The Student Room Group

A-Level Chemistry Energetics

Hi! I’m totally over complicating this question and need a little help~
“Use the table of mean bond theories shown below to determine which of the answers shows the standard enthalpy change for the addition reaction of bromine to 3-iodobut-1-ene
C=C 614 kJ/mol
C-H 413 kJ/mol
C-C 348 kJ/mol
Br-Br 193 kJ/mol
C-Br 276 kJ/mol”

+93 kJ/mol
-93 kJ/mol
-359 kJ/mol
the question cannot be answered due to insufficient data


Thank you!! :smile:
(edited 3 weeks ago)
Reply 1
Try writing the equation for the reaction, and then draw the displayed formula for the reactants/products. I'm guessing you're trying to calculate using every single bond energy (?), but you'll notice that a lot of them are the same between reactants and products. So it's only required to calculate just for the few bonds types that change.
Let me know if you're still confused, this isn't really a very good explanation 😅
Reply 2
Original post by Methene
Try writing the equation for the reaction, and then draw the displayed formula for the reactants/products. I'm guessing you're trying to calculate using every single bond energy (?), but you'll notice that a lot of them are the same between reactants and products. So it's only required to calculate just for the few bonds types that change.
Let me know if you're still confused, this isn't really a very good explanation 😅
I’ve tried doing that, I think the wording of the question isn’t all that clear really. I’m trying to find the overall enthalpy change. I’ve redone my calculations and gotten +596 so I think I’m a mile off :/
Reply 3
Original post by lolaconleyy
I’ve tried doing that, I think the wording of the question isn’t all that clear really. I’m trying to find the overall enthalpy change. I’ve redone my calculations and gotten +596 so I think I’m a mile off :/

When I did it, I found the difference in bond types were:

For reactants, one Br-Br bond (193) and one C=C bond (614)

For products, one C-C bond (348) and two C-Br bonds (2 x 276)


So enthalpy change for reaction is 193 + 614 - 348 - 2 x 276 = -93 kJ mol-1
Reply 4
Original post by Methene
When I did it, I found the difference in bond types were:

For reactants, one Br-Br bond (193) and one C=C bond (614)

For products, one C-C bond (348) and two C-Br bonds (2 x 276)


So enthalpy change for reaction is 193 + 614 - 348 - 2 x 276 = -93 kJ mol-1

Ohhh- I see where I was going wrong. I was drawing it as though H-Br as the electrophile haha. Thank you so much for your help 🙂 really appreciate it!
Reply 5
Original post by lolaconleyy
Ohhh- I see where I was going wrong. I was drawing it as though H-Br as the electrophile haha. Thank you so much for your help 🙂 really appreciate it!

Nws, glad you figured it out 😃
I’d honestly be tempted to say there is insufficient data to work out an answer.

You need to remember that bond enthalpies are measured in the gas phase and you haven’t been given any enthalpy data that would allow you to correct for this.

From your knowledge of the halogens, Br2 in it’s standard state is a liquid and we don’t actually know from the information given in the question what state the 3-iodobut-1-ene and 1,2-dibromo-3-butane are in under standard conditions - chances are that neither are gases, though.
Reply 7
Original post by UtterlyUseless69
I’d honestly be tempted to say there is insufficient data to work out an answer.
You need to remember that bond enthalpies are measured in the gas phase and you haven’t been given any enthalpy data that would allow you to correct for this.
From your knowledge of the halogens, Br2 in it’s standard state is a liquid and we don’t actually know from the information given in the question what state the 3-iodobut-1-ene and 1,2-dibromo-3-butane are in under standard conditions - chances are that neither are gases, though.
I had the same thought at first, I’ve just submitted it all and -93 was the correct answer. It’s externally set questions that we are now using and they aren’t worded very well at all :/
Original post by lolaconleyy
I had the same thought at first, I’ve just submitted it all and -93 was the correct answer. It’s externally set questions that we are now using and they aren’t worded very well at all :/
I see. It’s a very dodgy question imo, but well done for working through it and getting a right answer.
Reply 9
Totally fair point, and one that I completely overlooked... and so too did the question-setters it seems :P

Quick Reply

Latest