[sweetdaisy186]

What is the final concentration for the decomposition of 0.45 M X after 1.6 hours if the half-life is 35.0 minutes and the rate law is rate = k[X]²?


So, here is my theory. After 35 minutes, it would be .225MX
                                After 1.6 hours, it would be .196 MX

We know that 1.6 hours is equal to 96 minutes

And it we know that 105 minutes needed to reach half life.

Am I on the right track? Or am I making this way harder than it needs to be?

[madscientist]

How did you come to the conclusion that half-life is reached after 105 minutes??

[madscientist]

k= 0.693/ t_1/2
  = 0.693/35min
  =0.0198min-1

ln [x ]t / [x ]0 = -kt = ?

[x ]t / [x ]0=e^?= fraction remaining

(0.45mol)(fraction remaining)=?.??mol

I havnt included answers to the equation cause it will do you no good to just rote learn this problem,understand the equation and youl find it as easy as 1+1 :bigwink:
hope ive helped and not confused you with my rough explaination!

cheers,

madscientist :albert:

[xiankai]

in that case, where did 0.693 jump out from?  

[Donaldson Tan]

ln 2 = 0.693

[xiankai]

i get it... thanks.  

[sweetdaisy186]

Hmmm, I like that you didn't tell me word for word. I learn better that way. BTW, what does rote mean?

I think I might have messed up my question because I meant to say that the Rate law = k[ x ] ²

So, I would use the T_1/2 = 1/k[A]₀ right? Apparently, the X and the [ ] make a dot. So then I got that k=.06349. And then I thought that I could plug it into the 1/[A_t =kt + 1/[A₀ and when I did that I got .074 M. That seems sort of small if you ask me. My math was (0.06349*96) + (1/0.45)

[madscientist]

You have showed an honest attempt at the question so ive done the working with answers. If anyone thinks ive made a mistake with my working please point this person in the right direction. ( rote learning is just memorization without understanding the problem)

k= 0.693/ t1/2
  = 0.693/35min
  =0.0198min-1

ln [x ]t / [x ]0 = -kt = -(0.0198mins^-1)(96mins)= -1.9008

[x ]t / [x ]0=e^-1.9008= 0.15

(0.45mol)(0.15)= 0.0675mol

cheers,

madscientist :albert:

[sweetdaisy186]

Hmmm, I understand your work, but I don't understand why this is a first order reaction and not a second order. Thanks for showing me your work!!  

[madscientist]

I didnt realise that you had typed k=

• ²
  Sorry :blush: your right it is a second order reaction
  
  im workin on it and will post somthing in an hour or so.
  
  cheers,
  
  madscientist :albert:
  
  

[madscientist]

Ok youve got my brain workin overtome on this so here goes,

[x ]_t= [x ]_o / 1+[x ]_okt

your right to use: t½ = 1 / k [A ]_o, k=0.06349L/mol*min

therefore:

[x ]_t=0.45M / 1+0.45*0.06349*96min

       =0.45/3.742768

       =0.12M (after 96 mins)

Ive also just read that the length of half-life increases as concentration decreases in a second order reaction,so 70 and 105 mins (respectivly) arn't  2nd and 3rd half-lives.

I really hope this is right as my brain hurts from trying to figure it out!
thanks for the challenge,

madscientist :albert:

[sweetdaisy186]

Okay, that's what I got, I see what I did wrong now. Thanks sooo much! Sry, I didn't mean to make you work overtime.

THANKS AGAIN!