1. THERMOCHEMISTRY or Thermodynamics
To play the movies and simulations included, view the presentation in Slide Show Mode.

2. Geothermal power —Wairakei North Island, New Zealand

3. Energy & Chemistry
Burning peanuts supply sufficient energy to boil a cup of water.
Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)

4. Energy & Chemistry These reactions are PRODUCT FAVORED
They proceed almost completely from reactants to products, perhaps with some outside assistance.

5. Energy & Chemistry 2 H2(g) + O2(g) --> 2 H2O(g) + heat and light
This can be set up to provide ELECTRIC ENERGY in a fuel cell.
Oxidation:
2 H2 ---> 4 H e-
Reduction:
4 e- + O H2O ---> 4 OH-
CCR, page 845

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

7. Potential & Kinetic Energy
Potential energy — energy a motionless body has by virtue of its position.

8. Potential Energy on the Atomic Scale
Positive and negative particles (ions) attract one another.
Two atoms can bond
As the particles attract they have a lower potential energy
NaCl — composed of Na+ and Cl- ions.

9. Potential Energy on the Atomic Scale
Positive and negative particles (ions) attract one another.
Two atoms can bond
As the particles attract they have a lower potential energy

10. Potential & Kinetic Energy
Kinetic energy — energy of motion
• Translation

11. Potential & Kinetic Energy
Kinetic energy — energy of motion.

12. Internal Energy (E) PE + KE = Internal energy (E or U)
Int. E of a chemical system depends on
number of particles
type of particles
temperature

13. Internal Energy (E)
PE + KE = Internal energy (E or U)

14. Internal Energy (E) The higher the T the higher the internal energy
So, use changes in T (∆T) to monitor changes in E (∆E).

15. Heat transfers until thermal equilibrium is established.
Thermodynamics
Thermodynamics is the science of heat (energy) transfer.
Heat energy is associated with molecular motions.
Heat transfers until thermal equilibrium is established.

16. Directionality of Heat Transfer
Heat always transfer from hotter object to cooler one.
EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.
T(system) goes down
T(surr) goes up

17. Directionality of Heat Transfer
Heat always transfer from hotter object to cooler one.
ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.
T(system) goes up
T (surr) goes down

18. Energy & Chemistry All of thermodynamics depends on the law of
CONSERVATION OF ENERGY.
The total energy is unchanged in a chemical reaction.
If PE of products is less than reactants, the difference must be released as KE.

19. Energy Change in Chemical Processes
PE of system dropped. KE increased. Therefore, you often feel a T increase.

20. 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

21. HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C.
Which has the larger heat capacity?

22. Specific Heat Capacity
How much energy is transferred due to T difference?
The heat (q) “lost” or “gained” is related to
a) sample mass
b) change in T and
c) specific heat capacity

23. Specific Heat Capacity
Substance Spec. Heat (J/g•K)
H2O
Ethylene glycol 2.39 Al
glass
Aluminum

24. 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?

25. 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?
heat gain/lose = q = (sp. ht.)(mass)(∆T)
where ∆T = Tfinal - Tinitial
q = (0.897 J/g•K)(25.0 g)( )K
q = J
Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

26. Heat Transfer No Change in State
q transferred = (sp. ht.)(mass)(∆T)

27. Heat Transfer with Change of State
Changes of state involve energy (at constant T)
Ice J/g (heat of fusion) -----> Liquid water
q = (heat of fusion)(mass)

28. Heat Transfer and Changes of State
Liquid ---> Vapor
Requires energy (heat).
This is the reason
a) you cool down after swimming
b) you use water to put out a fire.
+ energy

29. Heating/Cooling Curve for Water
Evaporate water
Heat water
Note that T is constant as ice melts
Melt ice

30. Heat & Changes of State
What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC?
Heat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/g
+333 J/g
+2260 J/g

31. Heat & Changes of State 1. To melt ice
What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC?
1. To melt ice
q = (500. g)(333 J/g) = x 105 J
2. To raise water from 0 oC to 100 oC
q = (500. g)(4.2 J/g•K)( )K = 2.1 x 105 J
3. To evaporate water at 100 oC
q = (500. g)(2260 J/g) = x 106 J
4. Total heat energy = 1.51 x 106 J = 1510 kJ

32. Chemical Reactivity What drives chemical reactions? How do they occur?
The first is answered by THERMODYNAMICS and the second by KINETICS.
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

33. Chemical Reactivity
But energy transfer also allows us to predict reactivity.
In general, reactions that transfer energy to their surroundings are product-favored.
So, let us consider heat transfer in chemical processes.

34. Heat Energy Transfer in a Physical Process
CO2 (s, -78 oC) ---> CO2 (g, -78 oC)
Heat transfers from surroundings to system in endothermic process.

