# The total energy of the universe cannot change Spontaneity ...

of 20
/20

Embed Size (px)

### Transcript of The total energy of the universe cannot change Spontaneity ...

122_Ch-17_Thermodyn_NOTES_ZUMD_March_29_14.pptxbe created or
destroyed.

ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings

(though you can transfer it from one place to

another).

(system è surroundings)

two ways energy “lost” from a system: converted to heat, q and/or used to do work, w

Energy conservation requires that the energy change in the system equal the heat released plus the work done.

ΔE = q + w ΔE = ΔH + pΔV

Chapter 17 4

First Law of Thermodynamics

ΔE is a state function: internal energy change independent of how it was done

3/30/14

2

Nonspontaneous processes require energy input to go.

Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction

If the system after the rxn has less potential energy than before the rxn, the rxn is thermodynamically favorable.

spontaneity ≠ fast or slow

Chapter 17 6

The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

Comparing Potential Energy

Chapter 17 7

Reversibility of Process

Any spontaneous process is irreversible, i.e., it will proceed in only one direction.

A reversible process will proceed back and forth between the two end conditions.

- equilibrium rxns (they result in no change in free energy;)

If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction.

Chapter 17 8

Thermodynamics and Spontaneity

Thermodynamics and Spontaneity

diamond vs. graphite

Factors Affecting Whether a Rxn is Spontaneous

Reactions are spontaneous in the direction of lower chemical potential energy.

There are two factors that determine the thermodynamic favorability: the enthalpy change and the entropy change.

The enthalpy change, ΔH, is the difference in the sum of the internal energy and pV work energy of the reactants compared to the products.

The entropy change, ΔS, is the difference in randomness of the reactants compared to the products.

Chapter 17 11

Enthalpy (review)

ΔH is generally measured in kJ/mol. - stronger bonds = more stable molecules

exothermic rxn: if the bonds in the products are stronger than the bonds in the reactants

ΔHrxn < 0: exothermic rxn; ΔHrxn > 0: endothermic rxn

endothermic rxn: the bonds in the products are weaker than the bonds in the reactant;

ΔΗrxn = Σ (ΔH°f products) − Σ (ΔH°f reactants)

Chapter 17 12

Entropy, S, is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components (degree of freedom) increases. S is generally measured in J/mol. Random systems require less energy than ordered systems. (read section 17.1)

Entropy

3/30/14

4

Changes in Entropy, ΔS

Entropy change is favorable when the result is a more random system (ΔS = Sfinal – Sinit. > 0 )

ΔS > 0 if: - products are in a more random state - going from solid to less ordered liquid to less ordered gas

- rxns that have larger # of product molecules than reactant molecules

- increase in temperature

Chapter 17 14

Chapter 17 15

Changes in Entropy, ΔS

ΔSsystem > 0 for a process in which the final condition is more random than the initial condition (favorable entropy) ΔSsystem < 0 for a process in which the final condition is more orderly than the initial condition (unfavorable entropy)

ΔSsystem = ΔSreaction = Σ (S°prod) − Σ (S°react)

Chapter 17 16

Entropy Change and State Changes When a material changes physical state, the number of macrostates it can have changes as well.

- the more degrees of freedom the molecules have, the more macrostates are possible;

3/30/14

5

Chapter 17 17

Entropy Change and State Changes - solids have fewer macrostates than liquids, which have fewer macrostates than gases;

Chapter 17 18

•Water vapor condensing

•Dissolving sugar in tea

•2 NH3 (g) → N2 (g) + 3 H2 (g)

•Ag+ (aq) + Cl−(aq) → AgCl(s)

Predict whether ΔSsystem is + or − for each of the following:

Problem A

Chapter 17 19

• The total entropy change of the universe must be positive for a process to be spontaneous. - for reversible process, ΔSuniv = 0. - for irreversible (spontaneous) process, ΔSuniv >0

ΔSuniverse = ΔSsystem + ΔSsurroundings • If the entropy of the system decreases, then the entropy

of the surroundings must increase by a larger amount. - when ΔSsystem < 0 , ΔSsurroundings > 0 .

the 2nd Law of Thermodynamics

Chapter 17 20

Temperature Dependence of ΔSsurroundings

When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings.

When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings.

The amount the entropy of the surroundings changes depends on its initial temperature.

