1. Thermochemistry

2. A. Phase Changes

3. A. Phase Changes Evaporation
molecules at the surface gain enough energy to overcome IMF
Volatility
measure of evaporation rate
depends on temp & IMF

4. A. Phase Changes Equilibrium
trapped molecules reach a balance between evaporation & condensation

5. directly related to volatility
A. Phase Changes
Vapor Pressure
pressure of vapor above a liquid at equilibrium
v.p.
depends on temp & IMF
directly related to volatility
temp

6. A. Phase Changes Boiling Point
temp at which v.p. of liquid equals external pressure
depends on Patm & IMF
Normal B.P. - b.p. at 1 atm

7. A. Phase Changes Melting Point equal to freezing point
Which has a higher m.p.?
polar or nonpolar?
covalent or ionic?
polar
ionic

8. EX: dry ice, mothballs, solid air fresheners
A. Phase Changes
Sublimation
solid  gas
v.p. of solid equals external pressure
EX: dry ice, mothballs, solid air fresheners

9. B. Heating Curves Gas - KE  Boiling - PE  Liquid - KE 
Melting - PE 
Solid - KE 

10. energy required to raise the temp of 1 gram of a substance by 1°C
B. Heating Curves
Temperature Change
change in KE (molecular motion)
depends on heat capacity
Heat Capacity
energy required to raise the temp of 1 gram of a substance by 1°C

11. energy required to melt 1 gram of a substance at its m.p.
B. Heating Curves
Phase Change
change in PE (molecular arrangement)
temp remains constant
Heat of Fusion (Hfus)
energy required to melt 1 gram of a substance at its m.p.

12. EX: sweating, steam burns, the drinking bird
B. Heating Curves
Heat of Vaporization (Hvap)
energy required to boil 1 gram of a substance at its b.p.
usually larger than Hfus…why?
EX: sweating, steam burns, the drinking bird

13. C. Phase Diagrams
Show the phases of a substance at different temps and pressures.

14. Thermal Energy
A. Temperature & Heat
1. Temperature is related to the average kinetic energy of the particles in a substance.

15. 2. SI unit for temp. is the Kelvin
a. K = C (10C = 283K)
b. C = K – 273 (10K = -263C)
3. Thermal Energy – the total of all the kinetic and potential energy of all the particles in a substance.

16. 4. Thermal energy relationships
a. As temperature increases, so does thermal energy (because the kinetic energy of the particles increased).
b. Even if the temperature doesn’t change, the thermal energy in a more massive substance is higher (because it is a total measure of energy).

17. Cup gets cooler while hand gets warmer
5. Heat
a. The flow of thermal energy from one object to another.
b. Heat always flows from warmer to cooler objects.
Ice gets warmer while hand gets cooler

18. a. Some things heat up or cool down faster than others.
6. Specific Heat
a. Some things heat up or cool down faster than others.
Land heats up and cools down faster than water

19. b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K).
1) C water = 4184 J / kg C
2) C sand = 664 J / kg C
This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

20. Why does water have such a high specific heat?
water metal
Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

21. How to calculate changes in thermal energy
q = m x c x  T
q = change in thermal energy
m = mass of substance
T = change in temperature (Tf – Ti)
c = specific heat of substance
Heat loss (released) = heat gained (absorbed)
-q = + q

22. First, mass and temperature of water are measured
c. A calorimeter is used to help measure the specific heat of a substance.
First, mass and temperature of water are measured
Knowing its Q value, its mass, and its T, its Cp can be calculated
Then heated sample is put inside and heat flows into water
This gives the heat lost by the substance
T is measured for water to help get its heat gain

23. Project Description and Outline DUE FRIDAY! MAY 2nd .
HAPPY Monday!
PRE-AP
Take out your notebooks!!
Turn in any late work/check your boxes for returned quiz and returned redox homework.
Heat Capacity Homework DUE WEDNESDAY
REGULAR
Take out your Enthalpy Homework – we are going to go over some of the problems.
Enthalpy Hmk DUE TOMORROW.
Pre-TEST TOMORROW!!
TEST ON THURSDAY
REMINDERS
Project Description and Outline DUE FRIDAY! MAY 2nd .

24. STUDY FOR PRE-TEST LeChatelier’s Principle (Disney ride analogy!)
Add too much to L, move to right
Take away from L, move to left
Phase Changes
Heating Curve
Phase Change Diagram
Exothermic/Endothermic
Vocab for phase changes (sublimation etc)

25. 3. Basic Temperature Vocabulary
Kinetic Energy and Potential Energy
Thermal Energy
Intermolecular forces

26. 4. q and heat capacity problems
1. Q = amount of heat transferred (neg/pos)
2. C = specific heat
3. Units come from the C value
q = mc T

27. 5. Enthalpy Problems
1. H = (Products added) – (Reactants added)
2. Multiply by the coefficents
3. Elements in basic form are 0.

