15.1/R1.2.5 Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides [HL IB Chemistry]
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- เผยแพร่เมื่อ 29 เม.ย. 2012
- 15.1 Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change. Atomise both reactants then ionize the metal, and "electron affinity" the non-metal. Don't forget the electrons and the state symbols and to feed the cat. Students especially enjoy you calling it the "Yawn"-Haber Cycle - we have never heard that joke before....
![15.2/R1.4.1 Predict the entropy change for a given reaction or process [HL IB Chemistry]](http://i.ytimg.com/vi/ro3AnXdajAM/mqdefault.jpg)
the shade throwing at the beginning of this video is real...lol
I can see clearly now the pain is gone!
I missed this lesson in class, but watching this video I feel as if I was there with them, amazing video, very well done.
Just what i needed at 3:30 a.m cramming for my chemistry exam thats in exactly 5 hours, mucho gracias
Muchas
I just LOOVE your explanation👏🏽
Thank you so much! I finally got this after struggling for a year or so! I have an exam coming up and this has definitely boosted my confidence!!
Glad I can help
This is the only thing that helped me understand. Thank you so much!!!!
Bloody great video mate. Huge help!!!!
Great videos. Thank you so much!
Quite useful...Thank you!
The Mr Kelly part was hilarious. Just the way you said it.
In a previous video you said your daughter's second name would be molality. So I guess we know what your daughter will be called! Lattice Molality Thornley!
love your sense of humor!
this is so clear! thank you! :)
Thank you, I was struggling, but I can go to my test after watching your video, Thank you again
Something I struggled for two months to learn was taught to me in literally ten minutes thank you so much.
These are my favorite comments and mostly why I make these vieos - thank you.
Thank you for the explanation and the great chart! Professors here don't seem to explain every individual step as well as you do, I think we could really use you here in Florida! lol
Thank you so much for your videos they were really helpful in preparing for the exam! :D
Literally amazing thank you so much
Ooo Myy Godd, This is SO UNDERRATED, thanks, now I understand, I was stuck at the ionisation and electron affinity part
Thanks
You are a truly enthusiastic IB Chem teacher:D
Thank you this was very helpful!!!
Lol! The introduction was so perfect. I can't stop laughing.
fuckin love you man life saver for a missed class
Not the hero we deserve but the hero we need. Lord and Saviour Sir Richard Thornley!
Thank you so much your (sum of the clockwise - sum of the anticlockwise=0) has helped me a lot with the exams only next week I thought I would hate the born-haber calculation you just made it a lot easier :)
hahahaha "Lattice" killed me - thanks for the vid :)
Thanks! Helped me a ton with my homework!
Glad i could help.
Lattice Molality Thornley
That helped a lot. Thankyou so much 😁
Right off the bat the illustration is helpful! How was I so confused
this really helped out thankyou, fine work :)
so helpful!!! thank u sooooo much!
Amazing video tks mate
I loved seeing your frustration hahaha but thank you for this!!!! I get it now! I just have to learn that Bond Energy part...
This is a fabulous video with which I can actually understand Born-Haber Cycles...You're the best! Thank you so much! Also quick question, how does enthalpy of solution fit into the cycle? I'm not quite sure with that bit
Does it matter which order you do the left-handed ones in? For example, at 6:30, I did the two atomization steps first, then did the ionization, but you did atomization, ionization, atomization.
Thanks for your effort to make these videos. My lectures where sending me to sleep...
he makes chemistry sound so thick and sophisticated.....me likey
Okay studied it a bit more and I understand it all!!! All thanks to you :]
Thank you so much!!
u the greatest!!! #shoutout all the way fr jamaica
Oof it's my turn to prepare for Cape😣😣
Coolest vdo on born haber cycle.. Lvdd ir
Thank you so much 💕🙏
yes, you are right we need to use first as well as second Ionization energies for Mg.
Mg -------> Mg^+1 ----------> Mg^+2
Thanks a bunch sir, HL Chem exam in a little over a week!! :D
awesome,thank you very much
YOU ARE AMAZING
Thank you sir!
Thank you so much 💕
is it always 1/2X2 for the gas element on the reactant side?
Wouldn't the lattice value be positive due to bond making which is delta H positive?
