Write the expression for formation of AlCl3 using born haber cycle. My sir got the answer as delta Hsublimation +IE1 +IE2 +IE3 +3/2 bond energy-3 electron affinity +lattice energy. My doubt is why we wrote -3 EA directly? Why not like IE we wrote -EA1 -EA2 -EA3? Thanks
Dear User,
The Born Haber cycle of AlCl3 is shown below :
The Born Haber cycle of AlCl3 is shown below :
Now, ΔHsoln = ΔHf +ΔHatom(Al)+ (I1+I2+I3) - ΔHhyd(Al)+3 ΔHatom(Cl) -3EA -3ΔHhyd(Cl)
Here, I.E. is written as I1+I2+I3 because formation of Al3+ ion occur step wise, first one electron is removed as first I.E. is supplied for it then second electron is removed by supplying second I.E. to give Al2+ ion and then the third I.E. is supplied to finally produce Al3+. Therefore, the total I.E. for the formation of Al3+ is I1+I2+I3.
In case of E.A., it is not written as -EA1-EA2-EA3 because we are adding just one electron i.e. only -EA1 is produced. It is multiplied by 3 because 3 moles of Cl- ion are involved and the total electron affinity will be equal to the amount of energy released by these 3 ions.
Here, I.E. is written as I1+I2+I3 because formation of Al3+ ion occur step wise, first one electron is removed as first I.E. is supplied for it then second electron is removed by supplying second I.E. to give Al2+ ion and then the third I.E. is supplied to finally produce Al3+. Therefore, the total I.E. for the formation of Al3+ is I1+I2+I3.
In case of E.A., it is not written as -EA1-EA2-EA3 because we are adding just one electron i.e. only -EA1 is produced. It is multiplied by 3 because 3 moles of Cl- ion are involved and the total electron affinity will be equal to the amount of energy released by these 3 ions.