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)

Only through assuming that each reaction is an elementary reaction do we have

`A + A=> C`

With equal amounts of two disappearing reactants, and calling the appearing product `C`, you have the rate law:

`r(t) = k[A][A] = color(highlight)(k[A]^2 = -(d[A])/(dt)) = (d[C])/(dt)`

Note that despite the fact you have two `A` reactants, you use the stoichiometric coefficient `\mathbf1`, not `2`, because `A` is reacting with itself. It is not two `A` reactants reacting with a `B` reactant.

What we can do is a separation of variables:

`kdt = -1/([A]^2) d[A]`

Then, you can integrate each side with respect to the variable in question.

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

Where `X_0` means "initial `X`". Assuming `t_0 = 0`, we get:

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

Notice how for the half-life, `[A] = [A]_0/2`. Thus:

`kt = 2/[[A]_0] - 1/[[A]_0]`

`= 1/[[A]_0]`

Now, since `[A] = [A]_0/2`, `t = t_"1/2"` and we have:

`color(blue)(t_"1/2" = 1/(k[A]_0))`

(ii)

Using `[A]_0 ne [B]_0`, we get a second order reaction of two first-order components `A` and `B`:

`r(t) = k[A][B] = -(d[A])/(dt) = -(d[B])/(dt) = (d[C])/(dt)`

(Clearly if `A = B`, then we are just doing (i).)

Notice how we cannot just pick `[A]` and call it good. We have to consider both. When the concentrations of both compounds decrease, they decrease by an unknown amount, `x`, so we have:

`(dx)/(dt) = k[A][B]`

`= k([A]_0 - x)([B]_0 - x)`

(since `x = [A]_0 - [A] = [B]_0 - [B]`, `x` increases as `[A]` and `[B]` decrease, so `(dx)/(dt) > 0`)

Using separation of variables again, we have:

`int_(t_0)^t kdt = int_(0)^x 1/[([A]_0 - x)([B]_0 - x)] dx`

With this, we need to use partial fractions to integrate the right side. In the interest of time (and partial fractions is not the focus here):

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

Using the rules for logarithms, we get:

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

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

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

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

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

Even though we are doing half-life, we don't know their half-lives, so we can't assume that `[A] = [A]_0/2` or that `[B] = [B]_0/2`. So there is no general equation for the half-life for `[A]_0 ne [B]_0`!