Bond Energy or Bond Enthalpy Chemistry Tutorial
Key Concepts
- Bond energy is also known as:
⚛ bond enthalpy
⚛ average bond energy, or average bond enthalpy
⚛ mean bond energy, or mean bond enthalpy
- Bond energy, bond enthalpy, is the energy required per mole of gaseous compound to break a particular bond to produce gaseous fragments at 25°C and 1 atmosphere pressure.(1)
- Breaking chemical bonds requires an input of energy.
⚛ ΔH is positive for breaking bonds.
⚛ Breaking bonds is an endothermic reaction.
⚛ Bond energies, bond enthalpies, are positive.
⚛ Values of bond energy, bond enthalpy, can be found in tables.
- Making chemical bonds releases energy.
⚛ Bond formation releases energy.
⚛ ΔH is negative.
⚛ Bond formation is an exothermic reaction.
⚛ ΔHmaking bonds = -ΔHbreaking bonds = -bond energy (= -bond enthalpy)
- Bond energies, bond enthalpies, can be used to indicate how stable a compound is or how easy it is break a particular bond.
⚛ The more energy that is required to break a bond, the more stable the compound will be.
⚛ A larger bond energy implies the bond is harder to break, so the compound will be more stable.
- Bond energies, bond enthalpies, can be used to estimate the heat of a reaction (enthalpy change of a reaction, ΔH).
⚛ ΔH(reaction) = sum of the bond energies of bonds being broken - sum of the bond energies of the bonds being formed.
⚛ ΔH(reaction) = ΣH(reactant bonds broken) - ΣH(product bonds formed)
- Steps for calculating heat of reaction, enthalpy change of reaction, ΔH, from bond energies of reactants and products(2):
(i) Write the balanced chemical equation, with all reactants and products in the gaseous state.
If a reactant or product is NOT in the gaseous state, you will need to use Hess's Law to include the relevant energy (enthalpy) for the change of state.
(ii) Write a structural formula for each reactant and product
(iii) Write the general equation for the enthalpy change for reaction (heat of reaction), ΔH(reaction):
ΔH(reaction) = ΣH(reactant bonds broken) - ΣH(product bonds formed)
(iv) Substitute bond energy values into the equation and solve for ΔH(reaction)
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Bond Energy, Bond Enthalpy, Values
The amount of energy required to break particular bonds, values for bond energy or bond enthalpy, are often presented in a table like the one below:(3)
Single Bond Energies (kJ mol-1)
at 25°C |
|---|
| |
S |
Si |
I |
Br |
Cl |
F |
O |
N |
C |
H |
|---|
| H |
360 |
339 |
299 |
366 |
432 |
563 |
463 |
391 |
413 |
436 |
|---|
| C |
259 |
290 |
240 |
285 |
339 |
485 |
358 |
304 |
348 |
|
|---|
| N |
|
|
|
|
200 |
270 |
|
163 |
|
|
|---|
| O |
|
369 |
|
|
203 |
185 |
146 |
|
|
|
|---|
| F |
|
541 |
258 |
237 |
254 |
153 |
|
|
|
|
|---|
| Cl |
250 |
359 |
210 |
219 |
243 |
|
|
|
|
|
|---|
| Br |
|
289 |
178 |
193 |
|
|
|
|
|
|
|---|
| I |
|
213 |
151 |
|
|
|
|
|
|
|
|---|
| Si |
227 |
177 |
|
|
|
|
|
|
|
|
|---|
| S |
213 |
|
|
|
|
|
|
|
|
|
|---|
More compact tables of bond energies (bond enthalpies) are often provided in exams:
| Bond Energies at 25°C (kJ mol-1) |
|---|
| H-H |
436 |
C-C |
348 |
C-H |
413 |
O-H |
463 |
| H-Cl |
432 |
C=C |
610 |
C-O |
358 |
O-O |
146 |
| H-Br |
366 |
C≡C |
835 |
C=O |
736 |
O=O |
498 |
| H-I |
299 |
benzene |
518 |
C-Cl |
339 |
N-H |
391 |
| Cl-Cl |
243 |
|
|
C-Br |
285 |
S-H |
360 |
| Br-Br |
193 |
|
|
C-F |
485 |
N-N |
163 |
| I-I |
151 |
|
|
C-N |
304 |
N≡N |
945 |
Note that all values of bond energy, bond enthalpy, are positive because energy must be absorbed to break covalent bonds between atoms.
Breaking covalent bonds between atoms is an endothermic process.
When using these values it is important to remember that:
- the "reactant", the molecule containing the bonds to be broken, must be a gas
- the "product", the atoms produced after the bond is broken must also be in the gaseous state
- when the bond is broken, half the electrons from the bond go in one product fragment, and the other half of the electrons from the bond go in the other product fragment.
For example, from the table above we find the bond energy (bond enthalpy) for the H-H bond is 436 kJ mol-1.
If we have 1 mole of H-H bonds to break, we will need to supply 436 kJ of energy.
