From Thermochemistry to Thermodynamics

From Thermochemistry to Thermodynamics

Chapter 5 to Chapter 19

break out the clickers

11

Spontaneity• Every process has a preferred direction, under a

given set of conditions.

• This direction can be referred to as the spontaneous direction or the thermodynamically favored direction

• A spontaneous (thermodynamically favored) process is one that occurs without ongoing outside intervention

• A spontaneous (thermodynamically favored) process can be reversed only by some action, energy, or outside force.

2

2

Not to confuse spontaneous with spontaneous

• Most thermodynamically favored (spontaneous) reactions, will require an activation or ignition energy:✓ Water behind a dam won’t flow until the gate is opened✓ A mixture of CH4 and O2 will remain until sparked

• In the general language, spontaneous means: occurring as a result of a sudden inner impulse or inclination and without premeditation or external stimulus.

• In the thermodynamics, spontaneous means: occurring without continuous outside intervention.

3

3

The Common-Sense Law• Things tend to become dispersed.✓ Marbles will scatter, but not roll back in the bag.✓ Fences fall down, but piles of wood don’t assemble

into fences.✓ Jigsaw puzzles disassemble easily, but shaking the

box never result in reassembly.

• Each of the spontaneous (thermodynamically favored) processes above reduces the constraints on the objects making the system more disordered, more dispersed. 4

4

The Common-Sense Law applies at the molecular level as well.

• Salt dissolves• Gases disperse

5

5

The Common-Sense Law also Applies to Energy Dispersal

• A hot piece of metal will cool when dropped in water and the water will warm until the water and metal are in thermal equilibrium.

• Energy disperses.

6

6

Which process(es) shown below is(are) spontaneous (thermodynamically favored)?

7

situation #1 situation #2

7

Which process(es) shown below is(are) spontaneous? Both are spontaneous.1. Ice at room

temperature:• energy in, ice melts.

2. water in the freezer:• energy out, water

freezes.

8

situation #1 situation #2

8

The Common Sense Law• Situation #1 is endothermic.• Endothermic reactions seem to oppose the energy

dispersal Common Sense Law.• Most spontaneous reactions you are familiar with,

combustion reactions, are all exothermic, which obey the Common Sense Law, yet here we have an endothermic process, disobeying the Common Sense Law for energy, yet the reaction is spontaneous

• This tells us that analyzing just energy dispersal is not enough to determine if a process will be spontaneous or not...

• This reaction does obey the Common Sense Law for matter dispersal

9

#1

9

The Common Sense Law• Situation #2 is exothermic.• Exothermic reactions seem obey the energy

dispersal common sense law.• Yet this reaction seems to be disobeying the

Common Sense Law for matter − the water is freezing and becoming more ordered.

• This tells us that analyzing just matter dispersal is not enough to determine if a process will be spontaneous or not…

• So what should we consider to determine spontaneity?

10

10

What drives reactions, and determines spontaneity?

Dispersal of energy and matter

11

EnergyEntropyand

low

high

processes that move toward higher entropy and lower energy are thermodynamically favorable

11

EntropyEntropy, (S) is a measure of dispersal -

dispersal of matter and dispersal of energy

• In 1877, Austrian Ludwig Eduard Boltzmann defined the entropy of a substance as: ✓ the number of different ways to describe

a system✓ He called this the “degrees of freedom”

of the system• We can loosely describe entropy to be a

measure of the disorder of a substance. 12

1844-1906

12

Spontaneity and Kinetics and Reaction Direction

• Spontaneity is unrelated to the speed of a reaction - It’s kinetics that tells us about the rate of a reaction.✓ Some spontaneous processes may be extremely slow.✓ Some may be fast.

• Spontaneity tells us the direction in which a reaction proceeds, not the speed.

13

13

Entropycomparisons to help us predict entropy

1414

Which has the least amount of entropy?

1. solid2. liquid3. gas4. not enough information to

determine15

15

Which has the least amount of entropy?1. solid• phase is an important factor in

determining entropy2. liquid3. gas4. not really enough information to

determine16

16

Which has least entropy at room conditions1. 1 mole of gold2. 1 mole of mercury3. equal amounts of entropy since

there is 1 mole of each substance

4. not enough information to determine 17

17

Which has least entropy at room conditions1. 1 mole of gold, • since gold is solid, yet the mercury would

be liquid at room temp.• Yes, AP expects you to know the phase of

elements at room conditions.• Again, phase is a very important factor

when deciding on entropy2. 1 mole of mercury3. not enough information to determine 18

18

Which gas has more entropy?1. NO2. NO2

3. they have the same entropy4. not enough information to

determine5. I have absolutely no idea how

I would even venture a guess.19

These gases are at the same T, P, V conditions.

19

Which gas has more entropy?1. NO (211 J/molK)

2. NO2 (240 J/molK)• All else being the same, larger molecules have

higher entropy.There are more confirmations and more ways to distribute energy.

3. they have the same entropy4. not enough information to determine5. I have absolutely no idea how I

would even venture a guess.20

These gases are at the same T, P, V conditions.

20

Which has more entropy?

1. two moles of NO2(g)

2. one mole of N2O4(g)

3. the same4. not enough information to

determine21

at the same temp, same volume

21


• Hard to say...lets look it up on the Thermo Sheets

1. two moles of NO2 240.45 J/molK

2. one mole of N2O4 304.3 J/molK

3. the same4. not enough information to

determine22


22


1. two moles of NO2 2(240.45 J/molK)• since these molecules are not tied

together with a bond, they have more “degrees of freedom” or positional possibilities.

• more molecules means more entropy.2. one mole of N2O4 (304.3 J/molK)3. the same4. not enough information to determine 23


23

Which has more entropy?1. 10 ml of 1 M NaCl2. 10 ml of 2 M NaCl3. 1 and 2 have the same entropy

since they both represent the same molecule in solution.

24

24

Which has more entropy?1. 10 ml of 1 M NaCl2. 10 ml of 2 M NaCl• All other parameters being equal,

higher concentration will have more entropy since there are more molecules present.

3. 1 and 2 have the same entropy since they both represent the same molecule. 25

25

Which has the most entropy?

And can you name these three molecules?

26

1 2 3

All molecules at the same temperature.

HC

26

Which has the most entropy?• Propane is larger and thus has more

entropy than smaller molecules.

27

27


1. 1 mole HF(g) at 25ºC, 1 atm

2. 1 mole HCl(g) at 25ºC, 1 atm

3. not enough information to determine

4. 1 and 2 have the same entropy

28

28


1. 1 mole HF(g) at 25ºC, 1 atm• (174 J/mol K)

2. 1 mole HCl(g) at 25ºC, 1 atm • (187 J/mol K) • Likely, the more entropy is due to Cl’s larger

electron cloud.

3. not enough information to determine

4. 1 and 2 have the same entropy29

29

Which has the most entropy?(Choose more than one if there are molecules with the same entropy.)

30

1 2

3 45

1 2

3 45

1 2

3 45

12

3

30


• All three molecules are C5H12

• All other factors equal, straight chain molecules have more “degrees of freedom” than branched chain molecules.

• They are more “floppy.”

• What is the name of molecule #1?

