Read the passage carefully and answer the Questions.

The decomposition of $N_2O_5 (g)$ is given as

$N_2Os (g) → 2NO_2(g) +\frac{1}{2}O_2(g)$

The above reaction is found to be of first order.

The initial concentration of $N_2O_5$ was $1.24 × 10^{-2} M$ at 318 K. The concentration of $N_2O_5$ after 60 minutes was $0.20 × 10^{-2} M$. The rate constant of the reaction at 318 K. ......... ($\log 6.2= 0.792$)

$0.0304\, min^{-1}$

The correct answer is Option (1) → $0.0304\, min^{-1}$

For a first-order reaction, the rate constant is given by:

$k=\frac{2.303}{t}\log\frac{[A]_0}{[A]}$

Given:

• $[A]_0 = 1.24\times10^{-2}\,\text{M}$
• $[A] = 0.20\times10^{-2}\,\text{M}$
• $t = 60$ min
• [A]₀ / [A] = 1.24 / 0.20 = 6.2
• $\log 6.2 = 0.792$

$k=\frac{2.303}{60}\times 0.792$

$ k=\frac{1.824}{60}=0.0304\ \text{min}^{-1}$