If we have 1 mole of gaseous molecular hydrogen, H2(g), then we need to supply 436 kJ of energy for the following reaction to take place:
H2(g) → 2H(g)
H:H → H. + H.
If we had 2 moles of gaseous molecular hydrogen, H2(g), then we need to supply 2 × 436 kJ = 872 kJ of energy to break all the H-H bonds and form 4 moles of gaseous hydrogen atoms, H(g).
Similarly, if we had ½ mole of H2(g), we need to supply ½ × 436 = 218 kJ of energy in order to break all the H-H bonds and produce 1 mole of H(g).
Bond Energy and Chemical Stability
Breaking chemical bonds requires energy.
The more energy that is required to break a bond, the more chemically stable the compound will be.
When comparing two compounds, the compound containing the bond with the lowest bond energy will be the least stable compound, regardless of the strength of the other bonds present in the compound.
Consider these two molecules, water (H2O) and hydrogen peroxide (H2O2).
The electron dot diagram (Lewis Structure) for each molecule is shown below:
|
|
|
hydrogen
peroxide |
| | .. | | .. | |
| H: | O | : | O | :H |
| | .. | | .. | |
|
Both molecules contain O-H bonds that need to be broken if a chemical reaction is to occur.
Each mole of O-H bonds requires 463 kJ of energy (from the table of bond energies, or bond enthalpies, above)
However, in order for hydrogen peroxide to decompose, or react, another bond, the O-O bond, must be broken.
From the table of bond energies (bond enthalpies) above, we see that breaking 1 mole of O-O bonds requires only 146 kJ of energy.
It is easier, requires less energy, to break the O-O bond compared to an O-H bond.
Therefore, hydrogen peroxide is less stable than water because it contains a bond that requires less energy to break.
Hydrogen peroxide is therefore more chemically reactive than water.
Calculating Heat of Reaction (Enthalpy of Reaction) Using Bond Energy
Bond energies, bond enthalpies, can be used to estimate the heat of reaction (enthalpy change of a reaction), ΔH(reaction).
For the general chemical reaction in which reactants form products:
| chemical reaction: |
reactants |
→ |
products |
| bonds are |
broken |
→ |
made |
| energy is |
absorbed |
→ |
released |
A chemical reaction will be endothermic if the energy absorbed to break bonds in the reactant molecules is greater than the energy released when bonds are formed in the product molecules.
Hbreak bonds > Hmake bonds ΔHreaction is positive
A chemical reaction will be exothermic if the energy absorbed to break bonds in the reactant molecules is less than the energy released when bonds are formed in the product molecules.
Hbreak bonds < Hmake bonds ΔHreaction is negative
That is, if we add together the bond energies of all the bonds that need to be broken in the reactant molecules, and, add together all the bond energies of all the bonds we need to make in order to produce product molecules, then we can subtract one from the other to arrive at an estimate of the enthalpy change for the overall chemical reaction:
ΔH(reaction) = sum of the bond energies of bonds being broken - sum of the bond energies of the bonds being formed.
Or,
ΔH(reaction) = ΣH(reactant bonds broken) - ΣH(product bonds formed)
where Σ means "the sum of"
For example, we could use the bond energies provided in the table above to calculate the heat of reaction (enthalpy change for the reaction), ΔH, for the reaction:
| CH4(g) |
+ |
4Cl2(g) |
→ |
CCl4(g) |
+ |
4HCl(g) |
|
|
+ |
4 Cl−Cl |
→ |
|
+ |
4 H−Cl |
The bonds that need to be broken in the reactant molecules are:
- C-H bonds (there are 4 of these in a CH4 molecule so we need to break 4 lots of C-H bonds)
- Cl-Cl bonds (there is 1 of these in each Cl2 molecule, BUT, we need 4 lots of Cl2 molecules to balance the equation, so, 4 lots of Cl-Cl bonds need to be broken)
The bonds that need to be made when the products are formed are:
- C-Cl bonds (there are 4 lots of C-Cl bonds in each CCl4 molecule so we need to make 4 lots of C-Cl bonds)
- H-Cl bonds (there is only 1 H-Cl in each HCl molecule, but we need 4 lots of HCl molecules so we will need to make 4 lots of H-Cl bonds)
We will need to use published values of bond energies (bond enthalpies) to calculate the value of the heat of reaction (enthalpy change for the reaction).
Our solution to the problem is shown below:
- Write the balanced chemical equation, with all reactants and products in the gaseous state.
CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
- Write the structural formula of each reactant and product molecule (so we can identify which bonds are broken and which are formed)
- Write the general equation for the heat of reaction (enthalpy change for the reaction):
ΔH(reaction) = ΣH(reactant bonds broken) - ΣH(product bonds formed)
- Substitute bond energy values into the equation and solve for ΔHo(reaction)
| CH4(g) + 4Cl2(g) Energy absorbed to break bonds |
→ |
CCl4(g) + 4HCl(g) Energy released to form bonds |
|---|
| bond type |
bond energy |
|
bond type |
bond energy |
| 4 × C - H |
4 × 413 |
= 1652 |
|
4 × C - Cl |
4 × 339 |
= 1356 |
| 4 × Cl - Cl |
4 × 243 |
= 972 |
|
4 × H - Cl |
4 × 432 |
= 1728 |
|
| ΣH(reactant bonds broken) |
= 2624 |
|
ΣH(product bonds formed) |
= 3084 |
ΔH(reaction) = 2624 - 3084 = -460 kJ mol-1
Note that the reaction is exothermic, ΔH is negative (ΔH = -460 kJ mol-1).
The energy absorbed by reactant molecules to break bonds (2624 kJ) is less than the energy released when bonds are formed to make product molecules (3084 kJ).
Problem Solving: Using Bond Enthalpy to Calculate the Enthalpy Change for a Chemical Reaction
Question: Use the table of bond enthalpies to calculate the enthalpy change for the reaction in which gaseous nitrogen reacts with gaseous hydrogen to produce gaseous ammonia as shown in the balanced chemical equation below:
| N2(g) |
+ |
3H2(g) |
→ |
2NH3(g) |
| N≡N |
+ |
3 H−H |
→ |
2 |
|
Solution:
(using the StoPGoPS approach to problem solving)
| STOP |
STOP! State the Question.
|
| |
What is the question asking you to do?
Calculate ΔH for the reaction
|
| PAUSE |
PAUSE to Prepare a Game Plan
|
| |
(1) What information (data) have you been given in the question?
(a) Balanced chemical reaction:
| N2(g) |
+ |
3H2(g) |
→ |
2NH3(g) |
| N≡N |
+ |
3 H−H |
→ |
2 |
|
(b) Bond enthalpy data (in tutorial or from a data sheet):
| Type of bond |
Bond enthalpy (kJ mol-1) |
|---|
| N≡N |
945 |
| H−H |
436 |
| N−H |
391 |
(2) What is the relationship between what you know and what you need to find out?
ΔH = Σ(bonds broken) − Σ(bonds formed)
|
| GO |
GO with the Game Plan |
| |
Bond enthalpy for bonds broken:
| Type of bond broken |
Bond enthalpy |
Number of these bonds broken |
total bond enthalpy
(bond enthalpy × number) |
|---|
| N≡N |
945 |
1 |
945 |
| H−H |
436 |
3 |
1308 |
| Σ(bonds broken) = 945 + 1308 = 2253 |
Bond enthalpy for bonds formed:
| Type of bond formed |
Bond enthalpy |
Number of these bonds formed |
total bond enthalpy
(bond enthalpy × number) |
|---|
| N−H |
391 |
2×3 = 6 |
2346 |
| Σ(bonds formed) = 2346 |
Calculate ΔH (reaction)
| ΔH |
= |
Σ(bonds broken) |
− |
Σ(bonds formed) |
| |
= |
2253 |
− |
2346 |
| |
= |
−93 kJ mol-1 |
|
|
| PAUSE |
PAUSE to Ponder Plausibility |
| |
Have you answered the question?
Yes, we have calculated the enthalpy change, ΔH, for the chemical reaction given.
Is your answer plausible?
Work backwards using our calculated value for ΔH to find one of the bond enthalpies, eg, bond enthalpy for N−H bond:
ΔH = Σ(bonds broken) − Σ(bonds formed)
−93 kJ mol-1 = (1×945 + 3×436) − (2 × 3 × N-H)
−93 kJ mol-1 = (945+1308) − (6 × N-H)
−93 kJ mol-1 = 2253 − (6 × N-H)
−93 − 2253 kJ mol-1 = −6(N-H)
−2346 kJ mol-1 = −6(N-H)
(−2346 ÷ −6) kJ mol-1 = (N-H)
+391 kJ mol-1 = (N-H)
Since the value we calculate for the bond enthalpy of a N-H bond is the same as that given in the table in the tutorial, we are reasonably confident that our answer is plausible.
|
| STOP |
STOP! State the Solution |
| |
ΔH = −93 kJ mol-1
|
Footnotes
(1) The bonds must be broken homolytically, that is, half the electrons making up the bond go with one product fragment, and, the other half of the electrons making up the bond go to the other product fragment.
(2) If all the reactants and products are in their standard state (that is, if the standard state of all the reactants and products is the gaseous state), then this calculation results in the standard enthalpy change for the reaction, ΔHo
You can also use standard enthalpy of formation (standard heat of formation) to calculate the standard enthlpy change for a reacton (see Standard enthalpy of formation and reaction tutorial).
(3) Do not be concerned if the values for bond energy that you have been given differ from those shown here.
Published bond energies are an "average energy" since the same bond in different molecules can require different amounts of energy to break.
Also note that the energy required to break bonds will be dependent on the conditions of the reaction, such as the temperature, so check the conditions for your values.
Techniques for establishing the energy required to break bonds change, so the listed values will change.
Just use whatever values you have been given. If no values have been provided, you can use the ones given here.
