A first order reaction N2O5(g) → 2NO2(g) + ½ O2(g) started with 1.2 × 10−2 mol L−1 of N2O5 at 313 K. The concentration of N2O5 is found to be 0.3 × 10−2 mol L−1 after 23 minutes. The rate constant of the reaction at this temperature is
(Given: log 2 = 0.3010)
Answer & explanation
Correct answer: option 1
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.
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.”