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half life
« on: February 23, 2006, 10:11:57 AM »
Hi people,

I need to figure out how to derive an expression for the half life of a third order reaction and am a lttle unsure of my work so far, which is:

rate = k . [A]3

-(d[A] / dt) = k . [A]3

separating the variables "A" and "t" :

-(d[A] / [A]3) = k .dt

integrating both sides of the equation between the final and initial limits gives:

(1 / [A]t2 ) - (1 / [A]o2 ) = k . t

defining the half life:

t1/2 = 1 / k . [A]o2

as its probably clear to see ive guessed my way through a question that i have no idea how to answer and am probably wrong, if anyone could help out on this it would help immensly.

cheers,

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Re:half life
« Reply #1 on: February 23, 2006, 09:09:55 PM »
Ive attached a pdf. file to show my working a bit clearer,

cheers,

« Last Edit: February 23, 2006, 09:25:43 PM by madscientist »
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Donaldson Tan

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Re:half life
« Reply #2 on: February 23, 2006, 10:31:53 PM »
My answer is t1/2 = 3/(2.k.[[A]]2)

The equation for the concentration profile of reactant A is

where Ao is the initial concentration.
« Last Edit: February 23, 2006, 10:44:14 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:half life
« Reply #3 on: February 23, 2006, 10:52:11 PM »
ok, food for thought, thanks

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Re:half life
« Reply #4 on: February 23, 2006, 11:25:18 PM »
My answer is t1/2 = 3/(2.k.[[A]]2)

The equation for the concentration profile of reactant A is

where Ao is the initial concentration.

How do you get from:

rate=k.[A]^3

To the expression you have posted above?

sorry to bugg you on this but im absolutely stumped by it

cheers,

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Re:half life
« Reply #5 on: February 24, 2006, 12:13:32 AM »
I actually use a computer algebra system (Maple v10) to solve it analytically.

to find concentration profile
-dA/dt = k.A3 => dA/A3  = -k.dt
int A-3dA from Ao to A = -k.dt from 0 to t
1/2Ao2 - 1/2A2 = -k.t
=> 1/2A2 = 1/2Ao2 + k.t
=> 2A2 = (1/2Ao2 + k.t)-1 =2.Ao2/(2.K.Ao2.t+1)
A2 = Ao2/(2.K.Ao2.t + 1)

to find half-life
-dA/dt = k.A3 => dA/A3  = -k.dt
int A-3dA from A to 0.5A = -k.dt from t to (t + t1/2)
1/2A2 - 1/2(0.5A)2 = -k.t1/2
1/2A2 - 1/0.5A2 = -k.t1/2
1/2A2 - 2/A2 = -k.t1/2
k.t1/2 = 3/2A2
t1/2 = 3/(2.k.A2)
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:half life
« Reply #6 on: February 24, 2006, 04:51:06 PM »
so I was alsmost their with my working, thanks geodome, no wonder my head hurts im trying to calculate this with pen and paper!

P.S I absolutely love your picture, crack up!
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Re:half life
« Reply #7 on: February 24, 2006, 06:18:25 PM »
integrating both sides of the equation between the final and initial limits gives:

(1 / [A]t2 ) - (1 / [A]o2 ) = k . t

There is a 1/2 missing in there.  The integral of A-3 is *not* A-2.  Almost, but not quite.

I worked this out by hand and got the same answer geodome did.

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Re:half life
« Reply #8 on: February 24, 2006, 06:20:56 PM »
P.S I absolutely love your picture, crack up!

Super Mario gotta be the best console game in the world.

http://www.chemicalforums.com/users/geodome/mario.gif
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:half life
« Reply #9 on: February 24, 2006, 11:45:56 PM »
Thankyou so much for your help again geodome, and Mark.

I had a look at your answer and worked on it myself until i got the same answer, my working is very similar too.

rate=k.[A]3

-(d[A] / dt) = k.[A]3

separating variables:

-(d[A] / [A]3) = k.dt ----> -[A]-3.d[A] = k.dt

First working on the left hand side:

integrate -[A]-3d[A] from [A]o to [A]t = [-[A]-3+1 / -3+1][A]t[A]o

= [-[A]-2 / -2][A]t[A]o

= [1/2[A]-2 ][A]t[A]o

= [1/2[A]-2 t ] - [1/2[A]-2 o ]

= 1 / {(2). [A]2 t} - 1 / {(2). [A]2 o}

>Now working on the right hand side:

integrate k.dt from o to t,

this just ends up being  " k . (t - o) " which equals " k . t "

>after integrating both sides:

1 / {(2). [A]2 t} - 1 / {(2). [A]2 o} = k . t

>To find the half life:

[A]2 t = (10 mol/L)2 t

[A]2 o = (20 mol/L)2 o

{ 1 / ( (2)(10 mol/L)2 t ) } - { 1 / ( (2)(20 mol/L)2 o ) } = k . t1/2

>which simplifies to:

( 1 / 200 ) - ( 1 / 800 ) = ( 3 / 800)  = k . t1/2

>simplifies again:

(3)* { 1 / ( (2)(20 mol/L)2 o ) } = ( 3 / 800) = k . t1/2

>rearranging:

{ 3 / ( (2)(20 mol/L)2 o ) } = k . t1/2

>And now solving for t 1/2 :

k . t1/2 = { 3 / ( (2)(20 mol/L)2 o ) }

t1/2= { 3 / ( (2)(20 mol/L)2 o (k) ) }

Therefore the final expression for half life is:

t1/2 = { 3 / ( (2)[A]2 o (k) ) }

I hope ive made sense with my working, this is pretty full on stuff, for me anyway!

Cheers,