Read the passage carefully and answer the questions based on the passage:
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. In the rate equation Rate = \( k [A]^x [B]^y \)
\( x \) and \( y \) indicate how sensitive the rate is to the change in concentration of A and B, respectively. Sum of these exponents, i.e., \( x + y \) gives the overall order of a reaction where \( x \) and \( y \) represent the order with respect to the reactants A and B, respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. For a first order reaction, the concentration of the reactant varies as \( [R] = [R]_0e^{-kt} \)
The initial concentration of N2O5 in the following first order reaction \( N_2O_5(g) → 2 NO_2(g) + 1/2O_2 (g) \) was \( 1.2 × 10^{-2}\, mol\, L^{-1} \) at 298 K. The concentration of \( N_2O_5 \) after 30 minutes was \( 0.60 × 10^{-2}\, mol\, L^{-1} \). Calculate the rate constant of the reaction at 298 K.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.023\, min^{-1}\)
For a first order reaction the integrated rate equation is
\( \displaystyle k = \frac{2.303}{t}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} \)
The data: \( [\mathrm{R}]_0 = 1.2 \times 10^{-2}\ \mathrm{mol\,L^{-1}} \), \( [\mathrm{R}] = 0.60 \times 10^{-2}\ \mathrm{mol\,L^{-1}} \), \( t = 30\ \mathrm{min} \).
\( \displaystyle k = \frac{2.303}{30\ \mathrm{min}}\log\frac{1.2 \times 10^{-2}}{0.60 \times 10^{-2}} \)
\( \displaystyle k = \frac{2.303}{30\ \mathrm{min}} \times \log 2 = \frac{2.303 \times 0.3010}{30\ \mathrm{min}} \)
\( \displaystyle k = 0.0231\ \mathrm{min^{-1}} \)
So the rate constant is about 0.023 min−1.
Shortcut: The concentration falls from 1.2 to 0.60, that is to half, so 30 min is the half-life. For a first order reaction \( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30\ \mathrm{min}} = 0.0231\ \mathrm{min^{-1}} \).
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 N2O5(g) → 2 NO2(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.”