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Bond Enthalpy

Bond enthalpy measures the energy stored in a bond between two atoms. It is the energy required to symmetrically cleave a bond when a compound is in its gaseous phase. By symmetrical cleaving, we mean that the bond breaks so that each atom gains one electron and forms a radical [1-4].

Energy must be added to break a chemical bond. Therefore, bond enthalpy values are positive and are measured in the unit of kJ/mol. Higher values of bond enthalpy indicate strong bonds. It means that much energy is required to break the bond. The average values of bond enthalpy can be found in reference tables. These are only approximate values. The actual values depend on the neighboring atoms of the molecule under consideration.

Bond Enthalpy

Bond Enthalpy Table

The following chart gives the average bond enthalpy values of a few bonds [4].

BondBond Enthalpy (kJ/mol)BondBond Enthalpy (kJ/mol)
C – H413S – Cl253
C – C348S – Br218
C – N293S – S266
C – O358F – F155
C – F485Cl– F253
C – Cl328Cl – Cl242
C – Br276Br – F237
C – I240Br – Cl218
C – S259Br – Br193
N – H391I – Cl208
N – N163I – Br175
N – O201I – I151
N – F272Si – H323
N – Cl200Si – Si226
N – Br243Si – C301
H – H436Si – O368
H – F567C=C614
H – Cl431C≡C839
H – Br366C=N615
H – I299C≡N891
O – H463C=O799
O – O146C≡O1072
O – F190N=N418
O – Cl203N≡N941
O – I234O=O495
S – H338S=O523
S – F327S=S418

Examples

1. Hydrogen Chloride (HCl)

The energy required to dissociate one mole of gaseous HCl into gaseous H and Cl radicals is +431 kJ/mol.

2. Methane (CH4)

The energy required to cleave one C-H bond is +413 kJ/mol. In CH4, there are four C-H bonds. Therefore, the average energy required to dissociate one mole of CH4 into gaseous H and C radicals is 4 x 414 kJ/mol = +1656 kJ/mol.

How to Calculate Enthalpy of Reaction from Bond Enthalpy

To calculate the enthalpy of reaction from bond enthalpy, we apply the following steps [1-4]:

Step 1: Find out which bonds will break and their bond enthalpies.

Step 2: Add up the bond enthalpies of the broken bonds.

Step 3: Find the new bonds that will form in the products. Then, find their bond enthalpies and switch the sign from positive to negative. The reason is that energy is released when new bonds are formed, and the energy released is always negative.

Step 4: Add the bond enthalpies of the new bonds.

Step 5: Add the total values for breaking and forming bonds to get the enthalpy of reaction.

The following formula gives the heat of reaction (ΔHrxn) from bond enthalpy.

ΔHrxn = ΣΔHbonds broken + ΣΔHbonds formed

Examples

1. Consider the hydrogenation of ethene (C2H4) to give ethane (C2H6).

C2H4 (g)  +  H2 (g)  →  C6H6 (g)

We need to break the C=C bond in C2H6 and H-H bond. We need to form three new bonds in C2H6: two C – H bonds and one C – C bond. From the bond enthalpy table above, we have the following information.

BondBond Enthalpy (kJ/mol)Broken or Formed
C = C614Broken
H – H436Broken
C – H413Formed
C – C348Formed

The energy required to break the desired bonds is

ΣΔHbonds broken = 614 kJ/mol + 436 kJ/mol = 1050 kJ/mol

The energy liberated when new bonds form is

ΣΔHbonds formed = – 2 x 413 kJ/mol – 384 kJ/mol = – 1210 kJ/mol

Therefore, the enthalpy of reaction is

ΔHrxn = 1050 kJ/mol – 1210 kJ/mol = -160 kJ/mol

2. Consider the formation of water (H2O) from hydrogen (H2) and oxygen (O2).

2 H2 (g) + O2 (g) → 2 H2O (g)

We need to break two H-H bonds and one O=O bond. We need to form four OH bonds. From the bond enthalpy table above, we have the following values.

