Electronegativity Difference And Bond Character Page 3
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1
2
3
4Problem Set on Bonding
Bond energy and the Enthalpy of Reaction,
H
rxn
Bonds are potential energy and every bond therefore has a value of energy called bond energy.
H
n E
n E
rxn
bb
bm
The enthalpy or heat of reaction, H
, may be calculated by determining the difference in bond energy of the bonds broken,
rxn
E
, (reactants), and the bonds made, E
, (products). These values are supplied in a table and are not expected to be
bb
bm
memorized.
For example:
(g) CO
Consider the burning of natural gas: CH
(g) + 3O
(g) + 2H
O.
4
2
2
2
(a) Calculate the enthalpy of reaction.
solution:
H
n
E
n
E
rxn
bb
bm
H
1 [
4 {
C
H
}
3
1 {
O
O
}]
1 [
2 {
O
C
}
2
2 {
H
O
}]
rxn
kJ
kJ
kJ
kJ
1
mol
4
413
3
mol
495
1
mol
2
799
2
mol
2
467
rxn
mol
mol
mol
mol
H
329
kJ
rxn
(b) How much heat would be released when 0.00275 mol of methane, CH
, are burned?
4
solution:
From part-a, the balanced equation contains 1 mol CH
, 3 mol O
, 1 mol CO
and 2 mol H
O.
4
2
2
2
329
kJ
329
kJ
329
kJ
329
kJ
,
,
,
H
So,
1
mol CH
3
mol O
1
mol CO
2
mol H O
rxn
4
2
2
2
0.00275
mol CH
329
kJ
4
H
1
mol CH
rxn
4
0.905
kJ
H
rxn
Formal Charge
When drawing Lewis Structures, often more than one isomer or more than one resonance form (shift of -bonds, but
same -bond framework. The most likely structure is the one that has the electrons distributed most evenly – smallest formal
charge on each atom; and for nonzero formal charges, (-)’s on the more electronegative atoms and (+)’s on the least
electronegative atoms. Covalent molecules do not have charges embedded within them, but they do have localization of charge
density creating poles. These poles can be seen through the formal charges.
The formal charge of any atom in a molecule can be calculated by taking the difference between the number of valence
electrons and the number of electrons around the atom in the structure (lp’s +
bp’s)
1
2
bp e s
'
val e s
'
lp e s
'
FC
2
For example:
Determine the formal charge on the N in nitrate
bp e s
'
val e s
'
lp e s
'
FC
2
FC
8
5
0
2
FC
1
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