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# The half life period of a first order reaction is $20\text{ }minutes$ . The time required for the concentration of the reactant to change from $0.16\text{ }M\text{ }to\text{ }0.02\text{ }M$ is A) $\text{80 }minutes$B) $60\text{ }minutes$C) $40\text{ }minutes$D) $20\text{ }minutes$

Last updated date: 17th Apr 2024
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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}

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 $60\text{ }minutes$.