Chemical Kinetics Question 60
Question: The rate of a first order reaction is $ 0.04mol{L^{-1}}{S^{-1}} $ at 30 min and $ 0.03mol{L^{-1}}{S^{-1}} $ at 40 min. Thus, half-life of the reaction is
Options:
A) 32.4 mm
B) 51.2 mm
C) 46.8 mm
D) 24.1 min
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ k=\frac{2.303}{t}\log ( \frac{a}{a-x} ) $
$ \frac{kt}{2.303}=\log a-\log (a-x) $
$ \frac{kt_1}{2.303}=\log a-\log (a-x_1)at,time,t_1 $ …(i) $ \frac{kt_2}{2.303}=\log a-\log (a-x_2)at,time,t_2 $ …(ii) Subtracting (i) and (ii)
$ \therefore \frac{k}{2.303}(t_2-t_1)=log( \frac{a-x_1}{a-x_2} ) $ Rate, $ r_1=k(a-x_1);r_2=k(a-x_2) $
$ \therefore \frac{(a-x_1)}{(a-x_2)}=\frac{r_1}{r_2}\Rightarrow \frac{0.04}{0.03}=\frac{4}{3} $
$ \frac{k(40-30)}{2.303}=\log \frac{4}{3} $
$ \therefore k=\frac{2.303}{10}\log \frac{4}{3}=\frac{2.303}{10}\log 1.33 $
$ =\frac{2.303\times 0.125}{10}=0.02878{{\min }^{-1}} $
$ T_{50}=\frac{0.693}{0.02878}=24.1\min $
