Question

Two first order reactions have their half lives in the ratio $ 3:2 $ .Calculate the ratio of $ {t_1}:{t_2} $ . The time $ {t_1} $ and $ {t_2} $ are the time periods for $ 25\% $ and $ 75\% $ completion for the first and second reaction.
 $ \left( {a.} \right){\text{ 0}}{\text{.311 : 1}} $
 $ \left( {b.} \right){\text{ 0}}{\text{.420 : 1}} $
 $ \left( {c.} \right){\text{ 0}}{\text{.273 : 1}} $
 $ \left( {d.} \right){\text{ 0}}{\text{.119 : 1}} $

Answer 1

Hint : We will find the time taken for the reactions to complete $ 25\% $ and $ 75\% $ of its initial value. With the help of half-lives we can find out the reaction constant for the reaction respectively. We will find $ {t_1} $ and $ {t_2} $ separately for both the reactions. Then we will find the ratio of $ {t_1} $ and $ {t_2} $ .
 $ \left( 1 \right). $ $ t{\text{ = }}\dfrac{{2.303}}{k}{\text{log}}\left[ {\dfrac{A}{{A - {A_ \circ }}}} \right] $
Where,
 $ t{\text{ = }} $ Time period for completion of reaction
 $ {\text{k = }} $ Rate constant of reaction
 $ {\text{A = }} $ Initial concentration of reactant
 $ {\text{A - }}{{\text{A}}_ \circ }{\text{ = concentration at time , t}} $
 $ \left( 2 \right).{\text{ }}{{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{0.693}}{k} $ .

Complete Step By Step Answer:
The time taken by the first order reaction to complete is given by:
 $ t{\text{ = }}\dfrac{{2.303}}{{{k_1}}}{\text{log}}\left[ {\dfrac{A}{{A - {A_ \circ }}}} \right] $
Here at time $ {\text{ t}} $ the reaction is completed which means the reactants are completely converted into the products. For first reaction which takes times $ {t_1} $ to complete $ 25\% $ of its initial value we can write,
 $ {t_1}{\text{ = }}\dfrac{{2.303}}{{{k_1}}}{\text{log}}\left[ {\dfrac{{100}}{{100 - 25}}} \right] $
Here we consider the initial amount of reactant is $ 100\% $ , then according to the question at $ 25\% $ the left amount will be $ 100{\text{ - 25}} $ . Therefore it can be written as,
 $ {t_1}{\text{ = }}\dfrac{{2.303}}{{{k_1}}}{\text{log}}\left[ {\dfrac{{100}}{{75}}} \right] $ _______________ $ \left( 1 \right) $
Similarly for second reaction which takes time $ {t_2} $ for its $ 75\% $ completion, we can write,
 $ {t_1}{\text{ = }}\dfrac{{2.303}}{{{k_2}}}{\text{log}}\left[ {\dfrac{{100}}{{100 - 75}}} \right] $
 $ {t_2}{\text{ = }}\dfrac{{2.303}}{{{k_2}}}{\text{log}}\left[ {\dfrac{{100}}{{25}}} \right] $ _______________ $ \left( 2 \right) $
Now we have to find the ratio of $ {t_1} $ and $ {t_2} $ , therefore divide the equations $ \left( 1 \right) $ and $ \left( 2 \right) $ .
 $ \dfrac{{{t_1}}}{{{t_2}}}{\text{ = }}\dfrac{{{\text{ }}\dfrac{{2.303}}{{{k_1}}}{\text{log}}\left[ {\dfrac{{100}}{{75}}} \right]}}{{{\text{ }}\dfrac{{2.303}}{{{k_2}}}{\text{log}}\left[ {\dfrac{{100}}{{25}}} \right]}} $
On solving we get the result as:
 $ \dfrac{{{t_1}}}{{{t_2}}}{\text{ = }}\dfrac{{{{\text{k}}_2}{\text{ }} \times {\text{ 0}}{\text{.123}}}}{{{\text{ }}{{\text{k}}_1}{\text{ }} \times {\text{ 0}}{\text{.602}}}} $
For finding the ratio of $ {k_2}{\text{ and }}{{\text{k}}_1} $ we use the concept of half-lives for first order reaction. For first order reaction the half-life of a substance is given by,
 $ {\text{ }}{{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{0.693}}{k} $
Therefore for first reaction it is given by,
 $ {\text{ }}{{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{0.693}}{{{k_1}}} $ ___________ $ \left( 3 \right) $
Similarly for second reaction it can be written as,
 $ {\text{ }}{{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{0.693}}{{{k_2}}} $ ____________ $ \left( 4 \right) $
Also the ratio of half-life time is given as $ 3:2 $ . On dividing equation $ \left( 3 \right) $ and $ \left( 4 \right) $ we can write as,
 $ \dfrac{3}{2}{\text{ = }}\dfrac{{{\text{ }}\dfrac{{0.693}}{{{k_1}}}}}{{{\text{ }}\dfrac{{0.693}}{{{k_2}}}}} $
 $ \dfrac{3}{2}{\text{ = }}\dfrac{{{\text{ }}{{\text{k}}_2}}}{{{\text{ }}{{\text{k}}_1}}} $
Now using this relation we can find the ratio of time also.
 $ \dfrac{{{t_1}}}{{{t_2}}}{\text{ = }}\dfrac{{{{\text{k}}_2}{\text{ }} \times {\text{ 0}}{\text{.123}}}}{{{\text{ }}{{\text{k}}_1}{\text{ }} \times {\text{ 0}}{\text{.602}}}} $
 $ \dfrac{{{t_1}}}{{{t_2}}}{\text{ = }}\dfrac{{{\text{3 }} \times {\text{ 0}}{\text{.123}}}}{{2{\text{ }} \times {\text{ 0}}{\text{.602}}}}{\text{ }} $
 $ \dfrac{{{t_1}}}{{{t_2}}}{\text{ = }}\dfrac{{0.308}}{1}{\text{ }} $
Hence the ratio will be $ 0.308{\text{ }}:{\text{ }}1 $ . The nearest answer in the given option is $ \left( {a.} \right){\text{ 0}}{\text{.311 : 1}} $

Note :
The half-life is the time when the concentration of the reactants becomes half of initial value. The value of half time is different for different orders of reaction. The value of rate constant is easily determined by using the given formula for only first order reaction.