chemistry 102(01) spring 2009

Instructor: Dr. Upali Siriwardanee-mail: [email protected]

Office: CTH 311 Phone 257-4941

Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m.

Test Dates: March 25, April 26, and May 18; Comprehensive Final Exam: May 20,2009 9:30-10:45

am, CTH 328.March 30, 2009 (Test 1): Chapter 13 April 27, 2009 (Test 2): Chapters 14 & 15 May 18, 2009 (Test 3): Chapters 16, 17 & 18

Comprehensive Final Exam: May 20,2009 :Chapters 13, 14, 15, 16, 17 and 18

Chemistry 102(01) spring 2009

Review of Chapter 6. Energy and Chemical Review of Chapter 6. Energy and Chemical ReactionsReactions

6.1 The Nature of Energy

6.2 Conservation of Energy

6.3 Heat Capacity

6.4 Energy and Enthalpy

6.5 Thermochemical Equations

6.6 Enthalpy change for chemical Rections

6.7 Where does the Energy come from?

6.8 Measuring Enthalpy Changes: Calorimetry

6.9 Hess's Law

6.10 Standard Enthalpy of Formation

6.11 Chemical Fuels for Home and Industry

6.12 Food Fuels for Our Bodies

ThermochemistryThermochemistry

Heat changes during chemical reactions

Thermochemical equation. eg.

H2 (g) + O2 (g) ---> 2H2O(l) H =- 256 kJ;

is called the enthalpy of reaction.

if H is + reaction is called endothermic

if H is - reaction is called exothermic

Why is it necessary to divide Why is it necessary to divide Universe into System and Universe into System and

SurroundingSurroundingUniverse = System + Universe = System +

SurroundingSurrounding

system surroundings

universe

Types of SystemsTypes of SystemsIsolated systemIsolated system

no mass or energy exchange

Closed systemClosed systemonly energy exchange

Open systemOpen systemboth mass and energy exchange

Universe = System + SurroundingUniverse = System + Surrounding

Why is it necessary to divide Universe into Why is it necessary to divide Universe into System and SurroundingSystem and Surrounding

What is the internal energy change (What is the internal energy change (U) U) of a system?of a system?

U is associated with changes in atoms, molecules U is associated with changes in atoms, molecules and subatomic particlesand subatomic particles

Etotal = Eke + E pe + U

U = heat (q) + w (work)

U = q + w

U = q -P V; w =- P V

What forms of energy are found in What forms of energy are found in the Universe?the Universe?

mechanical mechanical

thermalthermal

electrical electrical

nuclearnuclear

mass: E = mcmass: E = mc22

others yet to discoverothers yet to discover

What is 1What is 1stst Law of Thermodynamics Law of Thermodynamics

Eenergy is conserved in the UniverseEenergy is conserved in the Universe

All forms of energy are inter-convertible All forms of energy are inter-convertible and conservedand conserved

Energy is neither created nor destroyed.Energy is neither created nor destroyed.

What exactly is What exactly is H?H?

Heat measured at constant pressure qp

Chemical reactions exposed to atmosphere and are held at a constant pressure.

Volume of materials or gases produced can change.

Volume expansion work = -PV

U = qp + w; U = qp -PV

qp = U + PV; w = -PV

H = U + PV; qp = H(enthalpy )

Heat measured at constant volume qv

Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -PV= 0 U = qv + wqv = U + o; w = 0U = qv = U(internal energy )

How do you measure How do you measure U?U?

What is Hess's Law of Summation of What is Hess's Law of Summation of Heat?Heat?

To heat of reaction for new reactions.

Two methodsTwo methods?

1st method: new H is calculated by adding Hs of other reactions.

2nd method2nd method: Where Hf ( H of formation) of reactants and products are used to calculate H of a reaction.

Method 1: Calculate Method 1: Calculate H for the reaction:H for the reaction:

SO2(g) + 1/2 O2(g) + H2O(g) ----> H2SO4(l) H = ?

