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
18-2CHEM 102, Fall 20010 LA TECH
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
18-3CHEM 102, Fall 20010 LA TECH
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
18-4CHEM 102, Fall 20010 LA TECH
Why is it necessary to divide Universe into System and SurroundingUniverse = System +
Surrounding
system surroundings
universe
18-5CHEM 102, Fall 20010 LA TECH
Types of Systems
Isolated system
no mass or energy exchange
Closed system
only energy exchange
Open system
both mass and energy exchange
18-6CHEM 102, Fall 20010 LA TECH
Universe = System + Surrounding
Why is it necessary to divide Universe into System and Surrounding
18-7CHEM 102, Fall 20010 LA TECH
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
18-8CHEM 102, Fall 20010 LA TECH
What forms of energy are found in the Universe?mechanical
thermal
electrical
nuclear
mass: E = mc2
others yet to discover
18-9CHEM 102, Fall 20010 LA TECH
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.
18-10CHEM 102, Fall 20010 LA TECH
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 )
18-11CHEM 102, Fall 20010 LA TECH
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?
18-12CHEM 102, Fall 20010 LA TECH
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.
18-13CHEM 102, Fall 20010 LA TECH
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
18-14CHEM 102, Fall 20010 LA TECH
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
18-15CHEM 102, Fall 20010 LA TECH
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
18-16CHEM 102, Fall 20010 LA TECH
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
18-17CHEM 102, Fall 20010 LA TECH
Why is DHof of elements is zero?
DHof, Heat formations are for compounds
Note: DHof of elements is zero
18-18CHEM 102, Fall 20010 LA TECH
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
18-19CHEM 102, Fall 20010 LA TECH
Chemical Thermodynamics
spontaneous reaction – reaction which proceed without external assistance once started
chemical thermodynamics helps predict which reactions are spontaneous
18-20CHEM 102, Fall 20010 LA TECH
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.
18-21CHEM 102, Fall 20010 LA TECH
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
18-22CHEM 102, Fall 20010 LA TECH
Bases on Energy: Product-Favored Reactions
In general, product-favored reactions are exothermic.
(Negative DH)
In general, reactant-favored reactions are endothermic.
(Positive DH)
18-23CHEM 102, Fall 20010 LA TECH
Product-Favored ReactionsBut many spontaneous reactions or processes
are endothermic or even have DH = 0.
NH4NO3(s) NH4NO3(aq); DH = +
18-24CHEM 102, Fall 20010 LA TECH
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)
18-25CHEM 102, Fall 20010 LA TECH
Expansion of a GasThe positional
probability is higher when particles are dispersed over a larger volume
Matter tends to expand unless it is restricted
18-26CHEM 102, Fall 20010 LA TECH
Gas Expansion and Probability
18-27CHEM 102, Fall 20010 LA TECH
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
18-28CHEM 102, Fall 20010 LA TECH
S[H2O(l)] > S[H2O(s)] at 0° C.
18-29CHEM 102, Fall 20010 LA TECH
Entropies of Solid, Liquidand Gas Phases
S (gases) > S (liquids) > S (solids)
18-30CHEM 102, Fall 20010 LA TECH
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
18-31CHEM 102, Fall 20010 LA TECH
Entropy and Molecular Structure
18-32CHEM 102, Fall 20010 LA TECH
Entropy and Dissolving
18-33CHEM 102, Fall 20010 LA TECH
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
18-34CHEM 102, Fall 20010 LA TECH
Entropy of a Solution of a Gas
18-35CHEM 102, Fall 20010 LA TECH
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
18-36CHEM 102, Fall 20010 LA TECH
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
18-37CHEM 102, Fall 20010 LA TECH
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
18-38CHEM 102, Fall 20010 LA TECH
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)
18-39CHEM 102, Fall 20010 LA TECH
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 +
18-40CHEM 102, Fall 20010 LA TECH
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)
18-41CHEM 102, Fall 20010 LA TECH
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)
18-42CHEM 102, Fall 20010 LA TECH
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.
18-43CHEM 102, Fall 20010 LA TECH
Standard Molar Entropy Values
18-44CHEM 102, Fall 20010 LA TECH
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
18-45CHEM 102, Fall 20010 LA TECH
Entropy & Spontaneity
How can water boil and freeze spontaneously?
Enthalpy change can not predict spontaneity!
Some endothermic processes are spontaneous
Need another thermodynamic property.