3/30/14

6

Temperature dependence of ΔSsurroundings

The higher the original temperature, the less effect addition or removal of heat has.

Chapter 17 22

Calculate the ΔSsurr. at 25ºC for the rxn: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔHrxn = − 2044 kJ.

Problem B

Chapter 17 23

The rxn below has ΔHrxn = +66.4 kJ at 25°C. 2 O2 (g) + N2 (g) → 2 NO2 (g)

i. Determine ΔSsurroundings.

The rxn below has ΔHrxn = +66.4 kJ at 25°C.

2 O2 (g) + N2 (g) → 2 NO2 (g)

ii. Determine the sign of ΔSsystem.

Problem C

Chapter 17 25

The rxn below has ΔHrxn = + 66.4 kJ at 25°C. iii. Determine whether the process is spontaneous 2 O2 (g) + N2 (g) → 2 NO2 (g)

Problem C

ΔGsys = ΔHsys− TΔSsys

ΔSuniv = ΔSsys + ΔSsurr

ΔSuniv = ΔSsys - ΔHsys

-Τ ΔSuniv = -T ΔSsys + ΔHsys

-Τ ΔSuniv = ΔHsys - Τ ΔSsys

-Τ ΔSuniv = ΔGsys

Gibbs Free Energy and Spontaneity

The Gibbs free energy is the maximum amount of work energy that can be released to the surroundings by a system.

- valid at a constant temperature and pressure - often called the chemical potential energy because it is analogous to the storing of energy in a mechanical system

ΔGsys = ΔHsys− T ΔSsys

- since ΔSuniv determines whether a process is spontaneous, ΔG also determines spontaneity;

-Τ ΔSuniv = ΔGsys Chapter 17 28

Gibbs Free Energy, ΔG

Possible ΔG values: ΔG < 0:

ΔH < 0 and ΔS > 0

ΔGsys = ΔHsys− T ΔSsys

rxn exothermic & more random

or high temperature

-Τ ΔSuniv = ΔGsys

Possible ΔG values: ΔG > 0 : ΔH > 0 and ΔS < 0

ΔGsys = ΔHsys− T ΔSsys

ΔG = 0 : the reaction is at equilibrium

Chapter 17 30

Chapter 17 31

The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. i. Calculate ΔG for the reaction.

Problem D

3/30/14

9

Chapter 17 33

The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. i. Calculate ΔG for the reaction.

Problem E

Chapter 17 35

The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. Calculate the minimum temperature for it to be spontaneous.

Problem F

Chapter 17 37

The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. Calculate the maximum temperature for it to be spontaneous.

Problem G

Read pages 656 - 659

Chapter 17 41

Changes in Entropy

the standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions;

ΔSºreaction = ( ∑ np Sºproducts) − ( ∑ nr Sºreactants)

Chapter 17 42

Calculate ΔS° for the rxn 4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g)

Problem H

Substance S°, J/mol⋅K NH3 (g) 192.8 O2 (g) 205.2 NO(g) 210.8 H2O(g) 188.8

Chapter 17 43

Problem H continued

Chapter 17 44

Problem H continued

Chapter 17 45

Substance S°, J/mol⋅K H2 (g) 130.7 O2 (g) 205.2 H2O(g) 188.8

Calculate ΔS° for the rxn 2 H2 (g) + O2 (g) → 2 H2O(g)

Problem I

at 25 °C at temperatures other than 25 °C:

assuming the change in ΔHo reaction and

ΔSo reaction negligible

1.

2.

3.

CO2 (g) −394.4 H2O(g) −228.6 O3 (g) 163.2

Calculate ΔG° at 25 °C for the reaction CH4 (g) + 8 O2 (g) → CO2 (g) + 2 H2O(g) + 4 O3 (g)

Problem J

Chapter 17 51

Determine the free energy change ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g)

Problem K

Chapter 17 53

Problem K continued

Chapter 17 54

Problem K continued

Chapter 17 55

- if a rxn can be expressed as a series of rxns, the sum of the ΔG values of the individual rxn is the ΔG of the total rxn (ΔG is a state function);

- if a rxn is reversed, the sign of its ΔG value reverses; - if the amount of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor; (the value of ΔG is an extensive property)

