Question

What is the derivation for the half life of a second order reaction with: (i) equal concentration of reactants for the reaction `2A -> "product"` (ii) unequal concentration of reactants for the reaction `A + B -> "product"`?

Answer 1

(i) `A + A -> P`

What you can start with is writing out the rate law.

`r(t) = k[A]^2 = (d[A])/(dt)`

Then, with a separation of variables (where `t_0 = "0 s"` and `[A] = [A]_0/2`)

`-kint_(t)^(t_0) dt = int_([A]_0)^([A]) 1/[A]^2 d[A]`

`-kt = (-1/([A])) - (-1/[A]_0)`

`-kt_"1/2" = -1/([A]) + 1/(2[A]) = -1/(2[A])`

`color(blue)(t_"1/2" = 1/(2k[A]))`

(ii) `A + B -> P`

This one is much harder, because `[A]_0 ne [B]_0`. So... let `[A]_0 = a`, `[B]_0 = b`, `[A] = a-x`, and `[B] = b-x`.

`-(dx)/(dt) = -k(a-x)(b-x)`

`int_0^x 1/((a-x)(b-x))dx = int_(t_0)^(t) kdt`

Using partial fraction decomposition (which you should know if you actually need to do this derivation for homework), and I'll skip to the answer, you get:

`1/(b-a) [ln(1/(a-x)) - ln(1/(b-x))] = kt`

If we plug values back in and normalize the natural logarithms such that `[A]_0 = 1`:

`1/([B]_0-[A]_0) [ln(([A]_0)/([A]_0-x)) - ln(([A]_0)/([B]_0-x))] = kt`

`color(blue)(ln(([A]_0[B])/([A][B]_0)) = k([B]_0-[A]_0)t)`

This is as far as you can go. You CANNOT get a half-life expression for this, because you don't know what `[A]` and `[B]` become. They have different half-lives, and so the half-life for the overall reaction is not able to be determined.