1 2

3 452

1 2

3 45

1

31

1 2

3 453

31


• All three molecules are C5H12

• All other factors equal, straight chain molecules have more “degrees of freedom” than branched chain molecules.

• What is the name of molecule #1?

• Pentane

1 2

3 452

1 2

3 45

1

32

1 2

3 453

32

Which has more entropy?1. 1 mol He(g) at 0ºC2. 1 mol He(g) at 10 K3. not enough information to

determine4. 1 and 2 have the same

entropy

33

33


1. He at 0ºC• All other parameters being equal,

higher temp, means higher entropy

2. He at 10 K3. not enough information to determine4. 1 and 2 have the same entropy

34

34

When predicting the entropy of one substance to another, consider★ Phase

• gases have significantly more entropy than liquids which have more entropy than solids★ Number of molecules

• more molecules have more positional possibilities, which means more entropy

• molecule size• significantly larger molecules have more entropy because they have more bending and

folding options

• molecule shape• long, straight chained molecules have more entropy than branched molecules

• larger atoms• for similar molecules, the one that contains larger atoms with larger electron cloud, will

have more entropy

• temperature• higher temperature causes greater entropy

If there are conflicting conditions, consider the change in the number of gas molecules as the determining factor.

35

35

Select the process(es) below in which entropy increases, ∆S = +

1. CaCO3(s) → CaO(s) + CO2(g)

2. 2SO2(g) + O2(g) → 2SO3(g)

3. H2O(g) → H2O(L)

4. Fe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(g)

36

Predict without using your thermo sheets

36

Select the processes below in which entropy increases.1. CaCO3(s) → CaO(s) + CO2(g)

• solid reactant to 1 mol gas product

2. 2SO2(g) + O2(g) → 2SO3(g)

• 3 mol gas reactant to 2 mol gas product

3. H2O(g) → H2O(L)

• 1 mol gas reactant to only liquid product

4. Fe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(g)

• 4 mol gas reactant to 4 mol gas product, but total 5 mol reactant to 7 mol product 37

37

Use your thermodynamic table to calculate ∆Sº forCaCO3(s) → CaO(s) + CO2(g)

• Report your answer to the nearest whole number.

38

∆ Srxn = ΣnSºProducts − ΣnSºReactants

38

Use your thermodynamic table to calculate ∆Sº forCaCO3(s) → CaO(s) + CO2(g)

• [(+39.75 J/molK + (+213.6 J/molK)] − (+92.88 J/molK) = +160.47 J/molK

• A positive value as you would predict because of the formation of a gas.

39

39

Use your thermodynamic table to calculate ∆Sº for2SO2(g) + O2(g) → 2SO3(g)


40

40

Use your thermodynamic table to calculate ∆Sº for2SO2(g) + O2(g) → 2SO3(g)

• 2(+256.2 J/molK) - [2(+248.5 J/molK + (+205.0 J/molK)]

• = -189.6 J/molK• A negative value as you would

predict because of the change of 3 moles of gas into only 2 moles of gas.

41

41

Use your thermodynamic table to calculate ∆Sº for

H2O(g) ⇋ H2O(L)


42

42

Use your thermodynamic table to calculate ∆Sº for H2O(g) ⇋ H2O(L)

• (+69.91 J/molK) + (+188.83 J/molK) = −118.92 J/molK

• A negative value as you would predict because of the formation of a liquid from a gas.

43

43

Use your thermodynamic table to calculate ∆Sº forFe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(g)


44

44

Use your thermodynamic table to calculate ∆Sº forFe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(g) [3(+27.15 J/molK + 4(+188.83 J/molK)] − [(+146.4 J/molK) + 4(+130.58 J/molK)]

• = +160.05 J/molK• A positive value as you would predict

because of the change of 5 moles of reactant into only 7 moles of product, while moles of gas is same on both side.

45

45

Select the process(es) below in which entropy increases, ∆S = +

1. 2O3(g) → 3O2(g)

2. 2CH4(g) → C2H6(g) + H2(g)

3. NaCl(s) → Na+(aq) + Cl−(aq)

4. MgCl2(s) → Mg2+(aq) + 2Cl−(aq)

46

Predict without using your thermo sheets

46

Select the processes below in which entropy increases.1. 2O3(g) ⇋ 3O2(g)

2. 2CH4(g) → C2H6(g) + H2(g)• It would be unlikely that you would be asked to distinguish something

like this as it is a VERY tough decision as demonstrated by the following slide showing that there is very small entropy decrease.

3. NaCl(s) → Na+(aq) + Cl−(aq)• Unless there is some other info in the problem, you can assume for

dissolution ∆S is positive

4. MgCl2(s) → Mg2+(aq) + 2Cl−(aq)• This is a case in which ∆S is negative, though there is nothing in the

problem that indicates this.• Let’s take a closer look to see why... 47

47

Use your thermodynamic table to calculate ∆Sº for 2O3(g) → 3O3(g)


48

48

Use your thermodynamic table to calculate ∆Sº for 2O3(g) → 3O2(g)

• 3(+205 J/molK) − 2(+237.6 J/molK)] = +139.8 J/molK

• This is a positive entropy change which makes sense since

• gas on both sides• # molecules increases during the reaction

49

49

Use your thermodynamic table to calculate ∆Sº for 2CH4(g) → C2H6(g) + H2(g)


50

50

Use your thermodynamic table to calculate ∆Sº for 2CH4(g) → C2H6(g) + H2(g)

• [(+229.5 J/molK) + (+130.58 J/molK)] − 2(+186.3 J/molK) = −12.52 J/molK

• This is a very small entropy change which makes sense since

• gas on both sides• same # molecules on both sides• It is possible that the very limited positional

possibilities for the tiny H2 molecule, contribute to the decrease in entropy. 51

51

Use your thermodynamic table to calculate ∆Sº for NaCl(s) → Na+(aq) + Cl−(aq)


52

52


NaCl(s) → Na+(aq) + Cl−(aq) (or NaCl(aq))• (+115J/molK) − (+72.33 J/molK) = +43 J/molK • This is a positive ∆S, thermodynamically

favorable, as you would expect from a salt breaking apart into its ions.

53

53

Use your thermodynamic table to calculate ∆Sº for MgCl2(s) → Mg2+(aq) + 2Cl−(aq)


54

54


MgCl2(s) → Mg2+(aq) + 2Cl−(aq)• [(+89.6 J/molK) + (−137 J/molK)] − 2(+56.1 J/molK) = −114.4 J/molK • This is a negative ∆S, thermodynamically unfavorable,

Whaaaat!!!???• The hydration shell makes the

water part of the system decreases the entropy more than the salt part of the system breaking apart increases the entropy .

• Nonethless, unless you have some other indicator in the problem, you should assume that dissolution increases entrogy

2+

55

55

56

The unit label on ∆Sº could be1. 2.

3.

4.

5.

6.

7.

8.

9.

JmolJ

mol ⋅KJgJºC

JgºC

kJmol

kJmol ⋅ oF

J

kJ

56

57

The unit label on ∆Sº could be1. 2.

3.

4.

5.

6.

7.

8.

9.

JmolJ

mol ⋅KJgJºC

JgºC

kJmol

kJmol ⋅F

J

kJ

57

Which graph below might best represent entropy changes as temperature increases

58

temp (K) →

Solid liquid gas

temp (K) →

S

Solid liquid gas

temp (K) →

S

Solid liquid gas

temp (K) →

S

Solid liquid gas

temp (K) →

S

1 5

4 6

Solid liquid gas

temp (K) →

S2

Solid liquid gas

temp (K) →

S

3

58

Which graph below might best represent entropy changes as temperature increases• #3• Note the larger

entropy change for the L-G phase change

59

temp (K) →

melting

boilingSolid liquid gas

temp (K) →

S

3

59

SpontaneityWhat drives reactions?

How can we mathematically combine energy and entropy?

What information do we need to know inorder to predict the spontaneity of a process?

6060

• Low energy is favored✓∆H decrease (−) is

thermodynamically favorable✓You are aware that exothermic

reactions are favorable reactions‣ combustion reactions

Two Forces Drive Reactions

61

61

LAD C.5Ammonium thiocyanate + barium hydroxide.

What drives reactions?

Caution, the products of this reaction have a potent smell. WAFT − do not put you nose into the beaker.

6262

The equation for the reaction between ammonium thiocyanate and barium hydroxide.2NH4SCN(s) + Ba(OH)2(s) → Ba(SCN)2(ppt) + 2NH3(g) + 2H2O(L)

2NH4SCN(s) + Ba(OH)2•8H2O(s) → Ba(SCN)2(ppt) + 2NH3(g) + 10H2O(L)

This reaction is spontaneous, yet very endothermic.

Since the energy factor is unfavorable and not helping the reaction be spontaneous.

What is it that is driving this reaction?

Entropy is thermodynamically favorable and helping the reaction to be spontaneous.

63

63

Energy AND Entropy drive reactions• High entropy is favored✓∆S increase (+) is favored

• For spontaneous processes with unfavorable energy change, the entropy change must be favorable.✓NH4NO3 → NH4+ + NO3− ✓H2O(s) → H2O(L)

✓ 2NH4SCN(s) + Ba(OH)2•8H2O(s) → Ba(SCN)2(ppt) + 2NH3(g) + 10H2O(L)

64

64

So how can we predict spontaneity?• Josaiah Willard Gibbs (USA) in 1873

combined enthalpy and entropy, the two driving forces into one mathematical relationship

• Gibbs Free Energy (G).✓∆G = ∆H - T∆S✓ ∆G is a measure of the spontaneity of a

process and the “useful” energy available from it.

✓ ∆H is enthalpy✓ ∆S is entropy✓ T is temperature in Kelvin 65

1839-1903

65

∆G = ∆H - T∆S• If ∆G < 0, a reaction will proceed in

the forward direction and is considered thermodynamically favorable (aka spontaneous).

• If ∆G > 0, a reaction will not proceed in forward direction and is considered thermodynamically unfavorable (aka not spontaneous).

66

66

Using ∆Gº• Standard conditions✓ 1 atm✓ 1 M concentration

• The ∆Gº values in the table are reported for temperatures at 25ºC

• ∆Gfº values have been established for many substances making the ∆Gº calculation at 1 M or 1 atm and for 25ºC.

• ∆GºRx = ∑n∆Gfºprod - ∑n∆Gfºreact

67

67

• q = c×m×∆T• q = C×∆T• q = ∆H×amount• ∆HºRx = ∑n∆Hfºprod - ∑n∆Hfºreact

• Hess’ Law• ∆SºRx = ∑n∆Sºprod - ∑n∆Sºreact

• ∆Gº = ∆Hº - T∆Sº• ∆GºRx = ∑n∆Gfºprod - ∑n∆Gfºreact

So what reactions do we know how to use in this unit?

68

13 Chemistry Practice Exam

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GASES, LIQUIDS, AND SOLUTIONS

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Ptotal = PA + PB + PC + . . .

n = mM

K = qC + 273

D = m

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KE per molecule = 12

mv2

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A = abc

1 1

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Gas constant, 8.314 J mol K

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68

Use the thermodynamic tables to calculate ∆Gº298 (at 25ºC) (º refers to standard conditions: 1atm and 1M)(Because temperature is not a standard condition like pressure and molarity, the subscript number gives the particular temperature. If no number is given, assume 298 K.)

2SO3(g) → 2SO2(g) + O2(g)

69

69

Use the thermodynamic tables to calculate ∆Gº298 (at 25ºC) (º refers to standard conditions: 1atm and 1M)(the subcript number give the temperature)

2SO3(g) → 2SO2(g) + O2(g)

∆Gºf SO2(g) = −300.4kJ∆Gºf SO3(g) = −370.4kJ∆Gºf O2(g) = 0 kJ

70

70

Use the thermodynamic tables to calculate ∆Gº298 at 25ºC (º refers to standard conditions: 1atm and 1M)

2SO3(g) → 2SO2(g) + O2(g)• ∆Grxº = ∑n∆Gfºprod - ∑n∆Gfºreact

• [2(-300.4kJ/mol) + 0kJ/mol] - [2(-370.4kJ/mol)]

• ∆Gºrx = +140 kJ/molrx (per reaction as written)

• A positive ∆Gº does not mean that nothing happens, it simply means that the reaction favors reactants NOT products.

• that is to say if you put 1 atm of each substance in a container, the reaction would not proceed in the forward direction, but it would proceed in the reverse direction.

71

71

Use the thermodynamic tables to calculate ∆Hº and ∆Sº at 25ºC2SO3(g) → 2SO2(g) + O2(g)

∆Hºf SO2(g) = −296.9 kJ∆Hºf SO3(g) = −395.2 kJ∆Hºf O2(g) = 0 kJ

Sº SO2(g)= 248.5 JSº SO3(g)= 256.2 JSº O2(g) = 205 J

72

72

Use the thermodynamic tables to calculate ∆Hº and ∆Sº at 25ºC2SO3(g) → 2SO2(g) + O2(g)

• ∆H = [2(-296.9)+ 0] - 2(-395.2)• ∆H = +196.6 kJ/molrx

✓ (which is to say per 2 mol of SO3, per 2 mol of SO2, per 1 mol of O2)

• ∆S = [2(+248.5)+ 205] - 2(256.2)• ∆S = +189.6 J/molrx K

✓ (which is to say per 2 mol of SO3, per 2 mol of SO2, per 1 mol of O2)

73

73

Using the ∆Hº and ∆Sº just calculated, use them to calculate ∆Gº with Gibbs Free Energy Formula at 25ºC.2SO3(g) → 2SO2(g) + O2(g)

• ∆Gº = ∆Hº − T∆Sº

• ∆H = [2(-296.9)+ 0] - 2(-395.2)• ∆H = +196.6 kJ/molrx

• ∆S = [2(+248.5)+ 205] - 2(256.2)• ∆S = +189.6 J/molrx K

74

74

Using the ∆Hº and ∆Sº just calculated, use them to calculate ∆Gº with Gibbs Free Energy Formula at 25ºC.2SO3(g) → 2SO2(g) + O2(g)

• ∆Gº = ∆Hº − T∆Sº• Watch out for the units!!• ∆G = +196.6kJ − [298K(0.1896kJ/K)] = +140 kJ/mol

• How does it compare to the previously calculated with ∆Gfº values?

• the same since it was all for 25º.75

• ∆S = +189.6 J/molrx K• ∆H = +196.6 kJ/molrx

75

Calculating ∆GºTemp at other than 25º• Conditions are often NOT at 25ºC.• ∆Gº is very temperature dependent✓ notice the T in the formula ∆Gº = ∆Hº − T∆Sº

• When not at 25º✓ You can not use ∆GRxº = ∑n∆Gfºprod − ∑n∆Gfºreact

‣ because ∆Gfº values are listed only for 25ºC✓ You must use ∆Gºtemp = ∆Hº − T∆Sº

• How can we get away with this? ✓ ∆Hº and ∆Sº are NOT very temp dependent,

thus a decent estimation of ∆Gºtemp can be determined at different temperatures using Gibbs Equation. 76

76

Using the ∆Hº and ∆Sº previously calculated and the Gibbs Free Energy Formula, to calculate free energy at 1127ºC, which is 1400K.2SO3(g) → 2SO2(g) + O2(g)

• ∆Gº1400. = ∆Hº − T∆Sº

• ∆Hº = +196.6 kJ/molrx

• ∆Sº = +189.6 J/molrxK77

77

Using the ∆Hº and ∆Sº previously calculated and the Gibbs Free Energy Formula, calculate free energy at 1127ºC2SO3(g) → 2SO2(g) + O2(g)

∆GºT,1400 at 1400 K?∆Gº = ∆Hº - T∆Sº

• Watch out for the units!!• ∆G = +196.6kJ/- [1400 K (0.1896kJ/K)] = -68.84kJ

• This is spontaneous! Say whaaaat!!• at 298 K the Rx was NOT spontaneous, +140 kJ/molrx

78


78

2SO3(g) → 2SO2(g) + O2(g)

∆Hº = +196.6 kJ and ∆Sº = 0.1896 kJ• This reaction is not spontaneous at room

temperature• (∆Gº298 = +140kJ),

• It is spontaneous at 1127ºC• (∆Gº1400 = −68.84kJ),

• at what temperature (in ºC) does the spontaneity change from spontaneous to not spontaneous?

79

79

2SO3(g) → 2SO2(g) + O2(g)

If this reaction is not spontaneous at room temperature (∆Gº298 = +140kJ), and it is spontaneous at 1127ºC (∆Gº1400 = −68.84kJ), at what temperature (in ºC) does the spontaneity change from spontaneous to not spontaneous?

• ∆Gº = ∆Hº − T∆Sº

• The sign of ∆G changes when ∆G = 0, solve for T

• 0 = +196.6kJ − T(0.1896kJ/K)

• T = 1037 K, which is 764ºC80


80

What is going on when

∆G = 0?

8181

2SO3(g) → 2SO2(g) + O2(g)

If this reaction is not spontaneous at room temperature, 25ºC (∆Gº298 = +140kJ), and it is spontaneous at 1127ºC (∆Gº1400 = −68.84kJ)What happens at 764ºC?1. The reaction “stops.”2. The reaction is at equilibrium.3. It depends on if you start above or

below 764ºC.4. Some other situation.

82

• ∆Hº = +196.6 kJ/for the reaction

• ∆Sº = +189.6 J/K for the reaction

82

2SO3(g) → 2SO2(g) + O2(g)

If this reaction is not spontaneous at room temperature, 25ºC (∆Gº298 = +140kJ), and it is spontaneous at 1127ºC (∆Gº1400 = −68.84kJ)What happens at 764ºC?1. The reaction “stops.”

• For now, we’ll define equlibrium as the moment at which a reversible reaction “stops.”

2. The reaction is at equilibrium, since ∆Gº1127 = 0

3. It depends on if you start above or below 764ºC

4. Some other situation. 83

83

Consider the effect of temperature on spontaneity. Remember Kelvin temps are always positive ∆H − T∆S = ∆G

84

∆H ∆S ∆G−

favorable+

favorableis always spontaneous

+ unfavorable

− unfavorable

is never spontaneous

− favorable

− unfavorable

Spontaneous only if |∆H| > |T∆S|lower T favors spontaneity

+unfavorable

+favorable

Spontaneous only if |∆H| < |T∆S|higher T favors spontaneity

84

What is ∆G = 0?• If ∆Gº < 0 the reaction is spontaneous• if ∆Gº > 0 the reaction is not spontaneous• What is ∆G = 0 ?✓Equilibrium✓For now, we’ll define equilibrium as the point

at which a reversible reaction “stops.”✓There is no “net” movement of the reaction

in the forward direction or the reverse direction − No changes of concentrations or amounts of reactants or products.

85

85

The reaction below is NOT spontaneous at 298 K. Without making any calculations, is there a temperature at which this reaction can become spontaneous?2NOCl(g) → 2NO(g) + Cl2(g)

For the reaction: ∆Hº = +74 kJ, ∆Sº = +117 J

1. Yes2. No3. Not enough information to determine.

86

86

The reaction below is NOT spontaneous at 298 K. Without making any calculations, is there a temperature at which this reaction can become spontaneous?2NOCl(g) → 2NO(g) + Cl2(g)

For the reaction: ∆Hº = +74 kJ, ∆Sº = +117 J

1. Yes2. Whenever ∆S is favorable (+), the

possibility exists for a large enough T to make |T∆S| large enough to counteract the unfavorable ∆H

3. No4. Not enough information to determine. 87

87

The reaction below is NOT spontaneous at 298 K. Without making any calculations, is there a temperature at which this reaction can become spontaneous?R → PFor the reaction: ∆Hº < 0 and ∆Sº < 0


88

88

The reaction below is NOT spontaneous at 298 K. Without making any calculations, is there a temperature at which this reaction can become spontaneous?X → YFor the reaction: ∆Hº < 0 and ∆Sº < 01. Yes• Whenever ∆S is unfavorable (−), the possibility

exists for a small enough T to make |T∆S| small enough to be overtaken by the favorable ∆H

• Remember, when the sign of ∆S and ∆H are the same, spontaneity can be changed with temperature changes.

2. No3. Not enough information to determine.

89

89

The reaction below is NOT spontaneous at 298 K. Without making any calculations, is there a temperature at which this reaction can become spontaneous?2 A → 3 BFor the reaction: ∆Hº > 0 and ∆Sº < 0


90

90

The reaction below is NOT spontaneous at 298 K. Without making any calculations, is there a temperature at which this reaction can become spontaneous?2 A → 3 BFor the reaction: ∆Hº > 0 and ∆Sº < 0

1. Yes2. No• Whenever ∆S(−) and ∆H(+) are both

unfavorable, ∆G can never be spontaneous (−).

3. Not enough information to determine.91

91

When a reaction is happening, we can describe the extent of the reaction with a ratio, Q

2NOCl(g) ⇋ 2NO(g) + Cl2(g)

• Q is called the reaction quotient.• If a reaction “finishes” (goes as far as it can, and

reaches equilibrium), we would now call the ratio K (instead of Q), the equilibrium konstant.✓ Kp in in atmospheres, Kc in molarity

• [ ] refers to molarity ( ) to pressure• Note that the substances are raised to the

power of their coefficient…(not to worry about why just yet.)92

Q = [Cl2 ][NO]2

[NOCl]2

Kp =PCl2( ) PNO( )2

PNOCl( )2Kc =

[Cl2 ][NO]2

[NOCl]2

92

Reversible Reactions

93

Q = Kfavors reactants

reverse reaction proceeds

Q is the reaction quotient. It is a ratio of extent.

