(C) For a first order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $90\%$ completion, $[A]_t = [A]_0

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(C) For a first order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $90\%$ completion, $[A]_t = [A]_0 - 0.9[A]_0 = 0.1[A]_0$ and $t = 1 \text{ hour}$.$k = \frac{2.303}{1} \log \frac{[A]_0}{0.1[A]_0} = 2.303 \log 10 = 2.303 \text{ h}^{-1}$.For $99.9\%$ completion, $[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.$t = \frac{2.303}{k} \log \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{2.303} \log 10^3 = 3 \log 10 = 3 \text{ hours}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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The time required for $90\%$ completion of a certain first order reaction is $1 \text{ hour}$. Calculate the time required for $99.9\%$ completion of the same reaction.

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A
$2 \text{ hours}$
B
$1 \text{ hour}$
C
$3 \text{ hours}$
D
$0.5 \text{ hour}$

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The time required for $90\%$ completion of a certain first order reaction is $1 \text{ hour}$. Calculate the time required for $99.9\%$ completion of the same reaction.

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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: $\log 2 = 0.3010$)

For a first-order reaction,the time required for $93.75\%$ completion is $4$ times the half-life of the reaction. What is the relationship between the time required for $93.75\%$ completion and the half-life $(t_{0.5})$?

The rate constant of the reaction $2H_2O_{2(aq)} \to 2H_2O_{(l)} + O_{2(g)}$ is $3 \times 10^{-3} \ min^{-1}$. At what concentration of $H_2O_2$,the rate of reaction will be $2 \times 10^{-4} \ M \ s^{-1}$? ............ $M$

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