Moderators: Chem_Mod, Chem_Admin

Jade Fosburgh Discussion 2C
Posts: 59
Joined: Fri Sep 29, 2017 7:07 am


Postby Jade Fosburgh Discussion 2C » Wed Jan 17, 2018 10:27 pm

Could someone explain the negative signs in this reaction? I was able to get up to the point where you have the enthalpy values to break the bonds, but from there, I don't understand why in the solutions manual it is -6*518, and +3*837. Do you add the two numbers after that?

Tanaisha Italia 1B
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.73a

Postby Tanaisha Italia 1B » Wed Jan 17, 2018 11:03 pm

It is -6*518 because you are forming six c-c bonds and the enthalpy of each one (with resonance) is 518. 3*837 refers to the enthalpy of c-c triple bonds broken. The enthalpy of bonds formed is subtracted from that of bonds broken.

Andrea ORiordan 1L
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

Re: 8.73a

Postby Andrea ORiordan 1L » Thu Jan 18, 2018 12:07 pm

Remember that it takes energy to break bonds, and energy is released when bonds are formed. Therefore, -6*518 + 3*837 = -597, which is the enthalpy of reaction.

Posts: 60
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time

Re: 8.73a

Postby RuchaDeshpande1L » Thu Jan 18, 2018 4:02 pm

Using bond enthalpies, we can calculate the enthalpy of the reaction. Breaking bonds requires energy (positive change) while creating bonds releases energy (negative change). Therefore, we can find the net energy change by adding all the bond enthalpies of the bonds broken, and subtracting from that value the bond enthalpies of the bonds formed. For question 8.73a, doing this method using the values in the book gives us a final answer of -597 kJ/mol.

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 2 guests