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Find units of rate constants and write reactions

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the units of k1, k2, k3, k4, and k5. Write rate reactions for dA/dt

    a. A[itex]\leftrightarrow[/itex]B
    k1 (forward)
    k2 (backward)

    b. A + B [itex]\rightarrow[/itex] C
    k3 (forward)

    c. A + R [itex]\rightarrow[/itex] 2R
    k4 (forward)

    d. A [itex]\rightarrow[/itex] B
    k5 (forward)


    2. Relevant equations



    3. The attempt at a solution

    a. [itex]\frac{d[A]}{dt}[/itex] = k1[A] - k2
    units of k1 & k2 are in s[itex]^{-1}[/itex]
    b. [itex]\frac{d[A]}{dt}[/itex] = -k3[C]
    units of k3 are in L/mol*s
    c. [itex]\frac{d[A]}{dt}[/itex] = -k4[R]^2
    units of k4 are in L/mol*s
    d. [itex]\frac{d[A]}{dt}[/itex] = -k5
    units of k5 are in s[itex]^{-1}[/itex]

    Are these correct? I'm a little confused since the problem specified finding dA/dt rather than each of the products. Thanks!
     
  2. jcsd
  3. Mar 14, 2013 #2

    epenguin

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    Looks all OK to me.

    Except for first order I never remember the units of the top of my head. But you shouldn't be confused, it is quite easy:

    [A] means moles.litre-1.

    /dt is s-1

    on the RHS, [A] is moles.litre-1.
    [A]2 is moles2.litre-2
    [A]ab[C]c... is molesa+b+c+....litre-(a+b+c+...)

    So e.g. for a 2nd-order reaction like b

    moles.litre-1.s-1 = k x moles2.litre-2


    Isolate k by dividing both sides by moles2.litre-2 and after any cancellations you have it.

    Do they insist nowadays, mnyah mnyah, you write out the cumbrous moles.litre-1 etc?

    M used to be good enough. Or rather, better.
     
  4. Mar 14, 2013 #3

    mfb

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    I would expect different signs for all "k", but that is probably just a convention issue.

    At b, why does the change of A depend on C (and not on A and B), if C does not react at all? The same applies to c and d.
    At b, how can you get units different from your answers at (a) and (d)?
     
  5. Mar 14, 2013 #4

    epenguin

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    It doesn't depend on C.

    Sorry I read your answer for the units which were all correct, not your equation which is not right. The answer for the units is right for b, c and d somehow, but your equations for those are wrong.

    I happen to have written a post recently about how simple it is to write the rate equations for a given mechanism.
    https://www.physicsforums.com/showpost.php?p=4304238&postcount=16

    Rate constants are always positive.
     
    Last edited: Mar 14, 2013
  6. Mar 14, 2013 #5
    Thanks for all the help! So, my equations are incorrect for b, c, d? I'm not sure how to write the reaction in terms of dA/dt, then. I know how to write them for dC/dt or dProduct/dt, but not for a reactant.
     
  7. Mar 14, 2013 #6
    Wait a minute...
    Am I just being an idiot?

    Would
    (b) simply be d[A]/dt = k3[A]

    (c) d[A]/dt = k4[A][R]
    actually I'm the most unsure about this one because of the 2R in the product. Would the rate be halved?

    (d) d[A]/dt = k5[A]?
     
  8. Mar 14, 2013 #7

    mfb

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    As A decreases in the reaction and the k-factors are always positive, those equations should have a minus sign. Apart from that, they are correct.
    The reaction product (as long as it is not A) does not matter, even in case of (c). The amount of R will change in a more complicated way, of course.
     
  9. Mar 14, 2013 #8
    Awesome! Thank you so much
     
  10. Mar 14, 2013 #9
    Another quick question: Are my units correct as epenguin said?
     
  11. Mar 14, 2013 #10

    mfb

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    With the fixed equation in (b), they are all correct now.
     
  12. Mar 14, 2013 #11

    epenguin

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    Given conservation of mass, or better, given that in b one molecule of A generates one of C what is the relation of dC/dt to dA/dt? Etc.

    If you don't have a doh! moment read through the first page or so of whatever book chapter your course is using.

    I initially thought your problems were with the dimensions or units, it did not occur to me that anyone would have a difficulty with formulating the equations so I did not even look at yours. (However their derivation is explained in the link I gave in last post.) In the same way I missed that your a is also wrong! - make it -d[A]/dt on the left. A is disappearing by reacting, the more A there is the faster it disappears! After you have corrected that, you have a correct first stage. But a second stage, for a more useful equation, is to express all in terms of A, without B appearing in the equation. Using the idea of my 1st line above.
     
  13. Mar 14, 2013 #12

    epenguin

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    b, c and d not quite, see my last post.
     
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