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]^{-3}d[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: