Question

The time elapsed between $ 33\;% $ and $ 67\;% $ completion of a first order reaction is $ 30\;minutes $ . What is the time needed for $ 25\;% $ completion?
(A) $ 15.5\;min $
(B) $ 12.5\;min $
(C) $ 18.5\;min $
(D) $ 16.5\;min $

Answer 1

Hint :The completion of the reaction represents the change in the concentration of the compound. If we take a reference initial concentration, we can find the change in the concentration and the concentration after some time interval. From the obtained date the rate constant can be calculated. Using rate constant, time required to reach a concentration can be calculated.
Rate constant $ k=\dfrac{2.303}{t}\log \dfrac{{{R}_{0}}}{{{R}_{t}}} $

Complete Step By Step Answer:
First of all, to simplify the given condition, we can take the initial concentration of the compound as $ R_0=100\;M $ .
Now, as per the given data, when the reaction gets $ 33\;% $ completed, we can say the concentration reduces by $ 33\;M $ . Hence, the final concentration after time interval $ t_1 $ is $ R_{t1}=100M-33M=67M $
From the equation of the rate constant for the first condition,
 $ k=\dfrac{2.303}{{{t}_{1}}}\log \dfrac{{{R}_{0}}}{{{R}_{t1}}} $
Substituting the obtained values, and making time intervals the subject of the equation.
 $ {{t}_{1}}=\dfrac{2.303}{k}\log \dfrac{100}{67} $ …… $ (1) $
Similarly when the reaction gets $ 67\;% $ completed, we understand that the concentration has reduced by $ 67\;M $ . Hence, the final concentration after time interval $ t_2 $ is $ R_{t2}=100M-67M=33M $
From the equation of the rate constant for the second condition,
 $ k=\dfrac{2.303}{{{t}_{2}}}\log \dfrac{{{R}_{0}}}{{{R}_{t2}}} $
Substituting the obtained values, and making time intervals the subject of the equation.
 $ {{t}_{2}}=\dfrac{2.303}{k}\log \dfrac{100}{33} $ …… $ (2) $
Now, we are given that the difference between the given time intervals is $ 30\;min $ .
 $ \therefore {{t}_{2}}-{{t}_{1}}=30\min $
Substituting the values from the equation $ (1) $ and $ (2) $
 $ \therefore \dfrac{2.303}{k}\log \dfrac{100}{33}-\dfrac{2.303}{k}\log \dfrac{100}{67}=30 $
 $ \therefore \dfrac{1.11}{k}-\dfrac{0.40}{k}=30 $
Rearranging the equation to find the value of rate constant
 $ \therefore k=\dfrac{1.11-0.40}{30} $
 $ \therefore k=0.0236{{\min }^{-1}} $
Now, we are required to find the time interval required for $ 25\;% $ completion i.e. for the concentration to reduce by $ 25\;M $ . Hence, the time required for the final concentration to be $ R_t=100-25=75M $ can be obtained as
 $ \therefore t=\dfrac{2.303}{k}\log \dfrac{{{R}_{0}}}{{{R}_{t}}} $
Substituting the given values,
 $ \therefore t=\dfrac{2.303}{0.0236}\log \dfrac{100}{75} $
 $ \therefore t\approx 12.5\min $
Hence, the correct answer is Option $ (B) $ .

Note :
The point to note is that, here we are given the change or say reduction in the concentration of the compound. For finding the rate constant, we need the concentration of the remaining compound after a particular time interval. Hence, we should remember to take the difference of the initial (reference) concentration and the change in the concentration, to get the final concentration after a time interval.