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# ln[A] vs Time graph - how do I work out the rate constant k?

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1. How would you work out the rate constant for this reaction?
I was given a table showing Time (min) and [A] mM.
The answer sheet says that [A] vs T should be used to determine the order through half lives and that Ln[A] vs T should be used to work out k, but there was no explanation of how. I know it's a 1st order reaction from the answer sheet but not sure how this is known.

(I've only plotted the first 5 points from the data)

2. Idk but if memory serves me right, I'd assume you deduce the order of the reaction with respect to [A]

Which i presume is 1st order? ( seen as the half lives are constant for A)

rate=k[A]

k= rate/[A]

[A] is the concentration of A at any given point

EDIT:
, assuming you know the rate, use the algorithm I gave above, or you can work out the initial rate at t=0 and input it onto your rate equation OR you can:

k=ln2/1/2t
where 1/2t= half life of A (which i think is 15 seconds looking at your graph)
3. (Original post by Petulia)
xyz
pree
4. (Original post by Petulia)
refresh
reload my comment, I updated it
5. (Original post by mercuryman)
Idk but if memory serves me right, I'd assume you deduce the order of the reaction with respect to [A]

Which i presume is 1st order? ( seen as the half lives are constant for A)

rate=k[A]

k= rate/[A]

[A] is the concentration of A at any given point

EDIT:
, assuming you know the rate, use the algorithm I gave above, or you can work out the initial rate at t=0 and input it onto your rate equation OR you can:

k=ln2/1/2t
where 1/2t= half life of A (which i think is 15 seconds looking at your graph)
Do you have a textbook or website where this is explained in more detail?

And is that the same method as this:
B The rate law for the reaction is thereforerate = k[N2O5]Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,
slope=ln[N2O5]3000−ln[N2O5]03000 s−0 s=(−4.756)−(−3.310)3000 s=−4.820×10−4 s−1

Thus k = 4.820 × 10−4 s−1.

That's what I found on this website: http://2012books.lardbucket.org/book...mine-rate.html

And what does it mean by the slope is -k for first orders?
6. (Original post by Petulia)
Do you have a textbook or website where this is explained in more detail?

And is that the same method as this:
B The rate law for the reaction is thereforerate = k[N2O5]Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,
slope=ln[N2O5]3000−ln[N2O5]03000 s−0 s=(−4.756)−(−3.310)3000 s=−4.820×10−4 s−1

Thus k = 4.820 × 10−4 s−1.

That's what I found on this website: http://2012books.lardbucket.org/book...mine-rate.html

And what does it mean by the slope is -k for first orders?

What i meant was that there is a mathematical function that can be used to derive k.

k= t1/2 ÷ ln(2)

as long as you know the half life of a reactant and its order, working out k can take like 2 seconds.

In your link it tells you how to work out k from 0 and 2nd order reactants mathematically too, in case you didn't know.

And Im not sure on ln[X] graphs and -k values. I'll have a look at it though.

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