Method 2 Example

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Xuan Kuang 2L
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Method 2 Example

Postby Xuan Kuang 2L » Sat Jan 26, 2019 10:54 pm

On Friday's Lecture, there was an example for Method 2 which considered the bond enthalpies regarding CH2=CH2 + H-Br --> CH3-CH2Br

Could someone reiterate why we utilized the value of the C=C bond on the reactants' side, and why we considered using the bond enthalpy of C-C on the products' side? I got kind of lost on that part, but the rest seems to make sense...

Thanks in advance! :)

Sophia Diaz - Dis 1B
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Re: Method 2 Example

Postby Sophia Diaz - Dis 1B » Sat Jan 26, 2019 11:37 pm

For what I understand, when you use bond enthalpies, you use them for the bonds that are being used in the reaction.

So for the reactants side, there is a C=C bond that breaks into a C-C bond when it forms into CH3-CH2Br. There is no C=C bond on the products side and there is no C-C bond on the reactants side.

The bonds that are actually present in the molecule, whether it is a reactant or product, are the bonds that you use the enthalpies for.

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Re: Method 2 Example

Postby GavinAleshire1L » Sat Jan 26, 2019 11:52 pm

When using bond enthalpies Dr. Lavelle said to only account for the bonds that change in the reaction so that less calculation is involved. In this case the double bond was broken and then a single bond was formed, in addition to the H-Br bond breaking and forming a C-Br bond. All the other hydrogens attached to the carbons remain the same in terms of net enthalpy. Another more arduous way to solve is to calculate every bond enthalpy on both sides of the equation, but that could lead to more room for error.

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Re: Method 2 Example

Postby 305113590 » Sun Jan 27, 2019 11:42 am

To add, the bond enthalpy calculation process can lead to inaccuracies in enthalpy values. This is because bond energy between two atoms may be different depending on the bond length/angle. The only guaranteed enthalpy value is with diatomic molecules.

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