Half-life decomposition, find final concentration

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Discussion Overview

The discussion revolves around a homework problem concerning first-order decomposition kinetics. Participants are trying to determine the final concentration of a compound after a specified time, given an initial concentration and a rate constant. The focus includes the appropriate equations to use for first-order reactions and the application of logarithmic functions in calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a first-order decomposition problem and expresses uncertainty about the equations being used, specifically questioning the appropriateness of two different equations.
  • Another participant points out that one of the equations is for zeroth-order reactions and asks which equation applies to first-order reactions.
  • A participant attempts to apply the formula ln[A] = -kt + ln[A]o but struggles with the calculation, leading to confusion about taking logarithms.
  • There is a discussion about the correct interpretation of logarithmic functions, with one participant suggesting that the issue lies in misunderstanding logarithms rather than the chemistry itself.
  • Another participant clarifies the relationship between natural logarithms and exponentials, indicating that the value sought is the antilogarithm of the calculated logarithm.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of the equations or the handling of logarithmic functions. There is ongoing confusion and debate about the appropriate mathematical approach to solve the problem.

Contextual Notes

Participants express uncertainty about the definitions and applications of logarithmic functions in the context of first-order kinetics, indicating potential gaps in understanding that affect their calculations.

dolpho
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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?
 
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dolpho said:

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.
 
Borek said:
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 :|
 
dolpho said:
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?
 
Borek said:
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 :|
 
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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