Chemical Kinetics Question 352
Question: $ {t_{1/4}} $ can be taken as the time taken for the concentration of a reactant to drop to $ \frac{3}{4} $ of its initial value. If the rate constant for a first order reaction is k, the $ {t_{1/4}} $ can be written as
Options:
A) $ 0.75/k $
B) $ 0.69/k $
C) $ 0.29/k $
D) $ 0.10/k $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ {t_{1/4}}=\frac{2.303}{k}\log \frac{1}{3/4}=\frac{2.303}{k}\log \frac{4}{3} $
$ =\frac{2.303}{k}(log4-\log 3)=\frac{2.303}{k}(2log2-log3) $
$ =\frac{2.303}{k}(2\times 0.301-0.4771)=\frac{0.29}{k} $
