Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: madscientist on February 23, 2006, 10:11:57 AM
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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,
madscientist
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Ive attached a pdf. file to show my working a bit clearer,
Can anyone please help?
cheers,
madscientist
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My answer is t1/2 = 3/(2.k.[[A]]2)
The equation for the concentration profile of reactant A is
(https://www.chemicalforums.com/users/geodome/rate_eqn.GIF)
where Ao is the initial concentration.
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ok, food for thought, thanks
madscientist :stupid:
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My answer is t1/2 = 3/(2.k.[[A]]2)
The equation for the concentration profile of reactant A is
(https://www.chemicalforums.com/users/geodome/rate_eqn.GIF)
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,
madscientist
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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)
(https://www.chemicalforums.com/users/geodome/rate_eqn.GIF)
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)
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so I was alsmost their with my working, thanks geodome, no wonder my head hurts im trying to calculate this with pen and paper!
madscientist :stupid:
P.S I absolutely love your picture, crack up!
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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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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 (http://www.chemicalforums.com/users/geodome/mario.gif)
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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,
madscientist :albert: