chapter 19: thermodynamics chemistry: the molecular nature of matter, 6e jespersen/brady/hyslop

Chapter 19: Thermodynamics

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Thermodynamics Study of energy changes and flow of energy Answers several fundamental questions:

Is it possible for a given reaction to occur? Will the reaction occur spontaneously (without

outside interference) at a given T ? Will reaction release or absorb heat?

Tells us nothing about time frame of reaction Kinetics

Two major considerations Enthalpy changes, H (heats of reaction)

Heat exchange between system and surroundings

Nature's trend to randomness or disorder Entropy

2


Review of First Law of Thermodynamics

Internal energy, E System's total energy Sum of KE and PE of all particles in system

or for chemical reaction

E + energy into system E – energy out of system

3


Two Methods of Energy Exchange Between System and

Surroundings Heat q Work w E = q + w Conventions of heat and work

q + Heat absorbed by system

Esystem increases

q – Heat released by system

Esystem decreases

w + Work done on system Esystem increases

w – Work done by system Esystem decreases

4


First Law of Thermodynamics Energy can neither be created nor destroyed It can only be converted from one form to

another Kinetic Potential Chemical Electrical Electrical Mechanical

E is a state function E is a change in a state function Path independent E = q + w

5


Work in Chemical Systems

1. Electrical2. Pressure-volume or PV

w = –PV Where P = external pressure

If PV only work in chemical system, then

6


Heat at Constant Volume Reaction done at constant V

V = 0 PV = 0, so E = qV

Entire energy change due to heat absorbed or lost

Rarely done, not too useful

7


Heat at Constant Pressure More common Reactions open to atmosphere

Constant P Enthalpy

H = E + PV Enthalpy change

H = E + PV Substituting in first law for E gives

H = (q – PV) + PV = qP H = qP

Heat of reaction at constant pressure8


Converting Between E and H For Chemical Reactions

H E Differ by H – E = PV Only differ significantly when gases

formed or consumed Assume gases are ideal

Since P and T are constant

9


Converting Between E and H For Chemical Reactions

When reaction occurs V caused by n of gas

Not all reactants and products are gases So redefine as ngas

Where ngas = (ngas)products – (ngas)reactants

Substituting into H = E + PV gives

or

10


Ex. 1 What is the difference between H and E for the following reaction at 25 °C?

2 N2O5(g) 4 NO2(g) + O2(g)

What is the % difference between H and E ?Step 1: Calculate H using data (Table 7.2) Recall

H ° = (4 mol)(33.8 kJ/mol) + (1 mol)(0.0 kJ/mol) – (2 mol)(11 kJ/mol)

H ° = 113 kJ 11


Ex. 1. H and E (cont.)Step 2: Calculate ngas

ngas = (ngas)products – (ngas)reactants

ngas = (4 + 1 – 2) mol = 3 mol

Step 3: Calculate E using R = 8.31451 J/K mol T = 298 K E = 113 kJ –

(3 mol)(8.314 J/K mol)(298 K)(1 kJ/1000 J)E = 113 kJ – 7.43 kJ = 106 kJ

12


Ex. 1. H and E (cont.)

Step 4: Calculate percent difference

Bigger than most, but still small

Note: Assumes that volumes of solids and liquids are negligible

Vsolid Vliquid << Vgas

13


Is Assumption that Vsolid Vliquid << Vgas Justified?

