Calculate the time in seconds required to reduce the concentration of reactant to half of initial concentration for a first order reaction if the rate constant is $1.386 \times 10^{-3} \ s^{-1}$. (in $s$)

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)

$A$ first order reaction is given as $A \rightarrow \text{products}$. Its integrated rate equation is:

The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation,$10 \%$ of the virus is inactivated. The rate constant for viral inactivation is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Use : $\ln 10 = 2.303; \log_{10} 3 = 0.477; \text{property of logarithm} : \log x^y = y \log x$]

$A$ first order reaction which is $30\%$ complete in $30 \ min$ has a half-life period of .............. $min.$ (in $.2$)

The decomposition of $N_2O_5$ is a first order reaction represented by $N_2O_5 \to N_2O_4 + \frac{1}{2} O_2$. After $15 \ min$,the volume of $O_2$ produced is $9 \ mL$ and at the end of the reaction,it is $35 \ mL$. The rate constant is equal to: