Chapter 19

1

Chapter 19

Chemical Thermodynamics Entropy, Free Energy and

Equilibrium

2

19.1 Thermodynamics

Thermo: heat Dynamics: power

State Functions: considers only initial and final states

Does not consider pathways or rate

3

19.1 Thermodynamics Study of energy flow and its transformations

(heat and energy flow) Determines direction of reactions

(spontaneous or nonspontaneous under given conditions)

Considers only initial and final states

Organized into three laws: 1st Law 2nd Law 3rd Law

4

19.1 Basic Definitions System Surrounding Open system Closed system Isolated system State of system: defined by values of

composition, pressure, T, V. State Function: defined only by initial and

final condition of the system (Enthalpy, Entropy, Gibbs Free Energy).

Energy change signaled by: accomplishment of work and/or appearance or disappearance of heat.

system surroundings

universe

system surroundings

universe

5

19.1 Reversible and Irreversible Processes

Reversible process: the system can be restored to its original state by exactly reversing the original change. Example: melting ice at its melting point

Irreversible process: we cannot restore the system to its original state by simple reversing the process. Different path has to be used. Example: expansion a gas into vacuum.

6

19.1 First Law of Thermodynamics 1st Law:

Energy can be neither created nor destroyed Energy of the universe is constant

Important concepts from thermochemistry Enthalpy Hess’s law

Purpose of 1st Law Energy bookkeeping

How much energy? Exothermic or endothermic? What type of energy? Δu = q + w q = heat; w = work system does on the

surroundings (-PΔV)

7

19.2 Spontaneous vs Nonspontaneous Spontaneous process

Occurs without outside assistance in the form of energy (occurs naturally) Product-favored at equilibrium May be fast or slow May be influenced by temperature

Spontaneous versus nonspontaneous processes. Expansion of gas: spontaneous. Reverse: not.

8

19.2 Spontaneous vs. NonspontaneousSpontaneous Processes: (exothermic or endothermic?) Gases expand into larger volumes at constant

temperature ________________ H2O(s) melts above 0C ____________ H2O(l) freezes below 0C ____________ NH4NO3 dissolves spontaneously in H2O ____________ Steel (iron) rusts in presence of O2 and H2O ________ Wood burns to form CO2 and H2O __________ CH4 gas burns to form CO2 and H2O __________

Evolution of Heat (Exothermicity) : not enough to predict spontaneity

9

19.2 Spontaneous vs Nonspontaneous

Nonspontaneous process Does not occur unless there is outside

assistance (energy?) Reactants-favored at equilibrium

All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction Spontaneous processes have a definite

direction Spontaneous processes are irreversible. Can

be reversed with considerable input of energy.

10

19.2 Factors That Favor Spontaneity

Spontaneous Processes driven by Enthalpy, H (Joules)

Many, but not all, spontaneous processes tend to be exothermic.

Entropy, S (Joules/K) Measure of the disorder of a system Many, but not all, spontaneous

processes tend to increase disorder of the system

Exothermicity favors spontaneity, but does not guarantee it.

11

19.2 Factors That Favor Spontaneity: Enthalpy

Examples of spontaneous reaction that is not exothermic:

NH4NO3(s) → NH4+ (aq) + NO3

-(aq) ΔH = 25 kJ/mol

Expansion of gas: energy neutral Phase changes: endothermic processes

that occurs spontaneously. Chemical system: H2(g) + I2(g) ↔ 2HI(g)

Equilibrium can be approached from both sides (spontaneous both ways) even though the forward reaction is endothermic and the reverse is exothermic.

12

19.2 Spontaneity: Examples

1. Based on your experience, predict whether the following processes are spontaneous, are spontaneous in reverse direction, or are in equilibrium:

(a) When a piece of metal heated to 150 ºC is added to water at 40 ºC, the water gets hotter.

(b) Water at room temperature decomposes into hydrogen and oxygen gases

(c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 ºC.