Other reactions:

SO2(g) ------> S(s) + O2(g) ; H = 297kJ

H2SO4(l)------> H2(g) + S(s) + 2O2(g); H = 814 kJ

H2(g) +1/2O2(g) -----> H2O(g); H = -242 kJ

SO2(g) ------> S(s) + O2(g); H1 = 297 kJ - 1

H2(g) + S(s) + 2O2(g) ------> H2SO4(l); H2 = -814 kJ - 2

H2O(g) ----->H2(g) + 1/2 O2(g) ; H3 = +242 kJ - 3

______________________________________

SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l); H = ?

H = H1+ H2+ H3

H = +297 - 814 + 242

H = -275 kJ

Calculate Calculate H for the reactionH for the reaction

Calculate Heat (Enthalpy) of Calculate Heat (Enthalpy) of Combustion: 2nd methodCombustion: 2nd method

C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; Ho = ?

Hf (C7H16) = -198.8 kJ/mol

Hf (CO2) = -393.5 kJ/mol

Hf (H2O) = -285.9 kJ/mol

Hf O2(g) = 0 (zero)

What method?Ho = n Hf

o products – n Hfo reactants

n = stoichiometric coefficients

2nd method

H = [n ( Hof) Products] - [n (Ho

f) reactants]

H = [ 7(- 393.5 + 8 (- 285.9)] - [-198.8 + 11 (0)]

= [-2754.5 - 2287.2] - [-198.8]

= -5041.7 + 198.8

= -4842.9 kJ = -4843 kJ

Calculate Calculate H for the reactionH for the reaction

Why is Why is HHoof f of elements is of elements is zero?zero?

Hof, Heat formations are for compounds

Note: Hof of elements is zero

Chapter 18. Thermodynamics: Directionality of Chemical Reactions

18.1 Reactant-Favored and Product-Favored Processes

18.2 Probability and Chemical Reactions18.3 Measuring Dispersal or Disorder: Entropy18.4 Calculating Entropy Changes18.5 Entropy and the Second Law of

Thermodynamics18.6 Gibbs Free Energy18.7 Gibbs Free Energy Changes and Equilibrium Constants18.8 Gibbs Free Energy, Maximum Work, and

Energy Resources18.9 Gibbs Free Energy and Biological Systems18.10 Conservation of Gibbs Free Energy18.11 Thermodynamic and Kinetic Stability

Chemical ThermodynamicsChemical Thermodynamics

spontaneous reaction – reaction which proceed without external assistance once started

chemical thermodynamics helps predict which reactions are spontaneous

ThermodynamicsThermodynamicsThermodynamicsThermodynamics

Will the rearrangement of a system decrease its Will the rearrangement of a system decrease its energy? energy?

If yes, system is favored to react — a If yes, system is favored to react — a product-product-favoredfavored system.system.

Most product-favored reactions are Most product-favored reactions are exothermic.exothermic.

Often referred to as Often referred to as spontaneousspontaneous reactions.reactions.

““Spontaneous” does not imply anything about time Spontaneous” does not imply anything about time for reaction to occur. Kinetic factors are more for reaction to occur. Kinetic factors are more important for certain reactions.important for certain reactions.

Thermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsDiamond Diamond graphite graphite

Thermodynamically Thermodynamically product favoredproduct favored

Slow KineticsSlow Kinetics

Paper burns

Thermodynamically Thermodynamically product favoredproduct favored

Fast KineticsFast Kinetics

In this chapter we only In this chapter we only look into look into thermodynamic thermodynamic factorsfactors

Bases on Energy: Product-Bases on Energy: Product-Favored ReactionsFavored Reactions

In general, product-favored reactions are In general, product-favored reactions are exothermicexothermic..

(Negative H) H)

In general, reactant-favored reactions are In general, reactant-favored reactions are endothermicendothermic..

(Positive H)H)

Product-Favored ReactionsProduct-Favored ReactionsBut many spontaneous reactions or processes But many spontaneous reactions or processes

are endothermic or even have are endothermic or even have H = 0.H = 0.

NHNH44NONO33(s) (s) NH NH44NONO33(aq); (aq); H = +H = +

Direction of ReactionDirection of ReactionProduct favored reactions are always a

transformation of a reactants favored reaction.

Product Favored ReactionProduct Favored Reaction

2Na(s) + 2Cl2(g) => 2NaCl(s)

Reactant Favored ReactionReactant Favored Reaction

2NaCl(s) => 2Na(s) + 2Cl2(g)

However, non spontaneous reactions could be made to take However, non spontaneous reactions could be made to take place by coupling with energy source: another reaction or place by coupling with energy source: another reaction or electric currentelectric current

ElectrolysisElectrolysis

2NaCl(l) => 2Na(s) + 2Cl2(g)

Expansion of a Gas

The positional probability is higher when particles are dispersed over a larger volume

Matter tends to expand unless it is restricted

Gas Expansion and Probability

Entropy, SEntropy, SEntropy, SEntropy, SThe thermodynamic property The thermodynamic property

related to randomness is related to randomness is ENTROPY, SENTROPY, S..

Product-favored processes: Product-favored processes: final state is more final state is more DISORDEREDDISORDERED or or RANDOMRANDOM than the original.than the original.

Spontaneity is related to Spontaneity is related to an increase in an increase in randomness.randomness.

Reaction of K Reaction of K with waterwith water

S[H2O(l)] > S[H2O(s)] at 0 C.

Entropies of Solid, Liquidand Gas Phases

S (gases) > S (liquids) > S (solids)S (gases) > S (liquids) > S (solids)

Entropy, SEntropy, SEntropy, SEntropy, S

Entropies of ionic solids depend on Entropies of ionic solids depend on coulombic attractions.coulombic attractions.

SSoo (J/K•mol) (J/K•mol)

MgOMgO 26.926.9

NaFNaF 51.551.5

SSoo (J/K•mol) (J/K•mol)

MgOMgO 26.926.9

NaFNaF 51.551.5

Entropy and Molecular Structure

Entropy and Dissolving

Qualitative Guidelines for Entropy Changes

Entropies of gases higher than liquids higher than solids

Entropies are higher for more complex structures than simpler structures

Entropies of ionic solids are inversely related to the strength of ionic forces

Entropy increases when making solutions of pure solids or pure liquids in a liquid solvent

Entropy decrease when making solutions of gases in a liquid

Entropy of a Solution of a

Gas

Phase TransitionsPhase Transitions

H2O(s) => H2O(l) H > 0; S > 0

H2O(l) => H2O(g) H > 0; S > 0

spontaneous at high temperatures

H2O(l) => H2O(s) H < 0; S < 0

H2O(g) => H2O(l) H < 0; S < 0

spontaneous at low temperatures

Entropy Changes for Phase Entropy Changes for Phase ChangesChanges

For a phase change, For a phase change, SSSYSSYS = q = qSYSSYS/T/T

(q = heat transferred)(q = heat transferred)

Boiling WaterBoiling Water

HH22O (liq) O (liq) H H22O(g)O(g)

H = q = +40,700 J/molH = q = +40,700 J/mol

mol•J/K 109+ = K 373.15

J/mol 40,700 =

T

q = S

Phase TransitionsPhase Transitions

Heat of Fusion

energy associated with phase transition solid-to-liquid or liquid-to-solid

Gfusion = 0 = Hfusion - T Sfusion

0 = Hfusion - T Sfusion

Hfusion = T Sfusion

Heat of Vaporization

energy associated with phase transition gas-to-liquid or liquid-to-gas

Hvaporization = T Svaporization

Qualitative prediction of Qualitative prediction of S of S of Chemical ReactionsChemical Reactions

• Look for (l) or (s) --> (g)

• If all are gases: calculate nn = n (gaseous prod.) - n(gaseous reac.)

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

n = 2 - 4 = -2

If n is - S is negative (decrease in S)

If n is + S is positive (increase in S)

Entropy ChangeEntropy Change

Entropy (Entropy (S) normally increase (+) for the following S) normally increase (+) for the following changes:changes:

i) Solid ---> liquid (melting) ++

ii) Liquid ---> gas ++

iii) Solid ----> gas most +most +

iv) Increase in temperature ++

v) Increasing in pressure(constant volume, and temperature) ++

vi) Increase in volume ++

Predict Predict S!S!

2 C2H6(g) + 7 O2(g)--> 4 CO2(g) + 6H2O(g)

2 CO(g) + O2(g)-->2 CO2(g)

HCl(g) + NH3(g)-->NH4Cl(s)

H2(g) + Br2(l) --> 2 HBr(g)

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq)O(liq)

SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]

SSoo = 2 mol (69.9 J/K•mol) – = 2 mol (69.9 J/K•mol) – [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)][2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]

SSoo = -326.9 J/K = -326.9 J/K

There is a There is a decrease in S decrease in S because 3 mol of gas give because 3 mol of gas give 2 mol of liquid.2 mol of liquid.

Calculating Calculating S for a ReactionS for a Reaction Based on Hess’s Law second method:

SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

Based on Hess’s Law second method:

SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

Third Law of ThermodynamicsThird Law of Thermodynamics

Provides reference point for absolute entropy

Entropy of a perfectly crystalline substance at absolute zero (T= 0 K) is zero.

Unlike H entropy values are positive above temperatures above absolute zero.

Standard Molar Entropy ValuesStandard Molar Entropy Values

Substance So (J/K.mol) Substance So (J/K.mol)

C (diamond) 2.37 HBr (g) 198.59

C (graphite) 5.69 HCl (g) 186.80

CaO (s) 39.75 HF (g) 193.67

CaCO3 (s) 92.9 HI (g) 206.33

C2H2 (g) ` 200.82 H2O (l) 69.91

C2H4 (g) 219.4 H2O (g) 188.72

C2H6 (g) 229.5 NaCl (s) 72.12

CH3OH (l) 127 O2 (g) 205.03

CH3OH (g) 238 SO2 (g) 248.12

CO (g) 197.91 SO3 (g) 256.72

Standard Entropies at 25Standard Entropies at 25ooCC

Entropy & SpontaneityEntropy & Spontaneity

• How can water boil and freeze spontaneously?

• Enthalpy change can not predict spontaneity!

• Some endothermic processes are spontaneous

• Need another thermodynamic property.

Laws of ThermodynamicsLaws of Thermodynamics

FirstFirst : The total energy of the universe is : The total energy of the universe is constantconstant

SecondSecond : The total entropy (S) of the : The total entropy (S) of the universe is always increasinguniverse is always increasing

Third Third : The entropy(S) of a pure, : The entropy(S) of a pure, perfectly formed crystalline substance at perfectly formed crystalline substance at absolute zero is zeroabsolute zero is zero

FirstFirst : The total energy of the universe is : The total energy of the universe is constantconstant

SecondSecond : The total entropy (S) of the : The total entropy (S) of the universe is always increasinguniverse is always increasing

Third Third : The entropy(S) of a pure, : The entropy(S) of a pure, perfectly formed crystalline substance at perfectly formed crystalline substance at absolute zero is zeroabsolute zero is zero

Dissolving NH4NO3 in water—an entropy driven process.

NHNH44NONO33(s) (s) NH NH44NONO33(aq); (aq); H = +H = +

2nd Law of Thermodynamics2nd Law of Thermodynamics

Second Law of ThermodynamicsSecond Law of ThermodynamicsIn the universe the ENTROPY cannot decrease for

any spontaneous process

The entropy of the universe strives for a maximum

in any spontaneous process, the entropy of the universe increases

for product-favored processfor product-favored process

Suniverse = ( Ssys + Ssurr) > 0

Suniv = entropy of the Universe

Ssys = entropy of the System

Ssurr = entropy of the Surrounding

Suniv = Ssys + Ssurr

Entropy of the UniverseEntropy of the Universe

Suniv = Ssys + Ssurr

Suniv Ssys Ssurr

+ + +

+ +(Ssys>Ssurr) -

+ - + (Ssurr>Ssys)

Can calc. that Can calc. that HHoorxnrxn = = HHoo

systemsystem = -571.7 kJ = -571.7 kJCan calc. that Can calc. that HHoorxnrxn = = HHoo

systemsystem = -571.7 kJ = -571.7 kJ


SSoosyssys = -326.9 J/K = -326.9 J/K

Entropy Changes in the Surroundings

T

H- =

T

q = systemsurr

gssurroundin

oS T

H- =

T

q = systemsurr

gssurroundin

oS

K 298.15

J/kJ) kJ)(1000 (-571.7 - = gssurroundin

oS K 298.15

J/kJ) kJ)(1000 (-571.7 - = gssurroundin

oS

= = +1917 J/K+1917 J/K= = +1917 J/K+1917 J/K



SSoosyssys = -326.9 J/K = -326.9 J/K

SSoosurrsurr = +1917 J/K = +1917 J/K

SSoouni uni = +1590. J/K= +1590. J/K

The entropy of the universe is increasing, so The entropy of the universe is increasing, so the reaction is product-favored. the reaction is product-favored.


Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G

SSunivuniv = = SSsurrsurr + + SSsyssys

Multiply through by (-T)Multiply through by (-T)

-T-TSSunivuniv = = HHsyssys - T - TSSsyssys

-T-TSSunivuniv = = GGsystemsystem

Under standard conditions —Under standard conditions —

GGoo = = HHoo - T - TSSoo

Suniv = Hsys

T + Ssys


GGoo = = HHoo - T - T SSoo

Gibbs free energy change = Gibbs free energy change = difference between the enthalpy of a system and

the product of its absolute temperature and entropy

predictor of spontaneity

Total energy change for system -Total energy change for system -energy lost in disordering the systemenergy lost in disordering the system

Predict the spontaneity of the Predict the spontaneity of the following processes from following processes from H and H and S S

at various temperatures.at various temperatures.

a)H = 30 kJ, S = 6 kJ, T = 300 Kb)H = 15 kJ,S = -45 kJ,T = 200 K

a)H = 30 kJ S = 6 kJ T = 300 K G = Hsys-TSsys or G = H - TS.

H = 30 kJS = 6 kJ T = 300 K

G = 30 kJ - (300 x 6 kJ) = 30 -1800 kJ

G = -1770 kJ

b) H = 15 kJ S = -45 kJ T = 200 KG = Hsys-TSsys or G = H - TS.

H = 15 kJS = -45 kJ T = 200 K

G = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJ

G = 15 + 9000 kJ = 9015 kJ

The sign ofGG indicates whether a reaction will occur spontaneously.

++ Not spontaneousNot spontaneous

0 0 At equilibriumAt equilibrium

-- SpontaneousSpontaneous

The fact that the effect of SS will vary as a function of temperature is important. This can result in changing the sign of GG.

Free energyFree energy

Predicting Whether a ReactionPredicting Whether a Reactionis Product Favored using is Product Favored using GG

Sign of Hsystem Sign of Ssystem Product-favored?

Negative (exothermic) Positive Yes

Negative (exothermic) Negative Yes at low T; no at

high T

Positive (endothermic) Positive No at low T;

yes at high T

Positive (endothermic) Negative No

Predict Predict G at different G at different H, H, S, TS, T

G = G = H - T H - T S.S.

- - all T T +

+ + all T T –

- /+ + high/low TT +

-/+ - low/high T T -



HHoo SSoo GGoo ReactionReaction

exo(-)exo(-) increase(+)increase(+) -- Prod-favoredProd-favored

endo(+)endo(+) decrease(-)decrease(-) ++ React-favoredReact-favored

exo(-)exo(-) decrease(-)decrease(-) ?? T dependentT dependent

endo(+)endo(+) increase(+)increase(+) ?? T dependentT dependent

GGffoo

Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states.

