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Chemical Kinetics - Concentration/time

  1. Oct 8, 2012 #1
    My textbook (Burdge, Chemistry, 2nd edition) is pretty vague mathematically when it comes to calculating the examples, so I'm a bit confused about solving this example problem.

    The equation is fairly straightforward:
    [itex]ln{[A]_{t}/[A]_{0}}=-kt[/itex]

    The 14.4 problem statement is (paraphrased):

    2a -> b
    [itex]k = 7.5 x 10^{-3}s^{-1}[/itex] (@ 110°c)
    [A] = 2.25M,
    What is [A] after 2.0 min?

    So, 2.0 min = 120 sec

    [itex]ln{[A]_{t}/2.25M}=-(7.5 x 10^{-3}s^{-1})(120s)[/itex]
    The answer given is 0.91M. I could really use the proper steps to get there.

    The thing that's really not clear is whether you take the natural log of each quotient value separately, or whether you divide the quotient and then take the natural log of that... and, then I'm getting my Properties of Natural Logs all confused while working the problem, so... I'm confused. Help, please?!
     
  2. jcsd
  3. Oct 9, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    To make it as unambiguous as possible:

    [tex]ln\Big(\frac{[A]_t}{[A]_0}\Big) = -kt[/tex]

    which is equivalent to

    [tex]\frac{[A]_t}{[A]_0} = e^{-kt}[/tex]

    You better work on your math, I guess the book assumes some proficiency in dealing with simple calculations. After all it is a chemistry textbook, not a math textbook.
     
  4. Oct 9, 2012 #3
    You have the same communication problem as the book does, apparently.
    I'm in Calculus 2, and I've gotten straight A's in all of my math classes, so please try not to be a condescending ***.

    [tex]ln\Big(\frac{X_1}{Y_1}\Big) \neq \Big(\frac{lnX_2}{lnY_2}\Big)[/tex]

    So, you're not really being clear at all.
     
  5. Oct 9, 2012 #4
    It appears to be a first-order reaction and therefore your equation is wrong as written. Look at this reference:

    http://en.wikipedia.org/wiki/Rate_equation

    for first-order reaction.

    Also, if you're a chem major, make sure and take DEs before p-Chem.
     
    Last edited: Oct 9, 2012
  6. Oct 9, 2012 #5

    Borek

    User Avatar

    Staff: Mentor

    No idea where you got it from, it is not something I have stated and I don't see how it is related to the problem - but then apparently I don't know what your problem is. From your description I could only guess you were confused by the fact your equations are written without parentheses so it is not clear whether they mean ln(At)/A0 or ln(At/A0). What I wrote is a correct formula for the first order reaction, if it doesn't help - please elaborate where the problem is.
     
  7. Oct 19, 2012 #6
    You gave:
    [tex]ln\Big(\frac{[A]_t}{[A]_0}\Big) = -kt[/tex]

    Which, if you look at my first post above, I had already provided.

    So, my question was, as I stated (if you had bothered to read instead of simply throwing insults about your assumption of my math skills), does that mean the natural log of the ratio, or the natural log of each rate and then dividing them. It turns out to be the latter, which you hinted at with your "which is equivalent to" response, but the insults and general obfuscation (nevermind the sloppy math notation) hid that from me.

    I ended up getting help from my professor and a fellow student on this, so... no big deal, I guess. You guys really disappointed me in this thread, though.
     
  8. Oct 20, 2012 #7

    AGNuke

    User Avatar
    Gold Member

    This forum is meant to ask the help to when you tried even last ditch effort in your power (asking teachers, fellow mates, brainstorming, google, wikipedia, etc.). I might say I am disappointed that you asked your doubt without investigating it yourself.

    Most of these people, excluding me, already had their share of problems, so they don't need yours to solve. But still they do it. So please at least don't disrespect them by saying "You guys disappointed me".
     
  9. Oct 20, 2012 #8
    you're confused about the properties of logarithms and you're disappointed in US?
     
  10. Oct 20, 2012 #9
    Nice attitude there.
     
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