1. Chapter 2 Elements of Chemical Thermodynamics Atkins: Ch6 & 7
Silberberg: Ch 6 & 20

2. Thermodynamics is the study of heat and its transformations.
Thermochemistry is a branch of thermodynamics that deals with
the heat involved with chemical and physical changes.
Fundamental premise
When energy is transferred from one object to another,
it appears as work and/or as heat.
For our work we must define a system to study; everything else
then becomes the surroundings.
The system is composed of particles with their own internal energies (U or E).
Therefore the system has an internal energy. When a change occurs, the
internal energy changes.

3. A chemical system and its surroundings.
the surroundings
the system

4. System & Surroundings
System - the specific part of the universe that is of interest in the study. often the reacting vessel in the chemistry lab
A beaker of NaOH and HCl
Surroundings - the rest of the universe
When heat is given off from the beaker, the energy is transferred out of the system

6. open
closed
isolated
Exchange:
mass & energy
energy
nothing

7. Exothermic & Endothermic Reactions
Exothermic - transfer of heat from the system to the surroundings
2H2(g) + O2(g)  2H2O(l) + energy
Endothermic - the transfer of heat from the surroundings to the system
energy + 2HgO(s)  2Hg(l) + O2(g)

8. Endothermic vs. Exothermic

9. Copyright McGraw-Hill 2009
Internal Energy
Internal energy: 2 components
Kinetic energy - molecular motion
Potential energy - attractive/repulsive interactions
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10. Copyright McGraw-Hill 2009
Internal Energy
We cannot calculate the changes in internal energy with any certainty
We can calculate the changes in energy of the system experimentally
U = U(products) - U(reactants)‏
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11. DU = Ufinal - Uinitial = Uproducts - Ureactants
Energy diagrams for the transfer of internal energy (U) between a system and its surroundings.
DU = Ufinal - Uinitial = Uproducts - Ureactants

12. DUuniverse = DUsystem + DUsurroundings
Units of Energy
Joule (J)
1 J = 1 kg*m2/s2
Calorie (cal)
1 cal = 4.18J
British Thermal Unit
1 Btu = 1055 J

13. First Law of Thermodynamics
Energy can be converted from one form to another but cannot be created nor destroyed
(the Law of Energy Conservation)‏
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14. Energy and the Universe
Usystem + Usurroundings = 0
When a system releases heat, some of the chemical energy is released as thermal energy to the surroundings but this does not change the total energy of the universe.
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15. Energy and the Universe
Usystem = - Usurroundings
When a system undergoes a change in energy, the surroundings must undergo a change in energy of equal magnitude but opposite in sign.
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16. Copyright McGraw-Hill 2009
Work and Heat
Usys = q + w
“q” = heat
“w” = work
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17. Pressure-volume work.

18. + = + + + + - - + - - - q w DU The Sign Conventions* for q, w and DU
depends on sizes of q and w - +
depends on sizes of q and w - - -
* For q: + means system gains heat; - means system loses heat.
* For w: + means work done on system; - means work done by system.

19. Copyright McGraw-Hill 2009
Work and Heat
Calculate the overall change in internal energy for a system that absorbs 125 J of heat and does 141 J of work on the system.
“q” = + (heat absorbed)
“w” = - (work done)
Usys = q + w = (+125 J) + (-141 J)‏
= -16 J
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20. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy
, pressure, volume, temperature
DU = Ufinal - Uinitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

21. Copyright McGraw-Hill 2009
Constant Volume
Constant volume: (V = 0)
U = q - PV
qv = U
(not easy to achieve in the chem lab)
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22. Copyright McGraw-Hill 2009
Constant Pressure
Constant pressure: (common in the chem lab)‏
U = q + w
U = qp - PV
qp = U + PV
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23. Enthalpy and Enthalpy Changes
Enthalpy (H) is a state function
Enthalpy measures heat content
Constant pressure process: (most are)‏
qp = H
Enthalpy of reaction:
H = H(products) - H(reactants)
H = - (exothermic) H = + (endothermic)
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24. Thermochemical Equations
Equations that represent mass and enthalpy relationships
H2O(s)  H2O(l) rH = kJ/mol
This is an endothermic process. It requires 6.01 kJ to melt one mole of ice.
(the “per mole” means per mole of reaction as written - respect stoichiometry)
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25. Thermochemical Equation Guidelines
Always specify state of reactants and products.
When multiplying an equation by a factor (n), multiply the H value by same factor.
Reversing an equation changes the sign but not the magnitude of the H.
Copyright McGraw-Hill 2009

26. Enthalpy and the First Law of Thermodynamics
DU = q + w
q = DH and w = -PDV
At constant pressure:
DU = DH - PDV
DH = DU + PDV

27. The Meaning of Enthalpy
w = - PDV
DH ≈ DU in
H = U + PV
1. Reactions that do not involve gases.
where H is enthalpy
2. Reactions in which the number of moles of gas does not change.
DH = DU + PDV
3. Reactions in which the number of moles of gas does change but q is >>> PDV.
qp = DU + PDV = DH

28. Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
H2O(l) H2O(g)
CH4 + 2O2
H2O(g)
Hfinal
Enthalpy, H
Hinitial
Enthalpy, H
heat out
heat in
DH < 0
DH > 0
CO2 + 2H2O
H2O(l)
Hfinal
Hinitial
A Exothermic process
B Endothermic process

30. The standard enthalpy of reaction (DrH0 ) is the enthalpy of a reaction carried out at 1 atm.
aA + bB cC + dD
DrH0
dDfH0 (D)
cDfH0 (C) = [ + ] -
bDfH0 (B)
aDfH0 (A)
DrH0
nDfH0 (products) = S
mDfH0 (reactants) -
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

31. Sample Problem 2.1
Using Hess’s Law to Calculate an Unknown DH
PROBLEM:
Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:
CO(g) + NO(g) CO2(g) + 1/2N2(g) DH = ?
Given the following information, calculate the unknown DH:
Equation A: CO(g) + 1/2O2(g) CO2(g) DHA = kJ/mol
Equation B: N2(g) + O2(g) NO(g) DHB = kJ/mol
PLAN:
Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g) CO2(g) DHA = kJ/mol
NO(g) /2N2(g) + 1/2O2(g)
DHB = kJ/mol
CO(g) + NO(g) CO2(g) + 1/2N2(g)
DHrxn = kJ/mol

32. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of formation (DfH0) as a reference point for all enthalpy expressions.
Standard enthalpy of formation (DfHm) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its most stable form is zero.
Df H m (O2) = 0
DfH0 (C, graphite) = 0
DfH0 (O3) = 142 kJ/mol
DfH0 (C, diamond) = 1.90 kJ/mol

33. Selected Standard Heats of Formation at 250C(298K)
Formula
DfHm (kJ/mol)
calcium
Ca(s)
CaO(s)
CaCO3(s)
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
chlorine
Cl(g)
-635.1
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
hydrogen
nitrogen
oxygen
Formula
DfHm (kJ/mol)
H(g)
H2(g)
N2(g)
NH3(g)
NO(g)
O2(g)
O3(g)
H2O(g)
H2O(l)
Cl2(g)
HCl(g)
-92.3
218
-45.9
90.3
143
-241.8
-285.8
107.8
Formula
DfHm (kJ/mol)
silver
Ag(s)
AgCl(s)
sodium
Na(s)
Na(g)
NaCl(s)
sulfur
S8(rhombic)
S8(monoclinic)
SO2(g)
SO3(g)
-127.0
-411.1 2
-296.8
-396.0

34. The general process for determining DrH0 from DfH0 values.
Elements
decomposition
formation
-DfH0
DfH0
Enthalpy, H
Reactants
Hinitial
DrH0
Products
Hfinal
DrH0 = S mDfH0 (products) - S nDfH0 (reactants)

35. Sample Problem 2.2
Calculating the Heat of Reaction from Heats of Formation
PROBLEM:
Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:
Calculate DH0rxn from DH0f values.
4NH3(g) + 5O2(g) NO(g) + 6H2O(g)
PLAN:
Look up the DH0f values and use Hess’s Law to find DHrxn.
SOLUTION:
DHrxn = S mDfH0 (products) - S nDfH0 (reactants)
DrH = [4(DfH0 NO(g) + 6(DfH0 H2O(g)]
- [4(DfH0 NH3(g) + 5(DfH0O2(g)]
= (4)(90.3 kJ/mol) + (6)( kJ/mol) -
[(4)(-45.9 kJ/mol) + (5)(0 kJ/mol)]
DrH = -906 kJ/mol

36. Some Important Types of Enthalpy Change
heat of combustion (DcombH)
C4H10(l) /2O2(g) CO2(g) + 5H2O(g)
heat of formation (DfH)
K(s) + 1/2Br2(l) KBr(s)
heat of fusion (DfusH)
NaCl(s) NaCl(l)
heat of vaporization (DvapH)
C6H6(l) C6H6(g)

37. Limitations of the First Law of Thermodynamics
DU = q + w
Uuniverse = Usystem + Usurroundings
DUsystem = -DUsurroundings
DUsystem + DUsurroundings = 0 = DUuniverse
The total energy-mass of the universe is constant.
However, this does not tell us anything about the direction of change in the universe.