In a first order reaction,$60 \%$ of the reactant converts into product in $45 \ minutes$. Calculate the rate constant of the reaction.

The half-life of a first-order reaction is $10 \ min$. If the initial concentration is $0.08 \ mol \ L^{-1}$ and the concentration at some instant is $0.01 \ mol \ L^{-1}$,then $t =$ ........... $\min$.

$N_{2}O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$
In the above first order reaction,the initial concentration of $N_{2}O_{5}$ is $2.40 \times 10^{-2} \ mol \ L^{-1}$ at $318 \ K$. The concentration of $N_{2}O_{5}$ after $1 \ hour$ was $1.60 \times 10^{-2} \ mol \ L^{-1}$. The rate constant of the reaction at $318 \ K$ is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Given: $\log 3 = 0.477, \log 5 = 0.699$]

The initial rate for a first order reaction is $0.6932 \times 10^{-2} \ mol \ L^{-1} \ min^{-1}$ and the initial concentration of the reactant is $0.1 \ M$. Then $t_{1/2}$ is equal to ...... $min$.

For a first order reaction $A \to B$,the reaction rate at reactant concentration of $0.01 \ M$ is found to be $2.0 \times 10^{-5} \ M \ sec^{-1}$. The half-life period of the reaction is .......... $sec$.

For a first order reaction,show that the time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of the reaction.