Consider CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)

37.0 mL 2×18.0 mL 18 mL 18 mL 24.4 L

Volumes assuming each coefficient equal number of moles

So V = Vprod – Vreac = 24.363 L 24.4 L Yes, assumption is justifiedNote: If no gases are present reduces

to 14


Learning Check Consider the following reaction for picric acid:8O2(g) + 2C6H2(NO2)3OH(l ) → 3 N2(g) + 12CO2(g) + 6H2O(l )

Calculate Η °, Ε °

8O2(g) + 2C6H2(NO2)3OH(l ) → 3 N2(g) + 12CO2(g) + 6H2O(l )

Η °f

(kJ/mol) 0.00 3862.94 0.00 –393.5 –241.83

Ε ° = H ° – ngasRT = H ° – (15 – 8) mol × 298 K × 8.314 × 10–3 kJ/(mol K)

H ° = 12 mol(–393.5 kJ/mol) + 6 mol(–241.83 kJ/mol) + 6 mol(0.00 kJ/mol) – 8 mol(0.00 kJ/mol) – 2 mol(3862.94 kJ/mol)

15

ΔH ° = –13,898.9 kJ

Ε ° = –13,898.9 kJ – 29.0 kJ = –13,927.9 kJ


Your Turn!Given the following:3H2(g) + N2(g) → 2NH3(g) Η °= –46.19 kJ

mol–1 Determine E for the reaction.A. –51.14 kJ mol–1

B. –41.23 kJ mol–1

C. –46.19 kJ mol–1

D. –46.60 kJ mol–1

Η = E + nRT E = Η – nRT E = –46.19 kJ mol –

(–2 mol)(8.314 J K–1mol–1)(298 K)(1 kJ/1000 J) E = –51.14 kJ mol–1

16


Enthalpy Changes and Spontaneity What are relationships among factors

that influence spontaneity? Spontaneous Change

Occurs by itself Without outside

assistance until finished e.g.

Water flowing over waterfall Melting of ice cubes in glass

on warm day

17


Nonspontaneous Change Occurs only with outside assistance Never occurs by itself:

Room gets straightened up Pile of bricks turns into a brick wall Decomposition of H2O by electrolysis

Continues only as long as outside assistance occurs: Person does work to clean up room Bricklayer layers mortar and bricks Electric current passed through H2O

18


Nonspontaneous Change Occur only when

accompanied by some spontaneous change You consume food,

spontaneous biochemical reactions occur to supply muscle power to tidy up room or to build wall

Spontaneous mechanical or chemical change to generate electricity

19


Direction of Spontaneous Change Many reactions which occur

spontaneously are exothermic: Iron rusting Fuel burning

H and E are negative Heat given off Energy leaving system

Thus, H is one factor that influences spontaneity

20


Direction of Spontaneous Change Some endothermic reactions occur

spontaneously: Ice melting Evaporation of water Expansion of CO2 gas into vacuum

H and E are positive Heat absorbed Energy entering system

Clearly other factors influence spontaneity

21


Your Turn!We can expect the combustion of

propane to be:A. spontaneousB. non-spontaneousC. neither

22


Entropy (Symbol S ) Thermodynamic quantity Describes number of equivalent ways

that energy can be distributed Quantity that describes randomness of

system Greater statistical probability of

particular state means greater the entropy! Larger S, means more possible ways to

distribute energy and that it is a more probable result

23


Fig. 19.6 - Entropy Distribution

24

Low Entropy – (a) A absorbs E in units of

10 Few ways to distribute

E ● represent E ’s of

molecules of A

High Entropy – (b) More ways to distribute E B absorbs E in units of 5 ●represent E ’s of

molecules of B


Entropy

If Energy = money Entropy (S ) describes number of different

ways of counting it25


Examples of Spontaneity Spontaneous reactions

Things get rusty spontaneously Don't get shiny again

Sugar dissolves in coffee Stir more—it doesn't undissolve

Ice liquid water at RT Opposite does NOT occur

Fire burns wood, smoke goes up chimney Can't regenerate wood

Common factor in all of these: Increase in randomness and disorder of system Something that brings about randomness more

likely to occur than something that brings order

26


Entropy, S State function Independent of path S = Change in entropy

For chemical reactions or physical processes

27


Effect of Volume on Entropy

For gases, entropy increases as volume increasesa) Gas separated from vacuum by partition b) Partition removed, more ways to distribute energy c) Gas expands to achieve more probable particle

distribution More random, higher probability, more positive S

28


Effect of Temperature on Entropy As T increases, entropy increases

(a) T = 0 K, particles in equilibrium lattice positions and S relatively low (b) T > 0 K, molecules vibrate, S increases(c) T increases further, more violent vibrations occur and S higher than in (b)

