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?

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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?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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

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

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

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

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:

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

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

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

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?

1 Answers