Calculate the rate constant of a first order reaction if the half-life of the reaction is $40 \ minute$.
- A$1.733 \times 10^{-2} \ minute^{-1}$
- B$1.951 \times 10^{-2} \ minute^{-1}$
- C$1.423 \times 10^{-2} \ minute^{-1}$
- D$1.256 \times 10^{-2} \ minute^{-1}$
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Similar Questions
$A$ first order reaction is half completed in $45 \ min$. How long does it need $99.9 \%$ of the reaction to be completed (in $Hours$)?
MediumKCET 2022
View SolutionThe following data were obtained during the first order thermal decomposition of a gas $A$ at constant volume:
$A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$
$S.No.$ $Time/s$ $Total Pressure/(atm)$ $1.$ $0$ $0.1$ $2.$ $115$ $0.28$
The rate constant of the reaction is . . . . . . $\times 10^{-2} \ s^{-1}$ (nearest integer).
$A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$
| $S.No.$ | $Time/s$ | $Total Pressure/(atm)$ |
| $1.$ | $0$ | $0.1$ |
| $2.$ | $115$ | $0.28$ |
The rate constant of the reaction is . . . . . . $\times 10^{-2} \ s^{-1}$ (nearest integer).
DifficultJEE MAIN 2024
View SolutionIf the rate law is $r = K[A]$,then the concentration of the reactant remaining after time $t = 1/k$ is: ($[A]_0$ is the concentration of the reactant at $t = 0$)
Medium
View SolutionFor a first order reaction,the half-life period is independent of
In a first order reaction,the concentration of the reactant is reduced to $(1/8)^{th}$ of its initial concentration in $23.03 \ min$. What is the half-life period of the reaction (in $min$)?
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