29


Effect of Physical State on Entropy Crystalline solid very low entropy

Liquid higher entropy, molecules can move freely More ways to distribute KE among them

Gas highest entropy, particles randomly distributed throughout container Many, many ways to distribute KE

30


Entropy Affected by Number of Particles Adding particles to system

Increase number of ways energy can be distributed in system

So all other things being equal Reaction that produces more particles will

have positive S

31


Your Turn!Which represents an increase in entropy?A. Water vapor condensing to liquidB. Carbon dioxide sublimingC. Liquefying helium gasD. Proteins forming from amino acids

32


Entropy Changes in Chemical Reactions

Reactions without gases Calculate number of mole molecules

n = nproducts – nreactants

If n is positive, entropy increases More molecules, means more disorder Usually the side with more molecules, has less

complex molecules

33

Reactions involving gases Calculate change in number of moles of

gas, ngas

If ngas is positive , S is positivengas is more important than nmolecules


Entropy Changes in Chemical ReactionsEx. N2(g) + 3H2(g) 2NH3(g)

nreactant = 4 nproduct = 2

n = 2 – 4 = –2Predict Srxn < 0

34

Higher positional probability

Lower positional probability


Ex. 2 Predict Sign of S for Following Reactions

CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)

ngas = 1 mol – 0 mol = 1 mol

since ngas is positive, S is positive

2 N2O5(g) 4 NO2(g) + O2(g)

ngas = 4 mol + 1 mol – 2 mol = 3 mol

since ngas is positive, S is positive

OH–(aq) + H+(aq) H2O

ngas = 0 mol n = 1 mol – 2 mol = –1 mol since ngas is negative, S is negative

35


Predict Sign of S in the Following:Dry ice → carbon dioxide gas

Moisture condenses on a cool window

AB → A + B

A drop of food coloring added to a glass of water disperses

2Al(s) + 3Br2(l ) → 2AlBr3(s)

positive

positive

positive

negative

CO2(s) → CO2(g)

H2O(g) → H2O(l )

negative36


Your Turn!Which of the following has the most entropy at standard conditions?A. H2O(l )

B. NaCl(aq)

C. AlCl3(s)

D. Can’t tell from the information

37


Your Turn!Which reaction would have a negative

entropy?A. Ag+(aq) + Cl–(aq) → AgCl(s)B. N2O4(g) → 2NO2(g)

C. C8H18(l ) + 25/2 O2(g) → 8CO2(g) + 9H2O(g)

D. CaCO3(s) → CaO(s) + CO2(g)

38


Both Entropy and Enthalpy Affect Reaction Spontaneity

Sometimes they work together Building collapses PE decreases H is negative Stones disordered S is positive

Sometimes work against each other Ice melting (ice/water mix) Endothermic

H is positive nonspontaneous Increase in disorder of molecules

S is positive spontaneous

39


Which Prevails? Hard to tell—depends on temperature!

At 25 °C, ice melts At –25 °C, water freezes

So three factors affect spontaneity: H S T Next few slides will develop the

relationship between H, S, and T that defines a spontaneous process

40


Second Law of Thermodynamics When a spontaneous event occurs, total

entropy of universe increases (Stotal > 0)

In a spontaneous process, Ssystem can decrease as long as total entropy of universe increases Stotal = Ssystem + Ssurroundings

It can be shown that

41


Law of Conservation of Energy

Says q lost by system must be gained by surroundings qsurroundings = –qsystem

If system at constant P, then qsystem = H

So qsurroundings = –Hsystem

and

42

Spontaneous Reactions (cont.)