18-46CHEM 102, Fall 20010 LA TECH
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
18-47CHEM 102, Fall 20010 LA TECH
Dissolving NH4NO3 in water—an entropy driven process.
2nd Law of Thermodynamics
NH4NO3(s) NH4NO3(aq); DH = +
18-48CHEM 102, Fall 20010 LA TECH
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
18-49CHEM 102, Fall 20010 LA TECH
Entropy of the UniverseDSuniv = DSsys + DSsurr
Dsuniv DSsys DSsurr
+ + +
+ +(DSsys>DSsurr) -
+ - + (DSsurr>DSsys)
18-50CHEM 102, Fall 20010 LA TECH
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
2nd Law of Thermodynamics
18-51CHEM 102, Fall 20010 LA TECH
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.
2nd Law of Thermodynamics
18-52CHEM 102, Fall 20010 LA TECH
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
18-53CHEM 102, Fall 20010 LA TECH
Gibbs Free Energy, GGibbs Free Energy, G
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
18-54CHEM 102, Fall 20010 LA TECH
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
18-55CHEM 102, Fall 20010 LA TECH
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
18-56CHEM 102, Fall 20010 LA TECH
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
18-57CHEM 102, Fall 20010 LA TECH
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
18-58CHEM 102, Fall 20010 LA TECH
Predict DG at different DH, DS, T
DG = DH - T DS.
- - all T +
+ + all T –
- /+ + high/low T +
-/+ - low/high T -
18-59CHEM 102, Fall 20010 LA TECH
Gibbs Free Energy, GGibbs Free Energy, G
DGo = DHo - TDSo
DHo DSo DGo Reaction
exo(-) increase(+) - Prod-favored
endo(+) decrease(-) + React-
favored
exo(-) decrease(-) ? T dependent
endo(+) increase(+) ? T
dependent
18-60CHEM 102, Fall 20010 LA TECH
DGo = DHo - TDSo
18-61CHEM 102, Fall 20010 LA TECH
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
18-62CHEM 102, Fall 20010 LA TECH
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
18-63CHEM 102, Fall 20010 LA TECH
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)
18-64CHEM 102, Fall 20010 LA TECH
Gibbs Free Energy, GGibbs Free Energy, G
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)
18-65CHEM 102, Fall 20010 LA TECH
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)
18-66CHEM 102, Fall 20010 LA TECH
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)
18-67CHEM 102, Fall 20010 LA TECH
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)
18-68CHEM 102, Fall 20010 LA TECH
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
18-69CHEM 102, Fall 20010 LA TECH
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
18-70CHEM 102, Fall 20010 LA TECH
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
18-71CHEM 102, Fall 20010 LA TECH
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
18-72CHEM 102, Fall 20010 LA TECH
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.
18-73CHEM 102, Fall 20010 LA TECH
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
18-74CHEM 102, Fall 20010 LA TECH
Effect of Temperature on Reaction Spontaneity
18-75CHEM 102, Fall 20010 LA TECH
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
18-76CHEM 102, Fall 20010 LA TECH
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
18-77CHEM 102, Fall 20010 LA TECH
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
18-78CHEM 102, Fall 20010 LA TECH
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
18-79CHEM 102, Fall 20010 LA TECH
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 = ?
18-80CHEM 102, Fall 20010 LA TECH
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
18-81CHEM 102, Fall 20010 LA TECH
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
18-82CHEM 102, Fall 20010 LA TECH
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
18-83CHEM 102, Fall 20010 LA TECH
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
18-84CHEM 102, Fall 20010 LA TECH
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
18-85CHEM 102, Fall 20010 LA TECH
DG as a Function of theExtent of the Reaction
18-86CHEM 102, Fall 20010 LA TECH
DG as a Function of theExtent of the Reactionwhen there is Mixing
18-87CHEM 102, Fall 20010 LA TECH
Maximum WorkDG = wsystem = - wmax
(work done on the surroundings)
18-88CHEM 102, Fall 20010 LA TECH
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)
18-89CHEM 102, Fall 20010 LA TECH
ADP and ATP
18-90CHEM 102, Fall 20010 LA TECH
Acetyl Coenzyme A
18-91CHEM 102, Fall 20010 LA TECH
Gibbs Free Energy and Nutrients
18-92CHEM 102, Fall 20010 LA TECH
Photosynthesis: Harnessing Light Energy
18-93CHEM 102, Fall 20010 LA TECH
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)