13

19.2 Entropy Direct measure of the randomness or disorder

of the system. Related to probability

describes # of ways the particles in a system can be arranged in a given state (position and/or energy levels)

The most likely state – the most random More possible arrangements, the higher

disorder, higher entropy Ordered state – low probability of occurring Disordered state: high probability of

occurring

14

19.2 Entropy versus Probability Systems tend to move spontaneously

towards increased entropy. Why? Entropy is related to probability

(positional) Disordered states are more probable

than ordered states

S = k (lnW) k (Boltzman’s constant) = 1.38 x 10-23 J/K W = # possible arrangements in system

15

19.2 Spontaneous Processes: Dispersal of Matter

Isothermal (constant temperature) expansion of gas

After opening stopcock the molecules could be in any arrangement shown (4 arrangements)Probability for each arrangement = (1/2)2

25% probability

25% probability

Two molecules present:

16

Gas Container = two bulbed flask

Gas Molecules

Ordered State

19.2 Spontaneous Process: Isothermal Gas Expansion

Consider why gases tend to isothermally expand into larger volumes.

17

Gas Container

S = k ln (W) = k (ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K

For 3 particles, probability = (1/2)3

For N particles, probability = (1/2)N

Ordered State

19.2 Spontaneous Process: Isothermal Gas Expansion

18

Disordered States

19

Disordered States

20

Disordered States

21

Disordered States

22

Disordered States

23

Disordered States

24

Disordered States

25

More probable that the gas molecules will disperse between two halves than remain on one side

Disordered States

26

Driving force for expansion is entropy (probability); gas molecules have a tendency to spread out

Disordered States

27S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K

Disordered States

28

Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K

Total Arrangements

29

19.2 Spontaneous Processes

With 1 mole of molecules, the number of possible arrangements increases dramatically. The probability that the gas molecules will be distributed between the two flasks increases greatly.

Spontaneous processes are those in which the disorder of the system increasesThe isothermal expansion of a gas is spontaneous because of the increase in randomness or “disorder” of the system.

30

19.2 Entropy Entropy, S (J/K) State Function S= (heat change)/T = q/T S = Sfinal – Sinitial

S > 0 represents increased randomness or disorder

Note: The magnitude of change in entropy depends on temperature.

31

19.2 Entropy: example

1. ΔS = q/TThe element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9 ºC, and its molar enthalpy of fusion is ΔHfusion = 2.331 kJ/mol. What is the entropy change when 50.0 g of Hg(l) freezes at the normal freezing point?

(-2.48 J/K)

32

19.2 Patterns of Entropy Change For the same or similar substances:

Ssolid < Sliquid < <Sgas

solidliquid vapor

33

19.2 Patterns of Entropy Change

solidliquid vapor

Rigidly held

particles; few

positions available

to particles

Particles free to flow; more

positions available for

particles

Particles farther apart, occupy

larger volume of space; even more positions available

to particles

34


solidliquid vapor

most ordere

d

least ordere

dless ordere

d

35

19.2 Patterns of Entropy Change Solution formation usually leads to increased

entropy for the system. Is it true for the surroundings (solvent)?

solute

solvent

solution

particles more disordered


36


Describe in words the entropy of the system

37

19.2 Patterns of Entropy ChangeDissolution of NaCl in water:1. The crystal breaks up2. The Na+ and Cl- ions are surrounded by hydrating

water molecules.NaCl(s) → Na+(aq) + Cl-(aq)

3. Each ion is surrounded by several water molecules.

4. NaCl becomes disordered (entropy increases)5. The water molecules become more ordered!

(entropy decreases)6. NaCl dissolves – net entropy increases.

38

19.3 Entropy and Temperature

(a) A substance at a higher temperature has greater molecular motion, more disorder, and greater entropy than (b) the same substance at a lower temperature.

39

Stan

dard

ent

ropy

, S(J

/K)

Temperature (K)0

20

10

30

40

50

10050 250 300150 200

What kind of changes are represented here?

40

Stan

dard

ent

ropy

, S°(

J/K)

Temperature (K)0

20

10

30

40

50

10050 250 300150 200

Solid

LiquidGas

What is the effect of temperature on entropy?

41

19.3 Entropy in Temperature

42

Substance

S (J/K) Substance

S (J/K)

H2O(l) 69.9 NaCl(s) 72.3H2O(g) 188.8 NaCl(aq) 115.5

I2(s) 116.7 Na2CO3(s) 136.0I2(g) 260.6 CH4(g) 186.3Na(s) 51.5 C2H6(g) 229.5K(s) 64.7 C3H8(g) 269.9Cs(s) 85.2 C4H10(g) 310.0

19.2 Standard Molar Entropies of Selected Substances at 298 K

43

19.2 Patterns of Entropy Change Chemical Reactions

#n gas molecules in product > #n molecules in reactants (Srxn > 0)

Physical Changes: cases where entropy increases Expansion of gas Formation of solutions Temperature changes. If only solids, ions and/or liquids involved, S

increases if total # particles increases.

