# Chemistry for medical professionals

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## Half-life in 1st order reactions

The half-life can be calculated from the law of speed. Again we consider a simple reaction:

$A.→B.$

These reactions take place in the 1st order:

$−d[A.]dt=k⋅[A.]$

By changing (variable separation) one obtains:

$−d[A.][A.]=k⋅dt$

The equation is now integrated, with the starting situation before the start of the reaction being chosen as the zero point. $[A.]0$ denotes the initial concentration of A at $t$ = 0.

$∫[A.]0[A.]d[A.][A.]=∫0t−k⋅dt$

It follows:

$ln([A.][A.]0)=−k⋅t$

This equation applies to the entire course of a first-order reaction. We now consider the half-life $t½$. After a half-life, the concentration of A has decreased by half: [A] = $½[A.]0$. This is now plugged into the equation, $[A.]0$ can be shortened.

$ln(12[A.]0[A.]0)=ln(12)=−k⋅t½$

It applies $ln(1/2)=-ln(2)$. By transformation we get an expression for the half-life:

Half-life
$t½=ln(2)k$

The half-life is therefore independent of the concentration of the educt. This is characteristic of first-order reactions. Radioactive decay also obeys a rate law 1. This is used when determining the age, for example using the C14 method.