35. Heat Energy Transfer in a Physical Process
CO2 (s, -78 oC) ---> CO2 (g, -78 oC)
A regular array of molecules in a solid > gas phase molecules.
Gas molecules have higher kinetic energy.

36. ∆E = E(final) - E(initial)
Energy Level Diagram for Heat Energy
Transfer
CO2 gas
∆E = E(final) - E(initial)
= E(gas) - E(solid)
CO2 solid

37. Heat Energy Transfer in Physical Change
CO2 (s, -78 oC) ---> CO2 (g, -78 oC)
Two things have happened!
Gas molecules have higher kinetic energy.
Also, WORK is done by the system in pushing aside the atmosphere.

38. FIRST LAW OF THERMODYNAMICS
heat energy transferred
∆E = q + w
work done
by the
system
energy
change
Energy is conserved!

39. ∆E = q + w SYSTEM heat transfer in (endothermic), +q heat transfer out
(exothermic), -q
SYSTEM
∆E = q + w
w transfer in
(+w)
w transfer out
(-w)

40. ENTHALPY ∆H = Hfinal - Hinitial
Most chemical reactions occur at constant P, so
Heat transferred at constant P = qp
qp = ∆H where H = enthalpy
and so ∆E = ∆H + w (and w is usually small)
∆H = heat transferred at constant P ≈ ∆E
∆H = change in heat content of the system
∆H = Hfinal - Hinitial

41. ENTHALPY ∆H = Hfinal - Hinitial Process is ENDOTHERMIC
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC

42. USING ENTHALPY Consider the formation of water
H2(g) + 1/2 O2(g) --> H2O(g) kJ
Exothermic reaction — heat is a “product” and ∆H = – kJ

43. USING ENTHALPY
Making liquid H2O from H2 + O2 involves two exothermic steps.
H2 + O2 gas
H2O vapor
Liquid H2O

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

45. Hess’s Law & Energy Level Diagrams
Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
Figure 6.18, page 227

46. Hess’s Law & Energy Level Diagrams
Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
Figure 6.18, page 227

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

48. Standard Enthalpy Values
Most ∆H values are labeled ∆Ho
Measured under standard conditions
P = 1 bar
Concentration = 1 mol/L
T = usually 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas

49. Enthalpy Values
Depend on how the reaction is written and on phases of reactants and products
H2(g) + 1/2 O2(g) --> H2O(g)
∆H˚ = -242 kJ
2 H2(g) + O2(g) --> 2 H2O(g)
∆H˚ = -484 kJ
H2O(g) ---> H2(g) + 1/2 O2(g)
∆H˚ = +242 kJ
H2(g) + 1/2 O2(g) --> H2O(liquid)
∆H˚ = -286 kJ

50. Standard Enthalpy Values
NIST (Nat’l Institute for Standards and Technology) gives values of
∆Hfo = standard molar enthalpy of formation
— the enthalpy change when 1 mol of compound is formed from elements under standard conditions.
See Table 6.2 and Appendix L

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

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

53. 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)
∆Hf˚ of H2O vapor = kJ/mol
C(s) + 1/2 O2(g) --> CO(g)
∆Hf˚ of CO = kJ/mol

54. Using Standard Enthalpy Values
H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ
C(s) + 1/2 O2(g) --> CO(g) ∆Ho = -111 kJ
H2O(g) + C(graphite) --> H2(g) + CO(g)
∆Honet = +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.

55. Using Standard Enthalpy Values
In general, when ALL enthalpies of formation are known,
Calculate ∆H of reaction?
∆Horxn =
 ∆Hfo (products)
-  ∆Hfo (reactants)
Remember that ∆ always = final – initial

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

57. Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
∆Horxn =  ∆Hfo (prod) -  ∆Hfo (react)
∆Horxn = ∆Hfo (CO2) + 2 ∆Hfo (H2O)
- {3/2 ∆Hfo (O2) + ∆Hfo (CH3OH)}
= ( kJ) + 2 ( kJ)
- {0 + ( kJ)}
∆Horxn = kJ per mol of methanol

58. Measuring Heats of Reaction
CALORIMETRY
Measuring Heats of Reaction
Constant Volume “Bomb” Calorimeter
Burn combustible sample.
Measure heat evolved in a reaction.
Derive ∆E for reaction.

59. Calorimetry Total heat evolved = qtotal = qwater + qbomb
Some heat from reaction warms water
qwater = (sp. ht.)(water mass)(∆T)
Some heat from reaction warms “bomb”
qbomb = (heat capacity, J/K)(∆T)
Total heat evolved = qtotal = qwater + qbomb

60. Measuring Heats of Reaction CALORIMETRY
Calculate heat of combustion of octane. C8H /2 O2 --> 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

61. 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 = (bomb heat capacity)(∆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