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Theloshua

  • one year ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. A + B yields products Trial [A] [B] Rate 1 0.10 M 0.20 M 1.2 × 10-2 M/min 2 0.10 M 0.40 M 4.8 × 10-2 M/min 3 0.20 M 0.40 M 9.6 × 10-2 M/min

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  1. aaronq
    • one year ago
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    For each of the reactants, isolate two trials where one of the concentrations is kept constant. You can now ignore the contribution of that reactant to the rate - that is ignore it from the rate law. Now examine how the rate changes and figure out the order for the reactant that IS changing in concentration (between the two trials). You can assign it's order (exponent) by using some simple math.

  2. Theloshua
    • one year ago
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    oh hey best friend!! :D

  3. Theloshua
    • one year ago
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    listen, idk if youll do this but im in a hurry, i kinda need this one done for me....

  4. Theloshua
    • one year ago
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    i needa turn it in.... can you maybe just do it for me?

  5. aaronq
    • one year ago
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    Can't man, sorry :S it's against the "rules", but more importantly, it's against my belief of doing your own work lol

  6. Theloshua
    • one year ago
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    okay *sigh* then walk me thru it please? i really really needa get this turned in...

  7. aaronq
    • one year ago
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    yeah, sure. did you understand what I said in the beginning?

  8. Theloshua
    • one year ago
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    nope

  9. aaronq
    • one year ago
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    Okay i'll do one, you do the other. Trial [A] [B] Rate 1 0.10 M 0.20 M 1.2 × 10-2 M/min 2 0.10 M 0.40 M 4.8 × 10-2 M/min 3 0.20 M 0.40 M 9.6 × 10-2 M/min The rate law is: \(\sf \large rate=k[A]^m[B]^n\) In trial 1 and 2, [A] is constant but [B] changes. It fact doubles 0.2 M to 0.4 M The rate changes in 4-fold, \(\sf \dfrac{4.8*10^{-2}}{1.2*10^{-2}}=4\) rate law: \(\sf rate=k[B]^n\) I'll use simple numbers since we know that the concentration doubles, and the rate quadruples: \([B]_1=1 ~and~ [B]_1=2\), \(rate_1=1 ~and ~rate_2=4\) \(\sf rate_1=k[B]^n=1=k[1]^n\) \(\sf rate_2=1=k[B]^n=4=k[2]^n\) Ignore the constant, k. We have \(\sf 1=[1]^n\) and \(\sf 4=[2]^n\) The only way for the above to be true is for n=2 so far, we have the rate law: \(\sf rate=k[A]^m[B]^2\) now figure out the exponent "m" in the same fashion and FINALLY, plug in any one set of data to figure out \(k\).

  10. aaronq
    • one year ago
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    Theres an error in "\(\sf rate_2=1=k[B]^n=4=k[2]^n\)" Ignore the 1, i mustve hit the key accidentally

  11. Theloshua
    • one year ago
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    confuzzled as heck

  12. aaronq
    • one year ago
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    Yeah, it's not a topic you can pick up in a few mins..

  13. aaronq
    • one year ago
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    I'll give you a hand. Trial [A] [B] Rate 1 0.10 M 0.20 M 1.2 × 10-2 M/min 3 0.20 M 0.40 M 9.6 × 10-2 M/min In trial 1 and 3, [B] is constant but [A] changes. How does the concentration of A change, and how does that affect the reaction (by what factor does the rate change)?

  14. Theloshua
    • one year ago
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    rate = k[h2]x[I2]y 1.2 × 10-2 M/min/4.8 × 10-2 M/min = [0.10]x[0.20]y/[0.10]x[0.40]y

  15. Theloshua
    • one year ago
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    this is hpw ive started out

  16. Theloshua
    • one year ago
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    the 0.10 can get crossed out right?

  17. aaronq
    • one year ago
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    Yeah, but I what are you achieving with this?

  18. Theloshua
    • one year ago
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    so its 0.20/0.40

  19. Theloshua
    • one year ago
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    apparently this is how to solve the problem according to my printed out notes *shrug*

  20. aaronq
    • one year ago
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    Hm, there are other ways to solve these types of problems, but the way i'm showing you simplifies everything a lot. Read what I wrote last about [B] being constant in trial 1 and 3

  21. Theloshua
    • one year ago
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    ok

  22. aaronq
    • one year ago
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    Crap it was trial 2 and 3, not 1 and 3.

  23. aaronq
    • one year ago
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    Trial [A] [B] Rate 2 0.10 M 0.40 M 4.8 × 10-2 M/min 3 0.20 M 0.40 M 9.6 × 10-2 M/min Do it with this

  24. aaronq
    • one year ago
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    In trial 2 and 3, [B] is constant but [A] changes. How does the concentration of A change, and how does that affect the reaction (by what factor does the rate change)?

  25. Theloshua
    • one year ago
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    2?

  26. aaronq
    • one year ago
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    2 what? lol

  27. Theloshua
    • one year ago
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    i dont knoooowwwwwww :/

  28. aaronq
    • one year ago
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    you're on the right track, the concentration doubles

  29. Theloshua
    • one year ago
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    ohhhh okay pfew, dont do that, i tought i was wrong :p

  30. aaronq
    • one year ago
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    haha you should elaborate on your answers, i know what you mean but if I didn't know the answer already I wouldnt know what youre talking about

  31. aaronq
    • one year ago
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    How does the rate change?

  32. Theloshua
    • one year ago
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    it doubles.... -_-

  33. aaronq
    • one year ago
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    Okay, so we can say \([A]_2=1\), \([A]_3=2\) and then \(rate_2=1\) \(rate_3=2\) Now ignoring k and ignoring [B] (because the concentration is constant) \(\sf rate_2=[A]^m\rightarrow 1=[1]^m \) \(\sf rate_2=[A]^m \rightarrow 2=[2]^m \) what is \(m\)?

  34. aaronq
    • one year ago
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    the second line is supposed to say \(\sf rate_3=[A]_3^m\) sorry, i'm just trying to answer fast

  35. Theloshua
    • one year ago
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    what the freak?? O.o idk...

  36. aaronq
    • one year ago
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    you could use logarithms, but using common sense, what number multiplied by 2 gives you 2?

  37. Theloshua
    • one year ago
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    1

  38. aaronq
    • one year ago
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    exponents work like this, btw \(3^1=3\) \(3^2=3*3=9\) \(3^3=3*3*3=27\)

  39. aaronq
    • one year ago
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    yes! so the rate law so far is: \(\sf rate=k[A]^1[B]^2\) now just plug in the rate and the concentrations from one of the trials to get \(k\)

  40. aaronq
    • one year ago
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    You need to do some algebra, it'll look like this: \(\sf k=\dfrac{rate}{[A]^1[B]^2}\)

  41. aaronq
    • one year ago
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    I have to go, good luck!

  42. Theloshua
    • one year ago
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    wait

  43. Theloshua
    • one year ago
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    is the answer .26833?

  44. aaronq
    • one year ago
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    the answer was supposed to be the rate law, not just a number. The last thing (solving for k) was to complete it. It should've looked like: \(\sf rate=k[A][B]^2\) with k as whatever the value was

  45. Theloshua
    • one year ago
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    oh pellet... guess I got the question wrong.... XD I was too tired. tanks anyway tho

  46. aaronq
    • one year ago
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    lol damn. no problem still

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