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Transcript of 17-1 CHEM 102, Fall 15 LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office:...
![Page 1: 17-1 CHEM 102, Fall 15 LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M.W &F, 8:00-9:00.](https://reader036.fdocuments.us/reader036/viewer/2022081513/5697c02f1a28abf838cda6e7/html5/thumbnails/1.jpg)
17-1CHEM 102, Fall 15 LA TECH
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: CTH 311
Phone 257-4941
Office Hours: M.W &F, 8:00-9:00 & 11:00-12:00 and Tu ,Th 8:00 - 10:00 am. am. or by appointment
Test Dates
Chemistry 102 Fall 2015
Sept. 29, 2015 (Test 1): Chapter 13
Oct. 27, 2015 (Test 2): Chapter 14 &15
Nove. 17, 2015 (Test 3): Chapter 16 &17
November 19, 2015 (Make-up test) comprehensive:
Chapters 13-17
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17-2CHEM 102, Fall 15 LA TECH
Chapter 6. Thermochemistry
6.1 Chemical Hand Warmers 231
6.2 The Nature of Energy: Key Definitions 232
6.3 The First Law of Thermodynamics: There Is No Free Lunch 234
6.4 Quantifying Heat and Work 240
6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry246
6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 249
6.7 Constant-Pressure Calorimetry: Measuring 253
6.8 Relationships Involving 255
6.9 Determining Enthalpies of Reaction from Standard Enthalpies
of Formation 257
6.1 0 Energy Use and the Environment 263
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17-3CHEM 102, Fall 15 LA TECH
Chapter 17. Free Energy and Thermodynamics17.1 Nature’s Heat Tax: You Can’t Win and You Can’t Break Even 769
17.2 Spontaneous and Nonspontaneous Processes 771
17.3 Entropy and the Second Law of Thermodynamics 773
17.4 Heat Transfer and Changes in the Entropy of the Surroundings 780
17.5 Gibbs Free Energy 784
17.6 Entropy Changes in Chemical Reactions: Calculating 788
17.7 Free Energy Changes in Chemical Reactions: Calculating 792
17.8 Free Energy Changes for Nonstandard States: The Relationship
between and 798
17.9 Free Energy and Equilibrium: Relating to the Equilibrium Constant
(K)
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17-4CHEM 102, Fall 15 LA TECH
Chapter 6. Thermochemistry
6.1 Chemical Hand Warmers 231
6.2 The Nature of Energy: Key Definitions 232
6.3 The First Law of Thermodynamics: There Is No Free Lunch 234
6.4 Quantifying Heat and Work 240
6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry 246
6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 249
6.7 Constant-Pressure Calorimetry: Measuring 253
6.8 Relationships Involving 255
6.9 Determining Enthalpies of Reaction from Standard Enthalpies
of Formation 257
6.1 0 Energy Use and the Environment 263
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17-5CHEM 102, Fall 15 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
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17-6CHEM 102, Fall 15 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
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17-7CHEM 102, Fall 15 LA TECH
1) Calculate entropy change for the reaction:
2 C(s) + 1/2 O2(g) + 3 H2(g) --> C2H6O(l); ∆H = ? (ANS 1493.2 kJ/mol)Given the following thermochemical equations:
C2H6O(l) + 3 O2(g) ---> 2 CO2(g) + 3 H2O(l); ∆H = - 1366.9 kJ/mol
1/2 O2(g) + H2(g) ----> H2O(l); ∆H = -285.8 kJ/mol
C(s) + O2(g) ----> CO2(g); ∆H = -393.3 kJ/mol
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17-8CHEM 102, Fall 15 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
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17-9CHEM 102, Fall 15 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
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17-10CHEM 102, Fall 15 LA TECH
Why is DHof of elements is zero?
DHof, Heat formations are for compounds
Note: DHof of elements is zero
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17-11CHEM 102, Fall 15 LA TECH
2) Calculate enthalpy change given the
∆Hfo[SO2(g)] = -297 kJ/mole and
∆Hfo [SO3(g)] = -396 kJ/mole
2SO2 (g) + O2 (g) -----> 2 SO3(g); ∆H= ? ANS -198 kJ/mole)
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17-12CHEM 102, Fall 15 LA TECH
What is relation of DH of a reaction to covalent bond energy?
DH = S½bonds broken½- S ½bonds formed½
How do you calculate bond energy from DH?
How do you calculate DH from bond energy?
