First Order Decomposition of Dinitrogen Pentoxide Gas Characteristics

To determine the correct statements for the given first-order reaction:
$2\text{N}_2\text{O}_5(\text{g}) \rightarrow 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$

Let us analyze each option individually:

1. Option (A): Variation of concentration with time
   For a first-order reaction, the integrated rate law is given by:
   $[\text{N}_2\text{O}_5]_t = [\text{N}_2\text{O}_5]_0 e^{-k t}$
   where:

   • $[\text{N}_2\text{O}_5]_0$ is the initial concentration of the reactant,
   • $[\text{N}_2\text{O}_5]_t$ is the concentration at time $t$,
   • $k$ is the rate constant of the reaction.

   This equation clearly shows that the concentration of the reactant decreases exponentially with time. Therefore, Option (A) is correct.

2. Option (B) and Option (C): Half-life characteristics
   The half-life ($t_{1/2}$) of a first-order reaction is expressed as:
   $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$

   • Dependence on initial concentration: The expression for $t_{1/2}$ is independent of the initial concentration $[\text{N}_2\text{O}_5]_0$. Therefore, Option (C) is incorrect.
   • Dependence on temperature: According to the Arrhenius equation, the rate constant $k$ increases with an increase in temperature:
     $k = A e^{-E_a / RT}$
     Since $t_{1/2} \propto \frac{1}{k}$, an increase in temperature increases $k$, which in turn causes the half-life ($t_{1/2}$) to decrease. Therefore, Option (B) is correct.

3. Option (D): Percentage completion after eight half-lives
   After $n$ half-lives, the amount of reactant remaining is given by:
   $[\text{N}_2\text{O}_5]_t = [\text{N}_2\text{O}_5]_0 \left(\frac{1}{2}\right)^n$
   For $n = 8$:
   $[\text{N}_2\text{O}_5]_t = [\text{N}_2\text{O}_5]_0 \left(\frac{1}{2}\right)^8 = \frac{[\text{N}_2\text{O}_5]_0}{256}$
   The fraction of the reaction completed is:
   $\text{Fraction completed} = 1 - \frac{1}{256} = \frac{255}{256}$
   Converting this into percentage:
   $\text{Percentage completion} = \left(\frac{255}{256}\right) \times 100\% \approx 99.609\% \approx 99.6\%$
   Therefore, Option (D) is correct.

Conclusion:
The correct statements are A, B, and D.