Answer
Verified
408.9k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE