Half-Life | ChemTalk – keiseronlineuniversity.com

Chemistry

Half-Life | ChemTalk

keiseronlineuniversity.com

18 August 2023

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Core Ideas

On this article, you’ll be taught what a half-life is, in 0th. 1st, and 2nd order reactions. Additionally, you will be taught what the half-life system is and how one can use it to search out the age of a substance.

Subjects Coated in Different Articles

What’s a Half-Life?

A half-life (t_1/2) is the time it takes half the focus of a pattern to decay. For instance, if the beginning focus of a pattern is 1M, then the half-life would be the time it takes for the focus of the pattern to decay to 0.5M.

First Order Half-Life

First-order reactions, together with radioactive decay, are reactions the place the speed of response is simply depending on the focus of 1 reactant. First-order reactions have fixed half-lives. We are able to show this utilizing the built-in first-order fee legislation:

$begin{gather*} ln[A] = -kt + ln[A]_0 end{gather*}$

[A] is the focus of a reactant at time t, [A]₀ is the preliminary focus of the reactant, and ok is the speed fixed. Right here, we are able to transfer all of the pure logs to the left aspect of the equation:

$begin{gather*} {text{ln}frac{[A]} {[A]_{text{0}}}} = -kt end{gather*}$

Since we’re looking for the half-life of the reactant, we are able to set t because the half-life and [A] as half of the preliminary focus:

$begin{gather*} {t} = {t_{text{1/2}}} end{gather*}$

$begin{gather*} [A] = frac {[A]_0}{2} end{gather*}$

Substituting these values into the equation provides us the next:

$begin{gather*} {text{ln}frac{[A]_0/2} {[A]_{text{0}}}} = -kt_{1/2} end{gather*}$

Lastly, dividing the ok and the adverse provides us an equation for the half-life of a first-order response:

$begin{gather*} {t_{text{1/2}} = frac{ln2}{k}} end{gather*}$

Radioactive decay of Carbon-14, which has a half-life of 5730 years. — Graph exhibiting the radioactive decay of Carbon-14. Carbon-14 has a half-life of 5730 years, so the quantity of Carbon-14 is lower in half after each 5730 years. Geologists usually use carbon courting to search out the age of rocks and organisms. Supply.

Second Order Half-Life

Second-order reactions are reactions the place the speed of response depends on the concentrations of two reactants. The half-lives of second-order reactions depend upon the reactant’s preliminary focus. The built-in fee legislation for second-order reactions is as follows:

$begin{gather*} frac{1}{[A]} = kt + frac{1}{[A]_0} end{gather*}$

Since we’re looking for the half-life of the reactant, we are able to set t because the half-life and [A] as half of the preliminary focus:

$begin{gather*} {t} = {t_{text{1/2}}} end{gather*}$

$begin{gather*} [A] = frac {[A]_0}{2} end{gather*}$

Substituting these values into the equation provides us the next:

$begin{gather*} frac{2}{[A]_0} = kt_{1/2} + frac{1}{[A]_0} end{gather*}$

We are able to then isolate the t to resolve for the half-life of a second-order response:

$begin{gather*} frac{1}{[A]_0} = kt_{1/2} end{gather*}$

$begin{gather*} t_{1/2} = frac{1}{k[A]_0} end{gather*}$

Zero Order Half-Life

In zero-order reactions, the speed of response doesn’t depend upon the reactant’s focus. The half-lives of zero-order reactions, nevertheless, do depend upon the preliminary focus of the reactant. The built-in fee legislation for zero-order reactions is as follows:

$begin{gather*} [A] = [A]_0 - kt end{gather*}$

Since we’re looking for the half-life of the reactant, we are able to set t because the half-life and [A] as half of the preliminary focus:

$begin{gather*} {t} = {t_{text{1/2}}} end{gather*}$

$begin{gather*} [A] = frac {[A]_0}{2} end{gather*}$

Substituting these values into the equation provides us the next:

$begin{gather*} frac{[A]_0}{2} = [A]_0 - kt_{1/2} end{gather*}$

We are able to then isolate the t to resolve for the half-life of a zero-order response:

$begin{gather*} kt_{1/2} = frac{[A]_0}{2} end{gather*}$

$begin{gather*} t_{1/2} = frac{[A]_0}{k2} end{gather*}$

Apply Issues

The half-life for a first-order response is 2768 years. Beginning with a focus of 0.345M, what is going to the focus be after 11072 years?
What’s the half-life of a compound if 75 p.c of a given pattern of the compound decomposes in 60 min? Assume first-order kinetics.
A substance hydrolyzes in water with a fee fixed of two.0 × 10⁻³ min⁻¹. Calculate the t_1/2 for the hydrolysis response. Assume first-order kinetics.

Options

0.0216M. First, learn the way many half-lives have handed. By dividing 11072 by 2768, we discover that 4 half-lives have handed. Due to this fact, we are able to discover the ultimate focus by dividing 0.345M by 2 4 occasions. This leaves us with 0.0216M.
30 minutes. 25 p.c of the pattern is left, which means 2 half-lives have handed. 60 minutes divided by 2 is 30 minutes, which is the half-life.
350 minutes. Utilizing the equation for the half-life of a first-order response, we are able to substitute ok with 2.0 × 10⁻³ min⁻¹.
$begin{gather*} {t_{text{1/2}} = frac{ln2}{2.0 times 10^{-3}}} end{gather*}$

We are able to then conclude that the half-life is 350 minutes.

Additional Studying

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Core Ideas

Subjects Coated in Different Articles

What’s a Half-Life?

First Order Half-Life

Second Order Half-Life

Zero Order Half-Life

Apply Issues

Options

Additional Studying

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