If ΔS > 0, , randomness increases and entropy increases

44

19.2 Entropy: Examples

2. Predict if ΔS increases, decreases or does not change

(a) Freezing liquid mercury(b) Condensing H2O(vapor)(c) Precipitating AgCl(d) Heating H2(g) from 60.0 ºC to 80 ºC (e) Subliming iodine crystals(f) Rusting iron nail

45

19.3 Entropy - Examples3. Predict which substance has the higher entropy:

a) NO2(g) or N2O4(g)b) I2(g) or l2(s)

4. Predict whether each of the following leads to increase or decrease in entropy of a system If in doubt, explain why.a) The synthesis of ammonia:

N2(g) + 3H2(g) ↔ 2NH3(g)

b) C12H22O11(s) C12H22O11(aq)

c) Evaporation to dryness of a solution of urea, CO(NH2)2 in vapor. CO(NH2)2(aq) → CO(NH2)2(s)

H2O (l)

46

19.3 Second Law of Thermodynamics: System

6. Predict the sign of ΔS0 for each of the following reactions:a) Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s)

b) MgCO3(s) → MgO(s) + CO2(g)

d) H2(g) + Br2(g) → 2HBr

47

19.3 Second Law of Thermodynamics

Expresses the connection between entropy and spontaneity.

In any spontaneous process there is always an increase in the entropy of the Universe. The entropy remains unchanged at equilibrium.

SPONTANEOUS PROCESS:ΔSuniv = Δ Ssys + Δ Ssurr > 0

NONSPONTANEOUS PROCESS:ΔS universe= ΔS syst. + ΔS surr. <0Reversible process is spontaneous

48

19.3 Second Law: Entropy Changes

EQUILIBRIUM PROCESSES (reversible)

♦ΔS universe= ΔS syst. + ΔS surr. =0

♦ΔS syst = ΔS surr

♦ΔS syst = ΔSº final - ΔS0 initial

49

19.3 Entropy Changes in a System (Reactions)

Entropy changes in a system aA + bB → cC + dD Standard entropy change ΔSº (25 ºC, 1atm). Only changes in entropy can be measured. Each element has an entropy value (compare

to enthalpy). Absolute value for each substance can be

determined. For a chemical system:

ΔSº rxn = ΔSº products - ΔS0 reactants

Standard Molar entropy, S0, is the entropy of one mole of a substance in its standard state (298 K)

50

5. Using standard molar entropies, calculate S°rxn for the following reaction at 25°C:

2SO2(g) + O2(g) → 2SO3(g)S° = 248.1 205.1 256.6 (J · K-

1mol-1)

(Ans.: -187.9 J/K)

19.3 Second Law: Example

51

19.3 Entropy of Reactions (System)6. Using thermodynamic tables, calculate the

standard entropy changes for the following reactions at 25 ºC

a) Evaporation of 1.00 mol of liquid ethanol to ethanol vapor.

b) The oxidation of one mole pf ethanol vapor (combustion reaction)c) Are the reactions spontaneous under the given conditions.

Answers:a) 122.0 J/K b) 96.09 J/K

52

19.3 Entropy Changes in the System

In a reaction More gas molecules produced: entropy

increases Less gas molecules produced: entropy

decreases No net change of # of gas molecules

produced: entropy changes, but slightly

Liquid, solid products: hard to estimate-needs calculations

53

19.3 Entropy Changes in Surrounding

Entropy of surroundings Increases in an exothermic process Decreases in an endothermic process Change depends on temperature ΔSsurr is then proportional to the change

in enthalpy in the system (negative sign needed to make entropy positive in an exothermic process)

54

19.3 Second Law of Thermodynamics (System)

7. Examples: H2O(s) melts above 0C (endothermic). What

about entropy? Steel (iron) rusts in presence of O2 and H2O

(exothermic). What about entropy? 4Fe(s) + 3O2(g) 2Fe2O3(s)

CH4 gas burns to form CO2 and H2O (exothermic). What about entropy?

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Each process increases entropy of the universe.

