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Bond Enthalpy which is also called bond energy, measures the strength of a chemical bond. Basically, it is the required amount of energy to break one mole of bond of any specific type between two atoms which are in a gaseous state. The SI unit used for the bond enthalpy is kJ mol-1.
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What is Bond enthalpy?
During any chemical reaction, the chemical bonds between the atoms may break or reform resulting to release or absorption of energy. This heat which is absorbed or released in this system under the constant pressure is known as enthalpy. Taking up an example: In the Hydrogen bond, H-H bond enthalpy of the molecule is 435.8kJ mol-1.
The chemical reaction is as follows:
H2 (g) → H (g) + H (g); Δa Ho = 435.8 kJ mol-1
N2 (N ≡ N) (g) → N (g) + N (g); Δa Ho = 946.0 kJ mol-1
According to bond enthalpy, the larger the dissociation, the stronger is the bond of the molecule.
In case of a Heteronuclear diatomic molecule, HCL, the bond enthalpy is:
HCl (g) → H (g) + Cl (g); Δa Ho = 431.0 kJ mol-1
For the Polyatomic molecules like H2O, the measurement of Bond strength is quite complicated. Therefore, the enthalpy required to break the O-H bonds is not as same as other molecules.
H2O (g) → H (g) + OH (g); Δa Ho1 = 502 kJ mol-1
OH (g) → H (g) + O (g); Δa Ho2 = 427 kJ mol-1
One can use the formula - ΔH = m x s x ΔT to calculate the bond enthalpy of any chemical molecule, where,
- m is the mass of reactant
- s is specific heat of the product
- ΔT is the change in temperature
Read : Bond Enthalpy Detailed Explaination and Examples
Average Bond Enthalpy
Average bond enthalpy refers to the Quantity of the energy stored within the chemical bonds of any two atoms in a specific molecule, that is, the amount of energy required in breaking the bonds of a molecule. This term is used for the polyatomic molecules.
Average bond enthalpy is calculated by:
ΔH = Sum of bond enthalpies – enthalpy of formation
Students can refer to the following video for more clarity:
What is Endothermic Process & Exothermic Process?
The breaking of any chemical bond is not necessarily an endothermic process every time as sometimes the energy is provided to the chemical molecule to break the chemical bonds. So, the change of enthalpy which is associated with the breaking of the chemical bond is always positive (ΔH >0). Also, in case of a chemical bond formation, it is almost always an endothermic process. The enthalpy associated in such cases will always have a negative value (H <0). So, the processes can be further explained:
Endothermic Process – Reaction products which has higher energy than the reactant products & in which the difference in energy between them is always positive (ΔH >0).
Exothermic Process – Reactions in which the reactant products are having much higher energy than the reaction products and the change in enthalpy remains always negative (H <0).
The following diagram will help us to understand the two processes:
Endothermic Reaction
Exothermic Reaction
Bond enthalpy: Hydrogenation of a Propene
According to the above diagram,
- The atom of one C= C bond & one H-H bond breaks
- The bond enthalpy found of C=C bond is 610kJ/mol ----(1)
- The bond enthalpy of H-H bond is 436kJ/mol -------(2)
- Combining (1) & (2), the energy is added to break the bonds that is – 610 kJ/mol + 436 kJ/mol = 1046 kJ/mol
- This results in formation of a new C-C & two new C-H bonds.
- So, the bond enthalpy of C-H bond is 413 kJ/mol & C-C bond is 346 kJ/mol
- Here the energy is released to form the new bonds: -346 kJ/mol + (2 x -413 kJ/mol) = -1172kJ/mol.
- Hence, the enthalpy we finally get from the reaction is:
ΔHrxn = energy added to break reactant bonds + energy released to make product bonds
= 1046 kJ/mol + (- 1172 kJ/mol)
= - 126kJ/mol
So, the hydrogenation of Propene is exothermic as the enthalpy of the chemical reaction resulted – negative (that is the energy is released).
To know more, refer to this presentation:
Bond Enthalpy: Sample Questions
Ques: What is the average bond enthalpy of O-H bond for a chemical molecule C2H5OH? (2 marks)
Ans: H2O (g) → H (g) + OH (g); Δa Ho1 = 502 kJ mol-1
OH (g) → H (g) + O (g); Δa Ho2 = 427 kJ mol-1
The average bond enthalpy of O-H bond in a polyatomic molecule- C2H5OH is:
Average bond enthalpy = 502 kJ/mol +427 kJ/mol2
= 464.5kJ/mol
Ques: What is a bond length? (1 mark)
Ans: The average distance between the two centres of nuclei of any two bonded atoms in a chemical molecule. This is known as a bond length. Bond length is measured by X-ray diffraction, electron- diffraction techniques.
