
A first order reaction $({{t}_{1/2}}=1\,day)$ completes $x%$ after 4 half-lives. Value of x is
(A)- 87.5
(B)- 75
(C)- 93.75
(D)- 50
Answer
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Hint: The first order reaction is a chemical reaction in which the rate of reaction is dependent on the concentration of one reactant. And at half-life, the time is independent of the concentration of the reactant.
Complete step by step answer:
Given the reaction is of first order. That is, it is linearly proportional to the concentration of only one reactant in the reaction process. We have the first order integrated rate law equation as: $\ln \left( \dfrac{\left[ A \right]}{{{\left[ A \right]}_{0}}} \right)=-kt$
where, $\left[ A \right]$is the concentration at time t and ${{\left[ A \right]}_{o}}$ is the initial concentration of the reaction. The time taken to consume half of the initial amount of reactant is known as its half-life. That is, $\left[ A \right]=\dfrac{1}{2}{{\left[ A \right]}_{o}}$
Then rate law equation becomes, $\ln \left( \dfrac{1}{2} \right)=-k{{t}_{1/2}}$ or ${{t}_{1/2}}=\dfrac{0.693}{k}$
where k is the rate constant. So, we have the half-life and the rate constant to be inversely proportional. It is given ${{t}_{1/2}}=1\,day$ for the given reaction. But as the half-life is independent of the initial concentration of the reactant, that is, the decrease in the concentration of half-life and any point of the reaction will be the same. So, after the first half-life the concentration is $\dfrac{{{\left[ A \right]}_{o}}}{2}$ . Then, the concentration after three consecutive half-lives will be $\dfrac{{{\left[ A \right]}_{o}}}{2\times 2}$, $\dfrac{{{\left[ A \right]}_{o}}}{{{2}^{3}}}$, $\dfrac{{{\left[ A \right]}_{o}}}{{{2}^{4}}}$ respectively. The percentage of the amount left is $=\dfrac{1}{{{2}^{4}}}\times 100=6.25%$
Then, the percentage of the reacted amount is $x%=100-6.26=93.75%$ .
Therefore, the first order reaction completes option (C)- 93.75 percent after 4 half-lives.
Note: For the first order reaction, the unit of the rate of reaction is mol/time and the rate constant is (1/time). The unit of rate constant is different for different order of reactions.
Complete step by step answer:
Given the reaction is of first order. That is, it is linearly proportional to the concentration of only one reactant in the reaction process. We have the first order integrated rate law equation as: $\ln \left( \dfrac{\left[ A \right]}{{{\left[ A \right]}_{0}}} \right)=-kt$
where, $\left[ A \right]$is the concentration at time t and ${{\left[ A \right]}_{o}}$ is the initial concentration of the reaction. The time taken to consume half of the initial amount of reactant is known as its half-life. That is, $\left[ A \right]=\dfrac{1}{2}{{\left[ A \right]}_{o}}$
Then rate law equation becomes, $\ln \left( \dfrac{1}{2} \right)=-k{{t}_{1/2}}$ or ${{t}_{1/2}}=\dfrac{0.693}{k}$
where k is the rate constant. So, we have the half-life and the rate constant to be inversely proportional. It is given ${{t}_{1/2}}=1\,day$ for the given reaction. But as the half-life is independent of the initial concentration of the reactant, that is, the decrease in the concentration of half-life and any point of the reaction will be the same. So, after the first half-life the concentration is $\dfrac{{{\left[ A \right]}_{o}}}{2}$ . Then, the concentration after three consecutive half-lives will be $\dfrac{{{\left[ A \right]}_{o}}}{2\times 2}$, $\dfrac{{{\left[ A \right]}_{o}}}{{{2}^{3}}}$, $\dfrac{{{\left[ A \right]}_{o}}}{{{2}^{4}}}$ respectively. The percentage of the amount left is $=\dfrac{1}{{{2}^{4}}}\times 100=6.25%$
Then, the percentage of the reacted amount is $x%=100-6.26=93.75%$ .
Therefore, the first order reaction completes option (C)- 93.75 percent after 4 half-lives.
Note: For the first order reaction, the unit of the rate of reaction is mol/time and the rate constant is (1/time). The unit of rate constant is different for different order of reactions.
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