Chemistry 102(01) Fall 2010

18-1CHEM 102, Fall 20010 LA TECH

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311 Phone 257-4941

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

Test Dates: September 23, October 21, and November 16; Comprehensive Final Exam: November 18, 2010

Exam: 10:0-10:15 am, CTH 328.

September 23, 2010 (Test 1): Chapter 13

October 21, 2010 (Test 2): Chapters 14 & 15

November 16, 2010 (Test 3): Chapters 16, 17 & 18

Comprehensive Final Exam: November 18, 2010 :Chapters 13, 14, 15, 16, 17 and 18

Chemistry 102(01) Fall 2010


Review of Chapter 6. Energy and Chemical Reactions 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


Thermochemistry

Heat changes during chemical reactions

Thermochemical equation. eg.

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

DH is called the enthalpy of reaction.

if DH is + reaction is called endothermic

if DH is - reaction is called exothermic


Why is it necessary to divide Universe into System and SurroundingUniverse = System +

Surrounding

system surroundings

universe


Types of Systems

Isolated system

no mass or energy exchange

Closed system

only energy exchange

Open system

both mass and energy exchange


Universe = System + Surrounding

Why is it necessary to divide Universe into System and Surrounding


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

DU is associated with changes in atoms, molecules and subatomic particles

Etotal = Eke + E pe + DU

DU = heat (q) + w (work)

DU = q + w

DU = q -P DV; w =- P DV


What forms of energy are found in the Universe?mechanical

thermal

electrical

nuclear

mass: E = mc2

others yet to discover


What is 1st Law of Thermodynamics

Eenergy is conserved in the Universe

All forms of energy are inter-convertible and conserved

Energy is neither created nor destroyed.


What exactly is DH?

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 = -PDV

DU = qp + w; DU = qp -PDV

qp = DU + PDV; w = -PDV

DH = DU + PDV; qp = DH(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 = -

PDV= 0

DU = qv + w

qv = DU + o; w = 0

DU = qv = DU(internal energy )

How do you measure DU?


What is Hess's Law of Summation of Heat?To heat of reaction for new reactions.

Two methods?

1st method: new DH is calculated by adding DHs of other reactions.

2nd method: Where DHf ( DH of formation) of reactants and products are used to calculate DH of a reaction.


Method 1: Calculate DH for the reaction:

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

Other reactions:

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

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

kJ

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


SO2(g) ------> S(s) + O2(g); DH1 = 297 kJ - 1

H2(g) + S(s) + 2O2(g) ------> H2SO4(l); DH2 = -814 kJ - 2

H2O(g) ----->H2(g) + 1/2 O2(g) ; DH3 = +242 kJ - 3

______________________________________

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

DH = DH1+ DH2+ DH3

DH = +297 - 814 + 242

DH = -275 kJ

Calculate DH for the reaction


Calculate Heat (Enthalpy) of Combustion: 2nd method C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; DHo = ?

DHf (C7H16) = -198.8 kJ/mol

DHf (CO2) = -393.5 kJ/mol

DHf (H2O) = -285.9 kJ/mol

DHf O2(g) = 0 (zero)

What method?DHo = S n DHf

o products – S n DHfo reactants

n = stoichiometric coefficients2nd method


DH = [Sn ( DHof) Products] - [Sn (DHo

f) reactants]

DH = [ 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 DH for the reaction


Why is DHof of elements is zero?

DHof, Heat formations are for compounds

Note: DHof 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 Thermodynamics

spontaneous reaction – reaction which proceed without external assistance once started

chemical thermodynamics helps predict which reactions are spontaneous


ThermodynamicsThermodynamics

Will the rearrangement of a system decrease its energy?

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

Most product-favored reactions are exothermic.

Often referred to as spontaneous reactions.

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


Thermodynamics and KineticsThermodynamics and KineticsDiamond graphite

Thermodynamically product favored

Slow Kinetics

Paper burns

Thermodynamically product favored

Fast Kinetics

In this chapter we only look into

thermodynamic factors


Bases on Energy: Product-Favored Reactions

In general, product-favored reactions are exothermic.

