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

The following two formulae are needed to solve this question.

& k=\dfrac{0.693}{{{t}_{1/2}}} \\\ & t=\dfrac{2.303}{k}\log \dfrac{a}{a-x} \\\ \end{aligned}$$ **Complete step by step answer:** The relationship between the rate constant k and the half life period for a first order reaction is as follows: $$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{aligned} & k=\dfrac{0.693}{{{t}_{1/2}}} \\\ & k=\dfrac{0.693}{20\text{ minutes}} \\\ & k=0.03465\text{ minut}{{\text{e}}^{-1}} \\\ \end{aligned}$$ The relationship between the time t required for a certain change in concentration of reactant and the rate constant k for the first order reaction is as follows: $$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{aligned} & 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{aligned}$$ The time required to decrease the concentration of the reactant from $$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{aligned} & \dfrac{0.02}{0.16}=\dfrac{1}{8}={{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{n}} \\\ & n=3 \\\ \end{aligned}$$ Three half life periods are needed. The half life period is $$20\text{ }minutes$$. For three half life periods, time will be $$60\text{ }minutes$$.