Kinda feels like captain J sparrow is teaching me chemistry... XD
Could you write the deltaH of atomization of chlorine next to deltaH of atomization of sodium (so there would be two arrows, and you would have be atomizing chlorine sooner), just to conserve some space? Or are the markers quite particular with the arrangement?
That's right, Mr. Kelly ...
why did we double the 1st IE of Na rather than adding the 1st and 2nd IE's ?
thank you really helps
Perfect!
Amazing. thankyou
What would be the Born-Haber cycle for aluminium iodide?
I searched online for the second ionisation energy for sodium because I remembered that it is much harder for Na to lose a second electron since Na+ is already a full octet, the 2nd IE energy of Na is actually 4563 kJ/mol, so the total IE energy of Na2+ should be 496+4563=5059, not 2*496.
please be my chemistry teacher
Hey its Kanna-chan! UwU
me too
Thank you for the video. I have one question however that always seems to confuse me: how do you set the born haber equation at the end to find the enthalpy of formation or lattice enthalpy? I'm always confused on what to set as positive or what to set as negative. Thanks!
+Vincent Prochoroff In the new syllabus the IB have decided that lattice energy is now positive. Which means the arrow should point up (+ve = endo)
The formation will always be negative and point down (exo)
Thanks!
What if you have a subscript of 3 like idk NaCl3 or anyother would it turn into 1/3 istead of 1/2?
Hi Richard,Thanks for the video!!In your second example for the ionisation of sodium you say 2 x the 1st ionisation energy. Should it not be the 1st ionisation plus the 2nd ionisation energy that you use?? Can you clarify im confused.
+oliver ingham I need to make Na+ and another Na+ = 2x1st IE of Na
I do not want to make Na2+ = 1st + 2nd IE of Na !
awesome :)
Question - How do get +657 KJ for the EA of O2? I keep getting 753 - 141 = 612 KJ
No one is coming to help you. :(
It is 612 kJ mol^-1, he made a mistake in the video.
Yeah absolutely - but what about in the case of something like MgO
Where Mg is 2+
Then you'd need 1st ionization energy plus 2nd.
But i cant find it in the data booklet!
Is it even there>?
Or will i just not be asked such a question?
THANKS!!!
Wow my chem teacher is good but this really good as well
How do you know whether the electron affinity arrow is up or down?
Thank you, I needed that, however sometimes, like in this example the arrow continued to go up.
Also, the reason it was upwards was because if you add up the first and second electron affinities of Oxygen, it turns out to be positive, thus an endothermic reaction.
@@aliarshad9739 The 1st electron affinity is bond between electron and nucleus. As we have learned, bond making is exothermic, thus negative (-).
For the 2nd electron affinity, it would be hard to add another electron as now you would be adding electron to a negative charged ion that was formed from the 1st electron affinity. Therefore there will be repulsion between the ion and the electron, so more energy is needed to overcome this repulsion. So, this is endothermic (+)
In conclusion, 1st electron affinity is always exothermic and 2nd electron affinity is always endothermic. The overall electron affinity can be positive or negative, depending on how exothermic or endothermic.
In this video, the electron affinity is positive as the value of the 2nd electron affinity is more positive. The value of the 1st electron affinity is not negative enough to make the overall value negative.
If anyone has a problem understanding my last paragraph, it just means that the magnitude of the 2nd electron affinity is larger than the 1st, so they add up to a positive number.
Thank you for telling me that if we only use one atom (In the Na2O example, O) we have to divide the bond enthalpy by number of atoms in the molecule... life=saved. Again
What stage is the bond dissociation enthalpy in the first born haber cycle?
There are 2 ways to account for this stage:
Atomisation of chlorine = making 1 chlorine atom from chlorine in its natural state = Cl2
This is the same energy change as breaking the bond in a Cl2 molecule thus making atomic chlorine
The fact that we only need 1 Cl but have to make it from Cl2 accounts for the 1/2 popping up.
thank you
Fantastic video had my first lesson of a2 chemistry and was baffled by this :( I got a couple of silly questions : 1 why did you have to times the enthalpy change atomisation of Na by 2 ?
+Sam Hanly Because you need 2 Na atoms to make the ionic crystal in question.
Great video! But isn't conversion of Na(s) to Na(g) delta H sublimation?
it is but the ib prefer atomisation. Sublimation of I2(s) would make I2(g) , which would not work since we need gaseous atoms, not gaseous molecules
it is very good video. thanks
ta
too good
I have a question. Say you have to get 2 Cl- ions from 2Cl to make say MgCl2, then how do you calculate the enthalpy change there?
