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 first order reaction is \[t _{1/2} = \frac{ \ln(2) }{ }\]

Answer 2

divided by k

Answer 3

for a first order reaction, use this equation to find the decay constant:

\[t _{1/2}=\frac{ \ln2 }{ k }\]

then use it in the exponent growth/decay equation (which I'm sure you know)

Answer 4

haha ok good

Answer 5

the thing is we have not worked with the exponent growth/decay in this class.

Answer 6

we've only had the integrated rate laws, half like formulas and arrhenius equation

Answer 7

ohh okay, do you know what the equation is? do you know how to solve logarithms?

Answer 8

okay, so since you've used the arrheius equation I'm guessing you know how to solve logarithms

Answer 9

yes.

Answer 10

so the equation is this:

\[A=A _{0}e ^{-kt}\]

A is the amount after whatever time elapsed
Ao is the initial amount
k is the constant
t is the time elapsed

Answer 11

so it's basically a plug and chug after that, but you have to take into account the differences in the time units

Answer 12

But I dont have the initial concentration...

Answer 13

you can assume the initial to be 1 or 100

Answer 14

because it's asking you for the difference (fraction)

Answer 15

ok

Answer 16

let me do it and Ill post what I get

Answer 17

k

Answer 18

alright, Im lost. I thoufht I had it.
this is what I did
\[\ln \frac{ x }{ 100 }=-0.046(120)\]

Answer 19

the answer should be 1/256

because 15 min is your half-life, then in 2 hours it has decayed 8 times

(1/2)^8 = 0.00390625 = 1/256

Answer 20

if you want il work it out the formal way

Answer 21

yeah, the answer is 1/256 but i didn't know how to get there without the initial concentration

Answer 22

okay.. so

k=ln2/(1/4)=4ln2=2.7725

A=100e^(-2.7725(2))

A=100e^(-5.545177444)

A=100(3.90625x10^-3)= 0.390625

100/0.390625 = 256

1/256

Answer 23

i took 15 mins as 1/4 of an hour just to simplify things

Answer 24

ok thanks! cuz I was jsut about to ask you where you got the 1/4 from

Answer 25

no problem!