Question

What is first order half life derivation?

Answer 1

Well, let us begin from the rate law... Is the FIRST-order half-life dependent on concentration?

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The first-order rate law for the reaction

`A -> B`

is

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

where:

• `r(t)` is the rate as a function of time, which we take to be the initial rate in `"M/s"`
• `k` is the rate constant. What are the units? You should know this.
• `[A]` is the concentration of a reactant `A` in `"M"`.
• `(d[A])/(dt)` is the rate of disappearance of reactant `A`, a NEGATIVE value. But the negative sign forces the rate to be positive, fitting for a FORWARD reaction.

By separation of variables:

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

Integrate on the left from time zero to time `t`, and the right from initial concentration `[A]_0` to current concentration `[A]`. We obtain:

`-kint_(0)^(t)dt = int_([A]_0)^([A])1/([A])d[A]`

`-kt = ln[A] - ln[A]_0`

Therefore, the first-order integrated rate law is:

`color(green)(ln[A] = -kt + ln[A]_0)`

As we should recognize, the half-life is when the concentration drops by half. Hence, we set `[A]_(1//2) -= 0.5[A]_0` to get:

`ln0.5[A]_0 = -kt_(1//2) + ln[A]_0`

`ln0.5[A]_0 - ln[A]_0 = -kt_(1//2)`

`ln\frac(0.5cancel([A]_0))(cancel([A]_0)) = -kt_(1//2)`

`-ln0.5 = ln2 = kt_(1//2)`

As a result, the half-life is given by...

`color(blue)(barul|stackrel(" ")(" "t_(1//2) = (ln2)/k" ")|)`