2N_2O_5 (g) longrightarrow 2N_2O_4 (g) + O_2 is a 1st order reaction.…

The correct answer is Option (3) → \(6.9 \times 10^{-3}\, \ s^{-1}\)

For a first order gas-phase reaction followed by the total pressure, the partial pressure of the reactant at time t is found from the stoichiometry, and

    k = (2.303/t) log (p_i/p_A)

Let the pressure of \( \ce{N2O5} \) fall by 2x atm. Two moles of \( \ce{N2O5} \) give two moles of \( \ce{N2O4} \) and one mole of \( \ce{O2} \):

    2N₂O₅(g) → 2N₂O₄(g) + O₂(g)
    t = 0: 0.8 atm, 0, 0
    t: (0.8 − 2x), 2x, x

    p_t = (0.8 − 2x) + 2x + x = 0.8 + x = 1 atm, so x = 0.2 atm
    p(N₂O₅) = 0.8 − 2(0.2) = 0.4 atm
    k = (2.303/100 s) log (0.8/0.4) = 0.02303 × 0.3010 s⁻¹
    = 6.93 × 10⁻³ s⁻¹

Shortcut: The pressure of \( \ce{N2O5} \) halves (0.8 → 0.4 atm) in 100 s, so t_1/2 = 100 s and k = 0.693/t_1/2 = 6.93 × 10⁻³ s⁻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 – “Example 3.6 The following data were obtained during the first order thermal decomposition of N₂O₅ (g) at constant volume:

    \( \ce{2N2O5(g) -> 2N2O4(g) + O2(g)} \)

Calculate the rate constant.”