G values can then be calculated from:

GGoo = = nnppGGffoo productsproducts – – nnrrGGff

oo reactantsreactants

Standard free energy of formation, Standard free energy of formation, GGffoo

SubstanceSubstance GGffoo SubstanceSubstance GGff

oo

C (diamond) 2.832 HBr (g) -53.43

CaO (s) -604.04 HF (g) -273.22

CaCO3 (s) -1128.84 HI (g) 1.30

C2H2 (g) 209 H2O (l) -237.18

C2H4 (g) 86.12 H2O (g) -228.59

C2H6 (g) -32.89 NaCl (s) -384.04

CH3OH (l) -166.3 O (g) 231.75

CH3OH (g) -161.9 SO2 (g) -300.19

CO (g) -137.27 SO3 (g) -371.08

All have units of kJ/mol and are for 25 All have units of kJ/mol and are for 25 ooCC

Standard free energy of formationStandard free energy of formation

How do you calculate How do you calculate GG

There are two ways to calculateG

for chemical reactions.

i) G = H - TS.

ii) Go = Gof (products) - G o

f (reactants)



Two methods of calculating Two methods of calculating GGoo

(a) Determine (a) Determine HHoorxnrxn and and SSoo

rxnrxn and use Gibbs and use Gibbs

equation.equation.

(b) Use tabulated values of free energies of (b) Use tabulated values of free energies of

formation, formation, GGffoo..

GGoorxnrxn = = GGff

oo (products) - (products) - GGffoo (reactants) (reactants)GGoo

rxnrxn = = GGffoo (products) - (products) - GGff

oo (reactants) (reactants)

Calculating Calculating GGoorxnrxn


Is the dissolution of ammonium nitrate product-Is the dissolution of ammonium nitrate product-favored? favored?

If so, is it enthalpy- or entropy-driven?If so, is it enthalpy- or entropy-driven?

NHNH44NONO33(s) + heat (s) + heat NH NH44NONO33(aq)(aq)



Method (a)Method (a) : From tables of thermodynamic data we find : From tables of thermodynamic data we find

HHoorxnrxn = +25.7 kJ = +25.7 kJ

SSoorxnrxn = +108.7 J/K or +0.1087 kJ/K = +108.7 J/K or +0.1087 kJ/K

GGoorxnrxn = +25.7 kJ - (298 K)(+0.1087 J/K) = +25.7 kJ - (298 K)(+0.1087 J/K)

= -6.7 kJ= -6.7 kJ

Reaction is Reaction is product-favoredproduct-favored in spite of negative in spite of negative HHoorxnrxn. .

Reaction is Reaction is “entropy driven”“entropy driven”

NHNH44NONO33(s) + heat (s) + heat NH NH44NONO33(aq)(aq)



Combustion of carbonCombustion of carbon

C(graphite) + OC(graphite) + O22(g) --> CO(g) --> CO22(g) (g) Method (b) :

GGoorxnrxn = = GGff

oo(CO(CO22) - [) - [GGffoo(graph) + (graph) + GGff

oo(O(O22)])]

GGoorxnrxn = -394.4 kJ - [ 0 + 0] = -394.4 kJ - [ 0 + 0]

Note that free energy of formation of an element in Note that free energy of formation of an element in its standard state is 0.its standard state is 0.

GGoorxnrxn = -394.4 kJ = -394.4 kJ

Reaction is Reaction is product-favoredproduct-favored

GGoorxnrxn = = GGff

oo (products) - (products) - GGffoo (reactants) (reactants)GGoo

rxnrxn = = GGffoo (products) - (products) - GGff

oo (reactants) (reactants)

We can calculate Go values from Ho and So values at a constant temperature and pressure.

Example.Example.

Determine Go for the following reaction at 25oC

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

Hfo, kJ/mol 0.00 0.00 -46.11

So, J/K.mol 191.50 130.68 192.3

Calculation of Go

Example.Example.

