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Topic: 1/2 order  (Read 16565 times)

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Offline madscientist

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1/2 order
« on: February 27, 2006, 10:08:50 AM »
Hi everyone,

Im asked to derive an expression for half life of a reaction whose order is 1/2,

Ive integrated and rearranged blah blah.... and have come to an answer that i cant seem to simplify any further, could you check if im right or wrong please?


rate = k.[A]1/2

t1/2 = { (-2)(sqrt([A]o/2) } - { (-2)(sqrt([A]o) } (k)

the only rationalisation that ive been able to make is that the units for k in a 1/2 order reaction must be M-1/2.s, or, 1/M 1/2.s

for eg,

rate = k.[A]1/2

M/s = 1/M-1/2.s * M1/2

and looking at my expression for half life, these units seem to fit, i think???

realy not sure on this one, please help

cheers,

madscientist
« Last Edit: February 27, 2006, 10:09:20 AM by madscientist »
The only stupid question is a question not asked.

Offline Winga

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Re:1/2 order
« Reply #1 on: February 27, 2006, 11:04:31 AM »
Quote
t1/2 = { (-2)(sqrt([A]o/2) } - { (-2)(sqrt([A]o) } (k)
But I got,

t1/2 = { (-1/2)(sqrt([A]o/2) } - { (-1/2)(sqrt([A]o) } / (k)

Offline madscientist

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Re:1/2 order
« Reply #2 on: February 27, 2006, 11:53:15 AM »
How sure are you of your answer cause i cant see how you arrived at

t1/2 = { (-1/2)(sqrt([A]o/2) } - { (-1/2)(sqrt([A]o) } / (k)

everything but the -1/2 is the same as my answer, but i cant see how you ended up with -1/2 ???  

you start with -[A]-1/2 and should end up with -[A](-1/2)+1 / 0.5 which is equal to -2 [A]1/2 which is then equal to -2 sqrt [A] .

anyone got any suggestions?

cheers,

madscientist
The only stupid question is a question not asked.

Offline pantone159

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Re:1/2 order
« Reply #3 on: February 27, 2006, 01:22:37 PM »
I got the same answer as Winga, but I wrote it as:

tH = sqrt(A0) * (1/k) * (2 - sqrt(2))

When dealing with factors of sqrt(2), they can usually be written in all kinds of different looking ways that aren't obviously the same until you do some calculations to check.

One hint, madscientist, that your first answer wasn't right, is that your half-life was proportional to k, the rate constant.  That meant that, the higher the rate, the longer the half-life.  It should be the other way around, so proportional to (1/k) makes more sense.

BTW - Since you seem to be doing this for multiple reaction orders, it isn't that much harder to calculate the result for an arbitrary reaction order n, and then keep re-using this result.  There is one 'special case', it is the usual one of n=1.

Offline Winga

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Re:1/2 order
« Reply #4 on: February 27, 2006, 01:38:26 PM »
d[A]/dt = -k[A]1/2

Int.(from [A] to [A]o) [A]-1/2 d[A] = -k Int.(from t1/2 to 0) dt

By substitution method,

Let u = [A]1/2
du/d[A] = ([A]-1/2)/2
du = ([A]-1/2)/2 d[A]

Int. 2 du = -k Int. dt (Oh, I made a mistake in this step, 2 du --> 1/2 du) ;D

2[A]1/2 - 2[A]o1/2 = -kt1/2
t1/2 = -2([A]o1/2/2 - [A]o1/2)/k

or

t1/2 = 2([A]o1/2 - [A]o1/2/2)/k

« Last Edit: February 27, 2006, 01:38:57 PM by Winga »

Offline madscientist

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Re:1/2 order
« Reply #5 on: February 28, 2006, 05:35:17 AM »
thanks winga and Mark, that makes sense.

 scooby snacks all round! ;D

cheers

madscientist :albert:
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Offline Winga

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Re:1/2 order
« Reply #6 on: February 28, 2006, 08:53:14 AM »
I think I should write it like that.

t1/2 = -2(([A]o/2)1/2 - [A]o1/2)/k

or

t1/2 = 2([A]o1/2 - ([A]o/2)1/2)/k
« Last Edit: February 28, 2006, 08:56:10 AM by Winga »

Offline madscientist

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Re:1/2 order
« Reply #7 on: February 28, 2006, 11:27:27 AM »
Yep i got caught on that too, having the 1/2 power inside the brackets doesnt work, took me 2 hours to figure that out lol

cheers

madscientist :albert:
The only stupid question is a question not asked.

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