Q =[NH 3]

2

[N2 ][H2 ]3

K =[NH 3]

2

[N2 ][H2 ]3

When at equilibrium, we call the reaction quotient, the equilibrium constant, K

favors productsforward reaction proceeds

reverse is spontaneous

forward is spontaneous

Gºreactants0

Gºproducts−33.32

G o

f the

reac

ting

syst

em

extent of the reaction

EquilibriumPosition

∆Gº of this reaction is −33.32 kJ/molrx

93

Reversible Reactions

94extent of the reaction

Q is the reaction quotient. It is a ratio of extent.

K = [SO2 ]2[O2 ]

[SO3]2

K = [SO2 ]2[O2 ]

[SO3]2

G o

f the

reac

ting

syst

em

Q = KEquilibrium

Position

2SO3(g) → 2SO2(g) + O2(g)∆Gº of this reaction is +140 kJ/molrx

Gºproducts−600.8

Gºreactants−740.8

Q > Kreverse reaction proceeds

Q < Kforward reaction proceeds

94

So we know how to calculate ∆Gº at temperatures other than 298 K, (using ∆Gºtemp = ∆Hº − T∆Sº)what if we want to know ∆G at non-standard conditions: pressures other than 1 atm or concentrations other than 1 M?

• We need a new equation….

95

95

So we know how to calculate ∆Gº at temperatures other than 298 K, using ∆Gºtemp = ∆Hº − T∆Sºwhat if we want to know ∆G at non-standard conditions: pressures other than 1 atm or concentrations other than 1 M?The relationship between ∆G and ∆Gº can be calculated with the following mathematical relationship:

∆Gtemp = ∆Gºtemp + RT lnQ9696

∆Gtemp at nonstandardpressure & concentration

• ∆Gtemp = ∆Gºtemp + RT lnQ✓ R = 8.314 J/mole K (the gas constant in energy units)✓ T is always in Kelvin✓ Q is the reaction quotient

‣ Right-side over left side raised to their equation coefficients.

• If temp = 298✓ calculate ∆Gº using ∆GºRx = ∑n∆Gfºprod − ∑n∆Gfºreact

✓ then use formula above to calculate ∆G (nonstandard pressure or concentration

• If temp is something other than 298✓ calculate ∆Gºtemp using ∆Gºtemp = ∆Hº − T∆Sº✓ then use formula above to calculate ∆G (nonstandard pressure or

concentration 97

97

What is ln? Natural Log• ∆Gtemp = ∆Gºtemp + RT lnQ• so what is lnQ?• Find the the ln (not log, base 10) key on your

calculator.• For now we will simply accept that taking the

natural log of Q is a mathematical tool to analyze the progress of the reaction i.e. the difference between the products and reactants.

• This is what we need to do in order to solve the mathematical equation above.

98

98

The following reaction is spontaneous: 2 NO(g) + O2(g) → 2 NO2(g) ∆Gº = −71.2 kJ/rx

∆G under the following pressure conditions: NO = 0.100 atm, O2 = 0.100 atm, NO2 = 2.00 atm, would be

1. more negative2. less negative3. stay the same

99

∆G =∆Gº+RT lnQ

99


∆G under the following pressure conditions: NO = 0.100 atm, O2 = 0.100 atm, NO2 = 2.00 atm, would

1. more negative2. less negative

• although still spontaneous, but less so3. stay the same

100

∆G =∆Gº+RT lnQ ∆G =∆Gº+RT lnPNO2( )2

PNO( )2 PO2( )⎛

⎝⎜⎜

⎞

⎠⎟⎟

∆G = −71.2 + 0.008314( )× 298( )ln 2( )2

0.1( )2 0.1( )⎛

⎝⎜⎞

⎠⎟

∆G = −50.7kJ

100

natural log Facts

• If products increase, Q >1, ln+, ∆G more positive

• If reactants increase, Q <1, ln−, ∆G more negative

101

ln(# > 1) +

ln(1) 0

ln(# < 1 & bigger than zero) -

∆G = ∆Gº+RT lnQ

101

LeChatelier’s PrincipleA general rule that will help us predict the direction in which an equilibrium reaction will move when the equlibrium is disturbed.

If an external “stress” is applied to a system at equilibrium, the system adjusts in such a way that the “stress” is partially offset as the system reaches a new equilibrium position.

102

102

LeChatelier’s PrincipleWhile the reaction just described, was not at equlibrium, we can use this same principle to guide the expected shift in ∆G change from standard conditions.

If the conditions are different than standard, the ∆G will adjusts in such a way to partially offset different conditions.

103

103


∆G under the following pressure conditions: NO = 0.100 atm, O2 = 0.100 atm, NO2 = 2.00 atm, would be larger or smaller.

• At standard conditions, Q = 1

• Since the products are greater than 1, and reactants are less than 1, Q will be greater than 1, and lnQ will be positive, then ∆G will end up more positve (less negative) and will “shift” left.

• That is to say be less spontaneous than at standard conditions.

104

∆G =∆Gº+RT lnQ

104

The following reaction is spontaneous: Zn + Cu2+ → Zn2+ + Cu ∆Gº = −212 kJ/mol, Eº = +1.10 VIf we increased the concentration of Cu2+ and decreased the concentration of Zn2+ in the cell, the spontaneity of the reaction would

1. more negative2. less negative3. stay the same

105

105

For the following reaction from a Galvanic cell, ∆Gº298 = −212 kJ/mol Zn + Cu2+ → Zn2+ + CuIf we increased the concentration of Cu2+ and decreased the concentration of Zn2+ in the cell, the spontaneity of the reaction would

1. increase• We can answer this question again by

invoking LeChatelier’s Principle. “Stressing” the reaction by increasing reactants (compared to standard) will increase the “push” to the right to try to relieve that “stress.” Let’s calculate….

2. decrease3. stay the same 106

106

For the following reaction from a Galvanic cell, ∆Gº298 = −212 kJ/mol Eº = 1.10 V

Zn + Cu2+ → Zn2+ + CuCaluculate ∆G if the Galvanic cell was set up with [Cu2+] = 2M and [Zn2+]= 0.1M

107

∆G = ∆Gº+RT lnQ

∆G = ∆Gº+RT ln [Zn2+ ]

[Cu2+ ]

107

For the following reaction from a Galvanic cell, ∆Gº298 = −212 kJ/mol Zn + Cu2+ → Zn2+ + CuCaluculate ∆G if the Galvanic cell was set up with [Cu2+] = 2M and [Zn2+]= 0.1M

• ∆G = −219 kJ more negative, thermodynamically favored108

∆G = ∆Gº+RT ln [Zn2+ ]

[Cu2+ ]∆G = −212kJ + (0.00831kJ )(298)ln [0.1]

[2]

Q <1, thus logQ negative

108

The Relationship between ∆G and K

• ∆Gtemp = ∆Gºtemp + RT lnQ• When at equilibrium,✓ ∆Gtemp = 0, and Q = Keq

• the equation at the top becomes✓ 0 = ∆Gºtemp + RT lnK

• and will reduce to, ✓ ∆Gºtemp = −RT lnK

109

109

Calculate ∆GºT, 300K in kJ/mole for a reaction at 300K in whichKeq = 2.2 x 10−4

Round to the nearest whole number - You may use your calculator.

