Question

Gen Chem weird question.

Answer 1

This is not an ochem question but rather a Gen Chem.
Here's the question.

CH3N2CH3(g)---> CH3CH3(g)+N2(g), the first order half is 1730 s
Calculate the time it takes for the concentration to go to 12.5% of its original concentration.

I forgot how to due these type of problems xD

Answer 2

did they give any initial concentrations?

Answer 3

Nope, they did not....

Answer 4

My sister is taking gen chem and she has this question in her homework.

Answer 5

Well i think first order half means half life. so I think we can figure this out.

well the rate of disappearance of a is equal to the rate constant times the concentration of A.

Translation

We start off with some amount of [A] and then as the reaction happens, we start to lose [A] some A. so it's represented like this.

\[-\frac{ d[A] }{ dT } = k[A] \]

Answer 6

We move our variables to solve this. we have to integrate this, to get our expression.

\[\frac{ d[A] }{ [A] } =- kdT\]

Rate of formation of product equals the the loss of k *t

\[\int\limits_{[A]_{0}}^{[A]} \frac{ d[A] }{ [A] } = -\int\limits_{T_{0}}^{T} kt\]

\[\ln([A])-Ln(A_{0}) = - k(t_{2}(\]

\[Ln\frac{ [A]_{t} }{ [A]_{0} } = -kt \]

but this if if you're given the rate constant which you're not

Answer 7

I think there's an easier way to do this it may just be the half life equation

Answer 8

because you have the half life already

Answer 9

Thank you!
I guess this question is weird. Yeah, it only gives us the half life which means there's a hint in to that.

Answer 10

The final concentration is 12.5% of its original concentration

\[(0.5)^{3}, or~3~half~lives \]

1 half life~0.5
2 half lives~0.25
3 half lives~0.125

we know that the half life is 1730s so i think we would multiply that by 3

Answer 11

to do the whole math intensive thing I did up there you would need to know what k is and we're not given that at all

Answer 12

We can still solve for k
\(\Large t_{1/2}=\frac{ln(2)}{k}\)
they already gave us the half life. So...
\(\Large 1730s=\frac{ln(2)}{k}\)

Answer 13

Lol would that work tho?

Answer 14

\[N_{0}*(\frac{1 }{ 2})^\frac{ 5190 }{ 1730 } = 0.125N\]

Answer 15

Like I thought this problem was really complicated, that's why i did all that math up there.

Well the reason why I don't think we need to find K is because they are asking for how long it would take to get to 12.5% of the original concentration, in english how long would it take for us to only have 12.5% left.

I think the simplest way to do this is probably to write out how many half lives it takes to get to 0.125.

that's how I got to 3

then multiply that by the half life to get the total time.

Answer 16

\[N_{0}(1/2)^\frac{ t }{ t_{1/2} } = 0.125N\]

\[N_{0}ln((1/2))^\frac{ t }{ t_{1/2} } = ln0.125N\]

\[\ln(0.5)*(\frac{ t }{ t_{1/2} }) = \ln(0.125)\]

\[ t = \frac{ \ln(0.125)*1730}{ \ln(0.5) } = 5190s\]

Answer 17

I also checked online and you can use this and the long math part above but, if you can use this method its probably better.

Answer 18

What formula did you use? Also, may i please get the links from the sites you used. I really want to dig this deeper. I was thinking of doing integrated rate law on this one, and use the first order based on the coefficient of the reactant.

Answer 19

well for the first part the most basic fact is that

you start off with a reactant/s right?

Answer 20

then they disappear at some rate k

Answer 21

well

Chemistry davidson edu
http://www.chm.davidson.edu/vce/kinetics/integratedratelaws.html

Radioactive Half life
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli.html

Answer 22

these are the sites I used.

Answer 23

Thank you!

Answer 24

anytime