Chemical compounds are groups of atoms or ions held together by chemical
bonds. The definition of a chemical reaction is 'a process in which
a new substance is formed'. For a chemical reaction to occur, these
bonds must be broken before others can form.
Syllabus reference R1.2.1
Reactivity 1.2.1 - Bond-breaking absorbs and bond-forming releases energy.
- Calculate the enthalpy change of a reaction from given average bond enthalpy data.
Guidance
- Include explanation of why bond enthalpy data are average values and may differ from those measured experimentally.
- Average bond enthalpy values are given in the data booklet.
Tools and links
- Structure 2.2 - How would you expect bond enthalpy data to relate to bond length and polarity?
- Reactivity 3.4 - How does the strength of a carbon–halogen bond affect the rate of a nucleophilic substitution reaction?
Bond dissociation enthalpy
This is the energy required to break one mole of specific bonds in a specific
molecule. It only applies to one particular type of bond in a unique molecule.
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Example: Methane has the structure:
The energy change for the process:
CH4 → CH3 + H ΔH = +435 kJ
This is different from the energy change required for the process:
CH3 → CH2
+ H ΔH = +444 kJ
which is different from:
CH2 → CH + H ΔH = +444 kJ
which is different from:
CH → C + H ΔH = +339 kJ
The average of the four bond dissociation enthalpies above is:
(435 + 444 + 444 + 339)/4 = +415.5 kJ
This average value is called the average bond enthalpy, or bond enthalpy
term for C-H.
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Bond dissociation enthalpies between non-identical atoms are of limited use,
as they do not apply to any other molecule.
However, they are useful in the case of specific dissociation of molecules.
For example the process:
Cl2(g) → 2Cl(g)
ΔH = +242 kJ mol-1
Represents the dissociation enthalpy of chlorine gas. We cannot measure this
value in any other molecule (we wouldn't want to!). The value is useful for
construction of Born Haber cycles, or other energetics calculations.
REM It is important to be able to distinguish
between bond dissociation enthalpy and average bond enthalpy, which is discussed
in the next section. It is often asked for in exams.
| Bond |
Bond dissociation enthalpy / kJ mol-1 |
| H-H |
+432 |
| O=O |
+494 |
| N≡N |
+942 |
| F-F |
+155 |
| Cl-Cl |
+242 |
| Br-Br |
+190 |
| I-I |
+148 |
| Ref: CRC Handbook of chemistry and physics
- Edition 44 |
Bond
dissociation enthalpies
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Average bond enthalpy
The average bond enthalpy term is the average amount of energy needed to
break a specific type of bond, measured over a wide variety of different gaseous
molecules. It is essentially the average of all of the bond dissociation enthalpies
for a specific type of bond.
Bond
enthalpy data table
|
Table of bond enthalpies / kJ mol-1
|
|
C-C
|
348 |
|
C-H
|
412 |
|
C-O
|
360 |
|
C=C
|
612 |
|
C=O
|
743 |
|
Cl-Cl
|
242 |
|
O-H
|
463 |
|
Cl-H
|
431 |
|
C-Cl
|
338 |
|
O=O
|
496 |
| Ref: CRC Handbook of chemistry and physics - Edition
44 |
Bond enthalpies are measured per mole of bonds. The units of the bond enthalpy
term is kJ mol-1
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Reaction enthalpy from bond energy terms
For any chemical reaction:
Reactants → products
There is a rearrangement of the bonding between the component atoms of the
reactants and products. Thanks to Hess' law, we can follow an alternative
route from the reactants to the products by breaking all of the bonds and
then reforming the new bonds.
Breaking the reactant's bonds is an endothermic process, it requires energy.
Making the product's bonds is an exothermic process. Using the alternative
route via the individual atoms we can state that the enthalpy change of the
reaction is the SUM of these two processes.
Enthalpy change = step 1 (bond breaking) + step 2 (bond formation)
However, forming the products releases energy, i.e. delta H is negative,
giving:
Σ the reactant bond enthalpies - Σ the product bond enthalpies.
