A first order reaction N2O5(g) → 2NO2(g) + ½ O2(g) started with 1.2 ×…

The correct answer is Option (1) → 0.0602 min⁻¹

For a first order reaction, the rate constant is found from the integrated rate equation, using the initial concentration [R]₀ and the concentration [R] left after time t:

    \( \displaystyle k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} \)

Only the ratio of the two concentrations is needed, so the unit mol L⁻¹ cancels, and k takes the unit of time⁻¹.

Here [R]₀ = 1.2 × 10⁻² mol L⁻¹, [R] = 0.3 × 10⁻² mol L⁻¹ and t = 23 min.

    \( \displaystyle \frac{[R]_0}{[R]}=\frac{1.2\times10^{-2}}{0.3\times10^{-2}}=4 \)

    \( \displaystyle \log 4=2\log 2=2\times0.3010=0.6020 \)

    \( \displaystyle k=\frac{2.303}{23}\times0.6020 \)

    \( \displaystyle k=\frac{1.386}{23}\approx0.060\ \mathrm{min^{-1}} \)

So the rate constant is 0.0602 min⁻¹.

Shortcut: The concentration falls to 1/4 of its initial value, which takes two half-lives (1 → 1/2 → 1/4). So t_1/2 = 23/2 = 11.5 min, and k = 0.693/11.5 ≈ 0.060 min⁻¹.

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
First Order Reactions – “… The initial concentration of N₂O₅ in the following first order reaction

    \( \ce{N2O5(g) -> 2NO2(g) + 1/2O2(g)} \)

was 1.24 × 10^–2 mol L^–1 at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10^–2 mol L^–1. Calculate the rate constant of the reaction at 318 K.”