Chemistry Question

[oreoeater82]

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hey everyone-

i had a question about how to determine reactivity of compounds. my question sprung from a question of DAT destroyer in which one of the answers says:

C.) The propagation steps of flouridation are highly endothermic

The author explains that Flourine is the most reactive member of the halogen family. And therefore, has a very high exothermicity in the propagation steps. He also says, Iodine has a very endothermic first propagating step and reacts much slower. How is it that we know this and what key principles am i not grasping? Does reactivity increase or decrease up and down a group or across a period like electronegativity or atomic radius? Someone please help!

10 days until the DAT!!! Pray for me

 

[BillytheEchidna]

hey everyone-

i had a question about how to determine reactivity of compounds. my question sprung from a question of DAT destroyer in which one of the answers says:

C.) The propagation steps of flouridation are highly endothermic

The author explains that Flourine is the most reactive member of the halogen family. And therefore, has a very high exothermicity in the propagation steps. He also says, Iodine has a very endothermic first propagating step and reacts much slower. How is it that we know this and what key principles am i not grasping? Does reactivity increase or decrease up and down a group or across a period like electronegativity or atomic radius? Someone please help!

10 days until the DAT!!! Pray for me

OK, chemistry is not my forte, but think of the size of these atoms and the bond lengths between diatomic fluorine (F2)... Fluorine is a very small and electronegative atom, so the bond is going to be shorter (stronger) than it is in something big like I2. There is more energy in the F-F bond than can be contained by the short bond length -- if it breaks, the F-F bond is going to release a lot of energy and it will be more energetically favorable to form a radical than to stay bonded to itself. The I2 requires UV light or heat before it forms a radical (hence, endothermic). Reactivity increases as you go up (due to increasing electronegativities) and across the table (due to unoccupied spots in the valence shell). I hope I haven't mangled this too badly... 🙂 Good luck!

 

[Danny289]

^ above explanation is correct but a little complicated.

inititation step is always endothermic because you are breaking bonds. for F2 you need lots of energy comparing for example I2 because
(F--F) bond is much stronger than (I---I).

some good points.
-delta H will calculate for all steps together not just one step.
- the rate in multi step reaction will be equal to slowest step.

hope it helped
good luck 👍

 

[oreoeater82]

one question, you say that propagation steps are always endothermic? why does the manual then explain that flourine is highly exothermic?

 

[Danny289]

one question, you say that propagation steps are always endothermic? why does the manual then explain that flourine is highly exothermic?

sorry my mistake, intitiation is always endothermic, not propragation because breaking bonds happen in intitiation step not in propragtion.

 

[oreoeater82]

ok, i get it... propogation can be either depending on reactants and initiation is endothermic b/c you are using energy to break bonds

thanks

 

[Danny289]

^ you got it. In DAT they don't go too depth like what destroyer goes. Kaplan has done nice job for covering what you need for DAT in Organic and general chemistry. GL 👍

 

[oreoeater82]

Thanks!

 

[oreoeater82]

are termination steps endothermic also? they require energy to form a final product right?