Calculate the Sorxn at 25 oC for the following

reaction.

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

SubstanceSubstance SSoo (J/K (J/K..mol)mol)

CH4 (g) 186.2

O2 (g) 205.03

CO2 (g) 213.64

H2O (g) 188.72

Calculation of standard entropy changesCalculation of standard entropy changes

Calculate the S for the following reactions usingSo

= So (products) - S o(reactants)a) 2SO2 (g) + O2 (g) ------> 2SO 3(g)

So [SO2(g)] = 248 J/K mole ; So [O2(g)] = 205 J/K mole; So [SO3(g)] = 257 J/K mole

b) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l)

So[ NH3(g)] = 193 J/K mole ; So [N2(g)] = 192 J/K mole;

So [N2O(g)] = 220 J/K mole; S[ H2O(l)] = 70 J/K mole

a)a) 2SO2SO22 (g) + O (g) + O22 (g------> 2SO (g------> 2SO 33(g)(g) So [SO2(g)] = 248 J/K mole ; So [O2(g)] = 205 J/K mole; So [SO3(g)] = 257 J/K moleSo 496 205 514

So = So (products) - S o(reactants)

So = [514] - [496 + 205]

So = 514 - 701

So = -187 J/K mole

Calculate the Calculate the G value for the following G value for the following reactions using:reactions using: Go = Go

f (products) - Gof (reactants)

N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; Go = ? Gf

o[ N2O5 (g) ] = 134 kJ/mole ; Gfo [H2O(g)] = -237 kJ/mole;

Gfo[ HNO3(l) ] = -81 kJ/mole

N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; Go = ?Gf

o 1 x 134 1 x (-237) 2 (-81) 134 -237 -162 Go = Go

f (products) - 3 Gof (reactants)

Go = [-162] - [134 + (-237)]Go = -162 + 103Go = -59 kJ/mole The reaction have a negative G and the reaction is spontaneous or will take place as written.

Free Energy and TemperatureFree Energy and Temperature

2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)

HHoorxnrxn = +467.9 kJ = +467.9 kJ SSoo

rxnrxn = +560.3 J/K = +560.3 J/K

GGoorxnrxn = +300.8 kJ = +300.8 kJ

Reaction is Reaction is reactant-favoredreactant-favored at 298 K at 298 K

At what T does At what T does GGoorxnrxn change from (+) to (-)? change from (+) to (-)?

Set Set GGoorxnrxn = 0 = = 0 = HHoo

rxnrxn - T - TSSoorxnrxn

K 835.1 = kJ/K 0.5603

kJ 467.9 =

S

H = T

rxn

rxn

Effect of Temperature on Reaction Spontaneity

How do you calculate How do you calculate G at G at different T and Pdifferent T and P

G = Go + RT ln Q

Q = reaction quotient

at equilibrium G = = Go + RT ln K

Go = - RT ln KIf you know Go you could calculate K

Concentrations, Free Energy, and Concentrations, Free Energy, and the Equilibrium Constantthe Equilibrium Constant

Equilibrium Constant and Free Energy

G = Go + RT ln Q

Q = reaction quotient

0 = Go + RT ln Keq

Go = - RT ln Keq

KKeqeq is related to reaction favorability and so to is related to reaction favorability and so to GGoorxnrxn..

The larger the (-) value of The larger the (-) value of GGoorxnrxn the larger the value the larger the value

of K.of K.

GGoorxnrxn = - RT lnK = - RT lnK

where R = 8.31 J/K•molwhere R = 8.31 J/K•mol

Thermodynamics and KThermodynamics and Keqeq

For gases, the equilibrium constant for a reaction can be related to Go by:

GGoo = - = -RTRT ln lnKK

For our earlier example,

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

At 25oC, Go was -32.91 kJ so K would be:

ln K = =

ln K = 13.27; K = 5.8 x 105

Go

-RT-32.91 kJ

-(0.008315 kJ.K-1mol-1)(298.2K)

Free energy and equilibriumFree energy and equilibrium

Calculate the Calculate the G for the following G for the following equilibrium reaction and predict the equilibrium reaction and predict the direction of the change using the equation:direction of the change using the equation: G = Go + RT ln Q ; [Gf

o[ NH3(g) ] = -17 kJ/mole

N2 (g) + 3 H2 (g) 2 NH3 (g); G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = 300

N2 (g) + 3 H2 (g) 2 NH3 (g); G = ?