55

19.3 Entropy of Surrounding

(a)When an exothermic reaction occurs in the system (ΔH < 0), the surroundings gain heat and their entropy increases (Δ Ssurr > 0).

(b) When an endothermic reaction occurs in the system (Δ H > 0), the surroundings lose heat and their entropy decreases (Δ Ssurr < 0).

62

19.3 Second Law of Thermodynamics To determine Suniv for a process, both Ssystem and

Ssurroundings need to be known: Ssystem

related to matter dispersal in system Ssurroundings

determined by heat exchange between system and surroundings and T at which it occurs

Sign of Ssurr depends on whether process in system is endothermic (Ssurr< 0) or exothermic (Ssurr >0)

Magnitude of Ssurr depends on T

Ssurr = -Hsystem/T

63

19.3 Second Law of Thermodynamics Consider transferring the same amount of heat to two

different systems, one at 298K and another at 500K.

More entropy (disorder) is created in the system at lower temp.

At high temperatures the system is already disordered

Tq

S rev

298K 500KQ Q

Kq

Kq

500298

64

19.3 Second Law of Thermodynamics

8. Reaction: N2(g) + 3H2(g) → 2NH3(g)ΔHº = -92.6kJ ΔSsys = -199 J/K at 25 ºC

ΔSsurr = -(-92.6 x1000)J/298K = 311 J/KΔSuniv = -199 J/K + 311 J/K = 112 J/K

Reaction spontaneous at 25 ºC

65

19.3 Third Law of Thermodynamics

Perfect crystal: its internal arrangement is absolutely regular. Nothing is in motion (vibrations, rotations and translations)

The entropy of a pure (perfect) crystal at 0K is 0.

66

Third Law of Thermo

Gives us a starting point, S at 0K is equal to zero.

All others must be >0.

Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed.

Products - reactants to find Sº (a state function)

More complex molecules higher Sº.

19.3 Entropy and Third Law

67

19.4 Gibbs Free Energy2nd law: Suniv = Ssys + Ssurr > 0 for spontaneousSuniv = Ssys - Hsys/T > 0Rearrange: multiply both sides by (-T)-TSuniv = -TSsys + Hsys < 0

-TSuniv = Hsys - TSsys < 0

GIBBS FREE ENERGY: G = -TSunivG = Gibbs free energy (J or kJ) = (G = -TSuniv ) G = H – TS < 0

andG = Hsys - TSsys < 0

G° = H°sys - TS° < 0 (if at standard state) for Spontaneous Process

68

19.4 Gibb's Free Energy at any Conditions

G=H-TS Never used this way. At constant temperature

G=Hsys –TSsys

G function eliminates the need to deal with entropy of the surroundings

If G is negative at constant T and P, the process is spontaneous.

We deal only with the SYSTEM.

69

19.4 Gibbs Free Energy Summary of Conditions for Spontaneity

G < 0 reaction is spontaneous in the forward

direction (Suniv > 0)

G > 0 reaction is nonspontaneous in the forward

direction(Suniv < 0)

G = 0 system is at equilibrium

(Suniv = 0)Remember- Spontaneity tells us nothing about rate.

70

19.4 Gibbs Free Energy The Gibbs Free Energy is a measure of the maximum

amount of work, at a given temperature and pressure, that can be done on the surroundings by a system.

Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.Wmax = G

For a spontaneous process: Maximum amount of energy released by the

system that can do useful work on the surroundings

Energy available from spontaneous process that can be used to drive non-spontaneous process.

For a nonspontaneous process: Minimum amount of work that must be done to

force the process to occur

71

19.4 Gibbs Standard Energy Free Energy

The standard free energies of formation, G°, are the free energy values for the formation of a substance under standard conditions.

The standard free energy of formation for any element in its standard state is zero. (Like enthalpy, but not entropy)

Compare: G = Hsys - TSsys (any conditions) G = H0

sys - TS0sys (standard conditions)

72

19.4 Convention for Standard States

State of Matter Standard StateGas 1 atm pressure

Liquid Pure liquidSolid Pure solidElements Gºf = 0Solution 1 molar

73

19.4 Free Energy in Reactions Gº = standard free energy change. Free energy change that will occur if reactants in their

standard state turn to products in their standard state. Can’t be measured directly, can be calculated from

other measurements. The reaction:

aA + bB → cC + dD

Gºrxn = ΣnGº (products) - ΣmGº (reactants)

or Gº=Hº-TSº

Use adapted Hess’s Law with known reactions.

