4E.7 in Textbook

[Julia Mizzi 2I]

Hi! I'm having a hard time figuring out how to solve 4E.7. For part A, I added up the bond enthalpies which I thought would be (944 + 3(158) -2(270) however I'm getting a vastly different answer than what's listed. Does anyone know where I'm going wrong?

[Justine Lim 3L]

Hi! I think you have the correct idea, I am just not sure about the values you are using. When using tables 4E2. and 4E3. for part A, I got my bond enthalpy equation to be 3(837) + (6)(412) - 6(518) - (6)(412) to equal = -597kJ/mol. I found the bonds by drawing out the lewis structure for C2H2 and C6H6.

[Daniel Lee 2J]

HI! I actually would say you are using the correct values (the person who responded might have looked at the wrong question). But I think what threw you off was the order of your calculations and the coefficients. Your equation should be [6(270)] — [3(158) + 1(944)] as the reaction enthalpy is essentially the sum of the product enthalpies times their coefficients minus the sum of the reactant enthalpies times their coefficients. The reason why 270 is multiplied by 6 is because we are using bonds. There are three NF bonds per NF3, and we have two moles of NF3, so this is then multiplied by 2.

[jhb123]

Hello!

You are using the correct values, yet your coefficients are slightly off. Reaction enthalpy is equal to the sun of product enthalpies multiplied by their respective, reaction coefficient minus the sum of the reactant enthalpies, multiplied by their reaction coefficients. This employs Hess’s law in overall enthalpy calculation.

This would give you:

6(270) - 3(158) - 1(944).

The coefficient of the first reaction would be 6, given the coefficient and the miner of NF bonds/NF3 molecule. (2x3=6)