I'm really confused about this aspect of Born Haber cycles. Why is the lattice enthalpy of CaCl3 more exothermic than CaCl2 when CaCl2 is actually more stable? I thought that the more exothermic a compound is the more stable it is? I hope I'm making sense.
I'm really confused about this aspect of Born Haber cycles. Why is the lattice enthalpy of CaCl3 more exothermic than CaCl2 when CaCl2 is actually more stable? I thought that the more exothermic a compound is the more stable it is? I hope I'm making sense.
No the more exothermic the less stable a compound is, think of benzene ring, one of the evidence for the delocalised structure was that its enthalpy of hydrogenation was less than expected, which means its bond are stronger and less energy released.
No the more exothermic the less stable a compound is, think of benzene ring, one of the evidence for the delocalised structure was that its enthalpy of hydrogenation was less than expected, which means its bond are stronger and less energy released.
Yes thank you that is very true. However, I read in my textbook that the more exothermic a compound is the more stable it is because it releases more energy and therefore the compound has less energy (more stability)? Does it differ in different cases?
Hmm maybe it's related to having more bonds, and no it doesn't exist - it's just referring to the theoretical lattice enthalpy.
Yes it is more exothermic, but its not enrrgetically favourable because in order to get Ca3+ ions you have to remove 3 electrons, amd from the trends of successive ionisation energies you should know that since Ca is in grp 2 there is a huge jump between the second and 3rd ie. Therefore even though the lattice enthalpy would be higher it takes so much energy to make the Ca3+ in the first place that the extra energy relaced by lattice formation is insignificant. See http://www.chemguide.co.uk/atoms/bonding/ionic.html and if you still have questions let me know
Yes it is more exothermic, but its not enrrgetically favourable because in order to get Ca3+ ions you have to remove 3 electrons, amd from the trends of successive ionisation energies you should know that since Ca is in grp 2 there is a huge jump between the second and 3rd ie. Therefore even though the lattice enthalpy would be higher it takes so much energy to make the Ca3+ in the first place that the extra energy relaced by lattice formation is insignificant. See http://www.chemguide.co.uk/atoms/bonding/ionic.html and if you still have questions let me know
Ohhh okay thank you so much - this really does make sense! Also on another note - this makes things a bit more complicated because on this link pg 6 it shows the lattice enthalpy of CaCl3 to be exothermic:
I'm really confused about this aspect of Born Haber cycles. Why is the lattice enthalpy of CaCl3 more exothermic than CaCl2 when CaCl2 is actually more stable? I thought that the more exothermic a compound is the more stable it is? I hope I'm making sense.
Struggling to make sense of what your asking as you haven't specified whether its LE. Form. or Dissoc. Assuming you are talking about LE. Form, to make CaCl3; Ca must be 3+ - the energy required to put in to achieve the third ionization energy for Ca is so large it will make the LE.Form more exothermic. If you draw out the B-H cycle it will make more sense as you can see more steps are required to form the CaCl3. The lattice dissociation enthalpy however will be more endothermic for the same reason (More steps in the cycle). CaCl3 is more stable as the energy released from the formation is greater than the energy released from the formation of CaCl3, the more energy we have to absorb to make formation happen the less stable it will be.
^ Can't remember at what point during the video, but E.Rintoul does explain exo/endothermic properties of ionic compounds and their stability very well.
Struggling to make sense of what your asking as you haven't specified whether its LE. Form. or Dissoc. Assuming you are talking about LE. Form, to make CaCl3; Ca must be 3+ - the energy required to put in to achieve the third ionization energy for Ca is so large it will make the LE.Form more exothermic. If you draw out the B-H cycle it will make more sense as you can see more steps are required to form the CaCl3. The lattice dissociation enthalpy however will be more endothermic for the same reason (More steps in the cycle). CaCl3 is more stable as the energy released from the formation is greater than the energy released from the formation of CaCl3, the more energy we have to absorb to make formation happen the less stable it will be.
^ Can't remember at what point during the video, but E.Rintoul does explain exo/endothermic properties of ionic compounds and their stability very well.
The ionisation energies do not change the lattice enthalpy, lattice enthalpy by definition is the energy released when gaseous constiuent ions form one mole of a lattice (or the reverse)
The ionisation energies do not change the lattice enthalpy, lattice enthalpy by definition is the energy released when gaseous constiuent ions form one mole of a lattice (or the reverse)
Yes, but the charge of the gaseous ions does - a more positively charged metal ion will yield a higher L.E.f.
The reason I mentioned I.E's as having an effect on your calcuated value of LE is that if you draw out a B-H cycle for both ionic compounds, you will see on the CaCl3 that the additional endothermic values in the cycle (owing to the extra ionization) will cause the calculated value for LE.f. to be more more negative (Exothermic) - whilst Ionization energies may not have a direct effect on the value of LE they will in your born-haber cycle.
Yes, but the charge of the gaseous ions does - a more positively charged metal ion will yield a higher L.E.f.
The reason I mentioned I.E's as having an effect on your calcuated value of LE is that if you draw out a B-H cycle for both ionic compounds, you will see on the CaCl3 that the additional endothermic values in the cycle (owing to the extra ionization) will cause the calculated value for LE.f. to be more more negative (Exothermic) - whilst Ionization energies may not have a direct effect on the value of LE they will in your born-haber cycle.
You are correct that a Ca3+ will indeed make the lattice enthalpy of formation more exothermic, but the reason for this is purely the increased charge and has nothing to do with the number of endothermic changes in your borne haber cycle.
You are correct that a Ca3+ will indeed make the lattice enthalpy of formation more exothermic, but the reason for this is purely the increased charge and has nothing to do with the number of endothermic changes in your borne haber cycle.
Yes, but the charge of the gaseous ions does - a more positively charged metal ion will yield a higher L.E.f. you will see on the CaCl3 that the additional endothermic values in the cycle (owing to the extra ionization) will cause the calculated value for LE.f. to be more more negative (Exothermic)
I'm so sorry - I'm not sure if you've seen my previous post above, but do you have any idea why the two websites show different values?
I haven't properly looked but most likely its due to the loose definition of lattice enthalpy. It can either be defined as the energy required to break a lattice, in which case it takes energy so is endothermic, or the enthalpy change when the lattice is formed, in which case it is releasing energy so is exothermic
I haven't properly looked but most likely its due to the loose definition of lattice enthalpy. It can either be defined as the energy required to break a lattice, in which case it takes energy so is endothermic, or the enthalpy change when the lattice is formed, in which case it is releasing energy so is exothermic
Ahh so sorry for the disturbance, haha that's fine take your time! And hmm that may be true you know about the different values...Thanks btw for the information you gave above - I understand much better now.