(Negative DH)

In general, reactant-favored reactions are endothermic.

(Positive DH)


Product-Favored ReactionsBut many spontaneous reactions or processes

are endothermic or even have DH = 0.

NH4NO3(s) NH4NO3(aq); DH = +


Direction of Chemical ReactionProduct favored reactions are always a

transformation of a reactants favored reaction.

Product Favored Reaction

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

Reactant Favored Reaction

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

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

Electrolysis

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


Expansion of a GasThe 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, SThe thermodynamic property

related to randomness is ENTROPY, S.

Product-favored processes: final state is more DISORDERED or RANDOM than the original.

Spontaneity is related to an increase in randomness. Reaction of K with water


S[H2O(l)] > S[H2O(s)] at 0° C.


Entropies of Solid, Liquidand Gas Phases

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


Entropy, SEntropy, S

Entropies of ionic solids depend on coulombic attractions.

So

(J/K•mol)

MgO 26.9

NaF 51.5

So

(J/K•mol)

MgO 26.9

NaF 51.5


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 TransitionsH2O(s) => H2O(l) DH > 0; DS > 0

H2O(l) => H2O(g) DH > 0; DS > 0

spontaneous at high temperatures

H2O(l) => H2O(s) DH < 0; DS < 0

H2O(g) => H2O(l) DH < 0; DS < 0

spontaneous at low temperatures


Entropy Changes for Phase Changes

For a phase change, DSSYS = qSYS/T

(q = heat transferred)

Boiling Water

H2O (liq) H2O(g)

DH = q = +40,700 J/mol

mol•J/K 109+ = K 373.15

J/mol 40,700 =

T

q = S


Phase Transitions

Heat of Fusion

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

DGfusion = 0 = DHfusion - T DSfusion

0 = DHfusion - T DSfusion

DHfusion = T DSfusion

Heat of Vaporization

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

DHvaporization = T DSvaporization


Qualitative prediction of DS of Chemical Reactions

Look for (l) or (s) --> (g)

If all are gases: calculate DnDn = Sn (gaseous prod.) - S n(gaseous reac.)

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

Dn = 2 - 4 = -2

If Dn is - DS is negative (decrease in S)

If Dn is + DS is positive (increase in S)


Entropy Change

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

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

ii) Liquid ---> gas +

iii) Solid ----> gas most +

iv) Increase in temperature +

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

vi) Increase in volume +


Predict DS!

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(g) + O2(g) 2 H2O(liq)

DSo = 2 So (H2O) - [2 So (H2) + So (O2)]

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

DSo = -326.9 J/K

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

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

DSo

= So

(products) - So

(reactants)

Based on Hess’s Law second method:

DSo

= So

(products) - So

(reactants)


Third Law of Thermodynamics

Provides reference point for absolute entropy

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

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


Standard 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 25oC


Entropy & 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 Thermodynamics

Zeroth: Thermal equilibrium and temperature

First : The total energy of the universe is constant

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

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

zero is zero

Zeroth: Thermal equilibrium and temperature

First : The total energy of the universe is constant

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

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

zero is zero


Dissolving NH4NO3 in water—an entropy driven process.

2nd Law of Thermodynamics

NH4NO3(s) NH4NO3(aq); DH = +


Second Law of Thermodynamics

In 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 processDSuniverse = ( Ssys + Ssurr) > 0 DSuniv = entropy of the UniverseDSsys = entropy of the SystemDSsurr = entropy of the SurroundingDSuniv = DSsys + DSsurr


Entropy of the UniverseDSuniv = DSsys + DSsurr

Dsuniv DSsys DSsurr

+ + +

+ +(DSsys>DSsurr) -

+ - + (DSsurr>DSsys)


Can calc. that DHo

rxn = DHo

system = -571.7 kJCan calc. that DHo

rxn = DHo

system = -571.7 kJ

2 H2(g) + O2(g) 2 H2O(liq)

DSosys = -326.9 J/K

Entropy Changes in the Surroundings

T

H- =

T

q = systemsurr

gssurroundin

oS

K 298.15

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

oS

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



2 H2(g) + O2(g) 2 H2O(liq)

DSosys = -326.9 J/K

DSosurr = +1917 J/K

DSouni = +1590. J/K

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



Gibbs Free Energy, GGibbs Free Energy, G

DSuniv = DSsurr + DSsys

Multiply through by (-T)

