8.73.b

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Jeremiah Samaniego 2C
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Joined: Fri Sep 29, 2017 7:05 am

8.73.b

Postby Jeremiah Samaniego 2C » Fri Jan 12, 2018 7:14 pm

For homework question 8.73b we are given the chemical equation CH4 (g) + 4 Cl2 (g) --> CCl4 (g) + 4 HCl (g). In the solutions manual, it shows that we are breaking 4 mols of a C-H bond and we are forming 4 mols of a H-Cl bond. Can someone explain to me why there isn't only one mol being broken and formed for both bonds?

Ashley Macabasco 2K
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Joined: Fri Sep 29, 2017 7:05 am

Re: 8.73.b

Postby Ashley Macabasco 2K » Fri Jan 12, 2018 7:28 pm

I believe that 4 moles are broken in the C-H bond because there are four hydrogens bonded to one carbon in CH4 (lewis structure) so there are four C-H bonds and not one C-H bond. There are 4 moles of the H-Cl bond formed because of the 4 coefficient in "4 HCl" indicating that four H-Cl bonds will be formed.

Cassandra Mullen 1E
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Re: 8.73.b

Postby Cassandra Mullen 1E » Sun Jan 14, 2018 4:37 pm

When determining the bond enthalpy, it helps to draw the molecules or at least visualize the individual bonds within each molecule. The bond enthalpies themselves are defined as the amount of energy to break one mole of the molecule. In this problem, there are four C-H bonds being broken and 4 H-Cl bonds being formed.

Yiwei Zhou 2I
Posts: 21
Joined: Fri Sep 29, 2017 7:07 am

Re: 8.73.b

Postby Yiwei Zhou 2I » Sun Jan 14, 2018 5:28 pm

Because 4 hydrogen atoms are bounded on C atom so the bond enthalpy should be multiplied by 4. You can also count every bond, broken or not, in the reactants and the products that it takes less time to think about the bond breaking and forming process.


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