Question

The half-life period of a first-order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be: (log2 = 0.301)
A. 230.3 minutes
B. 23.03 minutes
C. 46.06 minutes
D. 460.6 minutes

Answer 1

Hint: First order reactions are those reactions which depend on the concentration of only one reactant. Half-life period is the period in which the concentration of a substance reduces half of its initial value.

Complete answer:
-In the given question we have to calculate the time required in which 99% of the reaction will complete
-It is given that the half-life is equal to 6.93 minutes.
-So, firstly we will calculate the rate constant for the first-order reaction:
${{\text{k}}_{1}}\text{ = }\dfrac{\text{0}\text{.693}}{{{\text{t}}_{1/2}}}=\text{ }\dfrac{\text{0}\text{.693}}{6.93}\text{= 0}\text{.1 /min}\text{.}$
-Now, we will find the time that is required for 99% completion of the reaction by applying the first-order reaction:
$\text{k = }\dfrac{2.303}{t}\text{ log}\left( \dfrac{{{\text{A}}_{0}}}{\text{A}} \right)$
-Here the ${{\text{A}}_{0}}$represents the initial concentration which is equal to 100 and A represents the final concentration which is $\left( \text{100-99} \right)$that is 1.
-So,
$\text{t = }\dfrac{\text{2}\text{.303}}{\text{k}}\log \left( \dfrac{\text{100}}{100-99} \right)\text{ = }\dfrac{2.303}{0.1}\log \left( \dfrac{\text{1}{{\text{0}}^{2}}}{1} \right)\text{ = 23}\text{.03 }\cdot \text{ 2log10}$
$\text{= 23}\text{.03 }\cdot \text{ 2 }\cdot \text{ 1 = 46}\text{.06 minutes}$.

Therefore, option C is the correct answer.

Note: We can observe from the relation of rate constant and half-life period of first and second-order reaction that the first-order reaction is a constant but second-order reaction inversely depends on the initial concentration of the substance.