1. Submit your answer rounded to the nearest whole number.

110

110

Calculate ∆GºT, 300K in kJ/mole for a reaction at 300K in which Keq = 2.2x10−4

Round to the nearest whole number − You may use your calculator.

• ∆Gº = −RT lnK• ∆Gº = −(0.00831 kJ/K)(300K)(ln 2.2x10−4)• +21 kJ/mol or 21,000J/mol

111

111

Using the ∆Hº and ∆Sº previously calculated and the Gibbs Free Energy Formula, use those values to calculate free energy at 1127ºC2SO3(g) → 2SO2(g) + O2(g)

∆GºT=1400 = −69kJCalculate K for this reaction at this temp

112

112

Using the ∆Hº and ∆Sº previously calculated and the Gibbs Free Energy Formula, use those values to calculate free energy at 1127ºC2SO3(g) → 2SO2(g) + O2(g)

∆GºT,1400 = −69kJ∆GºT,1400 = −RT lnK-69 kJ = −(0.00831 kJ)(1400K)lnK

K = 377, large, thermodynamically favored as you would expect by the negative ∆G

113

−69− 0.00831( )(1400) = lnK e

−69− 0.00831( )(1400) = K

113

Need more practicewith your calculator? Calculate K for a reaction in which

1. ∆Gº = −4

2. ∆Gº = +12

3. ∆Gº = 0

114

∆Gº− 0.00831( )(298) = lnK

∆Gº= −RT lnK

e∆ G º

− 0.00831( )(298) = K

114

Need more practicewith your calculator? Calculate K for a reaction in which

1. ∆Gº = −4• K = 5

2. ∆Gº = +12

• K = 0.0079

3. ∆Gº = 0

• K = 1115

∆Gº− 0.00831( )(298) = lnK

∆Gº= −RT lnK

e∆ G º

− 0.00831( )(298) = K

115

Calculate the free energy at 25ºCC6H12O6 + 6O2 → 6CO2(g) + 6H2O(L)

∆GºT,298 = −2879 kJCalculate K for this reaction at 25ºC

116

116

Calculate the free energy at 25ºCC6H12O6 + 6O2 → 6CO2(g) + 6H2O(L)

∆GºT,298 = −2879 kJCalculate K for this reaction at 25ºCK = 8 x 10520 (this is ridiculously huge!!)

On most calculators, this will give you an error readout, because most calculators will only produce scientific notation up to 1099

117

117

Relationship of ∆G and K•∆Gºtemp = −RT ln Keq

• If ∆Gº < 0 then K > 1✓ natural log of any number larger than 1 is a positive #

• If ∆Gº = 0 then K = 1 (lnK = 0)✓ Since at standard conditions 1 M and 1 atm✓ and natural log of 1 = 0

• If ∆Gº > 0 then K < 1 ✓ natural log of any number smaller than 1 is a negative #

118

118

Free Energyand

Electrical EnergyThe relationship between

∆G and Ecell

Wrapping up the end of chapter 20

119119

Free Energy, ∆Gand Electrochem, E

• Remember, a reaction is thermodynamically favored when✓ Ecell > 0✓ ∆Grxn < 0✓ A more positive Ecell gives rise to a more negative ∆G

• Ecell is measured in V, volts✓ 1 volt = 1 Joule per coulomb✓ A coulomb is the charge on 6.242 x 1018 electrons

120

120

∆Gº = −nF Eº• Consider why the need for the negative sign✓ ∆G is negative = spontaneous✓ E is positive = spontaneous

• ∆G = free energy✓ usually kJ mol−1

• n = mole of electrons transferred• F = the Faraday, 96,485 coulombs/mol e−

• E = potential (voltage), joules/coulomb

121

121

Effect of Concentration on Cell Potential, EDon’t write….just watch.

• Remember

• and we now know

• substitute

• divide through by −nF

122

∆G = ∆Go + RT lnQ

∆Go = −nFEo

−nFE = −nFEo + RT lnQ

E = Eo −RTnFlnQ

E = Eo − 0.0592n

logQmany texts convert natural log to base 10 log: multiply by 2.303

E = Eo − 0.0257n

lnQE = Eo −8.314 J

molK× 298K

n × 96,485 coulmol e−

lnQ remember, a J/coul = Volt

122

For the following reaction from a Galvanic cell Zn + Cu2+ → Zn2+ + Cu Eº = 1.10 VCaluculate E if the Galvanic cell was set up with [Cu2+] = 2M and [Zn2+]= 0.1M

123

123

For the following reaction from a Galvanic cell Zn + Cu2+ → Zn2+ + Cu Eº = 1.10 VCaluculate E if the Galvanic cell was set up with [Cu2+] = 2M and [Zn2+]= 0.1M

124

E = Eo − 0.0257n

ln [Zn2+ ]

[Cu2+ ]

E = 1.10 − 0.02572

ln [0.1][2]

= 1.14V

• more positive, thermodynamically favored

Q <1, thus lnQ, negative

124

Electrochem and EquilibriumDon’t write….just watch.

• Now you know, ∆Gº = −nFEº• Remember ∆Gº = −RT lnKeq

• Now combine these two equations✓ −RT lnK = −nFEº then solve for Eº

• Units:‣ R = 8.314 J/molRxn K‣ T = temperature in Kelvin‣ n = number of electrons transferred, per moleRxn, unitless ratio‣ F = the Faraday, 96,485 C/mol‣ K = the equilibrium constant, unitless 125

Eo =RTnFlnK

125

• The reality is that many K values are very large or very small and expressed to the power of 10 so it is convenient to convert natural log to base 10 log

• ln K to log K: logx = 2.303 lnx

126

Eo =RTnFlnK

Eo = 2.3038.314 J

molRx iK⎛⎝⎜

⎞⎠⎟ 298K( )

n mol of e−

molRx⎛⎝⎜

⎞⎠⎟

96,500 coulmol of e−

⎛⎝⎜

⎞⎠⎟

logK

Eo =0.0592V

nlogK

by definition 1 V = 1J/coul

Eo = 0.0257Vn

lnK

Electrochem and EquilibriumDon’t write….just watch.

E = Eo − 0.0592n

logQ

• Further, at equilibrium, E = 0

thus

126

Consider a lead / silver voltaic cell made with each metal and their respective nitrate solutions at room temperature, 25ºC.

1. Write the balanced net ionic equation and calculate the voltage, Eº

2. Calculate free energy, ∆Gº

3. Calculate the equilibrium constant, K

127

half-reaction Eº (V)Ag+ + 1e− → Ag 0.80

Pb2+ + 2e− → Pb -0.13

127

Consider a lead / silver voltaic cell made with each metal and their respective nitrate solutions.


• Pb + 2Ag+ → Pb2+ + 2Ag• Eºcell = 0.93 V

2. Calculate free energy, ∆Gº• ∆Gº = −179 kJ/mol

128

∆Go = −nFEo

∆Go = −(2mole−) 96,485 coulmole−

⎛⎝⎜

⎞⎠⎟ 0.93 Joule

coul⎛⎝⎜

⎞⎠⎟

half-reaction Eº (V)Ag+ + 1e− → Ag 0.80Pb2+ + 2e− → Pb -0.13

128

Consider a lead / silver voltaic cell made with each metal and their respective nitrate solutions.


