Average Bond energies


Single Bonds

N-N 161
O-O 139
O-H 470
H-H 435
C-H 414
N-H 389
S-H 339
C-O 352
C-C 348
P-O 419
N-O 222
S-H 339
C-N 293
C-S 260
S-S 214
Si-O 369


Double Bonds

C=S 477
O=O 498
C=O 800
C=C 615
N=N 418


N=O 607

Triple Bonds

C ≡ C 812
N ≡ N 946
C ≡ N 890





Combustion enthalpy calculator

Just plug in how many bonds of each type are being broken and then formed (if a field is not needed leave it blank)

Bonds broken










Bonds formed



N ≡ N

Enthalpy of combustion =
0 - 0 = 0





Calculations involving enthalpy

There are 2 ways of calculating enthalpies of reactions.

1) using the bond energies (given on the left) as explained below

2) Using Hess's Law

3) Using enthalpies of formation of reagents and products

The second and third methods are more accurate, because the bond energies are only an average, e.g., they depend on the molecular environment.

1) Calculating enthalpy of reaction based on bond energies

The energy transferred in a chemical process originates on the formation of bonds. There is also a cost; in order to promote a reaction some bonds must be broken, and that takes energy in. So, the enthalpy of reaction can be calculated by subtracting the energy cost (breaking bonds) from the energy released (making bonds). The reaction enthalpy calculator on the right will facilitate these calculations.

Let's start with a simple reaction: the combustion of methane (this is a calculation of enthalpy of combustion)

CH4 + 2 O2 --> CO2 + 2 H2O

4 C-H bonds + 2 O=O must be broken

2 C=O bonds + 4 O-H bonds must be formed

Plugging these numbers in the Reaction enthalpy calculator (on the right) provides the answer directly: -828 kJ/mol

(a more accurate method, using Hess's Law, gives -891)

Another calculation of enthalpy of combustion: glucose

The structure of glucose is provided so that you can see what are the bonds that must be broken for the reaction to go.

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O

Answer: -2800 kJ/mol

That is pretty close to the experimental value of -2805.0 ± 1.3 (reference: http://webbook.nist.gov/cgi/cbook.cgi?ID=C492626&Mask=2 )

Another calculation of enthalpy of combustion: TNT

Notice that this is not a detonation, it is a combustion; if it was a detonation the extra oxygens needed to oxidize all the carbons in TNT wouldn't be able to get there in time, as lots of gases would be leaving fast (oxygen would have to be supplied via a solid substance, usually an oxidizing agent).

It is also interesting to calculate the volume of gas produced in this reaction. The production of large volumes of gases causes a substantial increase in the entropy of the system.This fact , allied to a high negative enthalpy, results in a very negative Gibb's energy, so that the reaction is very very favourable, as we know.

C7H2(NO2)3CH3 + 6.25 O2 --> 8 CO2 + 2.5 H2O + 1.5 N2

The bonds to be broken:

The bonding in NO2 is more complex because there is resonance. For simplicity, let's consider to exist one single and one double N-O bond in each nitro group. Hence:

group bonds
N-O 3
N=O 3

There is also resonance on the ring, so that there are 2 double bonds (C=C) and also 4 single bonds (C-C), when the methil group is also taken into consideration:

group bonds
C-C 4
C=C 3

And finally:

group bonds
C-H 5
C-N 3
O=O 6.25

The bonds formed are 16 C=O, 5 O-H and 1.5 N ≡ N


2) Calculate enthalpy of reaction using Hess's Law >>

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