$[A] \rightarrow [B]$. If the formation of compound $[B]$ follows first-order kinetics and after $70 \ min$ the concentration of $[A]$ was found to be half of its initial concentration,then the rate constant of the reaction is $x \times 10^{-6} \ s^{-1}$. The value of $x$ is $......$ (Nearest Integer).
- A$166$
- B$165$
- C$167$
- D$186$
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Similar Questions
Calculate the half-life of a first order reaction from the rate constants given below:
$(i) \ 200 \ s^{-1}$
$(ii) \ 2 \ min^{-1}$
$(iii) \ 4 \ years^{-1}$
$(i) \ 200 \ s^{-1}$
$(ii) \ 2 \ min^{-1}$
$(iii) \ 4 \ years^{-1}$
Difficult
View SolutionFor a first order reaction $A \rightarrow B$, the data is provided in the table below. Find $x$ in minutes. (Nearest integer)
$t/\text{min}$ $[A]/M$ $0$ $0.6500$ $x$ $0.0650$ $20$ $0.00065$
| $t/\text{min}$ | $[A]/M$ |
|---|---|
| $0$ | $0.6500$ |
| $x$ | $0.0650$ |
| $20$ | $0.00065$ |
DifficultJEE MAIN 2026
View SolutionThe integrated rate equation for a first-order reaction,$A \rightarrow \text{product}$,is
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View Solution$A$ reaction has a rate constant $k = 2.4 \times 10^{-4} \ s^{-1}$. Find the ratio of $t_{99.9}$ to $t_{50}$.
In a first order reaction,the concentration of the reactant decreases from $20 \ mmol \ dm^{-3}$ to $8 \ mmol \ dm^{-3}$ in $40 \ minute$. Find the rate constant of the reaction.
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