Thus Entropy for Entire Universe is

Multiplying both sides by T we getTStotal = TSsystem – Hsystem

orTStotal = – (Hsystem – TSsystem)

For reaction to be spontaneous TStotal > 0 (entropy must increase)

So, (Hsystem – TSsystem) < 0 must be negative for reaction to be

spontaneous 43


Gibbs Free Energy Would like one quantity that includes all

three factors that affect spontaneity of a reaction

Define new state function, G Gibbs Free Energy

Maximum energy in reaction that is "free" or available to do useful work

G H – TS At constant P and T, changes in free

energy G = H – TS

44


G = H – TS

G state function Made up of T, H and S = state

functions Has units of energy Extensive property

G = Gfinal – Ginitial

Gibbs Free Energy

G < 0 Spontaneous processG = 0 At equilibriumG > 0 Nonspontaneous

45


Criteria for Spontaneity? At constant P and T, process spontaneous

only if it is accompanied by decrease in free energy of system

H

S Spontaneous?

– + G = (–) – [T (+)] = –

Always, regardless of T

+ – G = (+) – [T (–)] = +

Never, regardless of T

+ + G = (+) – [T (+)] = ?

Depends; spontaneous

at high T, –G

– – G = (–) – [T (–)] = ? Depends; spontaneous

at low T, –G

46


Summary When H and S have

same sign, T determines whether spontaneous or nonspontaneous

Temperature-controlled reactions are spontaneous at one temperature and not at another

47


Your Turn!At what temperature (K) will a reaction become nonspontaneous when H = –50.2 kJ mol–1 and S = +20.5 J K–1 mol–1?A. 298 KB. 1200 KC. 2448 KD. The reaction cannot become non-spontaneous at any temperature

48


Third Law of Thermodynamics At absolute zero (0 K)

Entropy of perfectly ordered, pure crystalline substance is zero

S = 0 at T = 0 K Since S = 0 at T = 0 K

Define absolute entropy of substance at higher temperatures

Standard entropy, S ° Entropy of 1 mole of substance at 298 K

(25 °C) and 1 atm pressure S ° = S for warming substance from 0 K to

298 K (25 °C)49


Consequences of Third Law1. All substances have positive entropies as

they are more disordered than at 0 K Heating increases randomness S ° is biggest for gases—most disordered

2. For elements in their standard states S ° 0 (but Hf° = 0)

Units of S ° J/(mol K)Standard Entropy Change

To calculate S ° for reaction, do Hess's Law type calculation

Use S ° rather than entropies of formation

50


Learning Check

Calculate S ° for the following: CO2(s) → CO2(g)

S °: 187.6 213.7 J/mol KS ° = (213.7 – 187.6) J/mol KS ° = 26.1 J/mol K

CaCO3(s) → CO2(g) + CaO(s)

S °: 92.9 213.7 40 J/mol KS ° = (213.7 +40 – 92.9) J/mol KS ° = 161 J/mol K

51


Ex. 3. Calculate S° for reduction of aluminum oxide by hydrogen gasAl2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

Substance S ° (J/ K mol)

Al(s) 28.3

Al2O3(s) 51.00

H2(g) 130.6

H2O(g) 188.7

52


Ex. 3 Calculate ΔS° (cont.)

S ° = 56.5 J/K + 566.1 J/K – (51.00 J/K + 391.8 )

S ° = 179.9 J/K

53


Your Turn!What is the entropy change for the

following reaction?

