- 81 Views
- Uploaded on
- Presentation posted in: General

Chapter 19: Thermodynamics

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 19: Thermodynamics

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

- 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

- 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

- HeatqWorkw
- E = q + w
- Conventions of heat and work

- 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

- Electrical
- Pressure-volume orPV
- w = –PV
- Where P = external pressure

- If PV only work in chemical system, then

- w = –PV

- Reaction done at constantV
- V = 0
- PV = 0, so
- E = qV

- Entire energy change due to heat absorbed or lost
- Rarely done, not too useful

- 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 pressure

- H E
- Differ byH– E = PV
- Only differ significantly when gases formed or consumed
- Assume gases are ideal
- Since P and T are constant

- 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

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

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 molT = 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

Step 4: Calculate percent difference

Bigger than most, but still small

Note: Assumes that volumes of solids and liquids are negligible

VsolidVliquid << Vgas

- 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 justified
Note: If no gases are present reduces to

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

- Calculate Η°, Ε°

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)

ΔH° = –13,898.9 kJ

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

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

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 + nRTE = 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

- 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

- 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

- 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

- You consume food, spontaneous biochemical reactions occur to supply muscle power

- 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

- Some endothermicreactions 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

We can expect the combustion of propane

to be:

A. spontaneous

B. non-spontaneous

C. neither

- 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

- 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

- If Energy = money
- Entropy (S) describes number of different ways of counting it

- 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

- Things get rusty spontaneously
- 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

- State function
- Independent of path
- S = Change in entropy
- For chemical reactions or physical processes

- For gases, entropy increases as volume increases
- Gas separated from vacuum by partition
- Partition removed, more ways to distribute energy
- Gas expands to achieve more probable particle distribution
- More random, higher probability, more positive S

- As Tincreases, entropy increases
(a) T = 0 K, particles in equilibrium lattice positions and S relatively low

(b) T> 0 K, molecules vibrate, Sincreases

(c) Tincreases further, more violent vibrations occur and Shigher than in (b)

- 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

- 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

Which represents an increase in entropy?

A. Water vapor condensing to liquid

B. Carbon dioxide subliming

C. Liquefying helium gas

D. Proteins forming from amino acids

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

Reactions involving gases

- Calculate change in number of moles of gas, ngas
- If ngas is positive , S is positive
- ngas is more important than nmolecules

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

nreactant = 4nproduct = 2

n = 2 – 4 = –2

Predict Srxn < 0

Lower positional probability

Higher positional probability

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

- ngas = 1 mol – 0 mol = 1 mol
- sincengas is positive, S is positive
2 N2O5(g) 4 NO2(g) + O2(g)

- ngas = 4 mol + 1 mol – 2 mol = 3 mol
- sincengas is positive, S is positive
OH–(aq) + H+(aq) H2O

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

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)

CO2(s) → CO2(g)

positive

H2O(g) → H2O(l)

negative

positive

positive

negative

Which of the following has the most entropy at standard conditions?

- H2O(l)
- NaCl(aq)
- AlCl3(s)
- Can’t tell from the information

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)

- Sometimes they work together
- Building collapses
- PE decreases His 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

- 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

- 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

Spontaneous Reactions (cont.)

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

- If system at constant P, then
- qsystem = H

- So
- qsurroundings = –Hsystem

- and

Multiplying both sides by T we get

TStotal = TSsystem – Hsystem

or

TStotal = – (Hsystem– TSsystem)

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

(Hsystem– TSsystem) < 0

must be negative for reaction to be spontaneous

- TStotal > 0 (entropy must increase)

- 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
GH – TS

- Maximum energy in reaction that is "free" or available to do useful work
- At constant P and T, changes in free energy G = H – TS

G = H – TS

G state function

Made up of T, H and S = state functions

Has units of energy

Extensive property

G = Gfinal– Ginitial

- At constant P and T, process spontaneous only if it is accompanied by decrease in free energy of system

- When H and S have same sign, T determines whether spontaneous or nonspontaneous
- Temperature-controlledreactions are spontaneous at one temperature and not at another

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 K

B. 1200 K

C. 2448 K

D. The reaction cannot become non-spontaneous at any temperature

- 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)

- All substances have positive entropies as they are more disordered than at 0 K
- Heating increases randomness
- S° is biggest for gases—most disordered

- 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

Calculate S° for the following:

CO2(s)→CO2(g)

S°: 187.6213.7 J/molK

S° = (213.7 – 187.6) J/molK

S° = 26.1 J/molK

CaCO3(s)→ CO2(g) + CaO(s)

S°: 92.9 213.7 40 J/molK

S° = (213.7 +40 – 92.9) J/molK

S° = 161 J/molK

Ex. 3.Calculate S° for reduction of aluminum oxide by hydrogen gas

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

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

S° = 179.9 J/K

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

- 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:

Calculate G ° for reduction of aluminum oxide by hydrogen gas

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

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

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

H ° = 944.4 kJ

Step 2:Calculate S° see Example 3

S° = 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 isnot spontaneous

- 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

Calculate G° for reduction of aluminum oxide by hydrogen gas.

