1. Entropy, Enthalpy, and Gibb’s Free Energy
ENERGY OF REACTIONS
Entropy, Enthalpy, and Gibb’s Free Energy

2. ENERGY SO FAR Review: Energy is the ability to do work or produce heat
There is potential and kinetic energy
Energy can neither be created or destroyed
Every compound needs energy to increase temperature or to change from one state of matter to another

3. ENERGY SO FAR Units of energy: Joule (J) Calorie (cal) 1 cal = 4.184J
1kJ = 1000J
1kcal = 1000cal

4. REVIEW 255 cal = _____________ J 4200 J = ______________cal
Convert the following:
255 cal = _____________ J
4200 J = ______________cal
55kcal = ______________ J
6325 J = ______________ kJ
3.85 kJ = _______________kcal

5. OTHER ASPECTS OF ENERGY
Energy is also an important component of chemical reactions
Example:
4Fe(s) + 3O2(g)  2Fe2O3(s) kJ
In this example, you combine iron and oxygen to produce iron (III) oxide. This reaction also produces 1625kJ of energy/heat.

6. ENERGY AND REACTIONS
To better explain the energy changes in reactions, chemists have come up with:
Enthalpy (H): the heat content of substances under constant pressure
For a chemical reaction, we describe the change in enthalpy. This is called:
Enthalpy (or heat) of reaction (δHrxn): the change in heat or energy in a chemical reaction

7. HOW DO WE MEASURE δHrxn δHrxn = Hproducts - Hreactants
To measure the heat produced or used by a reaction, scientists again use calorimeters. (REVIEW: What is a calorimeter?)
To calculate, we have the following formula:
δHrxn = Hproducts - Hreactants

8. PREVIOUS EXAMPLE Previously, we looked at the following:
4Fe(s) + 3O2(g)  2Fe2O3(s) kJ
According to this equation, we produced (or lost) 1625kJ of energy to the environment (you feel this as hot) - EXOTHERMIC
Therefore Hreactants > Hproducts

9. WHAT THIS MEANS For the reaction:
Whenever Hreactants > Hproducts, the convention is to give a negative sign to the δHrxn
Therefore, exothermic reactions are always a negative “-” δHrxn
On the right side of the arrow

10. ENDOTHERMIC
Since we know that exothermic reactions have a negative heat of reaction:
WHAT IS THE SIGN FOR AN ENDOTHERMIC REACTION?

11. WHAT THIS MEANS Endothermic reactions are always a positive “+” δHrxn
On the left side of the arrow
You have to add energy to the reactant side to equal the products
Therefore we always have to add energy to an endothermic reaction:
Example:
27kJ + NH4NO3 (s)  NH4+(aq) + NO3-(aq)

12. CALCULATING HEATS OF REACTION
Chemists have measured different heats of reaction for combustion reactions.
In each case, they do this under a condition of standard pressure and temperature.
They call these heats of combustion:
δHcomb

13. What is the general format for a combustion reaction?
REVIEW
What is the general format for a combustion reaction?

14. COMBUSTION REACTION Molecule + O2  CO2 + H2O
Therefore, these are values for the combustion of 1 MOLE of the molecule
Example:
We will look at the combustion of 1 mole of glucose (sugar)
C6H12O6 + 6O2  6CO2 + 6H2O kJ
δHcomb = -2808kJ/mole

15. DIMENSIONAL ANALYSIS
How much energy is produced when you combust 2.5 moles of glucose
2.5 moles | kJ = -7020kJ
| 1mole

16. CALCULATING HEAT OF COMBUSTION
Example: You start with 3.55x103 g of glucose (C6H12O6). How much energy is released when glucose goes through a complete combustion reaction?
STEPS:
Convert to moles
Use the δHcomb for glucose to calculate energy
δHcomb = -2808kJ/mole

17. ANSWER δHcomb = -2808kJ/mole Molar mass of glucose: 180g/mole
3.55x103 g | 1 mole | -2808kJ
| 180g | 1 mole
-5.54x104 kJ

18. TRY THESE
The heat of combustion for octane (C8H18) is -5471kJ/mole. If you start with 1550g of octane, how much energy is released?
Sucrose (C12H22O11), or table sugar, has a heat of combustion of -5644kJ/mole. If you add 2.5g of sucrose to your cereal, how much energy will be added to your cereal?

19. ANSWERS
-7.44X104 kJ
-41 kJ

20. TRY THIS
You have a cup filled with 125mL of glucose (C6H12O6). If the density of glucose is 1.54g/mL, how many moles of glucose do you have?
If the heat of combustion for glucose is kJ/mole, what is the heat produced from the cup of glucose?

21. ANSWER
1.07 moles glucose
-3.00 x103 kJ

22. TRY THIS
A combustion reaction with octane (C8H18) releases a total of -5.55x104 kJ of energy. If the heat of combustion for octane is kJ/mole, how many grams of octane did you start with?

23. ANSWER
2.53X104 g

24. HEATS OF REACTION
As we said earlier, most heats of reaction or heats of combustion are measured using a calorimeter. Sometimes (because some reactions are toxic or unstable), we cannot use a calorimeter and must find an alternative way to measure heats of reaction.

25. HEATS OF REACTION
Hess’s law: If you add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction

26. EXAMPLE: FIND THE HEAT OF REACTION
2S + 3O2  2SO3
This reaction can occurs in 2 steps:
S + O2  SO2 ; δH = -297kJ
2SO3  2SO2 + O2 ; δH = 198kJ
How do these reactions go together to form the final reaction?

27. 2S + 3O2  2SO3
Step 1: reverse the second reaction because SO3 is a product. Therefore, you have to change the sign for the heat of reaction
S + O2  SO2 ; δH = -297kJ
2SO2 + O2  2SO3 ; δH =-198kJ
NOTE: Now the second reaction is exothermic

28. 2S + 3O2  2SO3
Step 2: there are 2 moles of S in the first reaction. This means we have to double everything in the first reaction (including the heat of reaction)
2(S + O2  SO2 ; δH = -297kJ)
2S + 2O2  2SO2; δH = -594kJ
2SO2 + O2  2SO3 ; δH =-198kJ

29. 2S + 3O2  2SO3
Step 3: you add the two reactions together to get the final product
2S + 2O2  2SO2; δH = -594kJ
2SO2 + O2  2SO3 ; δH = -198kJ__
2S + 3O2  2SO3; δH =-792kJ__
NOTE: If a compound is on opposite sides of a reaction, they cancel out

30. SUMMARY
You need to have known chemical reactions that you can combine to form the final chemical reaction
You need the δH for each reaction
If you need to reverse a reaction, you must also change the sign of δH
If you need to multiply a reaction by a number, you must also multiply the δH
When all reactions are completed, you then add them together to get the final δH

31. TRY THE FOLLOWING
Combine the two reactions to find the heat of reaction when hydrogen peroxide (H2O2) breaks apart to form water and oxygen)
2H2O2  2H2O + O2; ΔH = ?
2H2 + O2  2H2O; δH = -572kJ
H2 + O2  H2O2; δH = -188kJ

32. 2H2O2  2H2O + O2 Step 1: We need to reverse the second reaction
2H2 + O2  2H2O; δH = -572kJ
H2O2  H2 + O2; δH = 188kJ
Step 2: You need 2 moles of H2O2
2H2O2  2H2 + 2O2; δH = 376kJ

33. 2H2O2  2H2O + O2
Step 3: Combine the reactions
2H2 + O2  2H2O; δH = -572kJ
2H2O2  2H2 + 2O2; δH = 376kJ
2H2O2  2H2O + O2 δH = -196kJ
NOTE: When you cancel out reactants and products, 2H2 cancels out 2H2, but only 1O2 is cancelled out in the products

