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Second order and pseudo-first order reaction

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

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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
Reply 1
Avatar for shengoc
shengoc
12
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] and the conc. of A is effectively constant so rate=k'

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

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 then as [A] is effectively constant, then
rate=k'^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'
then rate(1/time) against will be linear

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

ln (rate) = ln (k') + 2ln

etc. very unlikely b would be 3, but b could also be zero (i dont know)
Reply 2
Avatar for DonnieBrasco
DonnieBrasco
OP
7
I understand what you're saying but how do I calculate the second order rate constant by plotting a graph of the data provided?
Reply 3
Avatar for DonnieBrasco
DonnieBrasco
OP
7
Actually I've got it now, thanks for the help :smile:

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