Method 2 example

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Method 2 example

Postby JustinHorriat_4f » Sun Jan 26, 2020 11:19 pm

In the method 2 example with CH2=Ch2, is the reason he doesn't include the bond enthalpy of the C-H bonds on the reactants side of the equation is because the products side has one more C-H bonds than the reactants side?

Ariana Iranmahboub1G
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Joined: Fri Aug 09, 2019 12:17 am

Re: Method 2 example

Postby Ariana Iranmahboub1G » Sun Jan 26, 2020 11:23 pm

Yes since the C-H bond does not break in the reactants, but the additional C-H in the products means that one C-H bond was formed.

Sofia Barker 2C
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Joined: Wed Sep 18, 2019 12:21 am

Re: Method 2 example

Postby Sofia Barker 2C » Sun Jan 26, 2020 11:35 pm

Will we be expected to know which bonds do not break when reactants react to form products? I wouldn't have known that the C-H bonds don't break unless someone were to tell me. Is it okay to assume that all bonds of reactants break?

christabellej 1F
Posts: 109
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Re: Method 2 example

Postby christabellej 1F » Mon Jan 27, 2020 8:34 am

I think that the reason only some of the C-H bonds were shown to break or form was based on the lewis structure. When he drew CH2=CH2, most of the C-H bonds remained in the "same" place, and so there was no need to show the bond breaking or forming. However, the = definitely broke as the product only had single bonds, which is why he calculated the double bond breaking but not other C-H bonds breaking.

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