Explain half-life for zero order and second order reaction.
Answer
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Hint: Half-life concept comes under the category of chemical kinetics. Half-life is the time that is required to reduce the quantity half times the initial values. Order of the reaction can be zero, first, second or third depending upon the reaction concentration.
Complete answer:
Now we will study the zero order reaction first and then second order reaction. In zero order reaction, the rate does not vary with the increase or decrease in the concentration of the reactants. Due to which, the rate of these reactions is always equal to the rate constant of the specific reactions.
The Differential form of a zero order reaction can be written as:
\[Rate = - \dfrac{{dA}}{{dt}}\]
\[ = k[A] = k\]
In zero order reaction, the rate is independent of reactant concentrations; it can be seen from the above differential equation.
Half-life for zero order can be defined as the timescale in which there is a \[50\% \] reduction in the initial population is referred to as half-life. Half-life is denoted by the symbol ‘ \[t\dfrac{1}{2}\] ’.
From the integral form, we have the following equation
\[[A] = [{A_0}]-kt\]
Replacing t with half-life \[t\dfrac{1}{2}\] we get:
\[\dfrac{1}{2}[A] = [{A_0}]-kt\dfrac{1}{2}\]
Therefore, t1/2 can be written as:
\[kt\dfrac{1}{2} = \dfrac{1}{2}[{A_0}]\]
And,
\[t\dfrac{1}{2} = \dfrac{1}{2}[{A_0}]\]
It can be noted from the equation given above that the half-life is dependent on the rate constant as well as the reactant’s initial concentration.
For second order reaction:
Differential form for second order reaction can be represented as:
\[ - d[R]dt = k{\left[ R \right]^2}\]
Half-life for second order reaction is:
Therefore, while attempting to calculate the half-life of a reaction, the following substitutions must be made:
\[\left[ R \right] = {\left[ {{R_0}} \right]^2}\]
And, \[t = t\dfrac{1}{2}\]
Now, substituting these values in the integral form of the rate equation of second order reactions, we get:
\[1{\left[ {{R_0}} \right]^2}-1\left[ {{R_0}} \right] = kt\dfrac{1}{2}\]
Therefore, the required equation for the half-life of second order reactions can be written as follows.
\[t\dfrac{1}{2} = 1k\left[ {{R_0}} \right]\]
This equation for the half-life implies that the half-life is inversely proportional to the concentration of the reactants.
Note:
We have to know that the half-life for zero order reaction is \[t\dfrac{1}{2} = \dfrac{1}{2}[{A_0}]\] whereas the half-life for second order reaction is \[t\dfrac{1}{2} = 1k\left[ {{R_0}} \right]\], and half-life can be defined as the timescale in which there is a \[50\% \]reduction in the initial population is referred to as half-life.
Complete answer:
Now we will study the zero order reaction first and then second order reaction. In zero order reaction, the rate does not vary with the increase or decrease in the concentration of the reactants. Due to which, the rate of these reactions is always equal to the rate constant of the specific reactions.
The Differential form of a zero order reaction can be written as:
\[Rate = - \dfrac{{dA}}{{dt}}\]
\[ = k[A] = k\]
In zero order reaction, the rate is independent of reactant concentrations; it can be seen from the above differential equation.
Half-life for zero order can be defined as the timescale in which there is a \[50\% \] reduction in the initial population is referred to as half-life. Half-life is denoted by the symbol ‘ \[t\dfrac{1}{2}\] ’.
From the integral form, we have the following equation
\[[A] = [{A_0}]-kt\]
Replacing t with half-life \[t\dfrac{1}{2}\] we get:
\[\dfrac{1}{2}[A] = [{A_0}]-kt\dfrac{1}{2}\]
Therefore, t1/2 can be written as:
\[kt\dfrac{1}{2} = \dfrac{1}{2}[{A_0}]\]
And,
\[t\dfrac{1}{2} = \dfrac{1}{2}[{A_0}]\]
It can be noted from the equation given above that the half-life is dependent on the rate constant as well as the reactant’s initial concentration.
For second order reaction:
Differential form for second order reaction can be represented as:
\[ - d[R]dt = k{\left[ R \right]^2}\]
Half-life for second order reaction is:
Therefore, while attempting to calculate the half-life of a reaction, the following substitutions must be made:
\[\left[ R \right] = {\left[ {{R_0}} \right]^2}\]
And, \[t = t\dfrac{1}{2}\]
Now, substituting these values in the integral form of the rate equation of second order reactions, we get:
\[1{\left[ {{R_0}} \right]^2}-1\left[ {{R_0}} \right] = kt\dfrac{1}{2}\]
Therefore, the required equation for the half-life of second order reactions can be written as follows.
\[t\dfrac{1}{2} = 1k\left[ {{R_0}} \right]\]
This equation for the half-life implies that the half-life is inversely proportional to the concentration of the reactants.
Note:
We have to know that the half-life for zero order reaction is \[t\dfrac{1}{2} = \dfrac{1}{2}[{A_0}]\] whereas the half-life for second order reaction is \[t\dfrac{1}{2} = 1k\left[ {{R_0}} \right]\], and half-life can be defined as the timescale in which there is a \[50\% \]reduction in the initial population is referred to as half-life.
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