Ag+(aq) + Cl–(aq) → AgCl(s)So 72.68 56.5 96.2 J K–1 mol–1

A. +32.88 J K–1 mol–1

B. –32.88 J K–1 mol–1

C. –32.88 J mol–1

D. +112.38 J K–1 mol–1 S ° = [96.2 – (72.68 + 56.5)] J K–1 mol–1 S ° = –32.88 J K–1 mol–1

54


Standard Free Energy Changes Standard Free Energy Change, G °

G measured at 25 °C (298 K) and 1 atm Two ways to calculate, depending on

what data is available Method 1: G ° = H ° – TS ° Method 2:

55


Ex. 4. Calculate G ° Method 1

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

Step 1: Calculate H° for reaction using heats of formation below

56

Substance (kJ/mol)

Al(s) 0.0

Al2O3(s) –1669.8

H2(g) 0.0

H2O(g) –241.8

Calculate G ° for reduction of aluminum oxide by hydrogen gas



H ° = 0.0 kJ – 725.4 kJ – 0.00 kJ – (– 1669.8 kJ) H ° = 944.4 kJ

57



Step 2: Calculate S ° see Example 3S ° = 179.9 J/K

Step 3: Calculate G ° = H ° – (298.15 K)S °

G ° = 944.4 kJ – (298 K)(179.9 J/K)(1 kJ/1000 J)G ° = 944.4 kJ – 53.6 kJ = 890.8 kJ

G ° is positive Indicates that the reaction is not

spontaneous

58


Ex. 4. Calculate G ° Method 2 Use Standard Free Energies of

Formation

Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 °C

59


Ex. 4. Calculate G ° Method 2Calculate G ° for reduction of aluminum

oxide by hydrogen gas. Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

60

Substance (kJ/mol)

Al(s) 0.0

Al2O3(s) –1576.4

H2(g) 0.0

H2O(g) –228.6



61

G° = 0.0 kJ – 685.8 kJ – 0.00 kJ – (– 1576.4 kJ) G° = 890.6 kJ

Both methods same within experimental error


Spontaneous Reactions Produce Useful Work

Fuels burned in engines to power cars or heavy machinery

Chemical reactions in batteries Start cars Run cellular phones, laptop computers, mp3 players

Energy not harnessed if reaction run in an open dish All energy lost as heat to surroundings

Engineers seek to capture energy to do work Maximize efficiency with which chemical energy is

converted to work Minimize amount of energy transformed to unproductive

heat62


Thermodynamically Reversible Process that can be reversed and is always

very close to equilibrium Change in quantities is infinitesimally small

Example - expansion of gas Done reversibly, it does most work on surroundings

63


G ° = Maximum Possible Work G ° is maximum amount of energy

produced during a reaction that can theoretically be harnessed as work Amount of work if reaction done under

reversible conditions Energy that need not be lost to

surroundings as heat Energy that is “free” or available to do work

64


Ex. 5 Calculate G ° Calculate G ° for reaction below at 1 atm and 25 °C, given H ° = –246.1 kJ/mol, S

° = 377.1 J/(mol K). H2C2O4(s) + ½O2(g) → 2CO2(g) + H2O(l )

G °25 = H – TS

G ° =(–246.1 – 112.4) kJ/mol G ° = –358.5 kJ/mol

65


Your Turn!Calculate G ° for the following reaction,

H2O2(l ) → H2O(l ) + O2(g)

given H °= –196.8 kJ mol–1 and S °= +125.72 J K–1

mol–1. A. –234.3 kJ mol–1

B. +234.3 kJ mol–1

C. 199.9 kJ mol–1

D. 3.7 × 105 kJ mol–1

G ° = –196.8 kJ mol–1 – 298 K (0.12572 kJ K–1 mol-1)

G ° = –234.3 kJ mol 66


G° and Position of Equilibrium When G ° > 0 (positive)

Position of equilibrium lies close to reactants Little reaction occurs by the time equilibrium is

reached Reaction appears nonspontaneous

When G ° < 0 (negative) Position of equilibrium lies close to products Mainly products exist by the time equilibrium is

reached Reaction appears spontaneous

67


G ° and Position of Equilibrium When G ° = 0

Position of equilibrium lies ~ halfway between products and reactants

Significant amount of both reactants and products present at time equilibrium is reached

