Question

What is the first-order half-life for `"0.50 M"` of a substance whose radioactive decay has a rate constant of `0.69 xx 10^(-2) "min"^(-1)`?

Answer 1

You don't need its concentration. (Why?)

`t_("1/2") = "100.5 min"`

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The Integrated Rate Law for the first-order process is:

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

where:

• `[A]` is the concentration of the reactant.
• `[A]_0` is its initial concentration.
• `k` is the rate constant in `"time"^(-1)`.
• `t` is of course the time in `"time"` units.

For a half-life, we take `t = t_"1/2"` such that `[A] = 1/2[A]_0`. Therefore:

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

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

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

But since `1/2 = 2^(-1)` and `ln(a^b) = blna`,

`-ln2 = -kt_"1/2"`,

and:

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

See, the concentration doesn't matter for first-order...

`color(blue)(t_"1/2") = (ln2)/(0.69 xx 10^(-2) "min"^(-1))`

`=` `color(blue)("100.5 min")`