Question

The reaction $ A \to B $ follows first order kinetics. The time taken for $ 0.8 $ mole of $ A $ to produce $ 0.6 $ mole of $ B $ is $ 1h $ . What is the time taken for conversion of $ 0.9 $ mole of $ A $ produce $ 0.675 $ mole of $ B $ ?
(A) $ 0.25h $
(B) $ 2h $
(C) 1h
(D) $ 0.5h $

Answer 1

Hint: Given that the time taken for $ 0.8 $ mole of $ A $ to produce $ 0.6 $ mole of $ B $ is $ 1h $ . Now we have to determine the time taken for conversion of $ 0.9 $ mole of $ A $ to produce $ 0.675 $ mole of $ B $ . by equating the both rate constants for the above moles gives the time required.
 $ k = \dfrac{{2.303}}{t}\log \dfrac{a}{{\left( {a - x} \right)}} $
 $ k $ is rate constant
 $ t $ is time required
 $ a $ is initial concentration
 $ x $ is the final concentration.

Complete answer:
Chemical reactions are classified into zero order, first order, and second order reactions. First order reactions are the chemical reactions that depend on the concentration of only one reactant.
The rate constant for the first order reaction is given as $ k = \dfrac{{2.303}}{t}\log \dfrac{a}{{\left( {a - x} \right)}} $
Given that the reaction $ A \to B $ follows first order kinetics.
The time taken for $ 0.8 $ mole of $ A $ to produce $ 0.6 $ mole of $ B $ is $ 1h $ . substitute these values and write the rate constant as $ {k_1} $
Thus, the rate constant can be written as $ {k_1} = \dfrac{{2.303}}{1}\log \dfrac{{0.8}}{{\left( {0.8 - 0.6} \right)}} $
Let the time taken for conversion of $ 0.9 $ mole of $ A $ produce $ 0.675 $ mole of $ B $ is $ t $ and has to be determined and rate constant be $ {k_2} $
The rate constant $ {k_2} $ can be written as $ {k_2} = \dfrac{{2.303}}{t}\log \dfrac{{0.9}}{{\left( {0.9 - 0.675} \right)}} $
By equation the above both obtained rate constants, $ {k_1} = {k_2} $
 $ \dfrac{{2.303}}{1}\log \dfrac{{0.8}}{{\left( {0.8 - 0.6} \right)}} = \dfrac{{2.303}}{t}\log \dfrac{{0.9}}{{\left( {0.9 - 0.675} \right)}} $
The time taken $ t $ will be $ 1h $
Option C is the correct one.

Note:
The initial concentrations and final concentrations were different in both the conversions. But the ratio of the initial concentration to change in concentration is the same, which results in equal time. The logarithmic values must be taken accurately while calculating the ratio.