1. CH Notes

2. So… about Thermal Energy What’s up with Temperature vs Heat?
Temperature is related to the average kinetic energy of the particles in a substance.

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

4. Thermal energy relationships
As temperature increases, so does thermal energy (because the kinetic energy of the particles increased).
If the temperature stays the same, the thermal energy in a more massive substance is higher (because it is a total measure of energy).

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

6. Things heat up or cool down at different rates.
Specific Heat (c)
Things heat up or cool down at different rates.
Land heats up and cools down faster than water, and aren’t we lucky for that!?

7. Specific heat is the amount of heat required to raise the temperature of 1 kg (but in Chem we use g) of a material by one degree (C or K, they’re the same size).
C water = 4184 J / kg C (“holds” its heat)
C sand = 664 J / kg C (less E to change)
This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

8. Why does water have such a high specific heat?
water metal
Water molecules form strong bonds with each other water molecule; (including H-bonds!)so it takes more heat energy to break the bonds. Metals have weak bonds (remember the “sea of e-) and do not need as much energy to break them.

9. Q = m x C x T WHERE’S THE MATH, MATE?! Q= change in thermal energy
m= mass of substance
C= specific heat of substance
ΔT= change in temperature (Tf – Ti)

10. Specific Heat Capacity
If 25.0 g of Al cool from oC to 37.0 oC, how many joules of heat energy are lost by the Al?
heat gain/lose = q = (mass)(c)(∆T)
where ∆T = Tfinal - Tinitial
q = (25.0 g)(0.897 J/g•°C)( )°C
q = J
Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

11. q transferred = (mass)(c)(∆T)
Heat can be Transferred even if there is No Change in State
q transferred = (mass)(c)(∆T)

12. Or… Heat Transfer can cause a Change of State
Changes of state involve energy (at constant T)
Ice J/g (heat of fusion) -----> Liquid water
Is there an equation? Of course!
q = (heat of fusion)(mass)

13. Heat Transfer and Changes of State
Liquid (l)  Vapor (g)
Requires energy (heat).
Why do you…
cool down after swimming ???
Evaporation leads to cooling!
use water to put out a fire???
1. Cools wood down by absorbing energy to vaporize the liquid water.
2. also, it creates a barrier between the wood and it’s fuel source (O2)

14. Note that T is constant as a phase changes
Remember this – it’s the Heating/Cooling Curve for Water!
Evaporate water
Note that T is constant as a phase changes

15. q transferred = (c)(mass)(∆T)
So, let’s look at this equation again…
q = (heat of fusion)(mass)
(There’s also q = (heat of vaporization)(mass), by the way, for when we are talking about vaporization)
WHY DO I NEED THIS WHEN I HAVE
q transferred = (c)(mass)(∆T)
HUH???
Well, when a phase changes THERE IS NO change in temperature… but there is definitely a change in energy!

16. So… if I want the total heat to take ice and turn it to steam I need 3 steps…
1) To melt the ice I need to multiply the heat of fusion with the mass…q = (heat of fusion)(mass)

17. 2) Then, there is moving the temperature from 0 C to 100C… for this there is a change in temperature so we can use… q transferred = (c)(mass)(∆T)
3) But wait, that just takes us to 100 C, what about vaporizing the molecules? Well, then we need q = (heat of vaporization)(mass)…

18. Add ‘em all up and there it is!
Now, lucky for us, just like there are tables for specific heats, there are also tables for heats of fusion and heats of vaporization.
Whew,
At least we don’t have to worry about that!

20. Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•°C
Heat & Changes of State
What quantity of heat is required to melt 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•°C
Heat of vaporization = 2260 J/g
+333 J/g
+2260 J/g

21. And now… More! Heat & Changes of State
How much heat is required to melt g of ice and heat the water to steam at oC?
1. To melt ice
  q  =  (333 J/g)(500.0 g)  =  1.67 x 105 J
2. To raise water from 0.0 oC to oC
  q = (500.0 g)(4.2 J/g•°C)(100.0 – 0.0)°C  =  2.1 x 105 J
3. To evaporate water at oC
  q  =  (2260 J/g)(500.0 g)  =  1.13 x 106 J
4. Total heat energy = 1.5 x 106 J = 1500 kJ

22. Aka… How we Measure Heats of Reaction
CALORIMETRY
Aka… How we Measure Heats of Reaction
In a Constant Volume or “Bomb” Calorimeter, we
burn a combustible sample.
From the heat change , we measure heat evolved in a reaction to get ∆E for reaction.

