For the second-order reaction 2A → B, time taken for the [A] to fall to one-fourth value is how many

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For the second-order reaction 2A → B, time taken for the [A] to fall to one-fourth value is how many times the time it takes for [A] to fall to half of its initial value?

Accepted Answer

For 2nd order reaction, $t_{1/2} = \frac{1}{K.\left\lbrack A_{0} \right\rbrack}$

Hence, second t_1/2 is double of first t_1/2.

$\left\lbrack A_{0} \right\rbrack\overset{\quad x\text{ min}\quad}{\rightarrow}\frac{\left\lbrack A_{0} \right\rbrack}{2}\overset{\quad 2x\text{ min}\quad}{\rightarrow}\frac{\left\lbrack A_{0} \right\rbrack}{4}$

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