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}}} $



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