74

19.4 Gibbs Free Energy: Spontaneity and Coupled Reactions

Conversion of rust to iron 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS)

To convert iron to rust, G must be provided from spontaneous reaction 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS) 4Fe + 3O2 2Fe2O3 G = - 1487 kJ (S) 6CO + 3O2 6CO2 G = -1543 kJ

(S)

75

19.4 Gibbs Free Energy: Coupled Reactions

Conversion of rust to iron 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS)

To convert iron to rust, G must be provided from spontaneous rxn 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS) 6CO + 3O2 6CO2 G = -1543 kJ (S) 2Fe2O3 + 6CO 4Fe + 6CO2 G = -56 kJ (S)

Reactions are “coupled”

76

19.4 Gibbs free Energy: Example

For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0?

Look at the equation G=H-TS Spontaneity can be predicted from the sign

of H and S.

(ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C)

77

ΔH ΔS -|TΔS| ΔG = ΔH-TΔS Reaction Characteristics

Example

- + - Always negative

Spontaneous at all temperatures

2O3(g) →3O2(g)

+ - +Always positive

Nonspontaneous at all temperatures

3O2(g) →2O3(g)

- - +

Negative at low T; positive at high T

Spontaneous at low T; nonspontaneous at high T. Enthalpy driven

H2O(l) →H2O(s)+

+ + -

Positive at low T; negative at high T

Nonspontaneous at low T; becomes spontaneous at high T. Entropy driven.

H2O(s) → H2O(l)

19.4 Gibbs Free Energy and SpontaneityEffect of Temperature on the Spontaneity of

Reactions

1

2

3

4

78Free energy change as a function

of temperature.

19.4 Gibbs Free Energy and Spontaneity

79

19.4 Gibbs Free Energy and Temperature

Spontaneous Processes H2O(s) melts above 0C (endothermic)

H > 0 (endo process), S > 0 ENTROPY DRIVEN

Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic) 4Fe(s) + 3O2(g) 2Fe2O3(s) H < 0 (Exo) , S < 0 (entropy decreases) ENTHALPY DRIVEN

G < 0 for each process T determines sign of G: G = H -TS

80

19.4 Gibbs Free Energy: Example (p. 748)

9. Predict which of the four cases in Table (slide #78) you expect to apply to the following reactions:a) C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(g) ΔH = -2540 kJ

b) Cl2(g) → 2Cl(g)

10. Calculate ΔG0 at 298 K for the reaction 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) ΔH0 = 114.4 kJ

a) using Gibbs free Energy equationb) from standard free energies of formation.

( -76.0 kJ)

81

19.4 Gibbs free Energy: Example

11. For the reaction SO2(g) + 2H2S(g) →3S(s) + 2H2O(g)

Calculate the temperature at which ΔG0 = 0Values of ΔH0, kJ/mol ΔS0 (kJ/(K mol) SO2 -296.8 0.2481H2S -20.6 0.2057S 0.0 0.0318H2O -241.8 +0.1887

Answer: 780K

82

19.4 Temperature and Chemical Reactions – Problem Set

14. Calculate the temperature at which the reaction: CaCO3(s) → CaO(s) + CO2(g) becomes spontaneous.ΔS º = 160.5 J/K; ΔH º = 177.8 kJ

Answer: 835 ºC

83

19.4 Phase Transitions: Temperature and Chemical Reactions

15. At its normal boiling point, the enthalpy of vaporization of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What should be its approximate normal boiling point temperature be? ΔS0 of vaporization is 87 J mol-1 K-1

(570 K)

84

19.5 Free Energy and Equilibrium

ΔG and ΔGº are not the same. ΔG = ΔH – T ΔS

ΔGº = ΔH º – T ΔS º ΔGº is only at standard conditions

(values obtained from Tables)ΔGº = Gº(products) - ΔGº

(reactants)

ΔG any conditions (no Tables of values available)

85

19.5 Free Energy and Equilibrium

Predicament: we start a reaction with all reactants in standard state (1 atm, 25ºC, 1M solution). Is the standard state preserved as the reaction progresses?

