1. Enthalpy, Entropy, and Gibb’s Free Energy
Unit theme: Energy

2. Collision Theory The reactant particles must collide with each other.
The collisions must be of enough energy to overcome the activation energy
The reactants must form new bonds to produce products.

3. Activation Energy Defined as:
The minimum energy required to start a chemical reaction.
Everything has an activation energy

4. Activation energy for a reaction is shown on reaction profile diagrams

5. Changing the Rate of a Chemical Reaction
To change the rate of a reaction one or more of the following things must happen:
Increase the number of collisions between the reactant particles
Increase the energy of the collisions.
Decrease the activation energy.

6. A catalyst reduces the activation energy of a reaction.
Effect of a Catalyst
A catalyst is a substance that increases the speed of a reaction, without being used up. A catalyst can be ‘recovered at the end of a reaction and used again.
A catalyst reduces the activation energy of a reaction.

7. Effect of a Catalyst Catalyst = faster reaction.
Activation energy without catalyst.
The lower activation energy in the presence of a catalyst means the reaction will be faster. More of the collisions have enough energy to react. There is a lower ‘energy barrier’.
energy
Activation energy with catalyst.
Catalyst = faster reaction.

8. Enthalpy of the reaction
The total energy change associated with a chemical or physical change
Given the symbol DHrxn
DHrxn = (energy of products) – (energy of reactants)

9. Exothermic reactions Energy is released to the surroundings
The temperature of the surroundings increases
The products have less energy than the reactants

10. Endothermic Reactions
Energy is absorbed from the surroundings
The temperature of the surroundings decreases
The products have MORE energy than the reactants

11. Thermochemical Equations
Exothermic reactions
Energy released by system
Can treat energy as a product
DH is negative
2 equivalent ways to write equation:
CaO + H2O  Ca(OH) kJ
CaO + H2O  Ca(OH) DH = kJ

12. Thermochemical Equations
Endothermic reactions
Energy absorbed by system
Can treat energy as a reactant
DH is positive
2 equivalent ways to write equation:
2 NaHCO kJ  Na2CO3 + H2O + CO2
2 NaHCO3  Na2CO3 + H2O + CO2 DH = kJ

13. 2 ways to manipulate thermochemical equations
1) Write equation backwards
Sign of DH must change
A + B  C DH = kJ
C  A + B DH = kJ
2) Multiply everything by a coefficient
Must multiply DH by coefficient, too!
3 A + 3B  3C DH = 3 (+123 kJ) = kJ

14. Problems with thermochemical equations
Consider the equation:
2 NaHCO3  Na2CO3 + H2O + CO2
DH = kJ
How many kJ would be required if 4.5 moles NaHCO3 reacted?

15. Hess’ Law
If you add two or more thermochemical equations to give a final equation, then add the heat changes to give the final enthalpy of reaction.

16. Example
What is the enthalpy change, DHrxn, for the decomposition of hydrogen peroxide?
Target: 2 H2O2(l)  2 H2O(l) + O2(g)
Given:
H2(g) + O2(g)  H2O2(l) DH = kJ
H2(g) + ½ O2(g)  H2O(l) DH = kJ

17. Standard Heats of Formation, DHof
The standard heat of formation of a compound is the change in enthalpy that accompanies the formation of one mole of a substance from its elements in their standard states.
The heat of formation of elements in their standard states is arbitrarily set to zero.

18. Using heats of formation
Thermodynamic stability: a measure of the energy required to decompose the compound
Compounds with large, negative enthalpies of formation are thermodynamically stable
Another way to do Hess’ Law!
DHrxn = SnDHf(products) – SmDHf(reactants)
where
n represents the coefficients for the products
m represents the coefficients for the reactants

19. Example Problems Compute DHrxn for the following reaction.
2NO(g) + O2(g)  2 NO2(g)
4 FeO(s) + O2(g)  2 Fe2O3(s)
Ans: kJ, kJ
Compound
Delta H kJ/mole
NO (g)
90.4
NO2 (g)
33.9
O2 (g)
FeO (s)
-226.5
Fe2O3 (s)
-822.2

20. Entropy A quantitative measure of the degree of disorder in a system
Symbol: S
A quantitative measure of the degree of disorder in a system
The greater the disorder, the larger the value of S
Solids have a high degree of order (low entropy)
Gases have a low degree of order (high entropy)
More particles (moles) results in higher entropy

21. Entropy, cont. Systems tend to proceed to higher disorder Examples
Stirring sugar into your coffee
The neatness of your room
The order of cards in a pack of playing cards after shuffling

22. Entropy, Cont
Can use entropy the same was as enthalply in Hess’ law. DSrxn = SnDS(products) – SmDS(reactants) C(graphite) (s) + O2 (g)  CO2 (g)
Determine S(rxn) from the equation above.
Compound
Delta S J/mole K
C (graphite) (s)
5.9
O2 (g)
205
CO2 (g)
213.8

23. Will a reaction occur spontaneously?
The answer depends on the balance between enthalpy and entropy
Gibb’s Free Energy
The energy available from the system to do useful work
DG = DH – TDS
Delta H is given in Kilojoules and Delta S is given in Joules so you MUST convert Delta S to kilojoules every time or you’ll get EVERY answer incorrect.
T must be in Kelven. Add 273 to the degrees C you’re given.

24. Gibb’s Free Energy DG = DH – TDS
If DG is negative, the reaction will occur spontaneously and can proceed on its own.
If DG is positive, the reaction is nonspontaneous and needs a sustained energy input to proceed.
If DG is zero, the reaction is at equilibrium (both the forward and reverse reactions take place!)

25. Gibb’s Free Energy