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The bonds are N2 one N≡N bond (triple bond) H2 one H-H bond NH3 three N-H bonds Hence: ΔH = D(N≡N) + 3·D(H-H) - 2·3·D(N-H) 942kJ/mol + 3·432kJ/mol - 6·386kJ/mol = ______________
ahh thank you SO much!!!! @abb0t !
@wcaprar thank you for coming! :) please help if you could :'(
@abb0t hi abb0i think you are like the only one who could actually do this problem. so please please please help :'(
Remember that breaking a bond gives off heat and forming a bond consumes heat there4 lets look at the equation. I had to look up those values, you should be able to look them up in your book, i think they are lsted in the back of your book. Do the same thing I did with the last problem.
I don't have all the values, but I have a few here in front of me: C=N is 891 C-H is 413 C≡N is 305 Please try it yourself first. I am not benefiting from giving you the answers, rather it is your duty to master this for your exam and or final.
@abb0t yes, i know. but what i actually need help is just to see how it is set up, and what the value of N triplebond C is. because it is not in the book.
I already gave you some values there.