Answer
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Hint: 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 answer:The first-order rate constant formula is as follows:
$\,\,{\text{k = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\,\dfrac{{{{\text{A}}_{\text{o}}}}}{{{{\text{A}}_{\text{x}}}}}$
Where,
k is the first-order rate constant. The unit of first-order rate constant is${\text{tim}}{{\text{e}}^{ - 1}}$.
it 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 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 $1/2$.
The first-order half-life formula 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.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{{\text{1/2}}}}\,\,}}$
We will rearrange the formula half-life as follows:
${{\text{t}}_{{\text{1/2}}}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
Now we will use the first-order half-life formula to determine the rate constant as follows:
On substituting $6.93\,{\text{s}}$ for ${{\text{t}}_{{\text{1/2}}}}$.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{\text{6}}{\text{.93}}\,{\text{s}}\,\,}}$
${\text{k}}\,\,{\text{ = }}\,{\text{0}}{\text{.1}}\,{{\text{s}}^{ - 1}}$
So, the rate constant is $\,0.1\,{{\text{s}}^{ - 1}}$.
In the question, it is given that the rate constant would be $\,10\,{{\text{s}}^{ - 1}}$ whereas the rate constant is $\,0.1\,{{\text{s}}^{ - 1}}$ so, the statement is not true.
Therefore, option (B) False, is correct.
Additional information: The first order half-life does not depend upon the initial concentration of the reactant. The first-order reaction is never complete. The completion of $99.9$% of a first-order reaction took $10$ half-life.
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 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 answer:The first-order rate constant formula is as follows:
$\,\,{\text{k = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\,\dfrac{{{{\text{A}}_{\text{o}}}}}{{{{\text{A}}_{\text{x}}}}}$
Where,
k is the first-order rate constant. The unit of first-order rate constant is${\text{tim}}{{\text{e}}^{ - 1}}$.
it 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 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 $1/2$.
The first-order half-life formula 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.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{{\text{1/2}}}}\,\,}}$
We will rearrange the formula half-life as follows:
${{\text{t}}_{{\text{1/2}}}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
Now we will use the first-order half-life formula to determine the rate constant as follows:
On substituting $6.93\,{\text{s}}$ for ${{\text{t}}_{{\text{1/2}}}}$.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{{{\text{6}}{\text{.93}}\,{\text{s}}\,\,}}$
${\text{k}}\,\,{\text{ = }}\,{\text{0}}{\text{.1}}\,{{\text{s}}^{ - 1}}$
So, the rate constant is $\,0.1\,{{\text{s}}^{ - 1}}$.
In the question, it is given that the rate constant would be $\,10\,{{\text{s}}^{ - 1}}$ whereas the rate constant is $\,0.1\,{{\text{s}}^{ - 1}}$ so, the statement is not true.
Therefore, option (B) False, is correct.
Additional information: The first order half-life does not depend upon the initial concentration of the reactant. The first-order reaction is never complete. The completion of $99.9$% of a first-order reaction took $10$ half-life.
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.
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