To work out the enthalpy change you do:
the sum of bond energies in the reactants - the sum of bond energies in the products
Given average bond energies, you can easily work out the sum of bond energies.
For example, if we are told that
C-H = 413
O=O= 498
C=O= 746
H-O=464
Work out the enthalpy change for this reaction: CH4 + 2O2 > CO2 + 2H2O
((4 x C-H) + (2 x O=O)) - ((2x C=O) + (4 x H-O))
Then we substitute in:
(4 x 413) + (2 x 498) - (2 x 746) + (4 x 464) = 2648KJ - 3348KJ = 700KJ enthalpy change
Credit to author of this source: https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&ved=0CFoQFjAE&url=http%3A%2F%2Fchemistryatdulwich.wikispaces.com%2Ffile%2Fview%2FSection%2B4%2Bb%2Benergetics.doc%2F183320535%2FSection%25204%2520b%2520energetics.doc&ei=a72UUZOaHYa_0QXMkoD4Bg&usg=AFQjCNF9yn2bAWyGOPYSO4W1NrJExx7HQw&sig2=_XxB8HEYh8MHVUe5abc6sA&bvm=bv.46471029,d.d2k
-700 kJ ?
ReplyDeleteIf you put that in a calculator it would give the wrong answer so the equation should be: ((4 x C-H) + (2 x O=O)) - ((2 x C=O) + (4 x H-O))
ReplyDeleteThat is correct
DeleteThanks, I changed it on the post now :)
DeleteThis comment has been removed by the author.
ReplyDeletewhy isn't it -700kJ?
Deletewhy isn't it -700kJ?
Deletewhy isn't it -700J?
ReplyDeleteit wold be -700KJ
ReplyDelete