Read the passage carefully and answer the Questions.
The decomposition of \( N_2O_5 (g) \) is given as
\( N_2O_5 (g) → 2NO_2(g) +\frac{1}{2}O_2(g) \)
The above reaction is found to be of first order.
The initial concentration of \( N_2O_5 \) was \( 1.24 × 10^{-2} M \) at 318 K. The concentration of \( N_2O_5 \) after 60 minutes was \( 0.20 × 10^{-2} M \). The rate constant of the reaction at 318 K is ......... (\( \log 6.2= 0.792 \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.0304\, min^{-1}\)
For a first order reaction the integrated rate equation gives the rate constant from the initial concentration [R]0 and the concentration [R] left after time t:
\( \displaystyle k = \frac{2.303}{t}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} \)
Here \( [\mathrm{R}]_0 = 1.24\times10^{-2}\,\mathrm{M} \), \( [\mathrm{R}] = 0.20\times10^{-2}\,\mathrm{M} \) and t = 60 min:
\( \displaystyle k = \frac{2.303}{60\,\mathrm{min}}\log\frac{1.24\times10^{-2}}{0.20\times10^{-2}} = \frac{2.303}{60\,\mathrm{min}}\log 6.2 \)
\( \displaystyle k = \frac{2.303\times0.792}{60}\,\mathrm{min^{-1}} = \frac{1.824}{60}\,\mathrm{min^{-1}} \)
\( \displaystyle k = 0.0304\,\mathrm{min^{-1}} \)
So the rate constant is 0.0304 min−1.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
First Order Reactions – “Example 3.5: 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.”