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Chemistry: Post your doubts here!

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hey...
C40H82 -------> C16H34 + C24H48

a good way of explaining this question is to see how many bonds u have before the reaction and how many bonds u have after the reaction for example...

before the reaction we have 39 C-C bonds and 82 C-H bonds...after the reaction we have 15+22 = 37 C-C bonds and one C=C bond and 48+34 = 82 C-H bonds

so we have broken 39-37 = 2 C-C bonds and formed 1 C=C bond
bond broken = 2 x 350 = +700
bond formed = -610 = -610
enthalpy = + 90 kjmol^-1 !!

thats it..the mark scheme got the answer as +180 because he has used a different equation in cracking instead of forming C24H48 as our alkene he formed 2 moles of C12H24 and in that way we will have 2 C=C bonds instead of one and 4 C-C bonds instead of 2 because one mole of C12H24 contains 1 C=C bond and 10 C-C bonds so the 2 moles will contain 20 c-c bond and 2 C=C bonds 20+15= 35 so 39-35 = 4 C-C bonds so we had broken 4 C-C bonds and formed 2 C=C bond..in that way u will get 180 kj per mole instead of 90..both are correct though depending on the equation u made ;)

i hope u got it!!

Oh my gosh, thanks! I thought in C24H48 all of them were C=C. How stupid of me. Thank you so much!! :)
 
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Can anyone please please tell how to calculate the enthalpy change for this reaction:
This is from October november 2008 paper 4 q4b
C40H82 : C16H34+2C12H24(it says considering the bonds broken and bonds formed use the data booklet to calculate the enthalpy change for the reaction)
 
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Guys can u xplain me q1(ii) in m/j 08 ....I dont understand isnt the E cell = (more +ve)-(less +ve) .....so why isnt it +1.36-0.77 instead it is 0.77-+1.36 ..pls help !
 
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