23. Complete worksheets

24. Le Chatelier’s Principle
Chemistry

25. Le Chatelier’s Principle
If a stress is applied to a system at equilibrium (this would include liquids and gases only), the system will change to relieve that stress and re –establish equilibrium

26. Factors that Affect Equilibrium
Concentration
Temperature
Pressure
For gaseous systems only!
The presence of a catalyst

27. Concentration Changes
Add more reactant  Shift to products
Remove reactants  Shift to reactants
Add more product  Shift to reactants
Remove products  Shift to products

28. Temperature Changes Exothermic Reactions
Consider heat as a product in exothermic reactions.
Add heat 
Shift to reactants
Remove heat 
Shift to products
A + B = AB + Heat

29. Temperature Changes Endothermic Reactions
Consider heat as a reactant in endothermic reactions.
Add heat 
Shift to products
Remove heat 
Shift to reactants
A + B + heat = AB

30. Pressure Changes
Only affects equilibrium systems with unequal moles of gaseous reactants and products.

31. N2(g) + 3H2(g) = 2NH3(g)
Increase Pressure
Stress of pressure is reduced by reducing the number of gas molecules in the container

32. N2(g) + 3H2(g) = 2NH3(g)
There are 4 molecules of reactants vs. 2 molecules of products.
Thus, the reaction shifts to the product ammonia.

33. PCl5(g) = PCl3(g) + Cl2(g) Decrease Pressure
Stress of decreased pressure is reduced by increasing the number of gas molecules in the container.

34. PCl5(g) = PCl3(g) + Cl2(g)
There are two product gas molecules vs. one reactant gas molecule.
Thus, the reaction shifts to the products.

35. Presence of a Catalyst
A Catalyst lowers the activation energy and increases the reaction rate.
Therefore, a catalyst has NO EFFECT on a system at equilibrium!
It just gets you to equilibrium faster!

36. Presence of an Inert Substance
An inert substance is a substance that is not- reactive with any species in the equilibrium system.
These will not affect the equilibrium system.
If the substance does react with a species at equilibrium, then there will be a shift!

37. S8(g) + 12O2(g)  8 SO3(g) + 808 kcals What will happen when ……
Given:
S8(g) O2(g)  8 SO3(g) kcals
What will happen when ……
Oxygen gas is added?
The reaction vessel is cooled?
The size of the container is increased?
Sulfur trioxide is removed?
A catalyst is added to make it faster?
Shifts to prodcuts 
Shifts to Products – to replace heat
V increases, Pressure decreases, shifts to more particles – to reactants!
Shift to products to replace it!
No change!

38. 2NaHCO3(s)  Na2CO3 (s) + H2O (g) + CO2(g)
Given
2NaHCO3(s)  Na2CO3 (s) + H2O (g) + CO2(g)
What will happen when
Carbon dioxide was removed?
Sodium carbonate was added?
Sodium bicarbonate was removed?
Shift to products – to replace it
No Change – solids do not affect equilibrium
No Change – solids do not affect equilibrium

39. Ca5(PO4)3OH(s)  5Ca2+(aq) + 3PO43-(aq) + OH- (aq)
Given
Ca5(PO4)3OH(s)    5Ca2+(aq)  + 3PO43-(aq) +  OH- (aq)
What will happen when
Calcium ions are added?
NaOH is added?
Na3PO4(aq) is added?
Shift to the reactants
Adding OH- , shifts to reactants
Adds PO43- ions, shifts to reactants

40. Complete worksheets

41. Topic: Potential Energy Curve

42. Spontaneous Processes
=physical or chemical change that occurs with
 no outside intervention
However, some energy may be supplied to get process started = activation energy
                    Iron rusting
4Fe(s) + 3O2(g)  2Fe2O3(s)
                                                                   H = kJ
Combustion CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -891kJ

43. Can you think of others?

44. Activation Energy & Reaction

45. Activation Energy = Energy needed to initiate reaction
Energy needed to overcome reaction barrier
difference between where reactants start & top of hill
Difference between reactants & activated complex