2X 1st Electron affinity - data booklet has this value
I wish I could just watch these instead of going to my chemistry classes; watching these, reading through one's textbook and doing lots of practice questions would be perfectly sufficient to score a good grade in IB chemistry I should think :)
Because these videos are absolutely brilliant :))
Just wondering if the order matters? Like if I do the atomization of sodium then followed by the atomization of oxygen, and then do the ionization energy of sodium and electron affinity of oxygen, because I have seen it like that in some books. Also, where does bond disassociation fit in and is that required? Thanks.
+Elishbah Amer Order does not matter for the maths but if you are not methodical you may miss a step. You might be given the bond energy for Cl2. So this much energy will produce 2 x Cl atoms = 2 x delta H atomisation. Its just another way to try to trip you up!
Richard Thornley Yes, I guess I should probably stick with this way to be on the safe side, thanks!
Yes this is OK - some people even write the sodium parallel to the chlorine - just clearly label it
Oh Mr. Kelley 😂👌🏾
Nice one
@richardthornley When do I use atomization and when bond enthalpy? are they always interchangeable?
I thought:
bond enthalpy was breaking covalent bonds
and
enthalpy of atomization was breaking intermolecular forces to create complete separation
normally in IB they use the same trick
atomisation of chlorine = 1/2 x bond energy of chlorine
since both make one chlorine atom (g)
atomisation is making atoms (g), either by breaking intermolecular forces or even intra if this will result in individual atoms.
Richard Thornley But for this example here, one atom of oxygen isn’t made by one mole of O2 but , 1/2 a mole of O2?
I could watch @7:44 all day on loop
will i be marked wrong if i inter-change the enthalpy change of atomisation for Na and the enthalpy change of Cl?
nope
for Na, why is the second ionization energy not used as 2 electrons are going out? why is the first ionization energy multiplied by 2?
it is not 2 electrons from one sodium (1st + 2nd IE)
it is one electron from TWO sodiums (1st IE x 2)
OHHHH I get it :) Thank you :)
Thanks sir
Your accent is gorgeous!
life saviourrrrrr
isn't the 1st ionization energy different from the 2nd? if so, then why did you double the 1st ionizatoin energy?
+MuchWow69 Assuming you're talking about ~6:48, he was saying that he was removing one electron from two moles, so double the first ionization energy (second ionization energy would be if he had removed two electrons). If you look at Na in that part, you'll notice it's missing one electron due to this.
Thanks
hi, how did you find the enthalpy change of atomisation for Na? at 6:37
it is not in the booklet - you are either given it or you must calculate it.
Hi, I have three questions. First, how do you know how many times to ionize? Second, why is it 1/2Cl2 (I understand that there is only 1 mole of the compound, but I don't understand why its 1/2Cl2. eg in MgCl2 how come it becomes Cl2 not 1/2Cl2)? Third in my data booklet the electron affinity is -141 and + 753 which becomes 612 rather than 657? Are they suppose to be different or am I doing something wrong? Thank you
1st - group 1 metals need to lose 1 electron, group 2 need to lose 2.
If the metal appear twice eg Na2O, you need to make sure you double your value. i.e. 2 x1st IE for Na
2nd MgCl2 needs Cl2, but NaCl needs only 1/2Cl2. You have to use the chlorine in its standard state = Cl2 AND you only need 1 chlorine atom, so it has to be half.
3rd in my data book I agree with you. The IB like to change the data in the data booklet - this is a pain in the arse.
ohh i see, thank you!
3:51 had to do a double take just to make sure that I heard what I really that I heard lmao
What did he say? I didnt quite hear/understand him
ea games challenge everything
lol oh ok thx
HAHA 'thank god for control z'
awesome video!
How come you divided the atomisation for 1/2O2(g) -> O(g) by 2, but in the previous example you did you did not do the same for the atomisation of 1/2Cl2(g) -> Cl(g) ?? I have a HL test tomorrow and need to know this please help!
Beacuse he wasnt doing it with numbers.
where in the data booklet can we find 2nd IE and 2nd EA?
they are not in there - they have to be given or calculated
I imagine Mr. Kelly wouldn't be too pleased with this video?
what about the dissociation step?
Jenny Lopez = atomization