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

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)

Thanks in advance

Reply 1

mercuryman

20

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:

input this into your calc

k=ln2/1/2t
where 1/2t= half life of A (which i think is 15 seconds looking at your graph)

(edited 7 years ago)

Reply 2

mercuryman

20

Original post by Petulia

xyz

pree

Reply 3

mercuryman

20

Original post by Petulia

refresh

reload my comment, I updated it

Reply 4

Petulia

OP

19

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:

input this into your calc

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/books/principles-of-general-chemistry-v1.0/s18-04-using-graphs-to-determine-rate.html

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

Reply 5

mercuryman

20

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/books/principles-of-general-chemistry-v1.0/s18-04-using-graphs-to-determine-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.

Hope I answered your query on k though.

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

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