The half-life for the first-order reaction $N_2O_5 \to 2NO_2 + \frac{1}{2} O_2$ is $2.4 \text{ hours}$. Starting with $10.8 \text{ g}$ of $N_2O_5$ at $STP$,how many liters of oxygen will be obtained after a period of $9.6 \text{ hours}$? (Given: $10^{0.2} = 1.584$)

For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

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 : $\ln 10 = 2.303; \log_{10} 3 = 0.477; \text{property of logarithm} : \log x^y = y \log x$]

It takes $1 \, h$ for a first order reaction to go to $50 \, \%$ completion. The total time required for the same reaction to reach $87.5 \, \%$ completion will be $..... \, h$.

The reaction $N_{2}O_{5} \longrightarrow 2NO_{2} + \frac{1}{2}O_{2}$ is first order in $N_{2}O_{5}$ having rate constant $6.2 \times 10^{-4} \ s^{-1}$. What is the value of rate of reaction when concentration of $N_{2}O_{5}$ is $1.25 \ mol \ L^{-1}$?

In a first order reaction,the concentration of the reactant is reduced from $0.6 \ mol \ L^{-1}$ to $0.2 \ mol \ L^{-1}$ in $5 \ min$. What is the rate constant of the reaction (in $min^{-1}$)? $(\log 3 = 0.4771)$