3. Born Haber Cycles, Magnesium Oxide
Uploader Comments (jamesmungall)
All Comments (22)
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If you don't get Born-Haber Cycles, this video is the one to watch. I never imagined that the concept could be explained in such concise a way. Thank you!
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@haxelex Same really, I wasn't in class for this and the textbook just has an extremely abstract and assumption-making way of explaining things. Thanks immensely for the video!
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I can't thank you enough for this video, its the only one that's been explained so clearly, I finally understand!! :D
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The only part I dont understand about born haber cycles are the fractions by the molecules. In my book, I have a 249.2 kj for elemental Oxygen in gas phase. so the whole fraction thing confuses me....
Nonetheless, this video definitely helped me become less confuse about the born haber cycle in all
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@jamesmungall oh, it's ok i got it . Thanx your videos are really helpfull today was my midyear exam and I did well . Thanks again , and by the way your notes are really really good it helped me alot .
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THANK YOU SO MUCH. YOU'RE SO MUCH MORE HELPFUL THAN MY TEACHER AND MY TEXTBOOK COMBINE. Lmao seriously though, thank you so much.
why didn' we break the bond of oxygne to get the free radical using bond dissociation energies. Would we than add electrons one and two to get oxygen with the 2- charge?
sschw006 4 months ago
@sschw006 In these examples I've used the atomisation energy of Oxygen (and other diatomic molecules). However, you could use the bond dissociation energy instead because atomisation energy = 1/2 dissociation energy.
jamesmungall 4 months ago
I have a question about the lattice energy , do we subtract it from the value of oxygen and Mg atomization and ionisation ? or it is already negative ? and thanks
universen1 1 year ago
@universen1 Usually it is already negative, and so doesn't need to be subtracted. In some older textbooks it's given as positive, in which case you do need to subtract it!
jamesmungall 1 year ago
the correct second EA for oxygen is 744kj/mol i looked it up in my descriptive inorganic book i did my calculations and i got exactly -602kj/mol, ur video really helped me in understanding the cycle thank u so much
naynay0x 2 years ago
That's very interesting...I've just looked up the value in 'Calculations in A-level Chemistry' by Jim Clark, and he uses a value of +844kJ/mol! In fact I think that EAs cannot be calculated independently but are actually calculated using Born Haber cycles. So it must depend upon which cycle you use as to what EA you get. Good luck with your studies, James
jamesmungall 2 years ago