Question

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) \[20\text{ }minutes\]

Answer 1

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 \[60\text{ }minutes\].