Slide 1 of 44 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics. ΔS total =...

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Slide 1 of 44 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics. ΔS total = ΔS universe = ΔS system + ΔS surroundings The Second Law of Thermodynamics: ΔS universe = ΔS system + ΔS surroundings > 0 All spontaneous processes produce an increase in the entropy of the universe.
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Transcript of Slide 1 of 44 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics. ΔS total =...

Slide 1 of 44

19-4 Criteria for Spontaneous Change:The Second Law of Thermodynamics.

ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings

The Second Law of Thermodynamics:

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

All spontaneous processes produce an increase in the entropy of the universe.

Slide 2 of 44

Free Energy and Free Energy Change

Hypothetical process: only pressure-volume work, at constant T and P.

qsurroundings = -qp = -ΔHsys

Make the enthalpy change reversible. large surroundings, infinitesimal change in temperature.

Under these conditions we can calculate entropy.

Slide 3 of 44

Free Energy and Free Energy Change

TΔSuniv. = TΔSsys – ΔHsys = -(ΔHsys – TΔSsys)

-TΔSuniv. = ΔHsys – TΔSsys

G = H - TS

ΔG = ΔH - TΔS

For the universe:

For the system:

ΔGsys = - TΔSuniverse

Slide 4 of 44

Criteria for Spontaneous Change

ΔGsys < 0 (negative), the process is spontaneous.

ΔGsys = 0 (zero), the process is at equilibrium.

ΔGsys > 0 (positive), the process is non-spontaneous.

Slide 5 of 44

Table 19.1 Criteria for Spontaneous Change

Slide 6 of 44

19-5 Standard Free Energy Change, ΔG°

The standard free energy of formation, ΔGf°. The change in free energy for a reaction in which a

substance in its standard state is formed from its elements in reference forms in their standard states.

The standard free energy of reaction, ΔG°.

ΔG° = [ p ΔGf°(products) - r ΔGf°(reactants)]

Slide 7 of 44

Worked Examples Follow:

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Slide 9 of 44

Slide 10 of 44

CRS Questions Follow:

Slide 11 of 44

At room temperature ~290 K the reaction of H2 and O2 to form water:

-1

2 2 2 -1 -1

H = 285.8 kJ mol2H (g) + O (g) 2H O(l)

S = 327 J K mol

1. is spontaneous because it is exothermic.

2. is non-spontaneous because S is negative.

3. is spontaneous because S is negative.

4. is spontaneous because G is negative.

5. is spontaneous because G is positive.

Slide 12 of 44

At room temperature ~290 K the reaction of H2 and O2 to form water:

-1

2 2 2 -1 -1

H = 285.8 kJ mol2H (g) + O (g) 2H O(l)

S = 327 J K mol

1. is spontaneous because it is exothermic.

2. is non-spontaneous because S is negative.

3. is spontaneous because S is negative.

4. is spontaneous because G is negative.

5. is spontaneous because G is positive.

Slide 13 of 44

Which of the following processes would you expect to be spontaneous at all temperatures?

4. None are spontaneous at all temperatures.

5. All are spontaneous irrespective of the temperature.

-12 2 32SO (g) + O (g) 2SO (g) H = -200 kJ mol 1.

-1H = 6 kJ mol2.

3. -1H = -5 kJ mol

Slide 14 of 44

Which of the following processes would you expect to be spontaneous at all temperatures?

4. None are spontaneous at all temperatures.

5. All are spontaneous irrespective of the temperature.

-12 2 32SO (g) + O (g) 2SO (g) H = -200 kJ mol 1.

-1H = 6 kJ mol2.

3. -1H = -5 kJ mol

Slide 15 of 44

A mixture of H2 and O2 can sit in a flask almost indefinitely at 298 K without reacting.

-1

2 2 2 -1 -1

H = 285.8 kJ mol2H (g) + O (g) 2H O(l)

S = 327 J K mol

What is the best explanation for the absence of observable reaction?

3. The reaction is entropically unfavorable.

2. The reaction is not spontaneous at this temperature.

1. A significant energy barrier hinders the start of the reaction.

4. All three of these factors contribute.

5. None of the above answers is correct.

Slide 16 of 44

A mixture of H2 and O2 can sit in a flask almost indefinitely at 298 K without reacting.

What is the best explanation for the absence of observable reaction?

3. The reaction is entropically unfavorable.

2. The reaction is not spontaneous at this temperature.

1. A significant energy barrier hinders the start of the reaction.

4. All three of these factors contribute.

5. None of the above answers is correct.

-1

2 2 2 -1 -1

H = 285.8 kJ mol2H (g) + O (g) 2H O(l)

S = 327 J K mol

Slide 17 of 44

There is enough energy in lightning bolts that O2 and N2 in the atmosphere are decomposed into N and O atoms. The reaction of N and O to form NO,

N(g) + O(g) NO(g)1. is spontaneous at all temperatures.

2. is non-spontaneous at all temperatures.

3. is spontaneous at high temperatures but non-spontaneous at low temperatures.

4. is spontaneous at low temperatures but non-spontaneous at high temperatures.

5. It is impossible to choose between the above responses without thermochemical data.

Slide 18 of 44

There is enough energy in lightning bolts that O2 and N2 in the atmosphere are decomposed into N and O atoms. The reaction of N and O to form NO,

N(g) + O(g) NO(g)1. is spontaneous at all temperatures.

2. is non-spontaneous at all temperatures.

3. is spontaneous at high temperatures but non-spontaneous at low temperatures.

4. is spontaneous at low temperatures but non-spontaneous at high temperatures.

5. It is impossible to choose between the above responses without thermochemical data.