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: For a first-order reaction the integrated rate law is \(k = \dfrac{2.303}{t} \log \dfrac{[A]_0}{[A]}\), where [A]₀ is the initial concentration and [A] the concentration after time t.
Step 2: Data: [A]₀ = \(1.24 \times 10^{-2}\) mol L^-1, [A] = \(0.20 \times 10^{-2}\) mol L^-1, t = 60 min.
Step 3: Ratio \(\dfrac{[A]_0}{[A]} = \dfrac{1.24}{0.20} = 6.2\).
Step 4: Substitute: \(k = \dfrac{2.303}{60} \log(6.2)\). Since \(\log 6.2 = 0.7924\), \(k = 0.03838 \times 0.7924 = 0.0304\) min^-1.
\[\boxed{k \approx 0.0304 \ \text{min}^{-1} = 5.07 \times 10^{-4}\ \text{s}^{-1}}\]