > 0 C favorable at T < 0 C

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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

answer

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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

answer

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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 Exothermic Surroundings Heat T +ΔT T System Endothermic Surroundings T-ΔT Heat 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.

work Gas Vacuum Movable partition Piston

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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

answer

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Entropy

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

T

Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it 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

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Entropy

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

ΔS =

qrev T

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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:

ΔSuniv = ΔSsystem

+ ΔSsurroundings = 0 For irreversible processes:

ΔSuniv = ΔSsystem

+ ΔSsurroundings > 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 enthalpy B internal energy C bond energy D entropy E heat flow

answer

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5 For an isothermal (constant temperature) process, ΔS = __________.

A q B qrev / T C qrev D Tqrev E q + w

answer

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6 Which one of the following is always positive when a thermodynamically favorable process

• ccurs?

A ΔSsystem B ΔSsurroundings C ΔSuniverse D ΔHuniverse E ΔHsurroundings

answer

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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

answer

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Entropy on the Molecular Scale

Ludwig Boltzmann described the concept of entropy

• n 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.

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 [**]

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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

[**]

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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 [**]

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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.

[**]

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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.

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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 **

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8 Which of the following would have 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.

answer

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Entropy and Physical States

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

*

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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

answer

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Solutions

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

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

*

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Entropy Changes

In general, entropy increases when · Gases are formed from liquids and solids; · Liquids or solutions are formed from solids; · The number of gas molecules increases; · The number of moles increases.

*

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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

*

The entropy is never zero because absolute zero is never achieved.

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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

*

The entropy is never zero because absolute zero is never achieved.

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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 S = Sfinal - Sinitial molar entropy values of substances in their standard states.

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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 Δ

C H H H H

Methane, CH

4

ΔS = 186.3 J/mol K

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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

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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: * Note that S is never zero; it can, however, be very small.

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10 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.

answer

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11 Of the following, the entropy of __________ is the largest. A

HCl (l)

B HCl (s)

C HCl (g) D

HBr (g) E

HI (g)

answer

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12 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

answer

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

True False

answer

be prepared to answer why!

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14 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

answer

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15 ΔS 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)

answer

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16 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

answer

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Calculating Entropy Changes

The entropy change, S°,for a reaction can be estimated in the same way we estimated enthalpy change, H: S° = Sum(nS°(products)) - Sum(mS°(reactants)) where n and m are the coefficients in the balanced chemical equation. However, all values must be looked up in a chart, the S° for elements in their standard state is NOT zero.

*

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17 The value of ΔSo 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) à C2H4(g) E

• 18.6

*

answer

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18 The value of ΔSo 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) à CO2(g) E +2.9

*

answer

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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 ΔHo for the system, so ΔSsurr = - qsys T ΔSsurr =

• ΔHsys

T

*

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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

*

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• T Suniverse = Hsystem - T Ssystem

Suniverse = Ssystem + Ssurroundings Suniverse = Ssystem +

• ΔHsys

T Multiply both sides by - T

• T Suniverse ≡ DG

G = Hsystem - T Ssystem ΔSsurr =

• ΔHsys

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.

*

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

• T Suniverse ≡ G

Gibbs Free Energy *

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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.

N2 + 3H2 < - - > 2NH3 spontaneous spontaneous

Free energy

Equilibrium Course of reaction

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

*

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19 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

answer

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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° = Sum(n G°f(products) - Sum(m G°f (reactants))

*

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20 The standard Gibbs free energy of formation of __________ is zero. (I) Al (s) (II) Br2 (l) (III) Hg (l)

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

*

answer

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21 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

*

answer

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22 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

*

answer

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23 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

*

answer

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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 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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24 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° = Sum(n G°f(products) - Sum(m G°f (reactants))

answer

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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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G and Thermodynamic Favorability

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.

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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)

G = H -T S

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

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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 = H -T S

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 the sun. Period.

move for answer G and Thermodynamic Favorability

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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 = H -T S

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

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H S

T

G

+

• all

+ + high low

• high

low

• +

all

Fill in the following summary chart

G = H -T S

G and Thermodynamic Favorability

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25 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

answer

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26 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

answer

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27 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

answer

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28 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

answer

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29 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

answer

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30

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

answer

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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

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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

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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. You can calculate the specific temperature at which G changes sign by setting G = 0.

Free Energy and Temperature

Below this temperature, the reaction will be unfavorable, while above it, the process is favorable.

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

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.

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31 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

answer

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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.

**

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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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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.

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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.

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