ΔHo(reaction) = ΣΔHo(bond enthalpy reactants) - ΣΔHo(bond
enthalpy products)
Bond
enthalpy animation
In reality, all of the bonds in the reactants are not broken before reforming
them into the products, this is another example Hess' law; the route may be
different, but the final answer must be the same.
Example: Use the bond enthalpy terms from the
table above to find the enthalpy change of the following reaction:
CH4 + 2O2 → CO2 + 2H2O
Sum of the bond enthalpy terms of the reactants:
- 4 x C-H = 4 x 412 = 1648 kJ
- 2 x O=O = 2 x 496 = 992 kJ
Total = 2640 kJ
Sum of the bond enthalpy terms of the products
- 2 x C=O = 2 x 743 = 1486 kJ
- 4 x H-O = 4 x 463 = 1852 kJ
Total = 3338 kJ
Reaction enthalpy change = 2640 - 3338 = -698
kJ
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Worked examples
Q442-01 Which of the following
is correct about the energy changes during bond breaking and bond formation?
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bond breaking
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bond formation
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| A. |
exothermic
|
endothermic
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| B. |
exothermic
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exothermic
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| C. |
endothermic
|
exothermic
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| D. |
endothermic
|
endothermic
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Answer
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Bond breaking requires energy - it is endothermic. Bond formation
releases energy - it is exothermic - Response
C
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Q442-02 Which statement about
bond enthalpies is correct?
- Bond enthalpies have positive values for strong bonds and negative values
for weak bonds
- Bond enthalpy values are greater for ionic bonds than for covalent bonds
- Bond breaking is endothermic and bond making is exothermic
- The carbon-carbon bond enthalpy values are the same in ethane and ethene.
Answer
|
Bond breaking is endothermic and bond making is exothermic Response
C
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Q442-03 The average bond enthalpy
of the C- H bond is 412 kJ mol
-1 . Which process has an enthalpy
change closest to this value?
- CH4(g) → C(s)
+ 2H2(g)
- CH4(g) → C(g)
+ 2H2(g)
- CH4(g) → C(s)
+ 4H(g)
- CH4(g) → CH3(g)
+ H(g)
Answer
|
The process involving the breaking of one C- H bond corresponds to
Response D
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Q442-04 What energy changes
occur when chemical bonds are formed and broken?
- Energy is absorbed when bonds are formed and when they are broken
- Energy is released when bonds are formed and when they are broken
- Energy is absorbed when bonds are formed and released when they are broken
- Energy is released when bonds are formed and absorbed when they are broken
Answer
|
Energy is released when bonds are formed and absorbed when they are
broken. Response D
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Q442-05 What is the value
of ΔH in kJ mol
-1 for the reaction below?
CH2=CH2 + H2 → CH3CH3
| Bond |
H-H |
C-C |
C=C |
C-H |
| bond energies / kJ mol-1 |
436 |
348 |
612 |
412 |
- 124
- 101
- -101
- -124
Answer
|
Bonds broken in the reactants (endothermic):
4 x C-H = 4 x 412 = 1648 kJ
1 x C=C = 1 x 612 = 612 kJ
1 x H-H = 1 x 436 = 436 kJ
total = +2696 kJ
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Bonds formed in the products
(exothermic):
6 x C-H = 6 x 412 = -2472 kJ
1 x C-C = 1 x 348 = -348 kJ
total = -2820 kJ
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Enthalpy of reaction = bonds broken (endo) + bonds formed (exo)
Enthalpy change of reaction = +2696 - 2820 = -124
kJ
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Q442-06 Some average bond
enthalpies in kJ mol
-1 are as follows: H - H = 436, Cl - Cl = 242,
H - Cl = 431
What is the enthalpy change (in kJ) for the decomposition of hydrogen chloride?
2HCl → H2 +
Cl2
- -184
- +184
- +247
- -247
Answer
|
Bonds broken in the reactants (endothermic):
2 x H-Cl = 2 x 431 = 862 kJ
total = +862 kJ
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Bonds formed in the products
(exothermic):
1 x H-H = 1 x 436 = -436 kJ
1 x Cl-Cl = 1 x 242 = -242 kJ
total = -678 kJ
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Enthalpy change for the decomposition = 862 + (-678) = +184
kJ
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Q442-07 Given the following
data:
- 1 C(s) + 2F2(g) → CF4(g) H1
= -680 kJmol-1
- 2 F2 (g) → 2F(g) H2= +158
kJmol-1
- 3 C(s) → C(g) H3 = +715
kJmol-1
Calculate the average bond enthalpy (in kJmol-1) for the C-F bond.
