Question

A certain first-order reaction has a rate of `"0.04 M/s"` after 10 seconds and `"0.03 M/s"` after 20 seconds. What is the half-life for this reaction?

Answer 1

`t_"1/2" = "24.1 s"`.

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Interesting question. The rates `r(t)` defined at each point in a reaction shall be given by the first order rate law:

`r(10) = k[A]_(10) = "0.04 M"cdot"s"^(-1)`

`r(20) = k[A]_(20) = "0.03 M"cdot"s"^(-1)`

where the subscript in front of `[A]` indicates the time in seconds. The rate constant `k` is the same for the same reaction at the same temperature.

Each would follow the same integrated rate law for first-order one-reactant reactions:

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

where `[A]_0` is the initial concentration of `A`.

Since the rate constants `k` are the same, we can write a ratio of the concentrations and relate that to the ratio of the rates:

`([A]_20)/([A]_10) = (r_20)/(r_10)`

So, using the integrated rate law for each time:

`ln[A]_10 = -10k + ln[A]_0`

`=> ln[A]_10 + 10k = ln[A]_0`

`ln[A]_20 = -20k + ln[A]_0`

`=> ln[A]_20 + 20k = ln[A]_0`

Since `ln[A]_0` for two times in a given reaction is the same for all time (i.e. the we start at the same initial concentration), we can thus say that:

`ln[A]_20 + 20k = ln[A]_10 + 10k`

`ln[A]_20 - ln[A]_10 = ln(([A]_20)/([A]_10)) = 10k - 20k`

Plug in the ratio of the rates:

`ln((r_20)/(r_10)) = -10k`

Lastly, the half-life `t_"1/2"` of a first-order reaction is given by

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

so `k = (ln2)/t_"1/2"`. Thus:

`ln((r_20)/(r_10)) = -(10ln2)/t_"1/2"`

`=> color(blue)(t_"1/2") = -(10ln2)/(ln(r_20//r_10)) = -(10ln2)/(ln(0.03//0.04))`

`=` `ulcolor(blue)("24.1 s")`