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Need HELP in AS Chemistry- Born Haber cycle

Why formation enthalpy and lattice enthalpy are exothermic?? and why second electron affinity is endothermic? and why the experimental values and theoretical values of lattice enthalpy differ?? what does it mean by more exothermic values gives stronger lattice enthalpy of the bonds??? .....Anyone out there could help me understanding these? Please!! :smile:
Formation Enthalpy can be either exothermic or endothermic. It depends on what you're forming.
Lattice Enthalpy is either exothermic or endothermic depending on how you define it. For some bizarre reason, A-level syllabuses define it as the formation of one mole of an ionic solid from constituent gaseous ions. This is exothermic because ionic bonds are being FORMED meaning energy is released.
Most real chemists, however, prefer to define Lattice Enthalpy as the energy required to rip a lattice apart into its constituent gaseous ions. This definition is endothermic because energy is required to break apart the lattice.

The values of Lattice Enthalpy for both definitions are the same; it's just the signs that differ because one definition is simply the reverse of the other.

A more exothermic value for Lattice Enthalpy means the ionic bonds between ions are stronger. If you use the other definition of Lattice Enthalpy (which is the same number, but endothermic) then the Lattice Enthalpy is more endothermic meaning more energy is needed to destroy the lattice (ie. more energy is required to break the ionic bonds)

The difference between experimental and theoretical values is because bonds are rarely 100% ionic. Some lattices may have a certain degree of covalent character in their bonding. This means that many of the assumptions underlying what is known as the "ionic model" don't always hold so well.

Second electron affinity is endothermic because you're sticking an electron on something which is already negatively charged so the two like charges are going to repel.
(edited 9 years ago)
Reply 2
Original post by TheWiseSalmon
Formation Enthalpy can be either exothermic or endothermic. It depends on what you're forming.
Lattice Enthalpy is either exothermic or endothermic depending on how you define it. For some bizarre reason, A-level syllabuses define it as the formation of one mole of an ionic solid from constituent gaseous ions. This is exothermic because ionic bonds are being FORMED meaning energy is released.
Most real chemists, however, prefer to define Lattice Enthalpy as the energy required to rip a lattice apart into its constituent gaseous ions. This definition is endothermic because energy is required to break apart the lattice.

The values of Lattice Enthalpy for both definitions are the same; it's just the signs that differ because one definition is simply the reverse of the other.

A more exothermic value for Lattice Enthalpy means the ionic bonds between ions are stronger. If you use the other definition of Lattice Enthalpy (which is the same number, but endothermic) then the Lattice Enthalpy is more endothermic meaning more energy is needed to destroy the lattice (ie. more energy is required to break the ionic bonds)

The difference between experimental and theoretical values is because bonds are rarely 100% ionic. Some lattices may have a certain degree of covalent character in their bonding. This means that many of the assumptions underlying what is known as the "ionic model" don't always hold so well.

Second electron affinity is endothermic because you're sticking an electron on something which is already negatively charged so the two like charges are going to repel.


Thanks :smile:,

But as a result calculation of experimental values (if exothermic) are more negative than theoretical values, why this is so??

Why ionic bonds like aluminium chloride have turned covalent bonds? do we get it by observing the strength of the charges of the cations in compare with the charges of anions??

any examples of formation enthalpy that are endothermic?
Original post by Pikee
But as a result calculation of experimental values (if exothermic) are more negative than theoretical values, why this is so??

Why ionic bonds like aluminium chloride have turned covalent bonds? do we get it by observing the strength of the charges of the cations in compare with the charges of anions??


It's to do with the difference in electronegativity between the ions. If the difference is relatively small (compared, to NaCl for example) then there is a higher degree of covalent character so the assumptions of the ionic model don't hold.
This is beyond what is required at A-level though so I don't know why you'd be expected to know this. Maybe I'm missing something obvious. You should discuss this with your teacher.

any examples of formation enthalpy that are endothermic?


Just look up some formation enthalpy tables. A lot of them are endothermic. The formation of most ionic solids, however, is exothermic.
Reply 4
Original post by TheWiseSalmon
It's to do with the difference in electronegativity between the ions. If the difference is relatively small (compared, to NaCl for example) then there is a higher degree of covalent character so the assumptions of the ionic model don't hold.
This is beyond what is required at A-level though so I don't know why you'd be expected to know this. Maybe I'm missing something obvious. You should discuss this with your teacher.



Just look up some formation enthalpy tables. A lot of them are endothermic. The formation of most ionic solids, however, is exothermic.


Thanks a lot :smile:

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