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# Second order and pseudo-first order reaction Watch

1. 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?

Thanks
2. (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?

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)
3. I understand what you're saying but how do I calculate the second order rate constant by plotting a graph of the data provided?
4. Actually I've got it now, thanks for the help

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Updated: April 10, 2013
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