Answer

Verified

381k+ 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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths