How do you derive first order half life?

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How do you derive first order half life?

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Answer 1 · 2017-02-05T21:14:21

You can always start from the first-order rate law:

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

for the reaction:

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

Separate variables:

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

Integrate from the initial to the final state. It is convenient to set $t_{0} = 0$ $t_0 = 0$ .

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

$- (k t - k \cdot 0) = ln [A] - {ln [A]}_{0}$ $-(kt - k*0) = ln[A] - ln[A]_0$

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

$\Rightarrow ln \frac{[A]}{{[A]}_{0}} = - k t$ $=> ln\frac([A])([A]_0) = -kt$

For the half-life, at time $t = t_{1/2}$ $t = t_"1/2"$ , the concentration of $A$ $A$ dropped to $\frac{1}{2} {[A]}_{0}$ $1/2[A]_0$ . Therefore:

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

$ln (1/2) = -kt_"1/2"$

$-ln (1/2) = kt_"1/2"$

$ln (1/2)^(-1) = kt_"1/2"$

$ln 2 = kt_"1/2"$

Therefore:

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

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How do you derive first order half life?

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