BondBond Enthalpy (kJ/mol)Broken or Formed
H – H436Broken
O = O495Broken
O – H463Formed

The energy required to break the desired bonds is

ΣΔHbonds broken = 2 x 436 kJ/mol + 495 kJ/mol = 1367 kJ/mol

The energy liberated when new bonds form is

ΣΔHbonds formed = – 4 x 463 kJ/mol = – 1852 kJ/mol

Therefore, the enthalpy of reaction is

ΔHrxn = 1367 kJ/mol – 1852 kJ/mol = -485 kJ/mol

When a Liquid is Present

So far, we have assumed that all substances involved in the chemical reaction are in their gaseous state. However, suppose one of the substances is in a liquid state. In that case, the heat of reaction cannot be directly calculated from the bond enthalpies. A new term, heat of vaporization (ΔHvap), must be introduced. The heat of vaporization is the enthalpy change when one mole of a substance is vaporized at its boiling point at standard pressure.

Example

Consider the combustion of methane (CH4) in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) at standard temperature and pressure.

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

We need to break four C – H bonds and two O = O bonds. We must form two C = O bonds and four O – H bonds. Aside, we need to convert liquid water into water vapor. From the bond enthalpy table above, we have the following:

BondBond Enthalpy (kJ/mol)Broken or Formed
C – H413Broken
O = O495Broken
C = O799Formed
O – H463Formed

The energy required to break the desired bonds is

ΣΔHbonds broken = 4 x 413 kJ/mol + 2 x 495 kJ/mol = 2642 kJ/mol

The energy liberated when new bonds form is

ΣΔHbonds formed = – 2 x 799 kJ/mol – 4 x 463 kJ/mol = – 3450 kJ/mol

The energy liberated when two moles of water vapor are transformed into liquid water is = – 2 x 41 kJ/mol = -82 kJ/mol

Therefore, the enthalpy of reaction is

ΔHrxn = 2642 kJ/mol – 3450 kJ/mol – 82 kJ/mol = -890 kJ/mol

Solved Problems

Problem 1: The complete combustion of propane can be represented by the following equation:

CH3CH2CH3 (g)  +  5 O2 (g)  →  3 CO2 (g)  +  4 H2O (g)

Find the enthalpy of reaction.

Solution

We use the following equation to calculate the heat of reaction (ΔHrxn).

ΔHrxn = ΣΔHbonds broken + ΣΔHbonds formed

The following bonds are broken:

BondBond Enthalpy (kJ/mol)Number of bonds broken
C – H4138
O = O4955
C – C3482

The total amount of energy required to break the desired bond is

ΣΔHbonds broken = 8 x 413 + 5 x 495 + 2 x 348 = 6475 kJ/mol

The following bonds are formed:

BondBond Enthalpy (kJ/mol)Number of bonds form
C = O7996
O – H4638

The total amount of energy released when new bonds form is

ΣΔHbonds formed = -6 x 799 – 8 x 463 = -8498 kJ/mol

Therefore, the enthalpy of reaction is

ΔHrxn = ΣΔHbonds broken + ΣΔHbonds formed

  • ΔHrxn = 6475 kJ/mol – 8498 kJ/mol
  • ΔHrxn = -2023 kJ/mol

Problem 2: Consider the oxidation reaction of methanol (CH3OH) to produce carbon dioxide (CO2) and water (H2O).

2 CH3OH (g) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (g)

Find the enthalpy of reaction.

Solution

We use the following equation to calculate the heat of reaction (ΔHrxn).

ΔHrxn = ΣΔHbonds broken + ΣΔHbonds formed

The following bonds are broken:

BondBond Enthalpy (kJ/mol)Number of bonds broken
C – H4136
O – H4632
C – O3582
O = O4953

The energy required to break the bonds is:

ΣΔHbonds broken = 6 x 413 + 2 x 463 + 2 x 358 + 3 x 495 = 5605 kJ/mol

The following bonds are formed:

BondBond Enthalpy (kJ/mol)Number of bonds broken
C = O7994
O – H4638

The total amount of energy released when new bonds form is

ΣΔHbonds formed = -4 x 799 – 8 x 463 = -6900 kJ/mol

Therefore, the enthalpy of reaction is

ΔHrxn = ΣΔHbonds broken + ΣΔHbonds formed

  • ΔHrxn = 5605 kJ/mol – 6900 kJ/mol
  • ΔHrxn = -1295 kJ/mol

References

  1. Khanacademy.org
  2. Chem.libretexts.org
  3. Ch301.cm.utexas.edu
  4. Kentchemistry.com

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