Answer
425.4k+ 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 \[60\text{ }minutes\].
\[\begin{align}
& k=\dfrac{0.693}{{{t}_{1/2}}} \\
& t=\dfrac{2.303}{k}\log \dfrac{a}{a-x} \\
\end{align}\]
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 \[60\text{ }minutes\].
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