Question

Given, ${{t}_{1/2}}$ of a first order reaction is 10 min. Starting with $10\,mol\,{{L}^{-1}}$, rate after 20 min is:
A.$0.0693\,mol\,{{L}^{-1}}\,{{\min }^{-1}}$
B.$0.0693\times 2.5\,mol\,{{L}^{-1}}\,{{\min }^{-1}}$
C.$0.0693\times 5\,mol\,{{L}^{-1}}\,{{\min }^{-1}}$
D.$0.0693\times 10\,mol\,{{L}^{-1}}\,{{\min }^{-1}}$

Answer 1

First calculate the rate constant of the reaction, as it is a first order reaction. Then multiply the rate constant with the concentration of the reactant to find the rate.

Complete answer:
In order to answer our question, we need to learn about kinetics and half life of a chemical reaction. Now, every reaction takes a certain amount of time to get completed. Moreover, the rates of reaction are different, for different reactions. More the rate of the equation, more is the speed and less is the time taken to complete the reaction. Now, let us come to the half life of a chemical reaction. Half life is defined as the time during which the concentration of the reactants is reduced to half of the initial concentration or it is the time required for the completion of half of the reaction. It is denoted by ${{t}_{1/2}}$. Now, let us calculate the half life for a first order reaction, in this case, decay of an isotope is an example of first order reaction. Now,

& k=\frac{2.303}{t}\log \frac{{{[R]}_{0}}}{[R]} \\\ & at\,t={{t}_{1/2}},[R]=\frac{{{[R]}_{0}}}{2} \\\ & \Rightarrow k=\frac{2.303}{{{t}_{1/2}}}\log \frac{{{[R]}_{0}}}{\frac{{{[R]}_{0}}}{2}} \\\ & \Rightarrow k=\frac{2.303}{{{t}_{1/2}}}\log 2 \\\ & \Rightarrow {{t}_{1/2}}=\frac{2.303}{k}\times \log 2=\frac{0.693}{k} \\\ & \\\ \end{aligned}$$ Now, if the initial concentration is $10mol\,{{L}^{-1}}$, then the concentration after 20 mins, which is equal to 2 half lives should be equal to $2.5mol\,{{L}^{-1}}$, now we will find the rate constant using the formula discussed above. So, substituting the data in the formula, we have the rate constant as: $$\begin{aligned} & k=\frac{0.693}{{{t}_{1/2}}} \\\ & \Rightarrow k=\frac{0.693}{10}=0.0693{{\min }^{-1}} \\\ \end{aligned}$$ Now, in order to calculate the rate, we will multiply the rate constant with the concentration of reactant. So, the rate is: $$\begin{aligned} & Rate=k\times [\operatorname{Re}ac\tan t] \\\ & Rate=0.0693\times 2.5mol\,{{L}^{-1}}{{\min }^{-1}} \\\ \end{aligned}$$ **So, we get the correct answer for this question as option B.** **Note:** In the expression of half life, does not carry conc. term hence the half life period or half change time for first order reaction does not depend upon initial concentration of the reactants. Similarly, the time required to reduce the concentration of the reactant to any fraction of the initial concentration for the first order reaction is also independent of the initial concentration.