HW 8.73 (Bond Enthalpies

[Palmquist_Sierra_2N]

I'm kind of confused with the concept of bond enthalpies.
Estimate the reaction enthalpy for
a) 3 C2H2(g)-->C6H6(g)
b) CH4(g)+4 Cl2(g)-->CCL4(g)+4 HCL(g)

[Ishita_Dubey_2D]

To calculate the enthalpy of a reaction using bond enthalpies you would have to add together all of the bond enthalpies of new bonds formed (which will be a negative value because formation of bonds releases energy) and add this value to the total of the bond enthalpies of all of the bonds broken (which will be a positive value because breaking a bond requires an input of energy). Personally, it helps me a little more to draw out the structures of the reactants and products to see exactly which bonds are broken and which ones are formed. For example, for part a, 3 carbon-carbon triple bonds (reactants) are broken to form 3 carbon-carbon single bonds and 3 carbon-carbon double bonds in the resulting hydrocarbon ring structure.

[Ann Zhang_1M]

use the sum of bond enthalpies of all products minus the sum of bond enthalpies of all reactants.
and do not forget to multiply each bond with their moles

[Zoe Robertson 2H]

Why don't we include the C-H bonds? Is it because there are 3 moles of 2 C-H bonds broken and 6 C-H formed in the benzene ring, so they cancel out?

[104781135]

I'm not getting the correct answer for part A. I get a -369kJ/mol but the answer is -597kJ/mol? Can anyone help me with this please?

[104781135]

I'm not getting the correct answer for part A. I get a -369kJ/mol but the answer is -597kJ/mol? Can anyone help me with this please?

Never mind, I figured it out. C6H6 is benzene, and all of benzene's carbon bonds are resonance bonds, which have a different bond enthalpy than a single carbon bond or triple carbon bond.