Chemistry 102(001) Spring 2015
Instructor: Dr. Upali Siriwardane
e-mail: upali@latech.edu
Office: CTH 311
Phone 257-4941
Office Hours: M,W 8:00-9:30 & 11:00-12:30 am; Tu,Th, F 8:00 10:00 am. or by appointment.;
Test Dates:
March 30,
April 27,
May 18,
May 20,
2015 (Test 1): Chapter 13
2015 (Test 2): Chapter 14 &15
2015 (Test 3): Chapter 17
2015 (Make-up test) comprehensive:
Chapters 13-17
CHEM 102, Spring 15 LA TECH
17-1
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
CHEM 102, Spring 15 LA TECH
17-2
Chapter 17. Free Energy and Thermodynamics
17.1 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)
CHEM 102, Spring 15 LA TECH
17-3
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
CHEM 102, Spring 15 LA TECH
17-4
Method 1: Calculate DH for the reaction:
SO2(g) + 1/2 O2(g) + H2O(g) ---->
DH = ?
H2SO4(l)
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
CHEM 102, Spring 15 LA TECH
17-5
Calculate DH for the reaction
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
CHEM 102, Spring 15 LA TECH
17-6
1) Calculate entropy change for the reaction:
2 C(s) + 1/2 O2(g) + 3 H2(g) --> C2H6O(l); ∆H = ? (ANS
-277.1 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
CHEM 102, Spring 15 LA TECH
17-7
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)
DHf (H2O)
DHf O2(g)
= -393.5 kJ/mol
= -285.9 kJ/mol
= 0 (zero)
What method?
DHo = S n  DHfo products – S n  DHfo reactants
n = stoichiometric coefficients
2nd method
CHEM 102, Spring 15 LA TECH
17-8
Calculate DH for the reaction
DH = [Sn ( DHof) Products] - [Sn (DHof) 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
CHEM 102, Spring 15 LA TECH
17-9
Why is DHof of elements is zero?
DHof, Heat formations are for compounds
Note: DHof of elements is zero
CHEM 102, Spring 15 LA TECH
17-10
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)
CHEM 102, Spring 15 LA TECH
17-11
What is relation of DH of a reaction to
covalent bond energy?
DH = Sbonds broken- S bonds formed
How do you calculate bond energy from DH?
How do you calculate DH from bond energy?
CHEM 102, Spring 15 LA TECH
17-12
CHEM 102, Spring 15 LA TECH
17-13
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
CHEM 102, Spring 15 LA TECH
17-14
Calculation of standard entropy changes
Example.
Calculate the DSorxn at 25 oC 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
CHEM 102, Spring 15 LA TECH
17-15
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) -------->
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
CHEM 102, Spring 15 LA TECH
4N2 (g) + 3 H2O (l)
17-16
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
CHEM 102, Spring 15 LA TECH
17-17
Calculating DS for a Reaction
Based on Hess’s Law second method:
DSo =
S So (products) - S So (reactants)
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.
CHEM 102, Spring 15 LA TECH
17-18
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
CHEM 102, Spring 15 LA TECH
17-19
Free energy, DG
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.
CHEM 102, Spring 15 LA TECH
17-20
Standard free energy of formation, DGf
o
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
CHEM 102, Spring 15 LA TECH
17-21
Standard free energy of formation
Substance
C (diamond)
CaO (s)
CaCO3 (s)
C2H2 (g)
C2H4 (g)
C2H6 (g)
CH3OH (l)
CH3OH (g)
CO (g)
DGfo
2.832
-604.04
-1128.84
209
86.12
-32.89
-166.3
-161.9
-137.27
Substance
HBr (g)
HF (g)
HI (g)
H2O (l)
H2O (g)
NaCl (s)
O (g)
SO2 (g)
SO3 (g)
DGfo
-53.43
-273.22
1.30
-237.18
-228.59
-384.04
231.75
-300.19
-371.08
All have units of kJ/mol and are for 25 oC
CHEM 102, Spring 15 LA TECH
17-22
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 of (reactants)
CHEM 102, Spring 15 LA TECH
17-23
Calculating DGorxn
NH4NO3(s) + heat  NH4NO3(aq)
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 kJ
Reaction is product-favored in spite of positive DHorxn.
Reaction is “entropy driven”
CHEM 102, Spring 15 LA TECH
17-24
Calculating DGorxn
DG
o
o
o
D
D
rxn = S Gf (products) - S Gf (reactants)
Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
Method (b) :
DGorxn = DGfo(CO2) - [DGfo(graph) + DGfo(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
17-25
CHEM 102, Spring 15 LA TECH
Calculation of DGo
We can calculate DGo values from DHo and DSo
values at a constant temperature and
pressure.
Example.
Determine DGo for the following reaction at 25oC
Equation
N2 (g) + 3H2 (g)
2NH3 (g)
DHfo, kJ/mol
So, J/K.mol
CHEM 102, Spring 15 LA TECH
0.00
191.50
0.00
130.68
-46.11
192.3
17-26
Predict the spontaneity of the
following processes from DH and DS
at various temperatures.
a)DH = 30 kJ, DS = 6 kJ, T = 300 K
b)DH = 15 kJ,DS = -45 kJ,T = 200 K
CHEM 102, Spring 15 LA TECH
17-27
a) DH = 30 kJ
DS = 6 kJ
b) DH = 15 kJ
DS = -45 kJ T = 200 K
DG = DHsys-TDSsys or DG = DH - TDS.
DH
= 30 kJ
DS
= 6 kJ
T
= 300 K
DG
= 30 kJ - (300 x 6 kJ)
= 30 -1800 kJ
DG
= -1770 kJ
T = 300 K
DG = DHsys-TDSsys or DG = DH - TDS.
DH
= 15 kJ
DS
= -45 kJ
T
= 200 K
DG
= 15 kJ -[200 (-45 kJ)]
= 15 kJ -(-9000) kJ
DG
= 15 + 9000 kJ = 9015 kJ
CHEM 102, Spring 15 LA TECH
17-28
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
CHEM 102, Spring 15 LA TECH
17-29
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
∆G o
system
∆H o
CHEM 102, Spring 15 LA TECH
system
∆S o
system
T
17-30
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
a)
a)
a)
a)
system
∆H
system
∆S system
Negative always
Negative at low T
Positive at high T
Positive at low T
Negative at high T
Negative
(exothermic)
Negative
(exothermic)
Positive
(endothermic)
Positive
Negative
Positive always
Positive
(endothermic)
Negative
CHEM 102, Spring 15 LA TECH
Positive
T
Yes
∆ G =- ,at low T;
∆ G= +, at high T
∆ G = + ,at low T;
∆ G= -, at high T
∆
G= +, at any T
17-31
Effect of Temperature on Reaction
Spontaneity
CHEM 102, Spring 15 LA TECH
17-32
o
DG
=
CHEM 102, Spring 15 LA TECH
o
DH
-
o
TDS
17-33
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:
CHEM 102, Spring 15 LA TECH
17-34
How do you calculate DG at different T
and P
DG = DGo + RT ln Q
Q = reaction quotient
at equilibrium DG = 0
0 = DGo + RT ln K
DGo = - RT ln K
If you know DGo you could calculate K
CHEM 102, Spring 15 LA TECH
17-35
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
CHEM 102, Spring 15 LA TECH
17-36
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
CHEM 102, Spring 15 LA TECH
17-37
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?
CHEM 102, Spring 15 LA TECH
17-38
Calculate the DG value for the following
reactions using:
D Go = SD Gof (products) - S D Gof (reactants)
N2O5 (g)
+ H2O(l) ------>
D Gfo[ N2O5 (g) ] = 134 kJ/mole ;
DGfo[ HNO3(l) ] = -81 kJ/mole
2 HNO3(l) ;
DGo = ?
D Gfo [H2O(g)] = -237 kJ/mole;
N2O5 (g)
+
H2O(l) ------> 2 HNO3(l) ; DGo = ?
