4E.5

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Jessica Esparza 2H
Posts: 130
Joined: Wed Sep 18, 2019 12:15 am

4E.5

Postby Jessica Esparza 2H » Tue Jan 28, 2020 11:16 am

C for 4E.5 stated that there enthalpy is zero, but I dont understand how if new bonds are forming?
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Kishan Shah 2G
Posts: 132
Joined: Thu Jul 11, 2019 12:15 am

Re: 4E.5

Postby Kishan Shah 2G » Tue Jan 28, 2020 12:48 pm

The way you calculate the enthalpy of these reactions is always (Delta H(Bonds Broken) - DeltaH(Bonds Formed)). So if you break a C-H bond lets just say, but then reform it, they will cancel out according to our equation. That is what is happening in this problem because the Bonds Form release energy so the values are negative and the bonds breaking the value is positive since we are investing energy .

Caitlyn Tran 2E
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am

Re: 4E.5

Postby Caitlyn Tran 2E » Tue Jan 28, 2020 10:51 pm

Another way you can think of this instead of trying to figure out which bonds are forming and which bonds are breaking is to break all the bonds in the reactants and to form all the bonds in the product. In this problem, you would be breaking 4 C-H bonds and 4 C-Cl bonds and then forming 4 C-H bonds and 4 C-Cl bonds. Since you are breaking and forming the exact same bonds, which require the exact same energies, they cancel and the overall enthalpy is 0 kJ/mol. The placement of each bond does not affect the enthalpy calculations. Only the types of bonds broken and formed affect the calculation. Hope this helps!


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