1. Assign.#6.6 – Enthalpies of Bonds, Phase Changes, and Reactions

2. Bond Strengths
The strength of a covalent bond is determined by the energy required to break it
Bond enthalpies is the ΔH for the breaking of a certain bond
Bond enthalpies are usually given as energy to break a given bond between two atoms
Always a positive number
Always requires energy to break a bond

4. Which Bond is Harder to Break? Why?#2 #1 #3

5. Single, Double, and Triple Bonds
The values of ΔH change based on whether the bond is single, double, or triple.
It takes more energy to break a double bond compared to a single bond and more energy to break a triple bond compared to a double bond.
Due to increased pi bonds that require more energy input in order to break
Also, bonds get shorter the more we have, making it harder for breaking to occur.

6. Heating Curves
Have students label what each of these regions indicate (solid, liquid, gas, etc.)

7. Enthalpies of Phase Change
When a substance changes phases, there is energy that is required in order to break intermolecular forces
The enthalpies of phase change are dependent upon the substance of interest:
ΔHvaporization, ΔHfusion, ΔHfreezing, etc.

8. Signs for ΔH for Phase Changes
To turn something from a solid to a liquid and/or a liquid to a gas requires energy. Therefore it is endothermic.
This means ΔH is always positive
To turn something from a gas to a liquid and/or a liquid to a solid releases energy. Therefore it is exothermic.
This means ΔH is always negative
The value for vaporization and condensation is the same, the only difference is the sign.

9. Enthalpy Cartoon
Create a cartoon creating of the different types of enthalpy that exist:
Enthalpy
Enthalpy of Bond Strengths
Enthalpy of Phase Changes
Enthalpy of Reactions

10. End Catalyst Why are bond enthalpies always positive values?
The ΔHvaporization = 890 kJ for a particular sample. What is the ΔHcondensation?
What is ΔH equivalent to at constant pressure?
End

11. Hydrogen Fire Ball

12. Justify – TPS
Why is there a large release of energy for this reaction? Use ΔH in your response.

13. Enthalpies of Reactions
Since ΔH means Hfinal – Hinitial, this means that for a reaction it is:
ΔHrxn = Hproducts – Hreactants
Every reaction has a enthalpy (or heat) of reaction ΔHrxn
-ΔHrxn = exothermic
+ΔHrxn = endothermic

14. Major Ideas of ΔHrxn
ΔHrxn is proportionate to amount of reactant consumed
When a reaction is run in reverse, the ΔHrxn is equal in magnitude, but opposite in sign
ΔHrxn depends on the state of the substance

15. CH4 (g) + 2 O2 (g)  CO2 (g) + 2H2O (l) ΔH =-890 kJ/mol
Class Example
Calculate the qrxn when 4.50 g of methane (CH4) is combusted at constant pressure. Note – For the reaction:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2H2O (l) ΔH =-890 kJ/mol

16. 2 H2O2 (l)  2 H2O (l) + O2 (g) ΔH = -196 kJ/mol
Table Talk
Hydrogen peroxide can decompose to water and oxygen by the reaction:
2 H2O2 (l)  2 H2O (l) + O2 (g) ΔH = -196 kJ/mol
Calculate the qrxn when 5.00 g of H2O2 decomposes at constant pressure.

17. ΔH as a State Function
All ΔH values are state functions, meaning it only depends on the amount of substance and the initial state of that substance.
The number of steps between initial and final state does not matter, only the difference between initial and final enthalpy values.
Example: Combustion of methane

18. Hess’s Law
If a reaction is carried out in a series of steps, ΔH for the reaction equations the sum of the individual steps
This law means that we do not need to measure ΔH for all known reactions, rather we only need to measure for a some reactions and then we can add equations together to get the equation of interest.

19. Class Example
The ΔHrxn for a combustion of C to CO2 is kJ/mol, and the enthalpy for the combustion of CO to CO2 is kJ/mol:
(1) C (s) + O2 (g)  CO2 (g) ΔH = kJ/mol
(2) CO (g) + ½ O2 (g)  CO2 (g) ΔH = -283 kJ/mol
Using these data, calculate the enthalpy for the combustion of C to CO:
(3) C (s) + ½ O2 (g)  CO (g) ΔH = ?

20. C (graphite)  C (diamond) ΔH = ?
Table Talk
(1) C (graphite) + O2 (g)  CO2 (g) ΔH = kJ/mol
(2) C (diamond) + O2 (g)  CO2 (g) ΔH = kJ/mol
Calculate ΔH for the conversion of graphite to diamond:
C (graphite)  C (diamond) ΔH = ?

21. Challenge Problem! Calculate ΔH for the reaction:
2 C (s) + H2 (g)  C2H2 (g)
Given the following chemical equations and their respective enthalpy changes:
C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (l) ΔH = kJ/mol
C (s) + O2 (g)  CO2 (g) ΔH = kJ/mol
H2 (g) + ½ O2 (g)  H2O (l) ΔH = kJ/mol

22. Around the World
Complete the 8 questions that are around the room

23. 2 C (s) + O2 (g) + 4H2 (g)  2CH3OH (g) ΔH = -402 kJ/mol
Exit Ticket
From the enthalpies of reaction:
2 C + O2  CO (g) ΔH = kJ/mol
2 C (s) + O2 (g) + 4H2 (g)  2CH3OH (g) ΔH = -402 kJ/mol
Calculate the ΔH for the reaction:
CO (g) + 2 H2 (g)  CH3OH (g)

24. Rate Yourself
Rate yourself 1 – 4 on LTs 5.6 to 5.8.

26. Closing Time
Read 5.4, 5.6, and 8.8
Do book problems: 5.36, 5.41, 5.43, 5.44, 5.62, 5.63, 5.64, 5.65, 5.66