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Half-life decomposition, find final concentration

  • Thread starter dolpho
  • Start date
  • #1
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Homework Statement



In a first order decomposition in which the rate constant is 0.03 sec-1, how much of the compound (in mol/L) is left after 39 sec, if there was 2.00 mol/L at the start?

I'm using a few equations and trying to plug it in but I don't know whether they are appropriate or not.

The answer is .621 but I'm not sure how to get there

Homework Equations



Ln(A(initial) /A(final) = -kt

A(final) = -KT + A(initial)


The Attempt at a Solution



I've tried plugging into these equations but haven't gotten the right answer yet. Are these the correct equations to use?
 

Answers and Replies

  • #2
Borek
Mentor
28,397
2,800

Homework Equations



Ln(A(initial) /A(final) = -kt

A(final) = -KT + A(initial)
They can't be both right at the same time - one is for the zeroth order reactions.

Which one is for first order reactions?

What is k?

Show details of what you did.
 
  • #3
66
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They can't be both right at the same time - one is for the zeroth order reactions.

Which one is for first order reactions?

What is k?

Show details of what you did.
I used the formula ln[A] = -kt + ln[A]o

Ao = unknown
Ao = 2 M
K = .03 sec^-1
t=39 seconds

Ln A = (-.03)(39 seconds) + Ln2
LnA = -.4767 which I can't take the natural log of. I think I'm doing something wrong in the equation but I'm really not sure what :|
 
  • #4
Borek
Mentor
28,397
2,800
I used the formula ln[A] = -kt + ln[A]o
OK

Ao = unknown
Ao = 2 M
So it is an unknown with a known value? But let's assume it was just a typo.

LnA = -.4767 which I can't take the natural log of.
That's were your thinking got derailed. Why do you want to take a logarithm of logarithm and not an antilogarithm?
 
  • #5
66
0
OK



So it is an unknown with a known value? But let's assume it was just a typo.



That's were your thinking got derailed. Why do you want to take a logarithm of logarithm and not an antilogarithm?
Yea to be honest logarithms have always been confusing to me. I found out how to get the answer but I used e^(-.4767)

Why would I use the Ln on the 2mol/l and then switch to e, even though the formula has LnA? I don't really get the concept of logs yet :|
 
  • #6
Borek
Mentor
28,397
2,800
Your problem is with math, not with the chemistry.

From the logarithm definitions, if b=ln(A), A=eb. You have calculated ln(A), so the value you are looking for is eln(A).
 

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