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t_{1/4}  can be taken as the time taken for the concentration of a reactant to drop to 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

Option: 1

0.10/k\;


Option: 2

\; \; 0.29/k\;


Option: 3

\; 0.69/k\;


Option: 4

\; 0.75/k


Answers (1)

best_answer

The formula for the first-order reaction is -

k = \frac{2.303}{t} \log \frac{a_0}{a_t}

Then,

 k = \frac{2.303}{t_{\frac{1}{4}}} \log \frac{1}{\frac{3}{4}}

\\t_\frac{1}{4} = \frac{2.303}{k} \log \frac{4}{3}\\ \\ t_\frac{1}{4} = \frac{2.303}{k}\times 0.125 \\ \\t_\frac{1}{4}= \frac{0.29}{k} 

The correct option is (2).

Posted by

Pankaj Sanodiya

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