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: $\log 2 = 0.3010$)

  • A
    $46$
  • B
    $92$
  • C
    $23$
  • D
    $69$

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Similar Questions

Identify $True$ $(T)$ and $False$ $(F)$ statements for the following equations related to a first-order reaction $R \rightarrow P$:
$(i) \ln [R] = -kt + \ln [R]_{0}$
$(ii) \ln [R] = +kt + \ln [R]_{0}$

What is the rate constant of a first-order reaction if $0.08 \ mol$ of reactant reduces to $0.02 \ mol$ in $23.03 \ min$ (in $min^{-1}$)?

$A$ reaction has a rate constant $k = 2.4 \times 10^{-4} \ s^{-1}$. Find the ratio of $t_{99.9}$ to $t_{50}$.

For the first order reaction $A \rightarrow 2B$,$1 \ mole$ of reactant $A$ gives $0.2 \ moles$ of $B$ after $100 \ minutes$. The half-life of the reaction is $..... \ min$. [Use: $\ln 2 = 0.69, \ln 10 = 2.3$]

Consider the two different first order reactions given below:
$A + B \rightarrow C$ (Reaction $1$)
$P \rightarrow Q$ (Reaction $2$)
The ratio of the half-life of Reaction $1$ : Reaction $2$ is $5 : 2$. If $t_1$ and $t_2$ represent the time taken to complete $2/3$ and $4/5$ of Reaction $1$ and Reaction $2$,respectively,then the value of the ratio $t_1 : t_2$ is $. . . . \times 10^{-1}$ (nearest integer).
[Given: $\log_{10}(3) = 0.477$ and $\log_{10}(5) = 0.699$]

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