Rate Law Integration: How to Overcome k-1[A]0 Term

[sparkle123]

How did this happen? I would separate the variables by dividing both sides by [A] but then the k-1[A]0 term is in the way :(

 

[Char. Limit]

That k_-1[A]₀ is just a constant, really. If you want, replace k₁+k_-1=a and k_-1[A]₀=b, and then you'll have an equation in the form:

\frac{d[A]}{dt} = a [A] - b

Which is easily solvable.

 

[sparkle123]

Sorry, but how do you solve this?

 

[Char. Limit]

Seperation of variables (and multiplication by dt) gives:

\int \frac{1}{a [A] - b} \frac{d[A]}{dt} dt = \int \frac{d[A]}{a [A] - b} = \int dt

And you can evaluate both of those integrals, right?

 

[sparkle123]

ln([A]-b/a)=at
:)
so ln([A] - k-1/(k1+k-1)[A]0)=(k1+k-1)t
could you possible provide some insight on how this turns into the equation in the solution?
Thanks!

 

[hunt_mat]

Not that isn't the only way of working it, you can treat it like an integrating factor problem. So multiply through by exp(-at) to find that:
<br /> e^{-at}\frac{d[ A]}{dt}-ae^{-at}[ A] =be^{-at}<br />
Then the LHS can be recognised as the derivative of exp(-at)[A] and you get:
<br /> \frac{d}{dt}(e^{-at}[ A]) =be^{-at}<br />
Integrating and using the boundary conditions will give the solution.