
Derive the relation between the half-life and rate constant for a first order reaction.
Answer
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Hint: First order reaction is the one which depends only on one reactant. Half-life \[{t_{\dfrac{1}{2}}}\] of the reaction is the time required for half of the reaction to be completed. Rate constant \[\left( k \right)\] is the proportionality constant in the equation expressing the relation between the rate of reaction and the concentration of reactant.
Complete step by step answer:
Let’s start with describing the meaning of the term half-life and rate constant along with first order reaction. First order reaction is one which depends only on one reactant. Half-life \[{t_{\dfrac{1}{2}}}\] of the reaction is the time required for half of the reaction to be completed. Rate constant \[\left( k \right)\] is the proportionality constant in the equation expressing the relation between the rate of reaction and the concentration of reactant.
Coming to the derivation,
The First order reaction rate equation is given by
$k = {\text{ }}\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\log _{10}}\dfrac{{[{{\text{A}}_0}]}}{{{{[{\text{A}}]}_{\text{t}}}}}$, where \[{{\text{A}}_{\text{0}}}\] is concentration at \[t = 0\].
Now, at ${{\text{t}}_{\dfrac{1}{2}}}{\text{A}} = \dfrac{{{{\text{A}}_0}}}{2}$, which gives us
$k = \dfrac{{2.303}}{{{{\text{t}}_{\dfrac{1}{2}}}}}{\text{lo}}{{\text{g}}_{10}}\dfrac{{\dfrac{{[{{\text{A}}_0}{\text{]}}}}{1}}}{{\dfrac{{[{{\text{A}}_0}]}}{2}}}$
Which gives us
$k = \dfrac{{2.303}}{{{{\text{t}}_{\dfrac{1}{2}}}}}{\text{lo}}{{\text{g}}_{10}}2$
Solving this we will get
$k = \dfrac{{0.693}}{{{{\text{t}}_{\dfrac{1}{2}}}}}$
Which can also be written as
${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$, \[k\] is rate constant and \[{t_{\dfrac{1}{2}}}\] is half-life of the reaction.
Hence the relation between the half-life and the rate of constant of the first order reaction has been established.
Note:
The relationship derived in this question is very important for both the experiment point of view as well as the theory point of view.
This relation helps us calculate an approximate rate constant for the reaction if we know about the time in which the reactant is reduced to half.
Complete step by step answer:
Let’s start with describing the meaning of the term half-life and rate constant along with first order reaction. First order reaction is one which depends only on one reactant. Half-life \[{t_{\dfrac{1}{2}}}\] of the reaction is the time required for half of the reaction to be completed. Rate constant \[\left( k \right)\] is the proportionality constant in the equation expressing the relation between the rate of reaction and the concentration of reactant.
Coming to the derivation,
The First order reaction rate equation is given by
$k = {\text{ }}\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\log _{10}}\dfrac{{[{{\text{A}}_0}]}}{{{{[{\text{A}}]}_{\text{t}}}}}$, where \[{{\text{A}}_{\text{0}}}\] is concentration at \[t = 0\].
Now, at ${{\text{t}}_{\dfrac{1}{2}}}{\text{A}} = \dfrac{{{{\text{A}}_0}}}{2}$, which gives us
$k = \dfrac{{2.303}}{{{{\text{t}}_{\dfrac{1}{2}}}}}{\text{lo}}{{\text{g}}_{10}}\dfrac{{\dfrac{{[{{\text{A}}_0}{\text{]}}}}{1}}}{{\dfrac{{[{{\text{A}}_0}]}}{2}}}$
Which gives us
$k = \dfrac{{2.303}}{{{{\text{t}}_{\dfrac{1}{2}}}}}{\text{lo}}{{\text{g}}_{10}}2$
Solving this we will get
$k = \dfrac{{0.693}}{{{{\text{t}}_{\dfrac{1}{2}}}}}$
Which can also be written as
${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$, \[k\] is rate constant and \[{t_{\dfrac{1}{2}}}\] is half-life of the reaction.
Hence the relation between the half-life and the rate of constant of the first order reaction has been established.
Note:
The relationship derived in this question is very important for both the experiment point of view as well as the theory point of view.
This relation helps us calculate an approximate rate constant for the reaction if we know about the time in which the reactant is reduced to half.
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