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    OH + H2 >> H2O + H

    H2 is in excess and so the concentration doesnt change noticeably during he reaction so its possible to determine a pseudo first order rate constant k'

    I have a table of data with the concentration of H2 and k' values for those concentrations (k' being the effective pseudo first order rate constant which is equal to k[A] )

    rate=k[A][B] and the conc. of A is effectively constant so rate=k'[B]

    ------------------

    So I have been asked to plot a graph to determine the second order rate constant but I am unsure what to plot. Usually with these things I would plot ln(concentration) against time but seeing as I dont have time I have k' (which has the units of per second) would I calculate 1/k' (giving me units of seconds/time) and then plot ln(concentration) against time?

    Sorry about the wordy question

    Thanks
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    (Original post by DonnieBrasco)
    OH + H2 >> H2O + H

    H2 is in excess and so the concentration doesnt change noticeably during he reaction so its possible to determine a pseudo first order rate constant k'

    I have a table of data with the concentration of H2 and k' values for those concentrations (k' being the effective pseudo first order rate constant which is equal to k[A] )

    rate=k[A][B] and the conc. of A is effectively constant so rate=k'[B]

    ------------------

    So I have been asked to plot a graph to determine the second order rate constant but I am unsure what to plot. Usually with these things I would plot ln(concentration) against time but seeing as I dont have time I have k' (which has the units of per second) would I calculate 1/k' (giving me units of seconds/time) and then plot ln(concentration) against time?

    Sorry about the wordy question

    Thanks
    If rate=k[A]^a[B]^b then as [A] is effectively constant, then
    rate=k'[B]^b

    you still dont know the order of the reactions with respect to each reactants, right, assign them a and b. then taking into account that k' variation is negligible, then you simply treat this as trial and error.

    if b = 1, meaning rate = k' [B]
    then rate(1/time) against [B] will be linear

    if b=2 meaning rate = k' [B] ^2
    then rate (1/time) against [B] will be quadratic, which you can convert to natural logarithm to show the straight line relationship, i.e.

    ln (rate) = ln (k') + 2ln[B]

    etc. very unlikely b would be 3, but b could also be zero (i dont know)
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    I understand what you're saying but how do I calculate the second order rate constant by plotting a graph of the data provided?
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    Actually I've got it now, thanks for the help
 
 
 
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