• Pb + 2Ag+ → Pb2+ + 2Ag

• Ag Eºred = 0.80 V Pb2+ Eºox = 0.13 V Eºcell = 0.93 V• ∆Gº = −179 kJ/mol

3. Calculate the equilibrium constant, K

4. solve for K = 2.97 x 1031

129

lnK = −180kJ mol0.00831kJ mol K −1( ) 298K( )

lnK = 72.5

∆Go = −RT lnK

129

Electrochem, Equilibrium, Free Energy

130

Eºcell K

∆Gº

Eo =0.0592n

logK

∆G o= −RT lnK∆G

o = −nFEo

Reaction Parameters at Standard StateReaction Parameters at Standard StateReaction Parameters at Standard StateReaction Parameters at Standard State∆Gº K Eºcell reaction at standard state is

130

Electrochem, Equilibrium, Free Energy

131

Eºcell K

∆Gº

∆G o= −RT lnK∆G

o = −nFEo

Reaction Parameters at Standard StateReaction Parameters at Standard StateReaction Parameters at Standard StateReaction Parameters at Standard State∆Gº K Eºcell reaction at standard state

− > 1 + spontaneous0 1 0 at Equlibrium+ < 1 − nonspontaneous

Eo =0.0592n

logK

131

In which of the following conditions can you be sure about the spontaneity of a reaction?

132

∆H ∆S1 + +2 + -3 - -4 - +

132

In which of the following conditions can you be sure about the spontaneity of a reaction? • For # 2 we know that it

will never be spontaneous since both ∆H and ∆S are unfavorable.

• For # 4 we know that it will always be spontaneous since both ∆H and ∆S are favorable. 133

∆H ∆S1 + +2 + -3 - -4 - +

133

Select the set of thermodynamic parameters that best describes the system below at room conditionsH2O(s) ⇋H2O(L)

134

∆G ∆H ∆S ∆Volume

1 0 > 0 > 0 > 02 0 > 0 < 0 03 < 0 > 0 < 0 < 04 < 0 > 0 > 0 < 05 < 0 < 0 > 0 > 06 < 0 > 0 > 0 > 0

134

Select the set of thermodynamic parameters that best describes the system at room conditionsH2O(s) ⇋H2O(L)

135

∆G ∆H ∆S ∆Volume

1 0 > 0 > 0 > 02 0 > 0 < 0 03 < 0 > 0 < 0 < 04 < 0 > 0 > 0 < 05 < 0 < 0 > 0 > 06 < 0 > 0 > 0 > 0

135

Calculating ∆Sfor phase changes

136136

At what temperature is this process at equilibrium? H2O(g) ⇋H2O(L)

• What temperature does ice melt?• What temperature does water

freeze?• At 0ºC, water can do either,

depending on whether or not energy is entering or exiting the system.

• If no heat is transfering, the system is at equilibrium

137

137

Using Boiling and Melting points to calculate ∆S

• When a substance is at its normal boiling or melting temperature it is at equilibrium.

• And at equilibrium ∆G = 0• Since ∆G = ∆Hº − T∆Sº• T∆Sº = ∆Hº• So the ∆S of melting or boiling = ∆H/T✓ This is one method of calculating entropy for some

substances.138

138

The molar heat of fusion, ∆Hfus, for water is 6.01 kJ mol-1. Which expression gives the molar entropy of fusion, ∆Sfus, in kJ K-1 mol-1 for ice at its normal melting point?

1. ∆Sfus = 6.01 x 4.182. ∆Sfus = 6.01 x 2733. ∆Sfus = 6.01 x 298

4. ∆Sfus =

5. ∆Sfus =

139

6.012736.01298

139

The molar heat of fusion, ∆Hfus, for water is 6.01 kJ mol-1. Which expression gives the molar entropy of fusion, ∆Sfus, in kJ K-1 mol-1 for ice at its normal melting point?

1. ∆Sfus = 6.01 x 4.182. ∆Sfus = 6.01 x 2733. ∆Sfus = 6.01 x 298

4. ∆Sfus =

5. ∆Sfus =

140

6.01298

6.01273

140

More Practice

141141

Consider the gas phase equilibrium 2CO + O2 ⇋ 2CO2 at 25ºCK = 11.1 Calculate ∆G

1. 500 kJ

2. 0.500 kJ

3. 5.96 kJ

4. 5960 kJ

5. −500 kJ

6. −0.500 kJ

7. −5.96 kJ

8. −5960 kJ142

142

Consider the gas phase equilibrium 2CO + O2 ⇋ 2CO2 at 25ºCK = 11.1 Calculate ∆G

1. 500 kJ2. 0.500 kJ3. 5.96 kJ4. 5960 kJ

5. −500 kJ6. −0.500 kJ7. −5.96 kJ8. −5960 kJ

143

∆G = −RT lnK ∆G = − 0.00831kJ / K( )(298K )(ln11.1)

143

Which apply to any reaction that proceeds spontaneously to form products from initial standard conditions.1. ∆G < 02. Keq > 13. ∆H < 04. ∆S > 0

144

144

Which apply to any reaction that proceeds spontaneously to form products from initial standard conditions.1. ∆G < 0

• because we know that if the reaction proceeds (spontaneous) it must have a negative ∆G.

2. Keq > 1• because we know at standard conditions of 1 atm, 1M K

would always = 1, so if the reaction proceeds K will end up greater than 1 since as the reaction proceeds products will increase and reactants will decrease resulting in K larger than 1.

3. ∆H < 04. ∆S > 0 145

145

Can every and any non spontaneous reaction be made spontaneous at some temperature?1. yes2. no3. I don’t know, and I would just be

guessing or following my neighbors, but I don’t really know how to analytically think through to an answer.

146

146

Can every and any non spontaneous reaction be made spontaneous at some temperature?1. yes2. no, remember: ∆G = ∆H − T∆S

• When ∆H is +, and ∆S is −, NO temperature will ever make the reaction spontaneous

• If ∆H is + and ∆S is +,✓ increasing the T will at some (even if theoretical)

temp make |T∆S| > |∆H| resulting in ∆G < 0

• If ∆H is − and ∆S is −,✓ decreasing the T will at some temp

make |T∆S| < |∆H| resulting in ∆G < 0147

147

For a spontaneous reaction, what happens to the value of ∆G as a reaction proceeds?1. gets more negative2. gets less negative3. stays the same

148

148

For a spontaneous reaction, what happens to the value of ∆G as a reaction proceeds?1. gets more negative2. gets less negative3. stays the same

149

149

The reaction below is not spontaneous at room condition but becomes spontaneous at higher temperatures. 2C(g) → A(g) + 3B(g) Which one of the following is true at 298 K (room temp)?

