The following is a first-order reaction:
$N_2O_5 \text{ (solution)} \rightarrow 2 NO_2 \text{ (solution)} + \frac{1}{2} O_2 \text{ (g)}$
In this reaction,$CCl_4$ is the solvent. The rate constant is $k = 5.0 \times 10^{-4} \ s^{-1}$. The initial concentration of $N_2O_5$ is $0.25 \ mol \ L^{-1}$.
$(i)$ What will be the initial rate of reaction?
$(ii)$ Calculate the half-life $(t_{1/2})$.
$(iii)$ How much time is required to complete $75\%$ of the reaction?
$(iv)$ Calculate the concentration of $N_2O_5$ and $NO_2$ after $30 \ min$.

(N/A) $(i)$ Initial rate $= k[N_2O_5]_0 = (5.0 \times 10^{-4} \ s^{-1}) \times (0.25 \ mol \ L^{-1}) = 1.25 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
$(ii)$ Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.0 \times 10^{-4} \ s^{-1}} = 1386 \ s$.
$(iii)$ For $75\%$ completion,$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} = \frac{2.303}{5.0 \times 10^{-4}} \log \frac{100}{25} = 4606 \times 0.6021 \approx 2773 \ s$.
$(iv)$ After $t = 30 \ min = 1800 \ s$,$[N_2O_5] = [N_2O_5]_0 e^{-kt} = 0.25 \times e^{-(5.0 \times 10^{-4} \times 1800)} = 0.25 \times e^{-0.9} \approx 0.25 \times 0.4066 \approx 0.1016 \ mol \ L^{-1}$.
Amount reacted $= 0.25 - 0.1016 = 0.1484 \ mol \ L^{-1}$.
Since $1 \ mol \ N_2O_5$ gives $2 \ mol \ NO_2$,$[NO_2] = 2 \times 0.1484 = 0.2968 \ mol \ L^{-1} \approx 0.30 \ mol \ L^{-1}$.