# Half-life decomposition, find final concentration

## 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?

Borek
Mentor

## 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.

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 :|

Borek
Mentor
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?

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 :|

Borek
Mentor
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).