Question

For the first-order reaction \(N_2O_5(g) \rightarrow 2NO_2(g) + O_2(g)\), the initial concentration of \(N_2O_5\) at 318 K was \(1.24 \times 10^{-2}\) mol L^-1, which decreased to \(0.20 \times 10^{-2}\) mol L^-1 after 60 minutes. Calculate the rate constant at 318 K.

Hint:

Use the first-order law \(k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\) with the ratio 6.2 over t = 60 min.

Answer 1

Step 1: A first-order reaction also obeys \(\ln \dfrac{[A]_0}{[A]} = kt\), so \(k = \dfrac{1}{t}\ln\dfrac{[A]_0}{[A]}\).
Step 2: Concentration ratio: \(\dfrac{[A]_0}{[A]} = \dfrac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} = 6.2\).
Step 3: Natural log: \(\ln 6.2 = 1.8245\).
Step 4: Rate constant: \(k = \dfrac{1.8245}{60} = 0.0304\) min^-1.
Step 5: Converting to SI units, divide by 60: \(k = 5.07 \times 10^{-4}\) s^-1.
\[\boxed{k \approx 0.0304\ \text{min}^{-1}}\]