1. THERMOCHEMISTRY OR THERMODYNAMICS
Chapter 6

2. Chemical Reactivity What drives chemical reactions? How do they occur?
The first is answered by THERMODYNAMICS and the second by KINETICS.
We have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
• electron transfer in a battery

3. Energy and Chemistry
ENERGY is the capacity to do work or transfer heat.
HEAT is the form of energy that flows between 2 samples because of their difference in temperature.
Other forms of energy —
light electrical nuclear
kinetic potential

4. Law of Conservation of Energy
Energy can be converted from one form to another but can neither be created nor destroyed.
(Euniverse is constant)

5. (m = mass, g = acceleration of gravity, and h = height)
The Two Types of Energy
Potential: due to position or composition - can be converted to work
PE = mgh
(m = mass, g = acceleration of gravity, and h = height)
Kinetic: due to motion of the object
KE = 1/2 mv2
(m = mass, v = velocity)

6. Kinetic and Potential Energy
Potential energy — energy a motionless body has by virtue of its position.

7. Kinetic and Potential Energy
Kinetic energy — energy of motion.
• Translation
• Rotation
• Vibration

8. Units of Energy
1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food “calorie”)
But we use the unit called the JOULE
1 cal = joules
James Joule

9. Temperature v. Heat
Temperature reflects random motions of particles, therefore related to kinetic energy of the system.
Heat involves a transfer of energy between 2 objects due to a temperature difference

10. Extensive & Intensive Properties
Extensive properties depends directly on the amount of substance present.
mass
volume
heat
heat capacity (C)
Intensive properties is not related to the amount of substance.
temperature
concentration
pressure
specific heat (s)

11. State Function
Depends only on the present state of the system - not how it arrived there.
It is independent of pathway.
Energy change is independent of the pathway (and, therefore, a state function), while work and heat are dependent on the pathway.

12. System and Surroundings
System: That on which we focus attention
Surroundings: Everything else in the universe
Universe = System + Surroundings

13. Exo and Endothermic Heat exchange accompanies chemical reactions.
Exothermic: Heat flows out of the system (to the surroundings).
Endothermic: Heat flows into the system (from the surroundings).

14. ENERGY DIAGRAMS
Exothermic
Endothermic

15. Endo- and Exothermic qsystem > 0 w > 0 ENDOTHERMIC heat
Surroundings
System
heat
qsystem > 0
w > 0
E goes up
ENDOTHERMIC

16. Endo- and Exothermic qsystem > 0 w > 0 qsystem < 0 w < 0
Surroundings
Surroundings
System
System
heat
heat
qsystem > 0
w > 0
qsystem < 0
w < 0
E(system) goes down
E(system) goes up
ENDOTHERMIC
EXOTHERMIC

17. First Law
First Law of Thermodynamics: The energy of the universe is constant.

18. Enthalpy DH = Hfinal - Hinitial
If Hfinal > Hinitial then DH is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then DH is negative
Process is EXOTHERMIC

19. First Law E = q + w E = change in system’s internal energy q = heat
w = work

20. Piston moving a distance against
a pressure does work.

21. Work
P = F/A
F = PA
w = F h
w = PA h
w = P V
w & V must have opposite signs, since work is being done by the system to expand the gas.
wsystem = -P V
1 Latm = J
1 J = kgm2/s2

22. Enthalpy Enthalpy = H = E + PV E = H  PV H = E + PV
At constant pressure,
qP = E + PV,
where qP = H at constant pressure
H = energy flow as heat (at constant pressure)

23. Some Heat Exchange Terms
specific heat capacity (s)
heat capacity per gram = J/°C g or J/K g
molar heat capacity (s)
heat capacity per mole = J/°C mol or J/K mol

24. Specific Heat Capacity
Substance Spec. Heat (J/g•K)
H2O Al
glass 0.84
Aluminum

25. Ho = - qp qp = mst q = heat (J) m = mass (g) Simple Calorimeter
s = specific heat (j/gCo)
t = “change” in temperature (Co)
Ho = “change in” enthalpy (kJ)
Simple Calorimeter

26. Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
where DT = Tfinal - Tinitial
heat gain/lost = q = m s DT

27. Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
where DT = Tfinal - Tinitial
q = (0.902 J/g•K)(25.0 g)( )K
q = J
heat gain/lost = q = m s DT

28. Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
q = J
Notice that the negative sign on q signals heat “lost by” or transferred out of Al.

29. Bomb Calorimeter

30. Heat Capacity E = qv & qv = -(C t + ms t)
E = “change in” internal energy (J)
qv = heat at constant volume (J)
C = heat capacity (J/Co)
t = “change”in temperature (Co)

31. In bomb calorimetry, volume is constant so:
REMEMBER!!!
In regular calorimetry pressure is constant, but the volume will change so:
qp = -H
qp = E + p  V
In bomb calorimetry, volume is constant so:
qv = E
since p  V = zero.

32. Measuring Heats of Reaction CALORIMETRY
Calculate heat of combustion of octane.
C8H /2 O2 --> 8 CO H2O
• Burn 1.00 g of octane
Temp rises from to oC
Calorimeter contains 1200 g water
Heat capacity of bomb = 837 J/K
Hcm = -(Ct + mst) where Hc is heat of combustion.

33. Measuring Heats of Reaction CALORIMETRY
Step 1 Calc. heat transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2 Calc. heat transferred from reaction to bomb.
q = C t
= (837 J/K)(8.20 K) = 6860 J
Step 3 Total heat evolved
41,170 J J = 48,030 J
Heat of combustion of 1.00 g of octane = kJ

34. Hess’s Law Reactants  Products
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

35. Standard States Compound For a gas, pressure is exactly 1 atmosphere.
For a solution, concentration is exactly 1 molar.
Pure substance (liquid or solid), it is the pure liquid or solid.
Element
The form [N2(g), K(s)] in which it exists at atm and 25°C.

36. Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g) H = 540 kJ

37. Using Enthalpy Consider the decomposition of water
H2O(g) kJ ---> H2(g) + 1/2 O2(g)
Endothermic reaction — heat is a “reactant”
DH = kJ

38. Using Enthalpy Making H2 from H2O involves two steps.
H2O(l) + 44 kJ ---> H2O(g)
H2O(g) kJ ---> H2(g) + 1/2 O2(g)
H2O(l) kJ --> H2(g) + 1/2 O2(g)
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others, the net DH is the sum of the DH’s of the other rxns.

39. Using Enthalpy Calc. DH for S(s) + 3/2 O2(g) --> SO3(g)
S(s) + O2(g) --> SO2(g) kJ
SO2(g) + 1/2 O2(g) --> SO3(g) kJ
_______________________________________
S(s) + 3/2 O2(g) --> SO3(g) kJ

40.  DH along one path =  DH along another path energy S solid
direct path DH = 1 +O 2 kJ
+ 3/2 O 2
DH = SO
gas 2 kJ
+ 1/2 O 2 SO
gas 3 DH
= kJ 2
 DH along one path =
 DH along another path

41. This equation is valid because DH is a STATE FUNCTION
DH along one path =
DH along another path
This equation is valid because DH is a STATE FUNCTION
These depend only on the state of the system and not how it got there.
Unlike V, T, and P, one cannot measure absolute H. Can only measure DH.

42. Standard Enthalpy Values
NIST (Nat’l Institute for Standards and Technology) gives values of
DHof = standard molar enthalpy of formation
This is the enthalpy change when 1 mol of compound is formed from elements under standard conditions. DHof is always stated in terms of moles of product formed.
See Appendix A21-A24.

43. DHof, standard molar enthalpy of formation
H2(g) + 1/2 O2(g) --> H2O(g)
DHof = kJ/mol
By definition, DHof = 0 for elements in their standard states.

44. Using Standard Enthalpy Values
Use DH°’s to calculate enthalpy change for
H2O(g) + C(graphite) --> H2(g) + CO(g)
(product is called “water gas”)

45. Using Standard Enthalpy Values
H2O(g) + C(graphite) --> H2(g) + CO(g)
From reference books we find
H2(g) + 1/2 O2(g) --> H2O(g)
DH°f of H2O vapor = kJ/mol
C(s) + 1/2 O2(g) --> CO(g)
DH° f of CO = kJ/mol

46. Using Standard Enthalpy Values
H2O(g) --> H2(g) + 1/2 O2(g) DHo = +242 kJ
C(s) + 1/2 O2(g) --> CO(g) DHo = -111 kJ
H2O(g) + C(graphite) --> H2(g) + CO(g)
DHonet = +131 kJ
To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy.
The “water gas” reaction is ENDOthermic.

47. Change in Enthalpy
Can be calculated from enthalpies of formation of reactants and products.
Hrxn° = npHf(products)  nrHf(reactants)
H is an extensive property--kJ/mol
For the reaction: 2H2 (g) + O2 (g) ---> 2H2O(g)
Enthalpy would be twice the H value for the combustion of hydrogen.

48. Using Standard Enthalpy Values
Calculate the heat of combustion of methanol, i.e., DHorxn for
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
DHorxn =  DHof (prod) -  DHof (react)

49. Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
DHorxn =  DHof (prod) -  DHof (react)
DHorxn = DHof (CO2) + 2 DHof (H2O)
- {3/2 DHof (O2) + DHof (CH3OH)}
= ( kJ) + 2 ( kJ)
- {0 + ( kJ)}
DHorxn = kJ per mol of methanol
DHorxn is always in terms of moles of reactant.

50. Pathway for the Combustion of Methane

51. Schematic diagram of the energy changes for the
combustion of methane.

52. Greenhouse Effect
-- a warming effect exerted by the earth’s atmosphere due to
thermal energy retained by absorption of infrared radiation.
Greenhouse Gases:
CO2 H2O
CH4 N2O