ΔGo Relationships

substance ΔG° (kJ/mol) H2 (g) + O2 (g) → H2O2 (g) −105.6

2 H2 (g) + O2 (g) → 2 H2O(g) −457.2

Determine ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g)

Problem L

Chapter 17 57

Problem L continued

Chapter 17 58

Find ΔGºrxn for the following rxn using the given equations: 3 C(s) + 4 H2 (g) → C3H8 (g)

Problem M

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔGº = −2074 kJ

C(s) + O2 (g) → CO2 (g) ΔGº = −394.4 kJ

2 H2 (g) + O2 (g) → 2 H2O(g) ΔGº = −457.1 kJ

Chapter 17 59

Problem M continued

Chapter 17 60

Problem M continued

Chapter 17 61

Problem M continued

Chapter 17 62

• The change in free energy is the theoretical limit as to the amount of work that can be done.

• If the reaction achieves its theoretical limit, it is a reversible reaction.

Free Energy and Reversible Reactions

Chapter 17 63

• In a real reaction, some of the free energy is “lost” as heat (if not most).

• Therefore, real reactions are irreversible.

Real Reactions

Chapter 17 64

Free Energy under Nonstandard Conditions ΔG = ΔG° only when the reactants and products are in their standard states;

- their normal state at that temperature - partial pressure of gas = 1 atm - concentration = 1 M

Under nonstandard conditions, ΔG = ΔG° + RT lnQ ( Q is the rxn quotient )

At equilibrium, ΔG = 0 , therefore ΔG° = −RTlnK

3/30/14

17

Chapter 17 66

NO2 (g) 2.00

Calculate ΔG at 298 K for the rxn under the given conditions: 2 NO(g) + O2 (g) → 2 NO2 (g) ΔGº = −71.2 kJ

Problem N

NH3 (g) 2.0

Calculate ΔGrxn for the given rxn at 700 K under the given conditions: N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ

Problem P

substance ΔG°f (kJ/mol) N2O4 (g) +99.8 NO2 (g) +51.3

Problem R Calculate K at 298 K for the reaction N2O4 (g) 2 NO2 (g)

ΔGo from Appendix IIB

Chapter 17 74

Problem R continued

Chapter 17 75

Problem R continued

Chapter 17 76

Problem S Estimate the equilibrium constant for the given rxn at 700 K: N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ

3/30/14

20

ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings

(though you can transfer it from one place to

another).

(system è surroundings)

two ways energy “lost” from a system: converted to heat, q and/or used to do work, w

Energy conservation requires that the energy change in the system equal the heat released plus the work done.

ΔE = q + w ΔE = ΔH + pΔV

Chapter 17 4

First Law of Thermodynamics

ΔE is a state function: internal energy change independent of how it was done

3/30/14

2

Nonspontaneous processes require energy input to go.

Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction

If the system after the rxn has less potential energy than before the rxn, the rxn is thermodynamically favorable.

spontaneity ≠ fast or slow

Chapter 17 6

The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

Comparing Potential Energy

Chapter 17 7

Reversibility of Process

Any spontaneous process is irreversible, i.e., it will proceed in only one direction.

A reversible process will proceed back and forth between the two end conditions.

- equilibrium rxns (they result in no change in free energy;)

If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction.

Chapter 17 8

Thermodynamics and Spontaneity

Thermodynamics and Spontaneity

diamond vs. graphite

Factors Affecting Whether a Rxn is Spontaneous

Reactions are spontaneous in the direction of lower chemical potential energy.

There are two factors that determine the thermodynamic favorability: the enthalpy change and the entropy change.

The enthalpy change, ΔH, is the difference in the sum of the internal energy and pV work energy of the reactants compared to the products.

The entropy change, ΔS, is the difference in randomness of the reactants compared to the products.