34. TRY THIS 2 Ca + 2C + 3O2  2CaCO3 Ca + 2C  CaC2; dHrxn = -62.8kJ
CO2  C + O2; dHrxn = -393kJ
2CaCO3 + 2CO2  2CaC2 + 5 O2; dHrxn = +1538kJ

35. OTHER PRACTICE (DID THIS ONE)
4 NH3 + 5 O2  4 NO + 6 H2O
N2 + O2  2 NO
∆H = kJ
N2 + 3 H2  2 NH3
∆H = kJ
2 H2 + O2  2 H2O
∆H = kJ

36. CH4 + NH3  HCN + 3 H2 TRY THIS (DID THIS ONE)
N2 + 3 H2  2 NH3; ∆H = kJ
C + 2 H2  CH4; ∆H = kJ
H2 + 2 C + N2  2 HCN; ∆H = kJ

37. TRY THIS (DID THIS ONE)
N2H4 + H2 → 2NH3 N2H4 + CH4O  CH2O + N2 + 3H2 ΔH = -37 kJ N2 + 3H2 → 2NH3 ΔH = -46 kJ CH4O → CH2O + H2 ΔH = -65 kJ

38. TRY THIS (DID THIS ONE) 2C + 2H2O  CH4 + CO2
C + H2O  CO + H2; ∆H = 131.3kJ
CO + H2O  CO2 + H2; ∆H = -41.2kJ
CH4 + H2O  3H2 + CO; ∆H = 206.1kJ

39. TRY THIS (DID THIS ONE) 5C + 6H2  C5H12 C + O2  CO2 ΔH = -393.5kJ
2H2 + O2  2H2O
ΔH = kJ
C5H12 +8O2  5CO2 + 6H2O
ΔH = -3536kJ

40. PREDICTING SPONTANEOUS CHEMICAL REACTIONS
Energy is an essential ingredient for a chemical reaction
Not all chemical reactions happen spontaneously
To determine if a reaction occurs spontaneously, we need to look at another concept: ENTROPY

41. ENTROPY
Entropy (S): a measure of the number of possible ways that the energy of a system can be distributed
In other words, entropy is the tendency for molecules to spread out as far as possible from each other
Since molecules spreading out is dependent on temperature, the unit for entropy is (J/K) – Joules/Kelvin

42. KELVIN TEMPERATURE
Before we go any further, we need to review Kelvins:
Kelvin (K)
A temperature scale based on absolute 0 (the coldest possible temperature)
The Celsius temperature converts to Kelvin:
K = °C + 273

43. THERMODYNAMICS
Thermodynamics: The laws of thermodynamics, in principle, describe the specifics for the transport of heat and work.
The only law we are interested in is the second law of thermodynamics

44. SECOND LAW OF THERMODYNAMICS
Second Law of Thermodynamics: spontaneous processes always proceed in such a way that the entropy of the universe increases.

45. PREDICTING CHANGES IN ENTROPY
Reminder:
ΔHrxn is the heat of reaction (enthalpy)
We have a similar symbol for entropy
Δ Srxn = the entropy in a reaction
If the entropy increases during a reaction, then Δ Srxn is positive
If the entropy decreases during a reaction, then Δ Srxn is negative

46. PREDICTING CHANGES IN ENTROPY
Changing states of matter: entropy changes when you go between solid, liquid or gas
H2O(l)  H2O (g); δSrxn > 0
Dissolving a gas in a solid or liquid always decreases entropy
When you increase the number of gas particles in a reaction, entropy tends to increase
Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g); δSrxn > 0

47. PREDICTING CHANGES IN ENTROPY
With some exceptions, entropy increases when a solid dissolves in a liquid
The solid tends to break apart
Random motion of particles increases as the temperature increases
As you increase temperature, the entropy increases

48. PREDICT THE FOLLOWING
Try to determine whether entropy increases or decreases for the following reactions:
CF(g) + F2(g)  CF3 (g)
NH3 (g)  NH3 (aq)
C10H8 (s)  C10H8 (l)
H2O (l)  H2 (g) + O2 (g)