To calculate To calculate GGoo

Using Go = Gof (products) - Go

f (reactants)

Gfo[ N2(g) ] = 0 kJ/mole; Gf

o[ H2(g) ] = 0

kJ/mole; Gfo[ NH3(g) ] = -17 kJ/mole

Notice elements have Gfo = 0.00 similar to Hf

o

N2 (g) + 3 H2 (g) 2 NH3 (g); G = ?Gf

o 0 0 2 x (-17) 0 0 -34 Go = Go

f (products) - Gof (reactants)

Go = [-34] - [0 +0]Go = -34Go = -34 kJ/mole

To calculate QTo calculate QEquilibrium expression for the reaction in terms of partial pressure:N2 (g) + 3 H2 (g) 2 NH3 (g) p2

NH3

K = _________ pN2 p3

H2

p2NH3

Q = _________ ; pN2 p3

H2

Q is when initial concentration is substituted into the equilibrium expression 752

Q = _________ ; p2NH3= 752; pN2 =300; p3

H2=3003

300 x 3003

Q = 6.94 x 10-7

To calculate To calculate GGoo

G = Go + RT ln Q

Go= -34 kJ/mole

R = 8.314 J/K mole or 8.314 x 10-3kJ/Kmole

T = 300 K

Q= 6.94 x 10-7

G = (-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln

6.94 x 10-7)

G = -34 + 2.49 ln 6.94 x 10-7

G = -34 + 2.49 x (-14.18)

G = -34 -35.37

G = -69.37 kJ/mole

Calculate K (from Calculate K (from GG00))

NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ

GGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K

GGoorxnrxn = - RT lnK = - RT lnK

1.94- = K)J/K)(298 (8.31

J 4800 - = lnK

K = 0.14K = 0.14GGoo

rxnrxn > 0 : K < 1 > 0 : K < 1GGoo

rxnrxn < 0 : K > 1 < 0 : K > 1

K = 0.14K = 0.14GGoo

rxnrxn > 0 : K < 1 > 0 : K < 1GGoo

rxnrxn < 0 : K > 1 < 0 : K > 1

Thermodynamics and KThermodynamics and Keqeq

Concentrations, Free Energy, and the Equilibrium Constant

The Influence of Temperature on Vapor Pressure

H2O(l) => H2O(g)

Keq = pwater vapor

pwater vapor = Keq = e- G'/RT

G as a Function of theExtent of the Reaction

G as a Function of theG as a Function of theExtent of the ReactionExtent of the Reactionwhen there is Mixingwhen there is Mixing

Maximum WorkMaximum Work

G = wsystem = - wmax

(work done on the surroundings)

Coupled ReactionsCoupled Reactions

How to do a reaction that is not thermodynamically favorable?

Find a reaction that offset the (+) G

Thermite ReactionThermite Reaction

Fe2O3(s) => 2Fe(s) + 3/2O2(g)

2Al(s) + 3/2O2(g) Al2O3(s)

ADP and ATPADP and ATP

Acetyl Coenzyme AAcetyl Coenzyme A

Gibbs Free Energy and Nutrients

Photosynthesis: Harnessing Light Energy

Using Electricity for reactions with (+) G: Electrolysis

Non spontaneous reactions could be Non spontaneous reactions could be made to take place by coupling with made to take place by coupling with energy source: another reaction or energy source: another reaction or electric currentelectric current

ElectrolysisElectrolysis

2NaCl(l) => 2Na(s) + 2Cl2(g)

chemistry 102(01) spring 2009

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