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Photon336

  • one year ago

Questions (rate(s) and rate laws) Enjoy

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  1. Photon336
    • one year ago
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    #1

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  2. Photon336
    • one year ago
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    #2

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  3. Photon336
    • one year ago
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    #3

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  4. Photon336
    • one year ago
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    @cuanchi @rushwr

  5. Rushwr
    • one year ago
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    #1 I haven't still done the environmental chemistry at school. But I'll go with D none of the above!

  6. Photon336
    • one year ago
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    #1 Graph I exothermic Graph II endothermic Agree/disagree?

  7. Rushwr
    • one year ago
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    oh sorry my answer goes to the second one !

  8. Rushwr
    • one year ago
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    that Ozone and NO reaction

  9. Rushwr
    • one year ago
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    for the exothermic endothermic question i'll go with C In an exothermic reaction the products are having less energy cuz in an exothermic reaction heat is lost to the surroundings !

  10. Rushwr
    • one year ago
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    the otherway round for the enothermic reaction.

  11. Photon336
    • one year ago
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    One thing you notice is that NO remains unchanged

  12. Photon336
    • one year ago
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    it's used in the first step of the reaction and then remains at the end of the reaction in the last step unchanged. so I would say that it's acting as a catalyst.

  13. Photon336
    • one year ago
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    #2 A is definitely wrong because it does change the concentration of O3

  14. Rushwr
    • one year ago
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    so its gonna e B for the #2 question ?

  15. Rushwr
    • one year ago
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    259. I'll go with enthalpy change of the reaction

  16. Photon336
    • one year ago
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    I think the issue is that NO is not present in the slow step so i thought it wouldn't cause any change. with that.

  17. Photon336
    • one year ago
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    I think i'll go with C. catalyst only lowers activation energy but doesn't do anything else in terms of overall energy. that stay the same.

  18. Photon336
    • one year ago
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    |dw:1438191437577:dw| the energy difference between products & reactants is the same

  19. Rushwr
    • one year ago
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    yeah I agree! that's what I was saying . and for the 260. is it the 3rd answer!

  20. Photon336
    • one year ago
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    For #1 we are correct.

  21. Photon336
    • one year ago
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    #2 i was between C and D and it was D. apparently NO does affect the overall energy change.

  22. Photon336
    • one year ago
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    @Ciarán95 thoughts on #2?

  23. Rushwr
    • one year ago
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    Then that was the 1st answer I gave is correct? Okai now this is how I argued to take that answer. Rate determining step is determined by the slowest reaction out of all right?

  24. Photon336
    • one year ago
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    right

  25. Rushwr
    • one year ago
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    There u can see that NO has no effect on it! Hence it says that even if we use NO as a catalyst its useless cuz it does not effect the rate determining step right?

  26. Rushwr
    • one year ago
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    Hence it won't lower the activation energy neither does it lower the overall energy !

  27. Photon336
    • one year ago
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    overall rate of reaction is dependent on the rate determining step..

  28. Rushwr
    • one year ago
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    yes ! So I think that is the way to approach it ?

  29. Photon336
    • one year ago
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    Thing is it does affect the overall energy choice C the one that says it doesn't is false.

  30. Rushwr
    • one year ago
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    The rate determining step is the slowest step . So if NO doesn't affect the rate of the slowest step that means it's useless as a catalyst cuz the rate of the reaction is not increased. if u want to increase the rate of the reaction u need to increase the rate of the slowest step. Since NO doesn't do that it doesn;t act as a catalyst.

  31. Photon336
    • one year ago
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    so.. it technically doesn't act as a catalyst at all.. so it does affect the overall energy change making C false..

  32. Photon336
    • one year ago
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    we still haven't done #3 (there's two questions for that)

  33. Rushwr
    • one year ago
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    for three 259. I have slight confusion with A and B But I would go for B

  34. Rushwr
    • one year ago
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    260. I would go with C

  35. Photon336
    • one year ago
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    #3 our rate constant is \[r = k[A]^{x}[B]^{y}\] \[Lnk =Ae ^{Ea/RT}\] if we look at the bottom equation if we change the temperature we also change the value of K , b/c RT is in the denominator. a catalyst lowers the activation energy and thus increases the rate of the reaction. so that would affect K as well. Ea would be smaller and RT would remain unchanged. so for this one I would say it's B. Rate constant is independent of the enthalpy change.

  36. Photon336
    • one year ago
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    any thoughts on #260 #3?

  37. Rushwr
    • one year ago
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    yey ! so it is the 2nd answer!

  38. Rushwr
    • one year ago
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    I though it would be that

  39. Photon336
    • one year ago
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    yeah

  40. Rushwr
    • one year ago
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    for carbonates to release CO2 at a higher rate the concentrations of the acids should be high so we can eliminate A and B I choose C over D cuz they say the concentration of H is is directly connected to the rate.

  41. Rushwr
    • one year ago
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    the no. of H in C is greater than in D so I choose C.

  42. Rushwr
    • one year ago
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    @Ciarán95 are we arguing right?

  43. Photon336
    • one year ago
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    \[H _{2}SO _{4}\] That's polyprotic? maybe more H+ So it could react faster.

  44. Photon336
    • one year ago
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    oh *** wait that's H\[H _{2}SO _{3}\] There are weak acids with the exception of HCL

  45. Photon336
    • one year ago
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    @Rushwr said the answer was HCl

  46. Rushwr
    • one year ago
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    NO I choose C

  47. Photon336
    • one year ago
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    Because it's a strong acid.

  48. Photon336
    • one year ago
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    I chose C too but i thought it was sulfuric acid

  49. Rushwr
    • one year ago
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    So they have said it as HCl?

  50. Photon336
    • one year ago
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    yeah because they said it' a strong acid so it will produce more H+

  51. Photon336
    • one year ago
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    \[Ka >>>> 1\]

  52. Rushwr
    • one year ago
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    okai then. But the concentration is small nah !

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