-TDSuniv = DHsys - TDSsys

-TDSuniv = DGsystem

Under standard conditions —

DGo = DHo - TDSo

D S univ = -D H sys

T + D S sys



DGo = DHo - T DSo

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 -energy lost in disordering the system


Predict the spontaneity of the following processes from DH and DS at various temperatures.

a)DH = 30 kJ, DS = 6 kJ, T = 300 Kb)DH = 15 kJ,DS = -45 kJ,T = 200 K


a) DH = 30 kJ DS = 6 kJ T = 300 K DG = DHsys-TDSsys or DG = DH - TDS.DH = 30 kJDS = 6 kJ T = 300 K DG = 30 kJ - (300 x 6 kJ) = 30 -1800 kJDG = -1770 kJ

b) DH = 15 kJ DS = -45 kJ T = 200 KDG = DHsys-TDSsys or DG = DH - TDS.DH = 15 kJDS = -45 kJ T = 200 K DG = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJDG = 15 + 9000 kJ = 9015 kJ


The sign of DG indicates whether a reaction will occur spontaneously.

+ Not spontaneous

0 At equilibrium

- Spontaneous

The fact that the effect of DS will vary as a function of temperature is important.

This can result in changing the sign of DG.

Free energy


Predicting Whether a Reactionis Product Favored using DG

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 DG at different DH, DS, T

DG = DH - T DS.

- - all T +

+ + all T –

- /+ + high/low T +

-/+ - low/high T -



DGo = DHo - TDSo

DHo DSo DGo Reaction

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

endo(+) decrease(-) + React-

favored

exo(-) decrease(-) ? T dependent

endo(+) increase(+) ? T

dependent


DGo = DHo - TDSo


DGfo

Free energy change that results when one mole of a substance if formed from

its elements will all substances in their standard states.

DG values can then be calculated from:

DGo

= S npDGfo

products – S nrDGfo

reactants

Standard free energy of formation, DGfo


Substance DGfo

Substance DGfo

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 oC

Standard free energy of formation


How do you calculate DG

There are two ways to calculate DG

for chemical reactions.

i) DG = DH - TDS.

ii) DGo = S DGof (products) - S DG o

f (reactants)



DGo = DHo - TDSo

Two methods of calculating DGo

(a) Determine DHorxn and DSo

rxn and use Gibbs

equation.

(b) Use tabulated values of free energies of

formation, DGfo.

Go

rxn = S Gfo

(products) - S Gfo

(reactants)Go

rxn = S Gfo

(products) - S Gfo

(reactants)


Calculating DGorxn

Calculating DGorxn

Is the dissolution of ammonium nitrate product-favored?

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

NH4NO3(s) + heat NH4NO3(aq)


Calculating DGorxn

Calculating DGorxn

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

DHorxn = +25.7 kJ

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

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

= -6.7 kJReaction is product-favored in spite of negative DHo

rxn.

Reaction is “entropy driven”

NH4NO3(s) + heat NH4NO3(aq)


Calculating DGorxn

Calculating DGorxn

Combustion of carbon

C(graphite) + O2(g) --> CO2(g)

Method (b) :

DGorxn = DGf

o(CO2) - [DGfo(graph) + DGf

o(O2)]

DGorxn = -394.4 kJ - [ 0 + 0]

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

DGorxn = -394.4 kJ

Reaction is product-favored

DGo

rxn = S DGfo

(products) - S DGfo

(reactants)DGo

rxn = S DGfo

(products) - S DGfo

(reactants)


We can calculate DGo

values from DHo and DSo

values at a constant temperature

and pressure.

Example.

Determine DGo

for the following reaction at 25o

C

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

DHfo

, kJ/mol 0.00 0.00 -46.11

So

, J/K.mol 191.50 130.68 192.3

Calculation of DGo


Example.

Calculate the DSo

rxn at 25 o

C for the following reaction.

