If the half-life of a first-order reaction is | vedclass

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(A) $1$. For a first-order reaction,the rate constant $k$ is given by $k = 0.693 / t_{1/2}$.Given $t_{1/2} = 6.93$ minutes,so $k = 0.693 / 6.93 = 0.1

Vedclass · Accepted Answer

$46$

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(A) $1$. For a first-order reaction,the rate constant $k$ is given by $k = 0.693 / t_{1/2}$.Given $t_{1/2} = 6.93$ minutes,so $k = 0.693 / 6.93 = 0.1 	ext{ min}^{-1}$.$2$. The time $t$ required for the completion of $99\%$ of the reaction is calculated using the formula: $t = (2.303 / k) \log([A]_0 / [A]_t)$.Here,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.$3$. Substituting the values: $t = (2.303 / 0.1) \log(100 / 1) = 23.03 	imes \log(10^2) = 23.03 	imes 2 = 46.06$ minutes.Rounding to the nearest integer,the time is $46$ minutes.

If the half-life of a first-order reaction is $6.93$ minutes,then the time required for the completion of $99\%$ of the reaction will be . . . . . . minutes. (Given: $\log 2 = 0.3010$)

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