Chemical Kinetics Question 64
Question: $ {t_{1/4}} $ can be taken as the time taken for concentration of reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is k, then $ {t_{1/4}} $ can be written as
Options:
A) 0.10/K
B) 0.29/K
C) 0.69/K
D) 0.15/K
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ k=\frac{2.303}{t}\log \frac{a}{a-x} $ or $ k=\frac{2.303}{{t_{1/4}}}\log \frac{4a}{3a}=\frac{2.303}{{t_{1/4}}}\log \frac{4}{3} $ or $ k=\frac{2.303\times 0.125}{{t_{1/4}}}=\frac{0.29}{{t_{1/4}}} $
