Method 2: Bond Enthalpies  [ENDORSED]

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John_Parks_3D
Posts: 6
Joined: Wed Sep 21, 2016 2:58 pm

Method 2: Bond Enthalpies

Postby John_Parks_3D » Sun Jan 15, 2017 10:23 pm

I understand that when bonds are form there is an exothermic reaction making delta H negative and when bonds are broken delta H is positive. However, in lecture we discussed that the delta H of the reaction can be found using the summation of the products minus the summation of the reactants. Therefore why is the answer -58kj from the example given in lecture. Shouldn't it be -2014 since it is -1036 for the products and 978 for the reactants?

April Z 1K
Posts: 10
Joined: Fri Jul 22, 2016 3:00 am

Re: Method 2: Bond Enthalpies  [ENDORSED]

Postby April Z 1K » Sun Jan 15, 2017 10:32 pm

When using bond enthalpies, it's just summation, not [summation of products] - [summation of reactants].
As stated in the example, double bond C=C and single bond H-Br is broken (+978kJ, requires energy). Single bonds C-C, C-H, and C-Br are formed (-1036kJ, releases energy). Adding the two together, -1036kJ + 978 kJ = -58kJ.
Hope this helps.

John_Parks_3D
Posts: 6
Joined: Wed Sep 21, 2016 2:58 pm

Re: Method 2: Bond Enthalpies

Postby John_Parks_3D » Mon Jan 16, 2017 12:01 am

Oh ok thank you.


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