agra,ahmedabad,ajmer,akola,aligarh,ambala,amravati,amritsar,aurangabad,ayodhya,bangalore,bareilly,bathinda,bhagalpur,bhilai,bhiwani,bhopal,bhubaneswar,bikaner,bilaspur,bokaro,chandigarh,chennai,coimbatore,cuttack,dehradun,delhi ncr,dhanbad,dibrugarh,durgapur,faridabad,ferozpur,gandhinagar,gaya,ghaziabad,goa,gorakhpur,greater noida,gurugram,guwahati,gwalior,haldwani,haridwar,hisar,hyderabad,indore,jabalpur,jaipur,jalandhar,jammu,jamshedpur,jhansi,jodhpur,jorhat,kaithal,kanpur,karimnagar,karnal,kashipur,khammam,kharagpur,kochi,kolhapur,kolkata,kota,kottayam,kozhikode,kurnool,kurukshetra,latur,lucknow,ludhiana,madurai,mangaluru,mathura,meerut,moradabad,mumbai,muzaffarpur,mysore,nagpur,nanded,narnaul,nashik,nellore,noida,palwal,panchkula,panipat,pathankot,patiala,patna,prayagraj,puducherry,pune,raipur,rajahmundry,ranchi,rewa,rewari,rohtak,rudrapur,saharanpur,salem,secunderabad,silchar,siliguri,sirsa,solapur,sri-ganganagar,srinagar,surat,thrissur,tinsukia,tiruchirapalli,tirupati,trivandrum,udaipur,udhampur,ujjain,vadodara,vapi,varanasi,vellore,vijayawada,visakhapatnam,warangal,yamuna-nagar

Bond Enthalpy- Bond Enthalpy, its Types, Applications, Practice Problems and FAQs

Bond Enthalpy- Bond Enthalpy, its Types, Applications, Practice Problems and FAQs

There is a family of five people- Krish, Sanya, Ruhi, Kelvin and Samayara. Kelvin and Samayara are the parents of Krish, Sanya and Ruhi. Now, everybody says that parents like their children equally but practically it is not possible, one of them is the favorite one.

Similarly, the force connecting two atoms called bond energy or bond enthalpy changes with the atoms bonded together. In a polyatomic molecule with similar bonds, the bond energy may not be the same for all of them. Let's study this in detail and learn an interesting concept about bond enthalpy.

Table of Contents

  • Bond Enthalpy
  • Application of Bond Enthalpy
  • Practice Problems
  • Frequently Asked Questions

Bond Enthalpy

If I ask why a ball is thrown up and comes down, you will say immediately, because the ball will be at rest or stable when the ball is on the ground than up in the air. So everything wants to lower their energy and attain stability.

Atoms also try to stabilize themselves by lowering their energy. But how? By reacting with other atoms. The reacting atoms lose some energy to form a bond between them. On the other, you have to supply energy to break them to become atoms once again.

Bonds between atoms in a molecule are either broken or formed or both during chemical reactions. A bond requires energy to break and releases energy during formation. The heat of a reaction can be linked to the strength of the bond to be made or broken during the chemical changes. In thermodynamics, the energy changes occurring at constant pressure conditions(like reactions in an open vessel) are called enthalpy. 

In chemical reactions, two enthalpy-one for the formation and another for breaking of a bond can be considered.

1. Bond formation enthalpy
2. Bond dissociation enthalpy

Bond formation enthalpy

It is the enthalpy change during the formation of one mole of bonds of a particular type. Since the formation of a bond is a stabilizing phenomenon, therefore, it is an exothermic process,


Bond dissociation enthalpy

It is the enthalpy change during dissociation of one mole of bonds of a particular type in a homodiatomic molecule. Since the dissociation of a bond is a destabilizing phenomenon, therefore, it is an endothermic process,


The bond enthalpy can be used to represent the strength of a single, specific bond in a molecule. Calculating the average value of all the bond dissociation energies of that type of bond in the molecule yields the mean bond energy of a chemical bond (in a molecule) or average bond enthalpy. As a result, the mean bond enthalpy differs from the bond dissociation energy (except for diatomic molecules).

Mean bond dissociation enthalpy or Average bond dissociation enthalpy (bondH)

Mean bond dissociation enthalpy is the average of the enthalpies required to break all identical bonds of a compound in one mole. Let’s understand this by taking the example of the dissociation of ammonia.

Standard mean bond enthalpy(bondHo)

Standard mean bond enthalpy is the average of enthalpies required to break all the identical bonds present in one mole of a compound at 1 bar and at 298 K.

Consider the succession dissociation of N−H bonds in NH3


In ammonia, all three N−H bonds are identical in bond length and energy. However, the energies required to break the individual N−H bonds in each successive step differ. This is due to the change in the electronic environment around the nitrogen with the removal of each hydrogen atom. As a result, different NH bond energies are produced in subsequent phases. Therefore, the average of all the enthalpies is taken as the bond enthalpy.

Mean bond dissociation enthalpy of NH3


Important Note

The stronger the bond and the more energy it takes to break it, the higher the value. 

Hess law

The overall change in enthalpy for the solution is the sum of all changes independent of the different phases or steps of a reaction, according to Hess' Constant Heat Summation Law (or simply Hess' Law).


Application of Bond Enthalpy

It is used to determine the enthalpy of any reaction. The formula to determine the enthalpy of any reaction is given as follows:

rH=Bond enthalpies(Reactant)-ΣBond enthalpies(Product)

Example: C2H4(g)+HCl(g)C2H5Cl(g)

The bond enthalpies(Reactant) required to split the reactants into gaseous atoms are as follows:


The following is the enthalpy necessary to split the products into gaseous atoms [Bond enthalpies(Product)]:


The following is the enthalpy change for the given reaction:

rH=Bond enthalpies(Reactant)-ΣBond enthalpies(Product)


Practice Problems

Q 1. From the following standard bond enthalpies:

H–H bond energy: 431.37 kJmol-1

C=C bond energy: 606.10 kJmol-1

C–C bond energy: 336.49 kJmol-1

C–H bond energy: 410.50 kJmol-1

Calculate the standard enthalpy of the reaction for C2H4(g)+H2(g)C2H6(g)

a. −243.6 kJmol-1
b. −120.02 kJmol-1
c. 553.0 kJmol-1
d. 153.26 kJmol-1



Q 2. If the standard bond enthalpies of H−H, Br−Br, and H−Br are 433, 192, and 364kJmol-1, respectively, then what is the value of rHofor the following reaction?


a. −261 kJmol-1
b. +103 kJmol-1
c. +261 kJmol-1
d. −103 kJmol-1



Q 3. The standard enthalpy change for the reaction, 4H(g)2H2(g) is −869.6 kJ.What is the dissociation energy of the H—H bond?

a. −434.8 kJ 
b. −869.6 kJ 
c. +434.8 kJ
d. +217.4 kJ

Answer: Given,


Q 4. Given that the standard bond enthalpies of H− H and Cl− Cl are 430 kJmol-1 and 240 kJmol-1respectively, and fHo for HCl is −90 kJmol-1, what is the standard bond enthalpy of HCl?

a. 380 kJmol-1
b. 425 kJmol-1
c. 245 kJmol-1
d. 290 kJmol-1



Frequently Asked Questions

Q 1. In basic terms, what is enthalpy?
Answer: Enthalpy is a term used in science and engineering to describe the quantity of heat and function are measured. Enthalpy tells how much heat and effort has been applied or taken from a substance when it is under constant strain. Enthalpy is similar to energy but not the same.

Q 2. Is bond enthalpy the same as atomisation enthalpy?
Answer: The change in enthalpy when one mole of gaseous atoms is created from atomic matter is known as atomization enthalpy. Bond enthalpy/bond energy/bond dissociation enthalpy, on the other hand, is the energy required to dissociate one mole of bond into separate atoms in the gaseous state.

Q 3. Can bond enthalpy be calculated in every state of matter?
Answer: The aspect about everything being in the gas state is crucial; you can't use bond enthalpies to calculate things straight from liquid or solid substances. Hence, bond enthalpy can be calculated in the gas state only.

Q 4. Can you tell the basic difference between Enthalpy and Energy?
Answer: The measurement of energy in a thermodynamic system is enthalpy. The overall content of heat in a system at constant pressure conditions is enthalpy, which is equal to the system's internal energy plus the product of volume and pressure.

Related Topics

Enthalpy of Neutralization

Enthalpy of Combustion

Enthalpy of solvation

Hess law

Talk to Our Expert Request Call Back
Resend OTP Timer =
By submitting up, I agree to receive all the Whatsapp communication on my registered number and Aakash terms and conditions and privacy policy