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

Substance So

(J/K.mol)

CH4 (g) 186.2

O2 (g) 205.03

CO2 (g) 213.64

H2O (g) 188.72

Calculation of standard entropy changes


Calculate the DS for the following reactions using D So

= S D So (products) - S D S o(reactants)a) 2SO2 (g) + O2 (g) ------> 2SO 3(g) D So [SO2(g)] = 248 J/K mole ; D So [O2(g)] = 205 J/K mole; D So [SO3(g)] = 257 J/K mole

b) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l) D So[ NH3(g)] = 193 J/K mole ; D So [N2(g)] = 192 J/K mole;

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


a) 2SO2 (g) + O2 (g------> 2SO 3(g) D So [SO2(g)] = 248 J/K mole ; D So [O2(g)] = 205 J/K mole; D So [SO3(g)] = 257 J/K mole

DSo 496 205 514

DSo = S DSo (products) - S DS o(reactants)

DSo = [514] - [496 + 205]

DSo = 514 - 701

DSo = -187 J/K mole


Calculate the DG value for the following reactions using: D Go = S D Go

f (products) - S D Gof (reactants)

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

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

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

N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; DGo = ?DGf

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

f (products) - 3 DGof (reactants)

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


Free Energy and Temperature

2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)

DHorxn = +467.9 kJ DSo

rxn = +560.3 J/K

DGorxn = +300.8 kJ

Reaction is reactant-favored at 298 K

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

Set DGorxn = 0 = DHo

rxn - TDSorxn

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 DG at different T and P

DG = DGo + RT ln Q

Q = reaction quotient

at equilibrium DG = 00 = DGo + RT ln K

DGo = - RT ln K

If you know DGo you could calculate K


Concentrations, Free Energy, and the Equilibrium Constant

Equilibrium Constant and Free Energy

DG = DGo + RT ln Q

Q = reaction quotient

0 = DGo + RT ln Keq

DGo = - RT ln Keq


Keq is related to reaction favorability and so to Gorxn.

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

of K.

DGorxn = - RT lnK

where R = 8.31 J/K•mol

Thermodynamics and Keq


For gases, the equilibrium constant for a reaction can be related to DGo

by:

DGo

= -RT lnK

For our earlier example,

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

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

ln K = =

ln K = 13.27; K = 5.8 x 105

DGo

-RT

-32.91 kJ

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

Free energy and equilibrium


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

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

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

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


To calculate DGo

Using DGo = S DGof (products) - S DGo

f (reactants)

DGfo[ N2(g)] = 0 kJ/mole; DGf

o[ H2(g)] = 0

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

Notice elements have DGfo = 0.00 similar to DHf

o

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

o 0 0 2 x (-17) 0 0 -34 DGo = S DGo

f (products) - S DGof (reactants)

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


To 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 DGo

DG = DGo + RT ln Q

DGo= -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

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

6.94 x 10-7)

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

DG = -34 + 2.49 x (-14.18)

DG = -34 -35.37

DG = -69.37 kJ/mole


Calculate K (from G0)

N2O4 --->2 NO2 DGorxn = +4.8 kJ

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

DGo

rxn = - RT lnK

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

J 4800 - = lnK

K = 0.14

DGorxn > 0 : K < 1

DGorxn < 0 : K > 1

K = 0.14

DGorxn > 0 : K < 1

DGorxn < 0 : K > 1

Thermodynamics and Keq


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


DG as a Function of theExtent of the Reaction


DG as a Function of theExtent of the Reactionwhen there is Mixing


Maximum WorkDG = wsystem = - wmax

(work done on the surroundings)


Coupled ReactionsHow to do a reaction that is not

thermodynamically favorable?

Find a reaction that offset the (+) DG

Thermite Reaction

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

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


ADP and ATP


Acetyl Coenzyme A


Gibbs Free Energy and Nutrients


Photosynthesis: Harnessing Light Energy


Using Electricity for reactions with (+) DG: Electrolysis

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

Electrolysis

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

Chemistry 102(01) Fall 2010

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Transcript of Chemistry 102(01) Fall 2010