1. Energy Thermodynamics
Chapter 6
Energy
Thermodynamics

2. Energy is... Conserved Made of heat and work.
Work is a force acting over a distance
Heat is energy transferred between objects because of temperature difference.
A state function which means that the result is independent of the path, or how you get from point A to B.

3. The universe
Is divided into two halves: the system and the surroundings.
The system is the part you are concerned with.
The surroundings are the rest.
Unfortunately, it is easier to measure the effect on the surroundings than the system directly

4. The universe Exothermic reactions release energy to the surroundings.
Endothermic reactions absorb energy from the surroundings.

5. Heat
Potential energy

6. Heat
Potential energy

7. Direction Every energy measurement has three parts.
A unit ( Joules or calories).
A number how many.
A sign to tell direction.
negative – exothermic
System to the surrounding
positive- endothermic
Surroundings to the system

8. Surroundings
System
Energy
DE <0

9. Surroundings
System
Energy
DE >0

10. Same rules for heat and work
Heat given off is negative.
Heat absorbed is positive.
Work done by system on surroundings is negative.
Work done on system by surroundings is positive.
Thermodynamics- The study of energy and the changes it undergoes.

11. First Law of Thermodynamics
The energy of the universe is constant.
Law of conservation of energy.
q = heat
w = work
DE = q + w
Take the systems point of view to decide signs.
Punch Line: DE = q (At constant P)

12. What is work? Work is a force acting over a distance. w= F x Dd
P = F/ area
d = V/area
w= (P x area) x D (V/area)= PDV
Work can be calculated by multiplying pressure by the change in volume at constant pressure.
units of liter - atm L-atm

13. Work needs a sign
If the volume of a gas increases, the system has done work on the surroundings.
work is negative
w = - PDV
Expanding work is negative.
Contracting, surroundings do work on the system w is positive.
1 L atm = J

14. Example #1
What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure?
volume increase – work is being done by the system on the surroundings (-)
w = - PDV
w = -(2.4atm)(25L-15L)
w = -24L atm
-24 L atm x (101.3J/Latm) = J = -2.4kJ

15. Example #2
If 2.36 J of heat are absorbed by the gas above. what is the change in energy? DE = q + w DE = 2.36 J J DE = J

16. Enthalpy (H) H = E + PV (that’s the definition) DH = DE + PDV
at constant pressure.
DH = DE + PDV
the heat at constant pressure qp can be calculated from
DE = qp + w = qp – PDV (w=-PDV)
qp = DE + P DV = DH
PUNCH LINE qp = DH

17. Calorimetry Measuring heat using a calorimeter.
Two kinds – the first kind being a
Constant pressure calorimeter (called a coffee cup calorimeter)
heat capacity for a material, C is calculated (heat required to change a substances temperature)
C= heat absorbed/ DT = DH/ DT

18. Calorimetry Specific heat capacity = C/mass
Molar heat capacity = C/moles
heat = specific heat x mass x DT
heat = molar heat x moles x DT
Make the units work and you’ve done the problem right.

19. Calorimetry A coffee cup calorimeter measures DH.
The specific heat of water is 1 cal/gºC or J/gC
Heat of reaction= DH = SH x mass x DT

20. Examples
The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.
DH = SH x mass x DT
DH = 0.71J/gC (75000g)(54C)
DH = 2876 kJ

21. Examples DHsurr = -DHsys .203 J/g C = SH
A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?
DHsurr = -DHsys
75.0g (4.184J/g C)(2.2C) = -[46.2g(SH)(-73.6C)]
690J = 3400g C (SH)
.203 J/g C = SH

22. Calorimetry Constant volume calorimeter is called a bomb calorimeter.
Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.
The heat capacity of the calorimeter is known and tested.
Since DV = 0, PDV = 0, DE = q

23. Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb
sample

24. Bomb Calorimeter
A bomb calorimeter works in a similar manner as the coffee cup calorimeter, but there is one significant difference.
In a coffee cup calorimeter, the reaction takes place in the water. In a bomb calorimeter, the reaction takes place in a sealed metal can, which is then placed in the water (contained in an insulated container).

25. Bomb Calorimeter
Analysis of the heat flow is a bit more complex than it was for the coffee cup calorimeter, because the heat flow absorbed by the metal parts of the calorimeter (the “bomb” part) must be taken into account:
qrxn = - (qwater + qbomb)
where qwater = 4.18 J/(g·°C) x mwater x ΔT
The heat flow of the bomb is:
qbomb = Ccal x ΔT

26. One More Twist – Phase Changes
Molar Heat of Fusion – Energy required to change 1 mol from a solid to a liquid (visa versa) 6kJ/mol for water Molar Heat of Vaporization – Energy required to change 1 mol from a liquid to a gas (vise versa) 40.7kJ/mol for water

27. Phase Changes qphasechange + qheating = -qwater
If 20g of ice at 0C is placed in a calorimeter containing 100ml of water at 60C, what is the final temperature of the system?
qphasechange + qheating = -qwater
20g (1mol/18.02g) = 1.10mol
1.10mol(6000J/mol) + 20g(4.184J/gC)(Tf-0) = -100g(4.184J/gC)(Tf-60)
Tf = Tf
Tf = 18504
Tf = 36.8 C

28. Problem Solving Strategies
The 1st step is to ascertain whether the process is constant pressure (open to the atmosphere) or constant volume.
If it’s constant pressure, use ΔH = −ΔEsurr;
for constant volume it’s ΔErxn = −ΔEsurr.
In many problems
ΔEsurr = (mass)(specific heat)ΔT.
To use this equation, you must determine the part of the overall system that is changing temperature.

29. Problem Solving Strategies
For example, if we add 1.0 g of Mg to mL of 1.0 M HCl, it is not the Mg that is changing temperature, but rather the mL of acidic solution in which the Mg is reacting.
In this example, then,
ΔEsurr = (100.0 g)(4.184 J/[g oC])ΔT,
where we have used the usual assumptions stated above. We would not use the mass of Mg (1.0 g) and the specific heat of Mg (1.02 J/[g oC]) in the ΔEsurr equation because it’s not the Mg that is changing temperature.

30. Problem Solving Strategies
Always ask yourself this question: “What is actually changing temperature in this process?”
Wherever the thermometer goes, that’s what is changing temperature.
*Note: in some problems – the heat capacity is given, so you’ll use ΔEsurr = CΔT, where C is heat capacity)

31. Problem Solving Strategies
In the equations ΔH = −ΔEsurr or ΔErxn = −ΔEsurr, the units on both sides are joules (J)!
Therefore, if the problem asks you to find an answer in J or kJ per mole (or g), you need to find J first and then divide by the appropriate moles (or g)
Likewise, if the problem gives you ΔH (or ΔErxn) and asks you to find one of the terms in ΔEsurr, you first need to make sure that the units on ΔH (or ΔErxn) are J and not J per mole or J per gram

32. Hess’s Law Enthalpy is a state function.
We can add equations to to come up with the desired final product, and add the DH
Two rules
If the reaction is reversed the sign of DH is changed
If the reaction is multiplied, so is DH

33. O2
NO2
-112 kJ
H (kJ)
180 kJ
NO2
68 kJ N2
2O2

35. Standard Enthalpy
The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions)
Symbol DHº
When using Hess’s Law, work by adding the equations up to make it look like the answer.
The other parts will cancel out.

36. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= kJ/mol
C(s) + O2(g) → CO2(g) ∆H= kJ/mol
H2(g) + ½O2(g) → H2O(g) ∆H= kJ/mol
Read through the whole question
Plan a Strategy
Evaluate the given equations.
Rearrange and manipulate the equations so that they will produce the overall equation.
Add the enthalpy terms. REWRITE THE CHANGES.

37. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= kJ/mol
C(s) + O2(g) → CO2(g) ∆H= kJ/mol
H2(g) + ½O2(g) → H2O(g) ∆H= kJ/mol

38. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
C(s) + O2(g) → CO2(g) ∆H= kJ/mol
H2(g) + ½O2(g) → H2O(g) ∆H= kJ/mol

39. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
4(C(s) + O2(g) → CO2(g)) ∆H= 4(-393.5kJ/mol)
H2(g) + ½O2(g) → H2O(g) ∆H= kJ/mol

40. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
4C(s) + 4O2(g) → 4CO2(g) distribute the ∆H= 4(-393.5kJ/mol)
H2(g) + ½O2(g) → H2O(g) ∆H= kJ/mol

41. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
4C(s) + 4O2(g) → 4CO2(g) distribute the ∆H= 4(-393.5kJ/mol)
5(H2(g) + ½O2(g) → H2O(g)) distribute the ∆H= 5(-241.8kJ/mol)

42. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol)
5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol)

43. Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol)
5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol)
_____________________________________________________
∆H = kJ/mol

44. Example Given calculate DHº for this reaction DHº= -1300. kJ

45. Example Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ
Calculate DHº for this reaction

46. Standard Enthalpies of Formation
Hess’s Law is much more useful if you know lots of reactions.
Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.
Standard states are 1 atm, 1M and 25ºC
For an element it is 0

47. Standard Enthalpies of Formation
Need to be able to write the equations.
What is the equation for the formation of NO2 ?
½N2 (g) + O2 (g) ® NO2 (g)
Have to make one mole to meet the definition.

