flexastexas
  • flexastexas
Can someone help me balance this chemical equation?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
flexastexas
  • flexastexas
http://imgur.com/Mpxpqju
flexastexas
  • flexastexas
I am supposed to balance this equation to reprsent the reaction aswell as draw and energy diagram which labels the energies of each step in the reaction and activation energy
anonymous
  • anonymous
Just refer this and try to solve it : Basic Tutorial on balancing chemical equations : http://openstudy.com/study#/updates/559f06b5e4b05bcb08a14634

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flexastexas
  • flexastexas
Will do. How would do the second part of the question? Label the the energies of each step of the reaction and activation energy.
Photon336
  • Photon336
Well, I think A would be your reactants, B would be the transition state, and C would be your products.. but do you know why?
flexastexas
  • flexastexas
Ohhh, OK I didn't look at the question that way. So what you are essentially saying is, A would be the process of using the catalyst to kick start the reaction. B would then be the peak of the reaction and C would be the reactions end results.
nincompoop
  • nincompoop
no
nincompoop
  • nincompoop
|dw:1436679303437:dw|
Photon336
  • Photon336
Catalyst lowers activation energy Ea of both the forward and reverse reactions. I don't remember what the colors/atoms represent for the ball stick model.
nincompoop
  • nincompoop
that is what essentially occurred, but what you really need to focus on are the reactants from A and the product in C balance those two
nincompoop
  • nincompoop
I am just assuming that it would be Chlorine, but it can be any halogen with either F or Cl
nincompoop
  • nincompoop
F, Cl, Br, I name your big ball
Photon336
  • Photon336
Seems like you have SN2 reaction going on there..
nincompoop
  • nincompoop
since the ball is much much bigger than C, it may be a halide on the third or 4th row
Photon336
  • Photon336
i see now , but that carbon at the bottom has 5 bonds.. that must be your Transition state.
nincompoop
  • nincompoop
it would be an \(\sf SN_2 \) and creating a true bond between the halide will violate the octet rule, so I am positing that the interaction was due to induction.
nincompoop
  • nincompoop
the carbon in methyl is slightly partially positively and that can provide a little attraction with the halide.
Photon336
  • Photon336
ok.. It looked like that was a bond there makes sense!
nincompoop
  • nincompoop
I didn't draw a curved arrow from halide to the partial positive carbon :)
nincompoop
  • nincompoop
in any case I will say that the products are \(\sf Me-OH + Cl^-\)
nincompoop
  • nincompoop
our charge was preserved (-)
Photon336
  • Photon336
I mean if that's a concerted mechanism.. practically happens in one step so .. like Cl- would be able to hold - charge good LG..
nincompoop
  • nincompoop
oh that did not look right let me type it again \(\sf Me–OH + Cl^- \) |dw:1436680471791:dw|
Photon336
  • Photon336
just curious.. how would you justify the opposite reaction
Photon336
  • Photon336
because i see two arrows, which implies the opposite reaction, I mean chlorine's pretty happy with that - charge
nincompoop
  • nincompoop
I shouldn't probably draw the last reversible arrow in the end because OH wouldn't be a good leaving group for a reversible reaction
Photon336
  • Photon336
OK. he asked about energy of the reactants vs products... technically if this process is favorable the nucleophilic attack on the on Methylchloride, then wouldnt this mean that the product is more stable, at a lower energy, than the reactant?
nincompoop
  • nincompoop
|dw:1436680676160:dw|
nincompoop
  • nincompoop
\(\sf Cl^- \) is a pretty stable base (weak), and initially \(\sf OH^- \) was a strong nucleophile
Photon336
  • Photon336
Ah.. I see.. that double dagger I think that's what they call it.. HO-C bond breaking, while the Cl- C bond is forming.
Photon336
  • Photon336
I guess if you think about it the opposite way you can't actually just kick off OH- unless you introduced some proton source.
nincompoop
  • nincompoop
and you are correct, the product is most likely more stable and our graph would look like this |dw:1436680983637:dw|
Photon336
  • Photon336
YOU sir NAILED this one
nincompoop
  • nincompoop
you are correct. OH is a bad leaving group so we're going to need a good source of energy to overcome the barrier, which translates to a good proton donor
nincompoop
  • nincompoop
my organic chemistry is pretty crappy so I am refreshing with some problems.
Photon336
  • Photon336
crappy is a relative. that was well explained
nincompoop
  • nincompoop
the double dagger that you mention is the opposite it is the concerted \(OH^- \) forming a bond to C whilst Cl breaking
Photon336
  • Photon336
mixed that up nucleophile was OH- so it's forming a bond haha
nincompoop
  • nincompoop
correct
flexastexas
  • flexastexas
Sorry Im late. It was really late when I posted. Question, how did you determine what kind of chemical this was from the picture?
flexastexas
  • flexastexas
And also, that last graph you posted would be what they asked for ?
flexastexas
  • flexastexas
And the first would be the balanced equation?
flexastexas
  • flexastexas
?

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