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A flask contains a mixture of compound A and B. Both compounds decompose by first order kinetics. The half lives are \[54.0\] min for A and \[18.0\] min for B. If the concentration of A and B are equal initially, how long (in a minute) will it take for the concentration of A to be four times that of B?
A.\[18.0\] min
B.\[18.0\] min
C.\[54.0\] min
D.None of these

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Last updated date: 17th Apr 2024
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Answer
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Hint: To answer this question we must have knowledge of the kinetic equation for the first order reaction and the relationship between half life and rate constant for first order. Start by assuming the initial and final amount as x and y respectively and solve the two linear equations formed.
Formula used: \[{{\text{t}}_{1/2}} = \dfrac{{0.693}}{{\text{K}}}\] and \[{{\text{A}}_{\text{t}}} = {{\text{A}}_{\text{0}}}{{\text{e}}^{ - {\text{Kt}}}}\]
Where \[{{\text{t}}_{1/2}}\] is the half life, \[{\text{K}}\] is rate constant, \[{{\text{A}}_{\text{t}}}{\text{ and }}{{\text{A}}_{\text{0}}}\] are the final and initial concentration respectively.

Complete step by step solution:
We have been given the half life of both the reactants A and B, so let us first calculate the rate constant for both these reaction because we will need them for the first order kinetic rate equation:
The relationship between half life and rate constant for a first order reaction is as follow:
\[{{\text{t}}_{1/2}} = \dfrac{{0.693}}{{\text{K}}}\], rearranging we will get \[{\text{K}} = \dfrac{{0.693}}{{{{\text{t}}_{1/2}}}}\].
Rate constant for reactant A with half life 54 min is
\[{{\text{K}}_{\text{A}}} = \dfrac{{0.693}}{{54{\text{ min}}}} = 0.0128{\text{ mi}}{{\text{n}}^{ - 1}}\]
Rate constant for reactant B with half life 18 min is
\[{{\text{K}}_{\text{B}}} = \dfrac{{0.693}}{{{\text{18 min}}}} = 0.0385{\text{ mi}}{{\text{n}}^{ - 1}}\]
Let the final and initial amount of reactant A \[{{\text{A}}_{\text{t}}}{\text{ and }}{{\text{A}}_{\text{0}}}\] and the final and initial amount of reactant B \[{{\text{B}}_{\text{t}}}{\text{ and }}{{\text{B}}_{\text{0}}}\] respectively.
Using the first order kinetic equation and substituting, the values we will get two equations as follow:
\[{{\text{A}}_{\text{t}}} = {{\text{A}}_{\text{0}}}{{\text{e}}^{ - 0.0128{\text{t}}}}\]
\[{{\text{B}}_{\text{t}}} = {{\text{B}}_{\text{0}}}{{\text{e}}^{ - 0.0385{\text{t}}}}\]
It is given to us that initial concentration of both A and B is same so \[{{\text{A}}_0}{\text{ = }}{{\text{B}}_{\text{0}}}\]
And e have to calculate the time when concentration of A becomes four times that of B. this give us \[{{\text{A}}_{\text{t}}}{\text{ = 4}}{{\text{B}}_{\text{t}}}\]
Putting the above values and dividing the two equations we will get:
\[\dfrac{{{\text{4}}{{\text{B}}_{\text{t}}}}}{{{{\text{B}}_{\text{t}}}}} = \dfrac{{{{\text{A}}_{\text{0}}}{{\text{e}}^{ - 0.0128{\text{t}}}}}}{{{{\text{B}}_{\text{0}}}{{\text{e}}^{ - 0.0385{\text{t}}}}}}\] Further solving and taking the exponential above we will get,
\[4 = {{\text{e}}^{ - 0.0128{\text{t + 0}}{\text{.0385t}}}}\]
\[4 = {{\text{e}}^{0.0257{\text{t}}}}\]
To cancel exponential we take Ln that is natural log both sides because natural log and exponential are inverse to each other.
\[\ln 4 = \ln {{\text{e}}^{0.0257{\text{t}}}}\]
\[\ln 4 = 0.0257{\text{t}}\]
The value of \[\ln 4\] is 1.3863. Substituting the values and rearranging.
\[\dfrac{{1.3863}}{{0.0257}} = {\text{t}}\] we will get \[{\text{t = 53}}{\text{.97}} \simeq 54{\text{ min}}\]
Hence, the correct option is option C.

Note: First order kinetics or first order reactions are those reactions in which rate of the reaction are directly proportional to the concentration of the reactant. The half life of the substance if the time in which reactant reduces to half of its initial amount, it is defined in terms of reactant and not product.