It can be shown mathematically that at non-standard conditions:

ΔG = ΔGº +RTlnQ

]tan[[products] quotient reaction ,

tsreacQ

86

The total free energy of a reaction mixture as a function of the progress of the reaction. Beginning with either pure reactants or pure products, the free energy decreases (ΔG is negative) as the system moves toward equilibrium. The graph is drawn assuming that the pure reactants and pure products are in their standard states and that Δ G° for the reaction is negative so the equilibrium composition is rich in products.

ΔGrxn < 0

ΔGrxn > 0

87

Spontaneous Spontaneous Reaction Reaction Products Products favoredfavored

equilibrium

position

Pure reactants

Pure products

extent of reaction

Greactants

Gproducts

Grxn < 0

89

19.5 Free Energy and the Equilibrium Constant

At equilibrium:

eq

eq

KRTG

KRTG

ln

ln0

From the above we can conclude:If G < 0, then K > 1 Product favoredIf G = 0, then K = 1 EquilibriumIf G > 0, then K < 1 Reactants favored

K = e-(ΔGº /RT)

Kc used for solutions and molarities, Kp used for gases

Useful equation to determine small

Keq

90

19.5 Temperature Dependence of K

Gº= -RTlnK = Hº - TSº

ln(K) = Hº/R(1/T)+ Sº/R

A straight line of lnK vs 1/T Slope?

Y-intercept?

19.5 Free Energy and Chemical Equilibrium

91

19.5 Reaction Path and ΔGº

ΔGº = -RTln K

Value of ΔGº Sign of K Path of reaction

Negative, <0 K>0 Products are favored

Positive, >0 K<0 Reactants are favored

zero, =0 K = 1 Equilibrium

92

19.5 Keq: Problem Set – 19.6 (p. 755)

16. Determine the value of Keq at 25 ºC for the reaction:2NO2(g) ↔ N2O4(g).

Note: calculate ΔG0 from tables Use ΔG0 = -RTln Kp

( 6.9)

93

19.5 Keq: Problem Set 19.7 (p. 755)

17. Using the solubility product of of silver iodide at 25 ºC (8.5 x 10-17), calculate ΔGº for the process:AgI(s) ↔ Ag+(aq) + I-(aq)

Answer:18. Estimate the value of ΔS0

298 for the dissociation of copper (II) oxide.CuO(s) ↔ 2Cu2O(s) + O2(g) ΔH0

298 = 283 kJ

( 0.203 kJ K-1

94

19.5 Gibbs and Equilibrium: Example

19. Calculate Grxn for the reaction below:2A(aq) + B(aq) C(aq) + D(g)

if G°rxn = 9.9 x 103 J/mol and (a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05 M, and PD = 0.05

atm, and (b) (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5 M, and PD = 0.5

atm. (c) Is the reaction spontaneous under these conditions?

(ans.: a) –2121 J, spontaneous; b) 17875 J, nonspontaneous)

95

Example 5

20. Calculate G°rxn for the ionization of acetic

acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is

this reaction spontaneous under standard

state conditions? (ans.: 27 kJ)


96

21. Calculate G° for the neutralization of a

strong acid with a strong base at 25°C. Isthis process spontaneous under theseconditions?For the reaction below, K = 1.0 x 1014.

H+ + OH- H2O

(ans.: -80 kJ)


97

22. Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored?

(ans.: a) 0.15; b) 216; 250C)


98

19.5 Summary

First Law of Thermodynamics: ΔU = Δq + Δw

Second Law: ΔSuniv = ΔS sys + ΔSsurr

ΔS sys= ΔH/TΔG = ΔH – T ΔSΔG = ΔGº +RTlnK

ΔGº = -RTlnK

99

19.5 Gibbs Free Energy Many biological reactions essential for life

are nonspontaneous Spontaneous reactions used to “drive”

the nonspontaneous biological reactions Example: photosynthesis

6CO2 + 6H2O C6H12O6 + 6O2 G > 0

What spontaneous reactions drive photosynthesis?

100

19.5 Entropy and Life Processes

If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible? growth of a complex life form

represents an increase in order (less randomness) lower entropy

101

19.5 Entropy and Life Processes

CO2H2Oheat

Organisms “pay” for their increased order by increasing Ssurr.

Over lifetime, Suniv > 0.

Chapter 19

Documents

Transcript of Chapter 19