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flexastexas

  • one year ago

Can someone help me balance this chemical equation?

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  1. flexastexas
    • one year ago
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    http://imgur.com/Mpxpqju

  2. flexastexas
    • one year ago
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    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

  3. anonymous
    • one year ago
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    Just refer this and try to solve it : Basic Tutorial on balancing chemical equations : http://openstudy.com/study#/updates/559f06b5e4b05bcb08a14634

  4. flexastexas
    • one year ago
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    Will do. How would do the second part of the question? Label the the energies of each step of the reaction and activation energy.

  5. Photon336
    • one year ago
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    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?

  6. flexastexas
    • one year ago
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    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.

  7. nincompoop
    • one year ago
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    no

  8. nincompoop
    • one year ago
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    |dw:1436679303437:dw|

  9. Photon336
    • one year ago
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    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.

  10. nincompoop
    • one year ago
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    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

  11. nincompoop
    • one year ago
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    I am just assuming that it would be Chlorine, but it can be any halogen with either F or Cl

  12. nincompoop
    • one year ago
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    F, Cl, Br, I name your big ball

  13. Photon336
    • one year ago
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    Seems like you have SN2 reaction going on there..

  14. nincompoop
    • one year ago
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    since the ball is much much bigger than C, it may be a halide on the third or 4th row

  15. Photon336
    • one year ago
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    i see now , but that carbon at the bottom has 5 bonds.. that must be your Transition state.

  16. nincompoop
    • one year ago
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    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.

  17. nincompoop
    • one year ago
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    the carbon in methyl is slightly partially positively and that can provide a little attraction with the halide.

  18. Photon336
    • one year ago
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    ok.. It looked like that was a bond there makes sense!

  19. nincompoop
    • one year ago
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    I didn't draw a curved arrow from halide to the partial positive carbon :)

  20. nincompoop
    • one year ago
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    in any case I will say that the products are \(\sf Me-OH + Cl^-\)

  21. nincompoop
    • one year ago
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    our charge was preserved (-)

  22. Photon336
    • one year ago
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    I mean if that's a concerted mechanism.. practically happens in one step so .. like Cl- would be able to hold - charge good LG..

  23. nincompoop
    • one year ago
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    oh that did not look right let me type it again \(\sf Me–OH + Cl^- \) |dw:1436680471791:dw|

  24. Photon336
    • one year ago
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    just curious.. how would you justify the opposite reaction

  25. Photon336
    • one year ago
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    because i see two arrows, which implies the opposite reaction, I mean chlorine's pretty happy with that - charge

  26. nincompoop
    • one year ago
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    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

  27. Photon336
    • one year ago
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    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?

  28. nincompoop
    • one year ago
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    |dw:1436680676160:dw|

  29. nincompoop
    • one year ago
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    \(\sf Cl^- \) is a pretty stable base (weak), and initially \(\sf OH^- \) was a strong nucleophile

  30. Photon336
    • one year ago
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    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.

  31. Photon336
    • one year ago
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    I guess if you think about it the opposite way you can't actually just kick off OH- unless you introduced some proton source.

  32. nincompoop
    • one year ago
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    and you are correct, the product is most likely more stable and our graph would look like this |dw:1436680983637:dw|

  33. Photon336
    • one year ago
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    YOU sir NAILED this one

  34. nincompoop
    • one year ago
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    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

  35. nincompoop
    • one year ago
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    my organic chemistry is pretty crappy so I am refreshing with some problems.

  36. Photon336
    • one year ago
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    crappy is a relative. that was well explained

  37. nincompoop
    • one year ago
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    the double dagger that you mention is the opposite it is the concerted \(OH^- \) forming a bond to C whilst Cl breaking

  38. Photon336
    • one year ago
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    mixed that up nucleophile was OH- so it's forming a bond haha

  39. nincompoop
    • one year ago
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    correct

  40. flexastexas
    • one year ago
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    Sorry Im late. It was really late when I posted. Question, how did you determine what kind of chemical this was from the picture?

  41. flexastexas
    • one year ago
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    And also, that last graph you posted would be what they asked for ?

  42. flexastexas
    • one year ago
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    And the first would be the balanced equation?

  43. flexastexas
    • one year ago
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    ?

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