[madscientist]

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:

t_1/2 = 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

[madscientist]

Ive attached a pdf. file to show my working a bit clearer,

Can anyone please help?

cheers,

madscientist

[Donaldson Tan]

My answer is t_1/2 = 3/(2.k.[[A]]²)

The equation for the concentration profile of reactant A is

where Ao is the initial concentration.

[madscientist]

ok, food for thought, thanks

madscientist :stupid:

[madscientist]

My answer is t_1/2 = 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,

madscientist

[Donaldson Tan]

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 + t_1/2)
1/2A² - 1/2(0.5A)² = -k.t_1/2
1/2A² - 1/0.5A² = -k.t_1/2
1/2A² - 2/A² = -k.t_1/2
k.t_1/2 = 3/2A²
t_1/2 = 3/(2.k.A²)

[madscientist]

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!

[pantone159]

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.

[Donaldson Tan]

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

[madscientist]

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 . t_1/2

>which simplifies to:

( 1 / 200 ) - ( 1 / 800 ) = ( 3 / 800)  = k . t_1/2

>simplifies again:

(3)* { 1 / ( (2)(20 mol/L)² _o ) } = ( 3 / 800) = k . t_1/2

>rearranging:

{ 3 / ( (2)(20 mol/L)² _o ) } = k . t_1/2

>And now solving for t _1/2 :

k . t_1/2 = { 3 / ( (2)(20 mol/L)² _o ) }
                           
     t_1/2= { 3 / ( (2)(20 mol/L)² _o (k) ) }


Therefore the final expression for half life is:

t_1/2 = { 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: