Answer

Verified

395.4k+ views

**Hint:**The following two formulae are needed to solve this question.

\[\begin{align}

& k=\dfrac{0.693}{{{t}_{1/2}}} \\

& t=\dfrac{2.303}{k}\log \dfrac{a}{a-x} \\

\end{align}\]

**Complete step by step answer:**

The relationship between the rate constant k and the half life period for a first order reaction is as follows:

\[k=\dfrac{0.693}{{{t}_{1/2}}}\]

The half life period of a first order reaction is 20 minutes.

\[{{t}_{1/2}}=20\text{ minutes}\]

Calculate the rate constant for the first order reaction:

\[\begin{align}

& k=\dfrac{0.693}{{{t}_{1/2}}} \\

& k=\dfrac{0.693}{20\text{ minutes}} \\

& k=0.03465\text{ minut}{{\text{e}}^{-1}} \\

\end{align}\]

The relationship between the time t required for a certain change in concentration of reactant and the rate constant k for the first order reaction is as follows:

\[t=\dfrac{2.303}{k}\log \dfrac{a}{a-x}\] ... ...(1)

Here, a is the initial concentration and a-x is the concentration at time t.

The initial concentration of the reactant is 0.16 M.

\[a=0.16M\]

The concentration of the reactant at time t is 0.02 M.

\[a-x=0.02M\]

Calculate the ratio of the initial concentration of the reactant to the concentration of the reactant at time t.

\[\dfrac{a}{a-x}=\dfrac{0.16M}{0.02M}=8\]

Calculate the logarithm of the ratio

\[\log \dfrac{a}{a-x}=\log 8=0.9031\]

Substitute values in equation (1)

\[\begin{align}

& t=\dfrac{2.303}{k}\log \dfrac{a}{a-x} \\

& t=\dfrac{2.303}{0.03465\text{ minut}{{\text{e}}^{-1}}}\times 0.9031 \\

& t=60\text{ minute} \\

\end{align}\]

The time required to decrease the concentration of the reactant from \[0.16\text{ }M\text{ }to\text{ }0.02\text{ }M\]is \[60\text{ }minutes\].

**So, the correct answer is “Option B”.**

**Note:**

Another approach to solve the same problem is to determine the number of half lives needed for the concentration of the reactant to change from \[0.16\text{ }M\text{ }to\text{ }0.02\text{ }M\].

\[\begin{align}

& \dfrac{0.02}{0.16}=\dfrac{1}{8}={{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{n}} \\

& n=3 \\

\end{align}\]

Three half life periods are needed. The half life period is \[20\text{ }minutes\]. For three half life periods, time will be \[60\text{ }minutes\].

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

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

Which are the major cities located on the river Ga class 10 social science CBSE

What is BLO What is the full form of BLO class 8 social science CBSE