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1. The question is asking me to find a value for the mean bond energy of the C-H bonds in Ethene and Ethane.

I am given the mean bond energies for the C=C, the H-H and the C-C bonds which are 612kj/mol, 436kj/mol and 346kj/mol respectively, and I am given a value for the Enthalpy of formation which is -136kj/mol but I don't have an answer and I am struggling to work it out.

So far I have

C2H4 + H2 -----> C2H6 = (-136kj/mol)

4C-H(?) + C=C(612) + H-H(436) ------> 6C-H(?) + C-C(348)

Therefore 4C-H(?) + 1038 ------> 6C-H(?) + 348

Therefore -2C-H = -348 + 1038

Therefore -2C-H = 690

Therefore C-H = -345kj/mol

2. link to image of question: https://www.flickr.com/photos/140547...posted-public/
3. hey there , i did not solve by numbers but the an easy way to solve it is by saying bond formed is endothermic oo bond breaking is exothermic
you can solve by subtracting the bond on right side from bond on left side
but one question what is the answer for part a ?
4. 612+ 436 + 4c-h ---> 348 + 6c-h + (-136)
612 + 436 ---> 348 + 2 C-H + (-136)
1048---> 212 + 2 c-h
836---> 2 C-H
418 ---> C-h

-136 = [612+ 436] - [2y + 348]
-136 = 1048 - 348 - 2C-H
-136 - 700 = -2C-H
-836/2 = C-H.

SO + 418 ...
5. (Original post by katenell)
hey there , i did not solve by numbers but the an easy way to solve it is by saying bond formed is endothermic oo bond breaking is exothermic
you can solve by subtracting the bond on right side from bond on left side
but one question what is the answer for part a ?
the answer to part A is 'Average Enthalpy required to break a covalent bond'

However I am only focussing on Question 5 in this instance, but just using some figures from the table in Q4
6. (Original post by rimstone)
612+ 436 + 4c-h ---> 348 + 6c-h + (-136)
612 + 436 ---> 348 + 2 C-H + (-138)
1048---> 210 + 2 c-h
838---> 2 C-H
419 ---> C-h
for the second part did you mean -136?
7. I rewrote it and yes i did, The answers, 418 anyways. basic algebra, with little chem knowledge needed, they could have made it a lot harder.
8. (Original post by rimstone)
612+ 436 + 4c-h ---> 348 + 6c-h + (-136)
612 + 436 ---> 348 + 2 C-H + (-138)
1048---> 210 + 2 c-h
838---> 2 C-H
419 ---> C-h
i said
4x +612+436 -(348+6x)= -136
-2x +1048-348 = -136
x= (-700-136)/2
x= 418
9. (Original post by Sam00)
the answer to part A is 'Average Enthalpy required to break a covalent bond'

However I am only focussing on Question 5 in this instance, but just using some figures from the table in Q4
thank you
10. (Original post by katenell)
i said
4x +612+436 -(348+6x)= -136
-2x +1048-348 = -136
x= (-700-136)/2
x= 418
Would it be 418 or -418
11. (Original post by Sam00)
Would it be 418 or -418
Bond enthalpies cannot be negative

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