Reaction appears spontaneous, whether start with reactants or products

Can Use G ° to Determine Reaction Outcome G ° large and positive

No observable reaction occurs G ° large and negative

Reaction goes to completion68


At EquilibriumG ° = –RT lnK and K = e–G

°/RT

Provides connection between G ° and K Can estimate K at various temperatures if G ° is

known Can get G ° if K is known

Relationship between K and G ° Keq G ° Reaction

> 1 – SpontaneousFavored

Energy released

< 1 +non-

spontaneousUnfavorable

Energy needed

= 1 0 At Equilibrium69


Learning Check

Ex. 6 Given that H ° = –97.6 kJ/mol, S ° = –122 J/(mol K), at 1 atm and 298 K, will the following reaction occur spontaneously?

MgO(s) + 2HCl(g) → H2O(l ) + MgCl2(s)

G ° = H ° – TS ° = –97.6 kJ/mol – 298 K(–0.122 kJ/mol K)G ° = –97.6 kJ/mol + 36.4 kJ/mol = –61.2 kJ/mol

70


Effect of Change in Pressure or Concentration on ∆G

G at nonstandard conditions is related to G ° at standard conditions by an expression that includes reaction quotient Q

This important expression allows for any concentration or pressure

Recall:

71


Ex. 7 Calculating G at Nonstandard Conditions

Calculate G at 298 K for the Haber process N2(g) + 3H2(g) 2NH3(g) G ° = –33.3 kJ

For a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2 and 0.5 atm NH3

Step 1 Calculate Q

72


Ex. 7 Calculating G at Nonstandard conditions

73


Free Energy Diagrams G is different than

G° G interpreted as the

slope of the free energy curve and tells which direction reaction proceeds to equilibrium

Minimum, at G = 0, indicates composition of equilibrium mixture

Because G ° is positive equilibrium lies closer to reactants

74

Equilibrium occurs here at Ptotal = 1 atm with

16.6% N2O4 decomposed

Non-spont. G ° > 0

N2O4(g) 2NO2(g)


Free Energy Diagrams

75

Spontaneous Rxn G ° < 0 Reaction proceeds toward equilibrium from either A or B where slope, G , is not zero

The curve minimum, where G = 0, lies closer to the product side for spontaneous reactions. This is determined by G °


How K is related to G°

Use relation G = G ° + RT lnQ to derive relationship between K and G °

At Equilibrium∆G = 0 and Q = K

So 0 = G ° + RT lnK

G ° = –RT lnK

Taking antilog (ex) of both sides gives

K = e–G ° /RT

76


Ex. 8 Calculating G ° from K Ksp for AgCl(s) at 25 °C is 1.8 10–10

Determine G ° for the process Ag+(aq) + Cl–(aq) AgCl(s) Reverse of Ksp equation, so

G ° = –RT lnK = –(8.3145 J/K mol)(298 K) × ln(5.6 109)(1

kJ/1000 J) G ° = –56 kJ/mol Negative G ° indicates precipitation will occur

77

910

106.5108.1

11

spKK


System at Equilibrium Neither spontaneous nor nonspontaneous In state of dynamic equilibrium Gproducts= Greactants

G = 0 Consider freezing of water at 0˚C H2O(l ) H2O(s)

System remains at equilibrium as long as no heat added or removed

Both phases can exist together indefinitely Below 0 °C, G < 0 freezing spontaneous Above 0 °C, G > 0 freezing nonspontaneous

78


No Work Done at Equilibrium G = 0 No “free” energy available to do work Consider fully charged battery

Initially All reactants, no products G large and negative Lots of energy available to do work

As battery discharges Reactants converted to products ∆G less negative Less energy available to do work

At equilibrium G = Gproducts – Greactants = 0 No further work can be done Dead battery

79


Phase Change = Equilibrium H2O(l ) H2O(g)