Chapter 17 11

Enthalpy (review)

ΔH is generally measured in kJ/mol. - stronger bonds = more stable molecules

exothermic rxn: if the bonds in the products are stronger than the bonds in the reactants

ΔHrxn < 0: exothermic rxn; ΔHrxn > 0: endothermic rxn

endothermic rxn: the bonds in the products are weaker than the bonds in the reactant;

ΔΗrxn = Σ (ΔH°f products) − Σ (ΔH°f reactants)

Chapter 17 12

Entropy, S, is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components (degree of freedom) increases. S is generally measured in J/mol. Random systems require less energy than ordered systems. (read section 17.1)

Entropy

3/30/14

4

Changes in Entropy, ΔS

Entropy change is favorable when the result is a more random system (ΔS = Sfinal – Sinit. > 0 )

ΔS > 0 if: - products are in a more random state - going from solid to less ordered liquid to less ordered gas

- rxns that have larger # of product molecules than reactant molecules

- increase in temperature

Chapter 17 14

Chapter 17 15

Changes in Entropy, ΔS

ΔSsystem > 0 for a process in which the final condition is more random than the initial condition (favorable entropy) ΔSsystem < 0 for a process in which the final condition is more orderly than the initial condition (unfavorable entropy)

ΔSsystem = ΔSreaction = Σ (S°prod) − Σ (S°react)

Chapter 17 16

Entropy Change and State Changes When a material changes physical state, the number of macrostates it can have changes as well.

- the more degrees of freedom the molecules have, the more macrostates are possible;

3/30/14

5

Chapter 17 17

Entropy Change and State Changes - solids have fewer macrostates than liquids, which have fewer macrostates than gases;

Chapter 17 18

•Water vapor condensing

•Dissolving sugar in tea

•2 NH3 (g) → N2 (g) + 3 H2 (g)

•Ag+ (aq) + Cl−(aq) → AgCl(s)

Predict whether ΔSsystem is + or − for each of the following:

Problem A

Chapter 17 19

• The total entropy change of the universe must be positive for a process to be spontaneous. - for reversible process, ΔSuniv = 0. - for irreversible (spontaneous) process, ΔSuniv >0

ΔSuniverse = ΔSsystem + ΔSsurroundings • If the entropy of the system decreases, then the entropy

of the surroundings must increase by a larger amount. - when ΔSsystem < 0 , ΔSsurroundings > 0 .

the 2nd Law of Thermodynamics

Chapter 17 20

Temperature Dependence of ΔSsurroundings

When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings.

When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings.

The amount the entropy of the surroundings changes depends on its initial temperature.

3/30/14

6

Temperature dependence of ΔSsurroundings

The higher the original temperature, the less effect addition or removal of heat has.

Chapter 17 22

Calculate the ΔSsurr. at 25ºC for the rxn: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔHrxn = − 2044 kJ.

Problem B

Chapter 17 23

The rxn below has ΔHrxn = +66.4 kJ at 25°C. 2 O2 (g) + N2 (g) → 2 NO2 (g)

i. Determine ΔSsurroundings.

The rxn below has ΔHrxn = +66.4 kJ at 25°C.

2 O2 (g) + N2 (g) → 2 NO2 (g)

ii. Determine the sign of ΔSsystem.

Problem C

Chapter 17 25

The rxn below has ΔHrxn = + 66.4 kJ at 25°C. iii. Determine whether the process is spontaneous 2 O2 (g) + N2 (g) → 2 NO2 (g)

Problem C

ΔGsys = ΔHsys− TΔSsys

ΔSuniv = ΔSsys + ΔSsurr

ΔSuniv = ΔSsys - ΔHsys

-Τ ΔSuniv = -T ΔSsys + ΔHsys

-Τ ΔSuniv = ΔHsys - Τ ΔSsys

-Τ ΔSuniv = ΔGsys

Gibbs Free Energy and Spontaneity

The Gibbs free energy is the maximum amount of work energy that can be released to the surroundings by a system.

- valid at a constant temperature and pressure - often called the chemical potential energy because it is analogous to the storing of energy in a mechanical system

ΔGsys = ΔHsys− T ΔSsys

- since ΔSuniv determines whether a process is spontaneous, ΔG also determines spontaneity;

-Τ ΔSuniv = ΔGsys Chapter 17 28

Gibbs Free Energy, ΔG

Possible ΔG values: ΔG < 0:

ΔH < 0 and ΔS > 0

ΔGsys = ΔHsys− T ΔSsys

rxn exothermic & more random

or high temperature

-Τ ΔSuniv = ΔGsys

Possible ΔG values: ΔG > 0 : ΔH > 0 and ΔS < 0

ΔGsys = ΔHsys− T ΔSsys

ΔG = 0 : the reaction is at equilibrium

Chapter 17 30

Chapter 17 31

The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. i. Calculate ΔG for the reaction.

Problem D

3/30/14

9

Chapter 17 33

The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. i. Calculate ΔG for the reaction.

