Ok I have many questions. Firstly, your value for the second election affinity of oxygen was wrong. You didn’t include the -602 from the enthalpy of formation. Secondly, at the start of the question, it said the first electron affinity of oxygen was -142kJmol^-1, when you put -147 in the born haber cycle. Ok, mistakes aside, i just wanted to know why the second electron affinity of oxygen is endothermic, when the first electron affinity was exothermic. Is it always like that or just specifically for oxygen? If not, how can we find out whether the enthalpy change is exothermic or not?
the second electron affinity for any atom is always endo as now in this case after its first electron affinity oxygen is negatively charged, there's a repulsion when u add the second electron to make O doubly charged. this repulsion makes it endo not exo
For the last question ..it is given ∆EA1H = 142.....but he used 147 in the answer for electron affinity....by using 142 we will get the answer as -784 ....by using 147 we will get -789
U forgot-602 while calculating
But the video was really helpful
I thought I was wrong but the comments reassured me :)
You missed the enthalpy of formation value (MgO) in calculating the enthalpy of 2nd electron affinity of oxygen at the end
Ok I have many questions. Firstly, your value for the second election affinity of oxygen was wrong. You didn’t include the -602 from the enthalpy of formation. Secondly, at the start of the question, it said the first electron affinity of oxygen was -142kJmol^-1, when you put -147 in the born haber cycle. Ok, mistakes aside, i just wanted to know why the second electron affinity of oxygen is endothermic, when the first electron affinity was exothermic. Is it always like that or just specifically for oxygen? If not, how can we find out whether the enthalpy change is exothermic or not?
first electron affinity is always exothermic for any reaction while the second and above are endothermic.
@@tadiwanashemutsayi9049 thanks
the second electron affinity for any atom is always endo as now in this case after its first electron affinity oxygen is negatively charged, there's a repulsion when u add the second electron to make O doubly charged. this repulsion makes it endo not exo
According to my textbook, its because the negative charge on the ion repels the second electron.
Yes, answer for last question is +789 according to my calculations. Cheers
Yeah, it's actually +784 KJ/mol
In the first electron affinity for oxygen the value given above is -142kj/mol and what u used was +147kj/mol........
Just proving how tricky these cycle calculations can be. You have to go really carefully!
I got -784. Some people are saying -789, WHY?
My final answer for the last question was 789kj mol-1, let me know what you guys get.
Hey bud you forgot the - 642 in the calculation before the lattic enthalpy. Well done anyway.
For the last question ..it is given ∆EA1H = 142.....but he used 147 in the answer for electron affinity....by using 142 we will get the answer as -784 ....by using 147 we will get -789
Thanks for the video
Thank you gr8 job on this video and the last
Wow great video ... This helped me a lot
Thank you so much!!
Very good Thanx
Padu bang