Slide 1

• Check your answers for Lewis Structures with
one another. Make sure you all come to a
consensus of the correct answers for each.
• Make sure you handed in all homework
assignments from previous days.
• Go through the rest of this Powerpoint.
• Try your best on the homework at the end of this
Powerpoint. I will go over it in more detail in
class on Monday. If you get stuck on any, save
them for Monday and we’ll work on them as a
group.


Slide 2

Last Topic of Chapter

Bond Enthalpy


Slide 3

C
C
C

C
C
C

Bond length
154 pm
134 pm
120 pm

Bond energy
346 kJ/mol
602 kJ/mol
835 kJ/mol

Since this bond is stronger, it requires
more energy to break this bond.

Bond Enthalpy
Energy needed to break a bond in 1
mole of a gaseous substance.


Slide 4

The strength of a bond is also related to it’s stability.
Consider N2 versus Cl2.

N

N

Cl

Cl

Bond Length
1.10 A

Bond Length
1.96 A

Bond Enthalpy
941 KJ / mol

Bond Enthalpy
242 KJ / mol

More Stable

Less Stable


Slide 5

*Bond enthalpies are determined experimentally using
calorimetry.

*Determining bond enthalpies for diatomic molecules is a
straight forward experimental process.

*When dealing with large molecules, individual bond
enthalpies can be calculated by considering the average of
each bond in the molecule
Example:


Slide 6

ΔH of breaking methane into it’s original atoms (atomization) is
experimentally determined to be 1660 KJ / mol

H
H

C

H

C (g) + 4H (g)

H
Average bond enthalpy of C-H bond is therefore
1660 / 4 = 415 KJ / mol
Experiments of other C-H containing molecules yield similar results.


Slide 7

A Return to Hess’s Law
*Bond enthalpies can be used to determine heats of reactions ΔHrxn.
*Allows us to see what causes a reaction to be endo. or exo.

Basic Concept
Reactant + Reactant  Product + Product
Energy required to
Break Bonds

Energy released when
new bonds form

The difference between the two determines
endothermic versus exothermic.
 H rxn   ( bond enthalpies

of bonds broken   ( b ond enthalpies

of bonds formed )


Slide 8

Bond Energy
Table 8.4 lists values of some bond energies.
(Handout)
Should be over on heater by window.

To illustrate, let’s estimate the heat of reaction for
reaction between methane and chlorine gas.

CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)


Slide 9

CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)
In this reaction, one C-H bond and one Cl-Cl
bond must be broken.
ΔH  1(C
 H)one
 1(Cl
 Cl)
and
1(Cone
 H-Cl
Cl) bond
1(Hare Cl)
In turn,
C-Cl
bond
formed.
H  (414  243  339  431) kJ / mol

H  113 kJ / mol


Slide 10

Notice: Only 1 C-H bond is broken on the left because
¾ C-H bonds appear as the same C-H bonds on the
product side. Since these stay the same, there is no
change in enthalpy associated with these.


Slide 11

Several Things to note:
1) Equation must be balanced.
2) Must draw Lewis Structures to see bonds between atoms.
3) To calculate bond enthalpy, simply determine how many
and of what type of bonds are broken on the left and how
many and of what type of bonds are formed on the right.
Then calculate: Enthalpy of bonds broken – enthalpy of
bonds formed.

4) A single C-C bond has a different enthalpy than a double
C=C bond. (true for all elements).
5) Must look enthalpies up in tables.

6) If a coefficient appears in the equation before a
compound, you must multiply that coefficient by the bonds
broken or formed.
Ex: 2CH4 = 8 CH bonds broken. = 6(414) KJ


Slide 12

Homework
8.61-8.63, 8.65a, 8.66a


Slide 13

Bond Enthalpies Can Be Found in Data Tables