Half-life period for first order reaction is $10$ sec. Calculate the rate constant for the reaction.
Answer
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Hint: To determine the answer we should know the half-life formula of first order reaction. The first-order reaction is the reaction in which the rate of reaction is directly proportional to the concentration of the reactant. The half-life of the first order reaction is inversely proportional to the rate constant.
Formula used:${{\text{t}}_{{\text{1/2}}}}\,\,{\text{ }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
Complete step-by-step solution:The first-order rate constant formula is given as follows:
$\,\,{\text{k = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\,\dfrac{{{{\text{A}}_{\text{o}}}}}{{\text{A}}}$
Where,
${\text{k}}$ is the first-order rate constant. The unit of first-order rate constant is ${\text{tim}}{{\text{e}}^{ - 1}}$.
${\text{t}}$ is the time.
${{\text{A}}_{\text{o}}}$ is the initial concentration of the reactant.
${{\text{A}}_{\text{x}}}$ is the concentration of the reactant left at time ${\text{t}}$.
Half-life is the time at which the concentration of the reactant becomes half of the initial concentration. So, if the initial concentration is $1$ at half-life the concentration will be $1/2$.
So, we can determine the half-life formula as is as follows:
\[\,\,{\text{k = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{{\text{t}}_{{\text{1/2}}}}}}{\text{log}}\,\dfrac{{\text{1}}}{{{\text{1/2}}}}\]
Where,
${{\text{t}}_{{\text{1/2}}}}$ is the half-life.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{{\text{1/2}}}}\,\,}}$
Now we will rearrange the formula of half-life as follows:
${{\text{t}}_{{\text{1/2}}}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
So, we will the above first-order half-life formula to determine the rate constant as follows:
On substituting $10$ sec for ${{\text{t}}_{{\text{1/2}}}}$.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{\text{10}}\,{\text{sec}}{\text{.}}\,\,}}$
$\therefore {\text{k}}\,\,{\text{ = }}\,0.0693\,{\text{se}}{{\text{c}}^{ - 1}}$
So, the value of rate constant of the reaction is $0.0693\,{\text{se}}{{\text{c}}^{ - 1}}$.
Note:The unit of half-life and rate constant should be noticed as both the units should be the same.
The unit of half-life is time and the unit of the rate constant is ${\text{tim}}{{\text{e}}^{ - 1}}$ and the time can be taken in second, minute, hour or year. The half-life time of a first order reaction is independent of reactant concentration. For other order reactions, the half-life is inversely proportional to the concentration of reactant.
Formula used:${{\text{t}}_{{\text{1/2}}}}\,\,{\text{ }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
Complete step-by-step solution:The first-order rate constant formula is given as follows:
$\,\,{\text{k = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\,\dfrac{{{{\text{A}}_{\text{o}}}}}{{\text{A}}}$
Where,
${\text{k}}$ is the first-order rate constant. The unit of first-order rate constant is ${\text{tim}}{{\text{e}}^{ - 1}}$.
${\text{t}}$ is the time.
${{\text{A}}_{\text{o}}}$ is the initial concentration of the reactant.
${{\text{A}}_{\text{x}}}$ is the concentration of the reactant left at time ${\text{t}}$.
Half-life is the time at which the concentration of the reactant becomes half of the initial concentration. So, if the initial concentration is $1$ at half-life the concentration will be $1/2$.
So, we can determine the half-life formula as is as follows:
\[\,\,{\text{k = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{{\text{t}}_{{\text{1/2}}}}}}{\text{log}}\,\dfrac{{\text{1}}}{{{\text{1/2}}}}\]
Where,
${{\text{t}}_{{\text{1/2}}}}$ is the half-life.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{{\text{1/2}}}}\,\,}}$
Now we will rearrange the formula of half-life as follows:
${{\text{t}}_{{\text{1/2}}}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
So, we will the above first-order half-life formula to determine the rate constant as follows:
On substituting $10$ sec for ${{\text{t}}_{{\text{1/2}}}}$.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{\text{10}}\,{\text{sec}}{\text{.}}\,\,}}$
$\therefore {\text{k}}\,\,{\text{ = }}\,0.0693\,{\text{se}}{{\text{c}}^{ - 1}}$
So, the value of rate constant of the reaction is $0.0693\,{\text{se}}{{\text{c}}^{ - 1}}$.
Note:The unit of half-life and rate constant should be noticed as both the units should be the same.
The unit of half-life is time and the unit of the rate constant is ${\text{tim}}{{\text{e}}^{ - 1}}$ and the time can be taken in second, minute, hour or year. The half-life time of a first order reaction is independent of reactant concentration. For other order reactions, the half-life is inversely proportional to the concentration of reactant.
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