Editing Half-life (section)

=== Half-life and reaction orders ===
In [[chemical kinetics]], the value of the half-life depends on the [[Rate equation|reaction order]]:

====Zero order kinetics====
The rate of this kind of reaction does not depend on the substrate [[concentration]], {{math|[A]}}. Thus the concentration decreases linearly.

:<math display="block" chem="">d[\ce A]/dt = - k</math>The integrated [[rate law]] of zero order kinetics is:

<math display="block" chem="">[\ce A] = [\ce A]_0 - kt</math>In order to find the half-life, we have to replace the concentration value for the initial concentration divided by 2: <math display="block" chem="">[\ce A]_{0}/2 = [\ce A]_0 - kt_{1/2}</math>and isolate the time:<math display="block" chem="">t_{1/2} = \frac{[\ce A]_0}{2k}</math>This {{math|''t''{{sub|½}}}} formula indicates that the half-life for a zero order reaction depends on the initial concentration and the rate constant.

====First order kinetics====
In first order reactions, the rate of reaction will be proportional to the concentration of the reactant. Thus the concentration will decrease exponentially.
<math display="block" chem="">[\ce A] = [\ce A]_0 \exp(-kt)</math>as time progresses until it reaches zero, and the half-life will be constant, independent of concentration.

The time {{math|''t''{{sub|½}}}} for {{math|[A]}} to decrease from {{math|[A]{{sub|0}}}} to {{math|{{sfrac|1|2}}[A]{{sub|0}}}} in a first-order reaction is given by the following equation:<math display="block" chem="">[\ce A]_0 /2 = [\ce A]_0 \exp(-kt_{1/2})</math>It can be solved for<math display="block" chem="">kt_{1/2} = -\ln \left(\frac{[\ce A]_0 /2}{[\ce A]_0}\right) = -\ln\frac{1}{2} = \ln 2</math>For a first-order reaction, the half-life of a reactant is independent of its initial concentration. Therefore, if the concentration of {{math|A}} at some arbitrary stage of the reaction is {{math|[A]}}, then it will have fallen to {{math|{{sfrac|1|2}}[A]}} after a further interval of {{tmath|\tfrac{\ln 2}{k}.}} Hence, the half-life of a first order reaction is given as the following:</p><math display="block">t_{1/2} = \frac{\ln 2}{k}</math>The half-life of a first order reaction is independent of its initial concentration and depends solely on the reaction rate constant, {{mvar|k}}.

====Second order kinetics====
In second order reactions, the rate of reaction is proportional to the square of the concentration. By integrating this rate, it can be shown that the concentration {{math|[A]}} of the reactant decreases following this formula:

<math display="block" chem>\frac{1}{[\ce A]} = kt + \frac{1}{[\ce A]_0}</math>We replace {{math|[A]}} for {{math|{{sfrac|1|2}}{{math|[A]}}{{sub|0}}}} in order to calculate the half-life of the reactant {{math|A}} <math display="block" chem="">\frac{1}{[\ce A]_0 /2} = kt_{1/2} + \frac{1}{[\ce A]_0}</math>and isolate the time of the half-life ({{math|''t''{{sub|½}}}}):<math display="block" chem="">t_{1/2} = \frac{1}{[\ce A]_0 k}</math>This shows that the half-life of second order reactions depends on the initial concentration and [[Reaction rate constant|rate constant]].