28. TWO Trends in Nature
Order  Disorder
 
High energy  Low energy 

29. 2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + H2O (s) H2O (l)
energy + 2HgO (s) Hg (l) + O2 (g)
6.2

30. DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0
DH > 0
6.4

31. Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0 and is positive
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DH = 6.01 kJ
6.4

32. Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0 and is negative
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DH = kJ
6.4

33. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ/mol ΔH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 mol x 6.01 kJ/mol = 12.0 kJ
6.4

34. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DHreaction = kJ
1 mol P4
123.9 g P4 x
3013 kJ
1 mol P4 x
266 g P4
= 6470 kJ
6.4

35. Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (O2) = 0 f
DH0 (C, graphite) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f
6.6

36. 6.6

37. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn = S
DH0 (products) f - S
DH0 (reactants) f
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.)
6.6

38. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
DH0 (products) f = S
DH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12 × × ] – [ 2 × ] = kJ
-6535 kJ
2 mol
= kJ/mol C6H6
6.6

39. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn
6.6

40. Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol
1 cal = J
1 Cal = 1000 cal = 4184 J

41. DHsoln = Hsoln - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
6.7

42. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7

43. Energy Diagrams Exothermic Endothermic
Activation energy (Ea) for the forward reaction
Activation energy (Ea) for the reverse reaction
(c) Delta H
50 kJ/mol
300 kJ/mol
150 kJ/mol
100 kJ/mol
-100 kJ/mol
+200 kJ/mol

44. Entropy (S) is a measure of the randomness or disorder of a system.
If the change from initial to final results in an increase in randomness
DS > 0
For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s) H2O (l)
DS > 0
18.3

45. First Law of Thermodynamics
Energy can be converted from one form to another but energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
18.4

46. Entropy Changes in the System (DSsys)
The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C.
rxn
aA + bB cC + dD
DS0
rxn
dS0(D)
cS0(C) = [ + ] -
bS0(B)
aS0(A)
DS0
rxn
S0(products) = S
S0(reactants) -
What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) CO2 (g)
S0(CO) = J/K•mol
S0(CO2) = J/K•mol
S0(O2) = J/K•mol
DS0
rxn
= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0
rxn
= – [ ] = J/K•mol
18.4

47. Entropy Changes in the System (DSsys)
When gases are produced (or consumed)
If a reaction produces more gas molecules than it consumes, DS0 > 0.
If the total number of gas molecules diminishes, DS0 < 0.
If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number.
What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) ZnO (s)
The total number of gas molecules goes down, DS is negative.
18.4

48. Spontaneous Physical and Chemical Processes
A waterfall runs downhill
A lump of sugar dissolves in a cup of coffee
At 1 atm, water freezes below 0 0C and ice melts above 0 0C
Heat flows from a hotter object to a colder object
A gas expands in an evacuated bulb
Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

49. For a constant-temperature process:
Gibbs Free Energy
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
For a constant-temperature process:
Gibbs free energy (G)
DG = DHsys -TDSsys
DG < The reaction is spontaneous in the forward direction.
DG > The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = The reaction is at equilibrium.
18.5

50. DG = DH - TDS
18.5

51. DG0 of any element in its stable form is zero.
The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.
rxn
aA + bB cC + dD
DG0
rxn
dDG0 (D) f
cDG0 (C) = [ + ] -
bDG0 (B)
aDG0 (A)
DG0
rxn
DG0 (products) f = S
DG0 (reactants) -
Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f
DG0 of any element in its stable form is zero. f
18.5

52. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DG0
rxn
DG0 (products) f = S
DG0 (reactants) -
DG0
rxn
6DG0 (H2O) f
12DG0 (CO2) = [ + ] -
2DG0 (C6H6)
DG0
rxn =
[ 12x– x–237.2 ] – [ 2x124.5 ] = kJ
Is the reaction spontaneous at 25 0C?
DG0 = kJ
< 0
spontaneous
18.5

53. Recap: Signs of Thermodynamic Values
Negative
Positive
Enthalpy (ΔH)
Exothermic
Endothermic
Entropy (ΔS)
Less disorder
More disorder
Gibbs Free Energy (ΔG)
Spontaneous
Not spontaneous

54. Heat (q) absorbed or released:
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = msDt
q = CDt
Dt = tfinal - tinitial
6.5

55. Dt = tfinal – tinitial = 50C – 940C = -890C
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt
= 869 g x J/g • 0C x –890C
= -34,000 J
6.5

56. Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = msDt
qcal = CcalDt
Reaction at Constant P
DH = qrxn
No heat enters or leaves!
6.5

57. Sample Problem
How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = J = 0.59 kJ
Step 2: Convert the solid to liquid ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid Q=mcΔT
Q = 36g x J/g deg C x 100 deg C = J = 15 kJ

58. Now, add all the steps together
Sample Problem
How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas ΔH vaporization
Q = 2.0 mol x kJ/mol = kJ
Step 5: Heat the gas Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ kJ = 118 kJ