48. Since we can manipulate the equations
We can use heats of formation to figure out the heat of reaction.
Lets do it with this equation:
ΔH = ∑ nΔHf (products) – ∑ nΔHf (reactants)  

49. Example
Use the table of standard enthalpies of formation at 25°C to calculate ΔH for the reaction
4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)
ΔH = ∑ nΔHf (products) – ∑ nΔHf (reactants)   
   = [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)]
       = 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0
       = – kJ mol– kJ mol– kJ mol–1
       = –905.2 kJ mol–1

50. Spontaneous Processes
Spontaneous processes are those that can proceed without any outside intervention.
The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously change.

51. Spontaneous Processes
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0C it is spontaneous for ice to melt.
Below 0C the reverse process is spontaneous.

52. Spontaneous Processes
Spontaneity depends on two important issues:
Entropy can be thought of as a measure of the randomness of a system.
Enthalpy is the heat change of a reaction.

53. Entropy is dependant on:
Temperature (phase changes) S(g) > S(l) > S(s)
The number of independently moving molecules
Ssolute < Ssolution
The change in phase usually has a larger affect than the number of molecules.
Polarity provides order to the particles because they align themselves like poles of magnets. Entropy decreases when polar molecules form.
The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes.

54. Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.

55. Standard Entropy
These are molar entropy values of substances in their standard states.
Standard entropies tend to increase with increasing molar mass.

56. Standard Entropies
Larger and more complex molecules have greater entropies.

57. Srxn° = Sf°(products) - Sf°(reactants)
Entropy
Like total energy and enthalpy, entropy is a state function (dependant on only final and initial conditions, not the changes it takes in between).
Entropy changes for a reaction can be estimated in the same manner by which H is estimated:
Srxn° = Sf°(products) - Sf°(reactants)

58. Entropy of Formation Example
Given the values below, calculate the ΔS°rxn of: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
C3H8 S° = J/mol
CO2 S° = J/mol
H2O S° = J/mol
O2 S° = J/mol
[3(213.6) + 4(69.91)] - [ (205.0)] = J/mol
Notice all numbers are nonzero and the units are J/mol
This only tells us that this reaction decreases in entropy or disorder which may be unfavorable.

59. Standard Free Energy Changes
Just like the standard enthalpies and entropies of formation, the standard Gibbs free-energy:
DG = SDG(products)  SG(reactants) f r f
This number can tell you if a reaction is spontaneous or not at a given temperature.
if G°rxn is negative it is spontaneous.
if G°rxn is positive it is not spontaneous.

60. Gibbs Free Energy of Formation Example
Given the values below, calculate the ΔG°rxn of: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
C3H8 ΔGf° = kJ/mol
CO2 ΔGf° = kJ/mol
H2O ΔGf° = kJ/mol
[3(-394.4) + 4( )] - [ (0)] = J/mol
Notice all elements are zero and the units are kJ/mol
This only tells us that this reaction is favorable and spontaneous at THIS temperature. But there is another way…

61. Gibbs in action Example
Recall from previous lessons:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
ΔH°rxn = KJ/mol
Δ S°rxn = J/mol
ΔG°rxn = kJ/mol
You can also calculate G using the equation ΔG= ΔH-TΔS
ΔG = – 298(-.3745) = kJ/mol
Notice the ΔS had to be modified for units.

62. Gibbs changes with Temperature G= H TDS

63. Free Energy and Equilibrium
Under any conditions, standard (298K and 1atm) or nonstandard, the free energy change can be found this way: G = G + RT lnQ

64. Free Energy and Equilibrium
G = G + RT lnQ
At equilibrium, Q = K, and G = 0.
The equation becomes
0 = G + RT lnK
Rearranging, this becomes
G = RT lnK

65. G and K
Recall that if K>1 the products are favored.. If K<1 the reactants are favored.
If G<1 (or negative) the reaction is spontaneous in the forward direction. If G>1 (or positive) the reaction is spontaneous in the reverse direction.
So, G and K are related. When G is negative, K is over 1. When G is positive, K is less than 1.
If K is 1, what is G and what does that mean?

66. Gibbs Final Example
The value of equilibrium constant for a reaction equals 45 at 298K. At the same temperature, the reaction not at equilibrium has Q=35. Determine the Gibbs value at 298K and show that the value of G indicates the same direction as Q states.
Q<K which predicts the reaction will proceed forward.
G = RT lnK = (298)ln(45) = -9.43KJ/mol
G is negative, which is spontaneous and favors the forward direction.
G = G + RT lnQ
= (298)ln(35)
= -620KJ/mol