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

G° = 0.0 kJ – 685.8 kJ – 0.00 kJ

– (– 1576.4 kJ)

G°= 890.6 kJ

Both methods same within experimental error

- 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 heat

- 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

- 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

CalculateG° 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

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

- 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

- 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 completion

- G° large and positive

- G°=–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°

Ex. 6Given 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

- Gat 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:

- 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

Step 2 Calculate G= G°+ RT lnQ

G= –33.3 kJ/mol + (8.314 J/K mol)(1 kJ/1000J)(298K)(ln(9.3 10–3))

= –33.3 kJ/mol + (2.479 kJ/mol)(ln(9.3 10–3))

= –33.3 kJ + (–11.6 kJ/mol) = – 44.9 kJ/mol

- At standard conditions all gases (N2, H2 and NH3) are at 1 atm of pressure
- Gbecomes more negative when we go to 1.0 atm N2, 3.0 atm H2 and 0.5 atm NH3
- Indicates larger driving force to form NH3
- Preactants > Pproducts

Non-spont. G° > 0

- 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

N2O4(g) 2NO2(g)

Equilibrium occurs here

at Ptotal = 1 atm with

16.6% N2O4 decomposed

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 °

- Use relation G= G° + RT lnQ to derive relationship between K and G°
- At Equilibrium
∆G = 0 andQ = K

- So 0 = G°+ RT lnK
G°=–RT lnK

Taking antilog (ex) of both sides gives

K = e–G°/RT

- 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

- 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

- 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

- Initially

- 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)

- Solid-liquid equilibrium
- Thus H = TS and
- or

CalculateTbp 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

What is the expected melting point for Cu?

Cu(s) Cu(l)

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

H = +341.1kJ

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

S = 133.19 J/K

- Reactions often run at temperatures other that 298 K
- Position of equilibrium can change as G° depends on temperature
G° = H° – TS°

- For temperatures near 298 K, expect only very small changes in H° and S°
- For reaction at temperature,we can write:

- 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 strategy
Step 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

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

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°
- H° = –92.38 kJ
- S° = –198.4 J/K

- Since both H° and S° are negative
- At lowtemperature
- G°will be negative and spontaneous

- At high temperature
- TS°will become a bigger positive number and
- G°will become more positiveand thuseventually, at high enough temperature, will become nonspontaneous

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 ex to obtain K

Large K indicates NH3 favored at room temp.

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

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

K =1.85 1011

- 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°

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

ln K = + 14.37 – 23.86 = –9.49

K = e–9.49 = 7.56 × 10–5

- 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

- Can be determined spectroscopically for simple diatomic molecules
- H2, O2, Cl2

- More complex molecules, calculate using thermochemical data and Hess’s Law
- UseH°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

- Ex. CH4(g) C(g) + 4H(g)
- Hatom = energy needed to break all bonds in molecule
- Hatom /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

- D = bond dissociation energy

- 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

- 2H2(g) 4H(g) H°1 = 4H°f (H,g)
- C(s) C(g)H°2 = H°f (C,g)
- 4H(g) +C(g) CH4(g) H°3 = –H°atom
2H2(g) +C(s) CH4(g) H° = H°f(CH4,g)

H°f(CH4,g) = 4H°f(H,g) + H°f(C,g) – Hatom

- Rearranging gives
Hatom= 4H°f(H,g) + H°f(C,g) – H°f(CH4,g)

- Look these up in Table 18.3, 6.2 or appendix C
Hatom= 4(217.9kJ/mol) + 716.7kJ/mol – (–74.8kJ/mol)

Hatom= 1663.1 kJ/mol of CH4

= 415.8 kJ/mol of C—H bonds

- 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

H°f(CH3OH,g) =

H˚f (C,g) + 4H˚f (H,g) + H˚f (O,g) – Hatom (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
- Hatom (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