Question

How do you find first order half life?

Answer 1

`t_"1/2" = (ln2)/k`

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Well, consider a general first-order rate law for a one-reactant reaction:

`A -> B`

`r(t) = k[A]`

where `k` is the rate constant and `[A]` is the concentration of `A` in `"M"`.

This is numerically equal to:

`= -(d[A])/(dt)`

where `(d[A])/(dt)` denotes an instantaneous rate of change in concentration of `A` over time.

By separation of variables:

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

Integrate the left-hand side from time zero to time `t`, and the right-hand side from initial to current concentration.

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

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

Thus, we obtain the first-order integrated rate law:

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

For a half-life, we have a current concentration of `[A]_0 = 1/2[A]`, so:

`ln (1/2[A]_0) = -kt_"1/2" + ln[A]_0`

where `t_"1/2"` is the half-life.

Using the properties of logarithms:

`=> ln((1/2[A]_0)/([A]_0)) = -kt_"1/2"`

`=> ln(1/2) = -kt_"1/2"`

`=> ln2 = kt_"1/2"`

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