Half-life decomposition, find final concentration

[dolpho]

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]

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.

 

[dolpho]

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]

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?

 

[dolpho]

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]

Your problem is with math, not with the chemistry.

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