DGfo 1 x 134
1 x (-237)
2 (-81)
134
-237
-162
DGo = DGof (products) - 3 DGof (reactants)
DGo = [-162] - [134 + (-237)]
DGo = -162 + 103
DGo = -59 kJ/mole
The reaction have a negative DG and the reaction is
spontaneous or will take place as written.
CHEM 102, Spring 15 LA TECH
17-39
Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
DHorxn = +467.9 kJ
DSorxn = +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 = DHorxn - TDSorxn
DH rxn
467.9 kJ
T=
=
= 835.1 K
DSrxn 0.5603 kJ/K
CHEM 102, Spring 15 LA TECH
17-40
Thermodynamics and 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
CHEM 102, Spring 15 LA TECH
17-41
Free energy and equilibrium
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 =
DGo
-RT
ln K = 13.27;
CHEM 102, Spring 15 LA TECH
=
-32.91 kJ
-(0.008315 kJ.K-1mol-1)(298.2K)
K = 5.8 x 105
17-42
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 Gfo[ NH3(g) ] = -17 kJ/mole
N2 (g) + 3 H2 (g)
2 NH3 (g); D G = ? at 300
K,
PN2 = 300,
N2 (g)
PNH3 = 75 and PH2 = 300
+ 3 H2 (g)
CHEM 102, Spring 15 LA TECH
2 NH3 (g); DG = ?
17-43
To calculate DGo
Using DGo = S DGof (products) - S DGof (reactants)
DGfo[ N2(g)] = 0 kJ/mole; DGfo[ H2(g)] = 0
kJ/mole; DGfo[ NH3(g)] = -17 kJ/mole
Notice elements have DGfo = 0.00 similar to DHfo
N2 (g) + 3 H2 (g)
2 NH3 (g); DG = ?
DGfo
0
0
2 x (-17)
0
0
-34
DGo = S DGof (products) - S DGof (reactants)
DGo = [-34] - [0 +0]
DGo = -34
DGo = -34 kJ/mole
CHEM 102, Spring 15 LA TECH
17-44
To calculate Q
Equilibrium expression for the reaction in terms of partial pressure:
N2 (g) +
3 H2 (g)
2 NH3 (g)
p2NH3
K
=
_________
pN2 p3H2
p2NH3
Q
=
_________ ;
pN2 p3H2
Q is when initial concentration is substituted into the equilibrium
expression
752
Q
=
_________ ; p2NH3= 752; pN2 =300; p3H2=3003
300 x 3003
Q=
6.94 x 10-7
CHEM 102, Spring 15 LA TECH
17-45
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
CHEM 102, Spring 15 LA TECH
17-46
Thermodynamics and Keq
o
DG rxn = - RT lnK
Calculate K (from DG0)
N2O4 --->2 NO2
DGorxn = +4.8 kJ
DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
ln K = = - 1.94
(8.31 J/K)(298K)
K =o 0.14
DGorxn > 0 : K < 1
DG rxn < 0 : K > 1
CHEM 102, Spring 15 LA TECH
17-47
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
CHEM 102, Spring 15 LA TECH
17-48
DG as a Function of the
Extent of the Reaction
CHEM 102, Spring 15 LA TECH
17-49
DG as a Function of the
Extent of the Reaction
when there is Mixing
CHEM 102, Spring 15 LA TECH
17-50
Maximum Work
DG = wsystem = - wmax
(work done on the surroundings)
CHEM 102, Spring 15 LA TECH
17-51
Coupled Reactions
How 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)
CHEM 102, Spring 15 LA TECH
17-52
ADP and ATP
CHEM 102, Spring 15 LA TECH
17-53
Acetyl Coenzyme A
CHEM 102, Spring 15 LA TECH
17-54
Gibbs Free Energy and Nutrients
CHEM 102, Spring 15 LA TECH
17-55
Photosynthesis: Harnessing
Light Energy
CHEM 102, Spring 15 LA TECH
17-56
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)
CHEM 102, Spring 15 LA TECH
=>
2Na(s) + 2Cl2(g)
17-57