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Topic: Intergrated rate law  (Read 5735 times)

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theunraveler

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Intergrated rate law
« on: June 04, 2006, 09:12:04 AM »
1st) this seems a fairly straight forward qn but the ans at the back of the book does not agree with my ans :(

we have a rate constant 5e-2 mol/L

inital concentration of 1e-3 M

calculate concentration after the time elapsed is 5e-3 s

my ans is 7.5e-4 but the ans at the back claims it is 2.5e-4

formula i using is the integrated rate law [A] = -kt + [A initial]

did i miss something?

2) for this reaction A ---> products, succesive half life is observed to be 10, 20 and 40 min for an experiment which A (initial) is 0.1, calculate the concentration of A at the following times

a) 80 min
b) 30 min

not sure where to start this qn, how to work out which order is it?

Offline Albert

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Re: Intergrated rate law
« Reply #1 on: June 04, 2006, 12:02:18 PM »
1st) this seems a fairly straight forward qn but the ans at the back of the book does not agree with my ans :(

we have a rate constant 5e-2 mol/L

inital concentration of 1e-3 M

calculate concentration after the time elapsed is 5e-3 s

my ans is 7.5e-4 but the ans at the back claims it is 2.5e-4

formula i using is the integrated rate law [A] = -kt + [A initial]

did i miss something?


I've never seen a rate constant expressed in mol/L and your integrated rate law seems incorrect too.

Can you post the whole text of the problem?
« Last Edit: June 04, 2006, 12:17:43 PM by Albert »

theunraveler

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Re: Intergrated rate law
« Reply #2 on: June 05, 2006, 11:13:16 AM »
the reaction A ---> B + C is known to be zero order in A and to have a rate constant of 5e-2 mol/L at 25 degrees celsius. an experiment was run at 25 degrees celsius at 25 degrees celsius where A (initial) is 1e-3 M

calculate the conc of B after 5e-3 has elapsed

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Re: Intergrated rate law
« Reply #3 on: June 05, 2006, 12:50:13 PM »
Now I've got it!  :)

Ok: you were right when you said the integrated rate law is [A] =[A]0 - kt
I apologize. ;)

Using this equation, however, you get the 'final' concentration of A, but you have to calculate the concentration of B.

I firmly believe there's a part of the text of the problem where it says that at the beginning (t=0), the concetration of B is 0=0.

Hence, given that [A] is 7.5*10-4 after 5*10-3 seconds, the concentration of B is:

= [A]0-[A] = 10-3 - 7.5*10-4 = 2.5*10-4
« Last Edit: June 05, 2006, 01:09:40 PM by Albert »

Offline lemonoman

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Re: Intergrated rate law
« Reply #4 on: June 05, 2006, 04:47:37 PM »
And as for your second problem, there's a couple ways to go about it.

1) Use the successive half-life data.

You know that half of whatever you started with will remain after the first half life (10 mins)
Half of THAT (or 1/4 the starting amount) will be present after the second half life (10 + 20 = 30 mins elapsed)
and Half of THAT (1/8 the starting amount) will be present after the third halflife (30 + 40 = 70 mins).

This gives you a good answer for 30 minutes...but other than that, we're stuck.

2) Figuring out the order of reaction (as you suggested).

Here, it may be easiest to make some data out of what you're given.

Let's say you start with a concentration of [c].

At t=0, [c] = [c]0
At t=10, [c] = (1/2)[c]0
At t=30, [c] = (1/4)[c]0
At t=70, [c] = (1/8)[c]0

From this, you should be able to get an expression for [c] based on [c]0 and t....then you can plug 80 in for t, and see what you get :D

Hope that helps.

P.S. There are other ways to do the question too (so many choices!)...if someone else wants to share, feel free..maybe your method is even cleaner (The second one is my preferred one)

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