46. Energy Diagram of a Reaction
Activated Complex = intermediate
formed during conversion from reactants to products
Reactants combine to form an unstable complex

47. Potential Energy Curve: Endothermic
Products have more P.E. than reactants
Start low, end high

48. Potential Energy Curve: Exothermic
Products have less P.E. than reactants
Start high, end low

49. Have to label 6 energies on curve: reactants & products
PE reactants
PE products
PE activated complex
Ea (activation Energy for forward reaction)
Ea (activation Energy for reverse reaction)
H (Enthalpy – heat flow of reaction)

50. What kind of reaction is represented?40 Ea
reverse rxn Ea
forward rxn 30
Enthalpy 
PE activated complex 20
PE products 10 PE
reactants
Time
What kind of reaction is represented?

51. 40 30
Enthalpy  20
H of reaction 10

52. Enthalpy  Time Based on the arrows, draw the curve! Ea Ea forward40 Ea
forward
rxn Ea
reverse
rxn 30
Enthalpy 
PE of activated complex 20
PE of reactants 10
P.E. of products
Time
Based on the arrows, draw the curve!

53. Enthalpy  Time What kind of reaction is represented? Ea Ea forward40 Ea
forward
rxn Ea
reverse
rxn 30
Enthalpy 
PE of activated complex 20
PE of reactants 10
P.E. of products
Time
What kind of reaction is represented?

54. 40 30
Enthalpy  20
H of reaction 10

55. Catalyst (in the body = enzyme)
Substance that increases rate of reaction without itself being consumed
does not participate in reaction
Lowers the activation energy for the reaction

56. DOES IT AFFECT ΔH?
Catalysts do not affect ΔH

57. DOES IT AFFECT ΔH? ΔH
Catalysts do not affect ΔH

58. Complete worksheets

59. Gibbs Free Energy

60. Gibbs free energy, denoted ΔG, combines enthalpy and entropy into a single value. The change in free energy, ΔG, is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system.

61. ΔG can predict the direction of the chemical reaction under two conditions:
constant temperature and
constant pressure.
If ΔG is positive, then the reaction is nonspontaneous (i.e., an the input of external energy is necessary for the reaction to occur) and if it is negative, then it is spontaneous (occurs without external energy input).

62. Gibbs Free Energy Equation ΔG = ΔH - TΔS
A reaction has a standard enthalpy change, ΔH, of kJ/mol at 298 K. The standard entropy change, ΔS, for the same reaction is J/molK. What is the value of ΔG for the reaction in kJ/mol?
ΔG = ? ΔH=+10.00kJ/mol T= 298K ΔS= J/molK = kJ/molK
ΔG = ΔH – TΔS = 10.00kJ/mol – (298K)( kJ/molK) = kJ/mol
POSITIVE ΔG So the reaction is non-spontaneous

63. Gibbs Free Energy Equation ΔG = ΔH - TΔS
The equation for the decomposition of calcium carbonate is given below:
CaCO3(s)  CaO(s) + CO2(g)
At 500K, ΔH for this reaction is kJ/mol. The standard entropy change, ΔS, for the same reaction is kJ/molK. What is the value of ΔG for the reaction in kJ/mol?
ΔG = ? ΔH=177.00kJ/mol T= 500K ΔS= kJ/molK
ΔG = ΔH – TΔS = kJ/mol – (500K)(0.161kJ/molK) = kJ/mol
POSITIVE ΔG So the reaction is non-spontaneous

64. Gibbs Free Energy Equation ΔG = ΔH - TΔS
The equation for the decomposition of calcium carbonate is given below:
N2(g) +3H2(g) ↔ CaO(s) + CO2(g)
At 298K, ΔH for this reaction is kJ/mol. The standard entropy change, ΔS, for the same reaction is kJ/molK. What is the value of ΔG for the reaction in kJ/mol?
ΔG = ? ΔH=-76.00kJ/mol T= 238K ΔS= kJ/molK
ΔG = ΔH – TΔS = kJ/mol – (298K)(-0.199kJ/molK) = kJ/mol
NEGATIVE ΔG So the reaction is spontaneous

65. Gibbs worksheet