Question

The half-life of a certain first order reaction is 15 minutes. What fraction of the original reactant concentration will remain after 2.0 hours??

Answer 1

I know the half life formula for a 1st order is t=ln(2)/k

Answer 2

The half-life is 15 minutes, so at time t = 15, you have half of the original amount left.
\[\frac{1}{2}=e^{15k}\]
(This comes from the half-life formula, which I'm assuming you know: \(y=Ce^{kt}\)).
Solving for k yields
\[k=\frac{\ln\left(\frac{1}{2}\right)}{15}\approx -0.046\]
So, now you have to find the amount left after time = 2.0 hours = 120 minutes:
\[y=Ce^{120k}\]
Of course, you don't know what y and C are, but that's fine, since you actually are looking for the fraction of the original amount that remains, which is given by
\[\frac{y}{C}=e^{120k}\]
Solve for the LHS.

Answer 3

what formula is the second one??

Answer 4

You could use \(1/2\) as the base instead of \(e\).

Answer 5

@Dama28 The second one isn't a formula, it's just solving for \(k\).

Answer 6

but where did that come from? Because I've only been taught the integrated rate laws, the half life formulas and that's it.

Answer 7

\[
\frac{1}{2}=e^{15k} \\
\ln\frac{1}{2}=\ln (e^{15k}) \\
\ln\frac{1}{2} = 15k
\]

Answer 8

I mean the y=Ce^120k

Answer 9

Basically, suppose the original concentration is \(C_0\)
We know that the concentration after one half-life is: \[
C_0(1/2)
\]After two half-lives it is \[
C_0(1/2)(1/2) = C_0(1/2)^2
\]After \(n\) half-lives \[
C_0(1/2)^n
\]Does this make sense so far?

Answer 10

yeah...

Answer 11

We also know that one half-life takes 15 minutes, so to find the number of half-lives in minutes. So that means \(n = t/15\).

Answer 12

So after \(t\) minutes, our concentration is: \[
C(t) = C_0(1/2)^{t/15}
\]

Answer 13

yes, I know how to get the 0.046, but I dont know what to do then

Answer 14

After \(t\) minutes, the fraction of the remaining concentration is: \[
f(t) = \frac{C(t)}{C_0} =(1/2)^{t/15}
\]

Answer 15

but what would be the initial concentration?? that's my problem

Answer 16

"What fraction of the original reactant concentration will remain after 2.0 hours??"

Well this is just saying: If \(2.0 \times 60 = t\), what is \(f(t)\)?

Answer 17

You don't need the initial concentration. They're asking for \(f(t)\) which fortunately doesn't contain the initial concentration.

Answer 18

They don't want the actual concentration \(C(t)\) they want the fraction of concentration \(f(t)\)\[

f(t) = \frac{C(t)}{C_0} =\frac{C_0(1/2)^{t/15}}{C_0} = (1/2)^{t/15}

\]

Answer 19

I really dont understand, and Im pretty sure this is very simple.

Answer 20

What don't you understand?

Answer 21

Would I just multiply the k value I got (0.046) times 120 minutes??

Answer 22

Are you expressing the equation as a base e or base 1/2?

Answer 23

This is the formula I know and what I did:
\[\ln \frac{ A _{t} }{ A _{0} }=-kt\]

Answer 24

I plugged in the values of k=0.046 and t=120 minutes and got 5.54

Answer 25

Okay...

Answer 26

now, should I find e^5.54??

Answer 27

Yes.

Answer 28

I understand the problem now... you didn't want to read anything that is longer than 2 sentences.

Answer 29

I get 255.9

Answer 30

http://www.wolframalpha.com/input/?i=e%5E%280.046*120%29

Answer 31

the answer is 1/256.

Answer 32

LOL

Answer 33

You would have gotten that if you had just used \[
(1/2)^{t/15}
\]Like I said....
http://www.wolframalpha.com/input/?i=%281%2F2%29%5E%28120%2F15%29

Answer 34

ok, thanks

Answer 35

The problem is that you forgot to put a negative sign on exponent.

Answer 36

are you kidding me? that's it?? wow!! I finally get it hahah

Answer 37

Did you understand the formula I got though...?

Answer 38

no, im sorry