8.75 part b

Moderators: Chem_Mod, Chem_Admin

Abigail Urbina 1K
Posts: 102
Joined: Thu Jul 27, 2017 3:01 am

8.75 part b

Postby Abigail Urbina 1K » Sun Jan 21, 2018 8:57 pm

CH3CHCH2(g) + H2O(g) --> CH3CH(OH)CH3(g)

I understand which bonds are broken and why (because I drew out the Lewis structures). However, for the bonds that are formed from this reaction, I am confused as to why the solution manual says that 1 mol of a C-C SINGLE bond is formed. Also, it says that 1 mol of a C-O SINGLE bond is formed when I thought that a C-O DOUBLE bond would be formed. I did, however, get 1 mol of the C-H single bond being formed. Can someone please clarify the other two bonds that were formed?

Posts: 20
Joined: Fri Sep 29, 2017 7:06 am

Re: 8.75 part b

Postby sahajgill » Sun Jan 21, 2018 9:27 pm

A C-C bond is formed because when the double bond C=C is broken, the C-C bond has to be reformed. It is not possible for one of the C=C bonds to be broken only. You are right about the C-H bond, but the other bond that forms is C-OH, which is not a double bond.

Ammarah 2H
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

Re: 8.75 part b

Postby Ammarah 2H » Wed Jan 24, 2018 5:17 pm

I don't know if this will help, but for these problems, I find it easier to include all of the bonds in the calculation! I draw the Lewis structures and then count all the bonds for both the products and reactants. The math works out because the repeated bonds are subtracted out anyways.

Adrian Lim 1G
Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

Re: 8.75 part b

Postby Adrian Lim 1G » Wed Jan 24, 2018 7:42 pm

I also thought that there would be a C=O double bond formed. Can anyone please clarify?

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 1 guest