Problem E

Chapter 17 35

The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. Calculate the minimum temperature for it to be spontaneous.

Problem F

Chapter 17 37

The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. Calculate the maximum temperature for it to be spontaneous.

Problem G

Read pages 656 - 659

Chapter 17 41

Changes in Entropy

the standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions;

ΔSºreaction = ( ∑ np Sºproducts) − ( ∑ nr Sºreactants)

Chapter 17 42

Calculate ΔS° for the rxn 4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g)

Problem H

Substance S°, J/mol⋅K NH3 (g) 192.8 O2 (g) 205.2 NO(g) 210.8 H2O(g) 188.8

Chapter 17 43

Problem H continued

Chapter 17 44

Problem H continued

Chapter 17 45

Substance S°, J/mol⋅K H2 (g) 130.7 O2 (g) 205.2 H2O(g) 188.8

Calculate ΔS° for the rxn 2 H2 (g) + O2 (g) → 2 H2O(g)

Problem I

at 25 °C at temperatures other than 25 °C:

assuming the change in ΔHo reaction and

ΔSo reaction negligible

1.

2.

3.

CO2 (g) −394.4 H2O(g) −228.6 O3 (g) 163.2

Calculate ΔG° at 25 °C for the reaction CH4 (g) + 8 O2 (g) → CO2 (g) + 2 H2O(g) + 4 O3 (g)

Problem J

Chapter 17 51

Determine the free energy change ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g)

Problem K

Chapter 17 53

Problem K continued

Chapter 17 54

Problem K continued

Chapter 17 55

- if a rxn can be expressed as a series of rxns, the sum of the ΔG values of the individual rxn is the ΔG of the total rxn (ΔG is a state function);

- if a rxn is reversed, the sign of its ΔG value reverses; - if the amount of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor; (the value of ΔG is an extensive property)

ΔGo Relationships

substance ΔG° (kJ/mol) H2 (g) + O2 (g) → H2O2 (g) −105.6

2 H2 (g) + O2 (g) → 2 H2O(g) −457.2

Determine ΔGo in the following rxn at 298 K: 2 H2O(g) + O2 (g) → 2 H2O2 (g)

Problem L

Chapter 17 57

Problem L continued

Chapter 17 58

Find ΔGºrxn for the following rxn using the given equations: 3 C(s) + 4 H2 (g) → C3H8 (g)

Problem M

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔGº = −2074 kJ

C(s) + O2 (g) → CO2 (g) ΔGº = −394.4 kJ

2 H2 (g) + O2 (g) → 2 H2O(g) ΔGº = −457.1 kJ

Chapter 17 59

Problem M continued

Chapter 17 60

Problem M continued

Chapter 17 61

Problem M continued

Chapter 17 62

• The change in free energy is the theoretical limit as to the amount of work that can be done.

• If the reaction achieves its theoretical limit, it is a reversible reaction.

Free Energy and Reversible Reactions

Chapter 17 63

• In a real reaction, some of the free energy is “lost” as heat (if not most).

• Therefore, real reactions are irreversible.

Real Reactions

Chapter 17 64

Free Energy under Nonstandard Conditions ΔG = ΔG° only when the reactants and products are in their standard states;

- their normal state at that temperature - partial pressure of gas = 1 atm - concentration = 1 M

Under nonstandard conditions, ΔG = ΔG° + RT lnQ ( Q is the rxn quotient )

At equilibrium, ΔG = 0 , therefore ΔG° = −RTlnK

3/30/14

17

Chapter 17 66

NO2 (g) 2.00

Calculate ΔG at 298 K for the rxn under the given conditions: 2 NO(g) + O2 (g) → 2 NO2 (g) ΔGº = −71.2 kJ

Problem N

NH3 (g) 2.0

Calculate ΔGrxn for the given rxn at 700 K under the given conditions: N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ

Problem P

substance ΔG°f (kJ/mol) N2O4 (g) +99.8 NO2 (g) +51.3

Problem R Calculate K at 298 K for the reaction N2O4 (g) 2 NO2 (g)

ΔGo from Appendix IIB

Chapter 17 74

Problem R continued

Chapter 17 75

Problem R continued

Chapter 17 76

Problem S Estimate the equilibrium constant for the given rxn at 700 K: N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ

3/30/14

20