G = 0 = H – TS Only one temperature possible for phase

change at equilibrium Solid-liquid equilibrium

Melting/freezing temperature (point) Liquid-vapor equilibrium

Boiling temperature (point) Thus H = TS and or

80

TH

S

SH

T


Ex. 9 Calculate Tbp

Calculate Tbp for reaction below at 1 atm and 25 °C, given H ° = 31.0 kJ/mol, S ° = 92.9 J/ mol K

Br2(l ) → Br2(g)

For T > 334 K, G < 0 and reaction is spontaneous (S ° dominates) For T < 334 K, G > 0 and reaction is nonspontaneous (H ° dominates) For T = 334 K, G = 0 and T = normal boiling point

81


Learning CheckWhat is the expected melting point for Cu?

Hf° (kJ/mol) Gf° (kJ/mol) S (J/mol K)

Cu(l ) 341.1 301.4 166.29

Cu(s) 0 0 33.1

82

H = 1mol(341.1 kJ/mol – 1mol(0 kJ/mol)

Cu(s) Cu(l )

H = +341.1kJ

S = 133.19 J/KS = 1mol(166.29 J/mol K – 1mol(33.1 J/mol K)


Effect of Temperature on G ° Reactions often run at temperatures

other that 298 K Position of equilibrium can change as

G ° depends on temperatureG ° = H ° – TS °

For temperatures near 298 K, expect only very small changes in H ° and S °

For reaction at temperature, we can write:

83


Ex. 10 Determining Effect of Temperature on Spontaneity

Calculate G ° at 25 °C and 500 °C for the Haber process

N2(g) + 3H2(g) 2NH3(g)

Assume that H ° and S ° do not change with temperature

Solving strategyStep 1. Using data in Tables 6.2 and 18.2

calculate H ° and S ° for the reaction at 25 °C H ° = – 92.38 kJ S ° = – 198.4 J/K 84



Step 2. Calculate G ° for the reaction at 25 °C using H ° and S ° N2(g) + 3H2(g) 2NH3(g)

H ° = –92.38 kJ S ° = –198.4 J/K

G ° = H ° – TS° G ° = –92.38 kJ – (298 K)(–198.4 J/K) G ° = –92.38 kJ + 59.1 kJ = –33.3 kJ So the reaction is spontaneous at 25 °C

85



Step 3. Calculate G ° for the reaction at 500 °C using H ° and S °. T = 500 °C + 273 = 773 K H ° = – 92.38 kJ S ° = – 198.4 J/K

G ° = H ° – TS ° G ° = –92.38 kJ – (773 K)(–198.4 J/K) G ° = –92.38 kJ + 153 kJ = 61 kJ So the reaction is NOT spontaneous at

500 °C 86


Ex. 10 Does this answer make sense? G ° = H ° – TS °

H ° = –92.38 kJ S ° = –198.4 J/K

Since both H ° and S ° are negative At low temperature

G ° will be negative and spontaneous At high temperature

TS ° will become a bigger positive number and

G ° will become more positive and thus eventually, at high enough temperature, will become nonspontaneous 87


Ex. 11 Calculating K from G° Calculate K at 25 °C for the Haber process

N2(g) + 3H2(g) 2NH3(g) G ° = –33.3 kJ/mol = –33,300 J/mol

Step 1 Solve for exponent

Step 2 Take e x to obtain K

Large K indicates NH3 favored at room temp. 88

54.13/ 107 eeK RTG


Your Turn!Calculate the equilibrium constant for the decomposition of hydrogen peroxide at 298 K given G ° = –234.3 kJ mol.A. 8.5 × 10-42

B. 1.0 × 10499

C. 3.4 × 10489

D. 1.17 × 1041

89


Ex. 12 Calculating K from G°, First Calculate G°

Calculate the equilibrium constant at 25 °C for the decarboxylation of liquid pyruvic acid to form gaseous acetaldehyde and CO2.

H3CC

C

O

OH

O

H3CC

O

H+ CO2

90


Ex. 12 First Calculate ∆G° from ∆Gf°

Compound ∆Gf°, kJ/mol

CH3COH –133.30

CH3COCOOH –463.38

CO2 –394.36

91


Ex. 12 Next Calculate Equilibrium Constant

92

K = 1.85 1011


Temperature Dependence of K G ° = –RT lnK = H ° – TS °

Rearranging gives

Equation for line Slope = –H °/RT Intercept = S °/R

Also way to determine K if you know H ° and S °

93


Ex. 12 Calculate K given H ° and S ° Calculate K at 500 °C for Haber process

N2(g) + 3H2(g) 2NH3(g)

Given H ° = –92.38 kJ and S ° = –198.4 J/K

Assume that H ° and S ° do not change with T

ln K = + 14.37 – 23.86 = –9.49 K = e–9.49 = 7.56 × 10–5

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Bond Energy Amount of energy needed to break chemical

bond into electrically neutral fragments Useful to know Within reaction

Bonds of reactants broken New bonds formed as products appear

Bond breaking First step in most reactions One of the factors that determines reaction rate e.g. N2 very unreactive due to strong NN bond

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Bond Energies Can be determined spectroscopically for

simple diatomic molecules H2, O2, Cl2

More complex molecules, calculate using thermochemical data and Hess’s Law Use H °formation enthalpy of formation

Need to define new term Enthalpy of atomization or atomization

energy, Hatom

Energy required to rupture chemical bonds of one mole of gaseous molecules to give gaseous atoms

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Determining Bond Energies Ex. CH4(g) C(g) + 4H(g)

H atom = energy needed to break all bonds in molecule

H atom /4 = average bond C—H dissociation energy in methane D = bond dissociation energy

Average bond energy to required to break all bonds in molecule

How do we calculate this? Use H °f for forming gaseous atoms from elements

in their standard states Hess’s Law

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Determining Bond Energies Path 1: Bottom

Formation of CH4 from its elements = H°f

Path 2: Top 3 step path Step 1: break H—H bonds Step 2: break C—C bonds Step 3: form 4 C—H bonds

1.2H2(g) 4H(g) H °1 = 4H °f (H,g)

2.C(s) C(g) H °2 = H °f (C,g)

3.4H(g) + C(g) CH4(g) H °3 = –H °atom

2H2(g) + C(s) CH4(g) H ° = H °f(CH4,g)

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Calculating H atom and Bond Energy H °f(CH4,g) = 4H °f(H,g) + H °f(C,g) – H atom

Rearranging givesH atom= 4H °f(H,g) + H °f(C,g) – H °f(CH4,g) Look these up in Table 18.3, 6.2 or

appendix CH atom= 4(217.9kJ/mol) + 716.7kJ/mol –

(–74.8kJ/mol)H atom= 1663.1 kJ/mol of CH4

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= 415.8 kJ/mol of C—H bonds


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Table 19.4 Some Bond Energies


Using Bond Energies to Estimate H

˚f Calculate H ˚f for CH3OH(g) (bottom reaction)

Use four step path Step 1: break C—C bonds Step 2: break H—H bonds Step 3: break O—O bond Step 4: form 4 C—H bonds

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Using Bond EnergiesH °f(CH3OH,g) =

H ˚f (C,g) + 4H ˚f (H,g) + H ˚f (O,g) – H atom (CH3OH,g)

H °f(C,g) + 4H ˚f (H,g) + H ˚f (O,g) = [716.7 +(4 × 217.9) + 249.2] kJ = +1837.5 kJ

H atom (CH3OH,g) = 3DC—H + DC—O + DO—H = (3 × 412) + 360 + 463 = 2059 kJ

H °f(CH3OH,g) = +1837.5 kJ – 2059 kJ = –222 kJ

Experimentally find H ˚f (CH3OH,g) = –201 kJ/mol So bond energies give estimate within 10% of actual

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chapter 19: thermodynamics chemistry: the molecular nature of matter, 6e jespersen/brady/hyslop

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