49. ANSWERS Entropy decreases: You go from 2 molecules of gas to 1
Entropy decreases: You are dissolving a gas into a liquid
Entropy increases: You have gone from a solid to a liquid
Entropy increases: You are going from no gas to 2 molecules of gas

50. PUTTING IT ALL TOGETHER
Generally, exothermic reactions are spontaneous (you do not need to add energy to make them work)
Generally, spontaneous reactions will increase in entropy
Yet, there is a way to know for sure if a reaction will occur spontaneously or not

51. GIBB’S FREE ENERGY
William Gibbs (an American) combined enthalpy and entropy to determine if a reaction was spontaneous
He called his equation Gibbs Free Energy equation
Free energy: the energy that is available to do work (δG)
The energy that holds a molecule together is called BOND ENERGY
Reactions are either making or breaking these bonds by releasing or adding energy

52. δGrxn = δHrxn – TδSrxn GIBB’S FREE ENERGY Gibbs Free Energy Equation:
δG (kJ): represents free energy
δH (kJ): represents change in enthalpy
T (K): temperature in kelvins
δS (J/K): represents change in entropy

53. EXAMPLE δHrxn = -91.8kJ δSrxn = -197J/K T = 25°C
N2(g) + 3H2(g)  2NH3(g)
δHrxn = -91.8kJ
δSrxn = -197J/K
T = 25°C
Why does the entropy decrease in this reaction?
What is the Gibb’s free energy?
Is this reaction spontaneous?

54. STEP 1: WRITE WHAT YOU KNOW
δG = ?
δH = -91.8kJ
T = 25°C +273 = 298K
δS = -197J/K = kJ/K
NOTE: Temperature must be in Kelvin and entropy is converted to kJ so that it matches the units for δG and δH.

55. STEP 2: PUT THE NUMBERS IN THE FORMULA
δGrxn = δHrxn – TδSrxn
δGrxn = -91.8kJ – (298K)(-0.197kJ/K)
δGrxn = -91.8kJ – (-58.7kJ)
δGrxn = -91.8kJ kJ
δGrxn = -33.1kJ

56. STEP 3: DETERMINE IF IT IS SPONTANEOUS
The reason why the entropy of this reaction decreased is that you are going from 4 moles of gas to 2 moles of gas. The amount of gas is reduced and lowers the entropy.
δGrxn = -33.1kJ
Since the Gibb’s free energy is negative, this is a spontaneous reaction.

57. TRY THIS δHrxn = 145 kJ δSrxn = 322 J/K T = 109°C
A Chemical reaction has the following information:
δHrxn = 145 kJ
δSrxn = 322 J/K
T = 109°C
What is the Gibb’s free energy?
Is this reaction spontaneous?

58. ANSWER
δG = +22kJ
Since the Gibb’s free energy is positive, this reaction is NOT spontaneous
Challenge: At what temperature would this reaction become spontaneous?

59. ANSWER We need to find what temperature δG<0
δHrxn = 145 kJ
δSrxn = 322 J/K
T = 109°C
We need to find what temperature δG<0
0kJ = 145kJ – T(0.322kJ/K)
T(0.322kJ/K) = 145kJ
T = 450K
the reaction has to be at least 450K to be spontaneous

60. TRY THIS
ΔH = 55kJ
ΔS = 225J/K
At what temperature does this reaction become spontaneous?

61. ANSWER REMEMBER: Set the ΔG = 0 0 = 55,000J – T(225J/K)
T(225J/K) = 55,000J
225J/K J/K
T = 244K

62. WHEN ARE REACTIONS SPONTANEOUS?
Reactions are spontaneous when:
δHrxn < 0; δSrxn > 0
δHrxn < 0; δSrxn < 0 (spontaneous at lower temperatures only)
δHrxn > 0; δSrxn > 0 (spontaneous at higher temperatures only
Reactions are not spontaneous when:
δHrxn > 0; δSrxn < 0

63. SPONTANEOUS?
Like entropy or enthalpy, Gibb’s free energy can be positive or negative
If positive (+): if the Gibb’s free energy is positive, the reaction is not spontaneous (you would need to add energy to make the reaction work) - ENDERGONIC
If negative (-): if the Gibb’s free energy is negative, the reaction is spontaneous -EXERGONIC

64. ENERGETICS
The graphs below show how energy is used as a chemical reaction proceeds

65. ENERGETICS AND REACTIONS
Even if spontaneous, reactions don’t just occur
They need energy to start (Energy of activation)
Many times this energy cost is very high and needs to be reduced to make the reaction work
A catalyst is a substance that lowers the amount of energy needed to start a chemical reaction
SEE GRAPH

66. CATALYST

67. HOW CATALYSTS WORK (END HERE)
Action of catalysts
When reactions occur it is because molecules are smashing into each other
Yet, they are like puzzle pieces and have to hit each other at an exact angle – EFFECTIVE COLLISION
This takes energy to move the molecules around enough to hit at this angle
A catalyst is a substance that “holds” molecules together so that it is easier to put them together

68. REACTION RATES
Just because a reaction is spontaneous, it doesn’t mean that you’ll be able to see anything happening
If a reaction is too slow, you won’t be able to see change for a long time
A measure of how fast a reaction goes is called the REACTION RATE
There are several factors that affect how fast a reaction goes and WHICH DIRECTION it goes

69. HOW DO REACTIONS WORK
REMEMBER: When we talked about reactions we mentioned that a reaction can go either direction:
Fe +O2  Fe2O3
The reaction can either go to the left or the right
Actually, it is continually going left AND right

70. HOW DO REACTIONS WORK Fe +O2  Fe2O3
In this reaction iron and oxygen are making iron (III) oxide, but iron (III) oxide is also breaking apart at the SAME TIME
The speed that this happens and which things are made or broken apart depends on many things
The fact that they are constantly doing this is called DYNAMIC EQUILIBRIUM or CHEMICAL EQUILIBRIUM

71. CHEMICAL/DNYAMIC EQUILIBRIUM
Reactions go to a certain ending that gives you a certain concentration of product
At that point you won’t get a higher concentration, but the chemicals are forming or breaking apart at a rate that makes sure that the concentration is the same
The reaction is still proceeding, but the concentrations are stay the same

72. HOW DO YOU KNOW WHICH WAY A REACTION GOES
Le Chatelier was a scientist that studies reaction rates and which direction reactions go
The things he found are called Le Chatelier’s Principles

73. EFFECTS OF CONCENTRATION
PCl5  PCl3 + Cl2
If you add more reactant (PCl5), the reaction shifts to the right
If you add more product (PCl3 OR Cl2), the reaction shifts to the left
SUMMARY: If you add a higher concentration to one side of the reaction it shifts the reaction to make more of the other side

74. PRESSURE/VOLUME N2(g) + H2(g)  NH3(g)
If you increase the pressure OR decrease the volume, the reaction shifts to the side with the fewest molecules (makes NH3)
If you decrease the pressure OR increase the volume, the reaction shifts to the side with the most molecules (N2 and H2)

75. 4Fe(s) + 3O2(g)  2Fe2O3(s) + heat
TEMPERATURE
4Fe(s) + 3O2(g)  2Fe2O3(s) + heat
EXOTHERMIC REACTION:
If the temperature is increased, the reaction will shift to the left (Fe and O2)
If the temperature is decreased, the reaction will shift to the right (Fe2O3)
THE OPPOSITE IS TRUE OF ENDOTHERMIC REACTIONS

76. EFFECTS OF CATLYST A catalyst will not shift the reaction either way
It will just make sure that the reaction reaches equilibrium much faster
Increased reaction rate