
${{\text{t}}_{{\text{1/2}}}}$ of a first order reaction is $6.93\,{\text{s}}$, the value of rate constant for the reaction would be $10\,{{\text{s}}^{ - 1}}$.
A. True
B. False
Answer
545.7k+ views
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.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Draw the diagram showing the germination of pollen class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

The computer jargonwwww stands for Aworld wide web class 12 physics CBSE