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17-13CHEM 102, Fall 15 LA TECH
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17-14CHEM 102, Fall 15 LA TECH
3) Use the table of bond energies to find the ∆Ho for the reaction:
H2(g) + Br2(g) 2 HBr(g);
H-H = 436 kJ, Br-Br= 193 kJ, H-Br = 366 kJ
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17-15CHEM 102, Fall 15 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
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17-16CHEM 102, Fall 15 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
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17-17CHEM 102, Fall 15 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
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17-18CHEM 102, Fall 15 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)
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17-19CHEM 102, Fall 15 LA TECH
4) Calculate the ∆S for the following reaction using:
a) 2SO2 (g) + O2 (g) ----> 2SO3(g)
So [SO2(g)] = 248 J/K mole ;
So [O2(g)] = 205 J/K mole;
So [SO3(g)] = 257 J/K mole
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17-20CHEM 102, Fall 15 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, DG
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17-21CHEM 102, Fall 15 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
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17-22CHEM 102, Fall 15 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
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17-23CHEM 102, Fall 15 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)
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17-24CHEM 102, Fall 15 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 positive DHo
rxn.
Reaction is “entropy driven”
NH4NO3(s) + heat NH4NO3(aq)
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17-25CHEM 102, Fall 15 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)
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17-26CHEM 102, Fall 15 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
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17-27CHEM 102, Fall 15 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
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17-28CHEM 102, Fall 15 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
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17-29CHEM 102, Fall 15 LA TECH
5) Predict the spontaneity of the following processes from ∆H and ∆S at various temperatures.
a) ∆H = 30 kJ ∆S = 6 kJ T = 300 K
b) ∆H = 15 kJ ∆S = -45 kJ T = 200 K
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17-30CHEM 102, Fall 15 LA TECH
6) Calculate the ∆Go for the following chemical reactions using given ∆Ho values, ∆So calculated above and the equation ∆G = ∆H - T∆S.2SO2 (g) + O2 (g) > 2 SO 3(g) ; ∆Go=∆Ho = -198 kJ/mole; ∆So = -187 J/K mole; T = 298 K
∆Go
system ∆Ho
system ∆So
system T
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17-31CHEM 102, Fall 15 LA TECH
7) Which of the following condition applies to a particular chemical reaction, the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K. This reaction is
∆G system ∆H system ∆S system T
a) Negative always Negative (exothermic) Positive Yes
a) Negative at low T Positive at high T
Negative (exothermic)
Negative ∆G =- ,at low T; ∆G= +, at high T
a) Positive at low T Negative at high T
Positive (endothermic)
Positive ∆G = + ,at low T; ∆G= -, at high T
a) Positive always Positive (endothermic)
Negative∆G= +, at any T
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17-32CHEM 102, Fall 15 LA TECH
Effect of Temperature on Reaction Spontaneity
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17-33CHEM 102, Fall 15 LA TECH
DGo = DHo - TDSo
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17-34CHEM 102, Fall 15 LA TECH
8) At what temperature a particular chemical reaction, with the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K becomes
a) Equilibrium:
b) Spontaneous:
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17-35CHEM 102, Fall 15 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
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17-36CHEM 102, Fall 15 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
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17-37CHEM 102, Fall 15 LA TECH
9) Calculate the non standard ∆G for the following equilibrium reaction and predict the direction of the change using the equation:
∆G= ∆Go + RT ln Q
Given ∆Gfo[NH3(g)] = -17 kJ/mole
N2 (g) + 3H2(g) → 2NH3(g); ∆G=? at 300K, PN2= 300, PNH3 = 75 and PH2 = 300
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17-38CHEM 102, Fall 15 LA TECH
10) The Ka expression for the dissociation of acetic acid in water is based on the following equilibrium at 25°C:
HC2H3O2(l) + H2O ⇄ H+(aq) + C2H3O2 -(aq)
What is ∆G° if Ka=1.8 x 10-5?
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17-39CHEM 102, Fall 15 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.
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17-40CHEM 102, Fall 15 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
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17-41CHEM 102, Fall 15 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
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17-42CHEM 102, Fall 15 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
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17-43CHEM 102, Fall 15 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 = ?
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17-44CHEM 102, Fall 15 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
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17-45CHEM 102, Fall 15 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
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17-46CHEM 102, Fall 15 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
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17-47CHEM 102, Fall 15 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 - = ln K
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
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17-48CHEM 102, Fall 15 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
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17-49CHEM 102, Fall 15 LA TECH
DG as a Function of theExtent of the Reaction
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17-50CHEM 102, Fall 15 LA TECH
DG as a Function of theExtent of the Reactionwhen there is Mixing
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17-51CHEM 102, Fall 15 LA TECH
Maximum WorkDG = wsystem = - wmax
(work done on the surroundings)
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17-52CHEM 102, Fall 15 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)
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17-53CHEM 102, Fall 15 LA TECH
ADP and ATP
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17-54CHEM 102, Fall 15 LA TECH
Acetyl Coenzyme A
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17-55CHEM 102, Fall 15 LA TECH
Gibbs Free Energy and Nutrients
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17-56CHEM 102, Fall 15 LA TECH
Photosynthesis: Harnessing Light Energy
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17-57CHEM 102, Fall 15 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)