Ques: What are the factors responsible that affect the bond enthalpy? (2 marks)
Ans: The factors which are responsible to affect the bond energy or enthalpy are:
- Size of the atoms which are involved in the bond
- The electrons affinities of atoms
- Difference in the atomic size & in their electronegativities
Ques: Differentiate between bond dissociation enthalpy and bond enthalpy. (2 marks)
Ans: The basic difference between bond dissociation enthalpy and bond enthalpy is that:
Bond enthalpy: it is the average amount of required energy to break the bonds in between two similar types of atoms present in a chemical compound.
Bond dissociation enthalpy: bond association enthalpy is used to describe the energy used for breaking down a bond of a specific type via homolytic cleavage.
Ques: How to use specific heat & temperature in an equation for bond enthalpy? (2 marks)
Ans: The bond enthalpy for any chemical equation can be calculated with the formula - ΔH = m x s x ΔT. We can simply multiply the heat value and mass of a reactant with the temperature changes of a product.
Ques: Calculate the enthalpy change for the following reaction:
H2(g) + 1⁄2O2(g) ---> H2O(g)Given that enthalpies (in kJ/mol) of specific bonds are: H−H (432); O=O (496); H−O (463) (3 Marks)
Ans: ΔH = Σ Hreactant - Σ Hproduct
| S.No. | Reactant | Product |
|---|---|---|
| 1 | H-H Bond: 432 | O-H Bond: 463 |
| 2 | O=O Bond: 496x0.5 | O-H Bond: 463 |
| Total | =432+248=680 | =2x463=926 |
Hence, overall enthalpy change = Σ Hreactant - Σ Hproduct = 680-926 = -246 kJ/mol
Ques. Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows:
4NH3(g) + 7O2(g) ---> 4NO2(g) + 6H2O(g)
Use the following data for bond energies to determine the bond energy of the N−O bond of NO2, given that the overall heat of reaction is -1135 kJ, using the follow values (given in kJ/mol): (4 marks)
| Bond | Energy |
|---|---|
| O−H | 464 |
| N−H | 389 |
| O=O | 498 |
Ans: The table below lists out the bonds involved in both reactant and product side:
| S.No. | Reactant | Product |
|---|---|---|
| 1 | N-H : 12 (389x12 kJ) | N-O : 8 (Let bond energy of one bond = x) |
| 2 | O=O : 7 (498x7 kJ) | O-H : 12 (12X464 kJ) |
| Total | 4776+3486 = 8262 kJ | 5568+8x |
overall enthalpy change = Σ Hreactant - Σ Hproduct
=> -1135 = 8262 – (5568+8x)
=> x = 465 kJ/mol
Ques: Calculate the enthalpy of the reaction below, using the bond enthalpies listed in the table: (5 Marks)
CH3CH=CH2 + 4.5O=O ---> 3O=C=O + 3H−O−H
| Bond | Energy |
|---|---|
| C−C | 347 |
| C=C | 611 |
| C−H | 414 |
| C=O | 736 |
| O=O | 498 |
| O−H | 464 |
Ans: The bonds present in the reactant and product side are summarised in the table below:
| S.No. | Reactant | Product |
|---|---|---|
| 1 | C-H : 6 (6x414 kJ) | C=O : 6 (6x736 kJ) |
| 2 | C=C : 1 (611 kJ) | O-H : 6 (6x464 kJ) |
| 3 | C-C : 1 (347 kJ) | - |
| 4 | O=O : 4.5 (4.5x498 kJ) | - |
| Total | 2484+611+347+2241 = 5683 kJ | 4416+2784 = 7200 kJ |
overall enthalpy change = Σ Hreactant - Σ Hproduct
=5683-7200 = -1517 kJ
Things to remember in Bond Enthalpy
- The amount of energy required in the formation or breaking of the bond is known as bond enthalpy.
- ΔH = m x s x ΔT is the formula used to calculate the bond enthalpy of a chemical molecule
- Δ H in a chemical reaction determines whether the reaction absorbs heat (endothermic) or releases heat (exothermic)
- For a reaction, the overall enthalpy change is energy added to break reactant bonds + energy released to make product bonds, or overall enthalpy change = Σ Hreactant - Σ Hproduct
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This particular topic from Chapter 4 carries 3-5 marks according to the current syllabus pattern of CBSE. Chapter 4 will have a weightage of total of 14 marks in the Examination.
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