$N_2O_5 \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$ is a first order reaction. $k = 5 \times 10^{-4} \ s^{-1}$. The initial concentration of $N_2O_5 = 0.2 \ mol \ L^{-1}$. After how much time will its concentration be $25 \%$ of the initial concentration (in $s$)?

The integrated rate equation is $Rt = \log C_0 - \log C_t$. The straight line graph is obtained by plotting

The rate constant $k$,for the reaction ${N_2}{O_5}_{(g)} \to 2N{O_2}_{(g)} + \frac{1}{2}{O_2}_{(g)}$ is $2.3 \times 10^{-2} \ s^{-1}$. Which equation given below describes the change of $[{N_2}{O_5}]$ with time? $[{N_2}{O_5}]_0$ and $[{N_2}{O_5}]_t$ correspond to concentration of ${N_2}{O_5}$ initially and at time $t$.

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 $A \rightarrow \text{Product}$,the rate of reaction is $1 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1}$ when $[A] = 0.2 \, \text{M}$. What is the half-life $(t_{1/2})$ of the reaction?

The ratio of $t_{0.75}$ and $t_{0.5}$ for a first-order reaction is: