How do you calculate half life decay?

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How do you calculate half life decay?

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Answer 1 · 2017-06-16T18:16:59

It depends on the order of the reaction with respect to the reactant undergoing half-life decay.

The common half-life expressions for one reactant $A$ $A$ are:

$t_{1/2} = \frac{{[A]}_{0}}{2 k}$ $t_"1/2" = ([A]_0)/(2k)$ (zero order)

$t_{1/2} = \frac{ln 2}{k}$ $t_"1/2" = (ln2)/k$ (first order)

$t_{1/2} = \frac{1}{k {[A]}_{0}}$ $t_"1/2" = 1/(k[A]_0)$ (second order)

$t_{1/2} = \frac{3}{2 k {[A]}_{0}^{2}}$ $t_"1/2" = 3/(2k[A]_0^2)$ (third order)

where ${[A]}_{0}$ $[A]_0$ is the initial concentration in $M$ $"M"$ and $k$ $k$ is the rate constant in the appropriate units to obtain time units of $s$ $"s"$ .

If you wish to derive them, that requires some calculus. I can do first order as an example, as that is the hardest one. For the general rate law

$r (t) = k [A] = - \frac{d [A]}{d t}$ $r(t) = k[A] = -(d[A])/(dt)$ ,

of the first-order reaction

$A \to B$ $A -> B$ ,

where the negative sign indicates disappearance of reactant,

separation of variables gives:

$- k d t = \frac{1}{[A]} d [A]$ $-kdt = 1/([A]) d[A]$

Now, integrate from $0$ $0$ to $t$ $t$ on the left-hand side and ${[A]}_{0}$ $[A]_0$ to $[A]$ $[A]$ for the right-hand side.

$- \int_{0}^{t} k d t = \int_{{[A]}_{0}}^{[A]} \frac{1}{[A]} d [A]$ $-int_(0)^(t) kdt = int_([A]_0)^([A]) 1/([A])d[A]$

The integral of $\frac{1}{x}$ $1/x$ is $ln | x |$ $ln|x|$ , and $[A] \geq 0$ $[A] >= 0$ , so:

$- k t = ln [A] - {ln [A]}_{0}$ $-kt = ln[A] - ln[A]_0$

For a half-life, $[A] = \frac{1}{2} {[A]}_{0}$ $[A] = 1/2[A]_0$ , i.e. the concentration of $A$ $A$ has halved. So:

$- k t_{1/2} = ln (\frac{1}{2} {[A]}_{0}) - {ln [A]}_{0}$ $-kt_"1/2" = ln(1/2 [A]_0) - ln[A]_0$

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

$=> color(blue)(t_"1/2 " ("1st order") = (ln2)/k)$

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How do you calculate half life decay?

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