[4]
Answer
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The average bond enthalpy (in kJ mol-1) for the C-F bond
is given by the process:
C-F → C(g) + F(g)
However, as C-F cannot exist on its own then we can calculate the
average bond energy by dividing the energy change for the following
process by four:
CF4(g) → C(g) + 4F(g)
To construct this equation from the equations given:
CF4(g) appears in equation 1
on the RHS so we must reverse equation 1
and change the sign of the energy.
CF4(g) → C(s) + 2F2(g)
ΔH = +680 kJ
Now we need C(g). This appears in equation 3
on the RHS, which is required. So we can add equation 3
directly to the equation above:
CF4(g) → C(s) + 2F2(g)
ΔH = +680 kJ
C(s) → C(g); H3
= +715 kJmol-1
CF4(g) + C(s) → C(s) + 2F2(g) + C(g)
ΔH = +1395 kJ
Finally, to finish constructing the required equation we need F(g).
This appears in equation 2 on the
RHS as 2F(g), BUT we must double the equation to get 4F(g) and then
add it to the result of the equations addition above:
CF4(g) → 2F2(g)
+ C(g) ΔH = +1395
kJ
2F2(g) → 4F(g)
ΔH=
+316kJmol-1
CF4(g) + 2F2(g) → 2F2(g) + C(g) + 4F(g)
ΔH = +1711 kJ
Cancelling terms gives us the required equation. The average bond
enthalpy is this enthalpy change divided by 4 = +428
kJ
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Q442-08 The average bond enthalpies
for O-O and O=O are 146 and 496 kJ mol
-1 respectively. What is the
enthalpy change in kJ for the reaction below?
H-O-O-H (g) → H-O-H(g)
+ ½O=O(g)
- -102
- +102
- +350
- +394
Answer
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Bonds broken in the reactants (endothermic):
2 x O-H = 2 x z = 2z kJ
1 x O-O = 146 kJ
total = +146 kJ
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Bonds formed in the products
(exothermic):
2 x O-H = 2 x -z = -2z kJ
½ x O=O = ½ x -496 = -248 kJ
total = -248 kJ
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Notice the 2z terms cancel out on either side.
The enthalpy change for the reaction = -102
kJ
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Q442-09 Consider the following
reaction:
N2(g) + 3H2(g) → 2NH3(g) ΔHo = ?
Bond enthalpies (in kJ mol-1) involved in the reaction are:
Which calculation will give the value of ΔHo?
- x + 3y - 6z
- 6z - x +3y
- x - 3y + 6z
- x + 3y - 2z
Answer
|
Bonds broken in the reactants (endothermic):
1 x N ≡ N = x
kJ
3 x H-H = 3y kJ
total = +x + 3y kJ
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Bonds formed in the products
(exothermic):
6 x N-H = -6z kJ
total = -6z kJ
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Enthalpy change of reaction = x + 3y -6z
kJ
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Q442-10 Enthalpy change may
be calculated using bond enthalpies, some values of which ( in kJmol
-1
) are provided below:
C=C 612; C-H 412; O-H 463; C=O 743; O=O 496.
The balanced equation for the complete combustion of one mole of ethene,
C2H4, in oxygen is shown below. Use this data to calculate
the enthalpy of combustion of ethene:
C2H4(g) + 3O2(g) → 2CO2 + 2H2O(g)
Answer
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Bonds broken in the reactants (endothermic):
1 x C=C = 612 kJ
4 x C-H = 1648 kJ
3 x O=O = 1488 kJ
total = +3748 kJ
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Bonds formed in the products
(exothermic):
4 x C=O = -2972 kJ
4 x O-H = -1852 kJ
total = -4824 kJ
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Enthalpy change for the combustion of ethene = -1076
kJ
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