1. ∆G, ∆H, and ∆S are all positive2. ∆G, ∆H, and ∆S are all negative3. ∆G and ∆H are positive, but ∆S is negative4. ∆G and ∆S are positive, but ∆H is negative5. ∆G and ∆H are negative, but ∆S is positive6. ∆G and ∆S are negative, but ∆H is positive7. ∆G is positive, but we can not know about the sign

of ∆S or ∆H 8. ∆G is negative, but we can not know about the sign

of ∆S or ∆H 150

150

The reaction below is not spontaneous at standard conditions but becomes spontaneous at higher temperatures. 2C(g) → A(g) + 3B(g) Which of the following is true at 298 K?1. ∆G, ∆H, and ∆S are all positive

• at 298 K, not spontaneous means G +• when ∆S is positive, the |T∆S| factor will increase at higher

temperatures and since the equation is, − T∆S, at some temp the |T∆S| factor will be greater than the |∆H| and the value of ∆G will become negative.

2. ∆G, ∆H, and ∆S are all negative3. ∆G and ∆H are positive, but ∆S is negative4. ∆G and ∆S are positive, but ∆H is negative5. ∆G and ∆H are negative, but ∆S is positive6. ∆G and ∆S are negative, but ∆H is positive7. ∆G is positive, but we can not know about the sign of ∆S or ∆H 8. ∆G is negative, but we can not know about the sign of ∆S or ∆H 151

151

Which parameters below correctly describe the condensation of water on the inside of your car windshield on a cold morning. Select all that apply.

1. ∆G < 02. ∆H < 03. ∆S < 0 4. ∆V < 0

152

152

Which parameters below correctly describe the condensation of water on the inside of your car windshield on a cold morning. Select all that apply.1. ∆G < 0• indeed this is a spontaneous reaction, you’ve seen it

happen as you are trying to see through the windshield while driving to school.

2. ∆H < 0• vaporization is endothermic, condensation is exothermic3. ∆S < 0 • liquid has less entropy than the gas4. ∆V < 0• the liquid takes up less space than its gas

153

153

∆H for the reaction below is negativeN2(g) + 3H2(g) → 2NH3(g) ∆H = −97kJWhich is larger, the bond dissociation value of the reactants or products?

1. reactants2. products3. depends on other factors and cannot

be determined with the given information

154

154

∆H for the reaction below is negativeN2(g) + 3H2(g) → 2NH3(g) ∆H = −97kJWhich is larger, the bond dissociation value of the reactants or products?1. reactants2. products• Without looking up and dissocation values,

we know that bond forming, exothermic must be greater than bond breaking in order for the ∆H of the reaction to be negative, exothermic.

3. depends on other factors and cannot be determined with the given information

155

155

Consider the disorder of the following reactions and select the choice for which the sign of ∆S is INCORRECT.Select all that apply.1. H2O(s) → H2O(L) ∆S − 2. Ag+(aq) + Cl−(aq) → AgCl(s) ∆S − 3. CaO(s) + CO2(g) → CaCO3(s) ∆S − 4. 2SO2(g) + O2(g) → 2SO3(g) ∆S +5. 2Fe2O3(s) → 4Fe(s) + 3O2(g) ∆S +

156

156

Consider the disorder of the following reactions and select the choice for which the sign of ∆S is INCORRECT.Select all that apply.1. H2O(s) → H2O(L) ∆S − 2. Ag+(aq) + Cl−(aq) → AgCl(s) ∆S − 3. CaO(s) + CO2(g) → CaCO3(s) ∆S − 4. 2SO2(g) + O2(g) → 2SO3(g) ∆S +5. 2Fe2O3(s) → 4Fe(s) + 3O2(g) ∆S +

157

157

Choose the pair of samples of matter that correctly compares their entropy values. Select all that apply.

158

The entropy of is the entropy of1 HCl(g) at 25º < Na2SO4(s) at 25º

2 H2O(L) at 25º < H2O(L) at 50º

3 Ne(g) at 25º < Ar(g) at 25º

4 NO2(g) at STP < N2O4(g) at STP

5 N2(g) + 3H2(g) at STP < 2NH3(g) at STP

158

Choose the pair of samples of matter that correctly compares their entropy values. Select all that apply.

159

The entropy of is the entropy of1 HCl(g) at 25º < Na2SO4(s) at 25º

2 H2O(L) at 25º < H2O(L) at 50º

3 Ne(g) at 25º < Ar(g) at 25º

4 NO2(g) at STP < N2O4(g) at STP

5 N2(g) + 3H2(g) at STP < 2NH3(g) at STP

159

Select the thermodynamic quantity below that is negative for any exothermic reaction. Select as many as apply.

1. Keq

2. ∆G 3. ∆H 4. ∆S 5. ∆V (volume)

160

160

Select the thermodynamic quantity below that is negative for any exothermic reaction. Select as many as apply.

1. Keq

2. ∆G 3. ∆H 4. ∆S 5. ∆V (volume)

161

161

Select the thermodynamic quantity below that is always positive for all reactions. Select as many as apply.

1. Keq

2. ∆G 3. ∆H 4. ∆S 5. ∆V (volume)

162

162

Select the thermodynamic quantity below that is always positive for all reactions. Select as many as apply.

1. Keq

2. ∆G 3. ∆H 4. ∆S 5. ∆V (volume)

163

163

Select the thermodynamic quantity below that can be negative for spontaneous reactions. Select as many as apply.

1. Keq

2. ∆G 3. ∆H 4. ∆S 5. ∆V (volume)

164

164

Select the thermodynamic quantity below that can be negative for spontaneous reactions. Select as many as apply.

1. Keq• Keq is never a negative value

2. ∆G (must be)3. ∆H4. ∆S5. ∆V (volume)

165

165

∆Gfº for NO = +87 kJ/mol∆Gfº for NO2 = +52 kJ/molSelect the true statement(s) below.1. The formation of NO will occur faster

than the formation of NO2.2. NO2 is more stable than NO.3. NO + ½O2 → NO2 is a spontaneous

reaction.4. The formation of NO2 is spontaneous

at all reaction temperatures.5. NO → N2 + ½O2 is spontaneous 166

166

∆Gfº for NO = +87 kJ/mol∆Gfº for NO2 = +52 kJ/molSelect the true statement(s) below.1. The formation of NO will occur faster

than the formation of NO2.2. NO2 is more stable than NO.3. NO + ½O2 → NO2 is a spontaneous

reaction.4. The formation of NO2 is spontaneous

at all reaction temperatures.5. NO → N2 + ½O2 is spontaneous 167

167

When two solutions are combined and a precipitate forms, the solution cools. Which describes the value of ∆G, and its components?1. ∆G < 0, |∆H| > |T∆S|2. ∆G < 0, |∆H| < |T∆S|3. ∆G > 0, |∆H| > |T∆S|4. ∆G > 0, |∆H| < |T∆S|5. ∆G = 0, |T∆S| = |∆H|

168

168

When two solutions are combined and a precipitate forms while the solution cools. Which describes the value of ∆G, and its components?1. ∆G < 0, |∆H| > |T∆S|2. ∆G < 0, |∆H| < |T∆S|• Since we know that the solution cools, we know that ∆H is

positive, and since that is an unfavorable factor the S must be driving the reaction, thus ∆S must be positive.

• although it is completely counter intuitive, the driving force for this reaction must be an increase in entropy,

• and for the reaction to be spontaneous (∆G<0), the |T∆S| factor must be larger than the |∆H| factor.

3. ∆G > 0, |∆H| > |T∆S|4. ∆G > 0, |∆H| < |T∆S|5. ∆G = 0, |T∆S| = |∆H|

169169

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