For the reaction \( \ce{2N2O5(g) -> 4NO2(g) + O2(g)} \), the initial concentration of N2O5 is 1.24 mol L−1. Over the first 20 minutes, O2 is formed at an average rate of 1.0 × 10−4 mol L−1 s−1. What is the concentration of N2O5 at the end of 20 minutes?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.00 mol L−1
The coefficients are not equal, so each rate is divided by its stoichiometric coefficient to give one rate of reaction:
\( \displaystyle \text{Rate} = -\frac{1}{2}\frac{\Delta[\ce{N2O5}]}{\Delta t} = \frac{1}{4}\frac{\Delta[\ce{NO2}]}{\Delta t} = \frac{\Delta[\ce{O2}]}{\Delta t} \)
So N2O5 is used up twice as fast as O2 is formed.
\( \displaystyle -\frac{\Delta[\ce{N2O5}]}{\Delta t} = 2 \times 1.0 \times 10^{-4} = 2.0 \times 10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}} \)
\( \displaystyle \Delta t = 20\ \mathrm{min} = 20 \times 60\ \mathrm{s} = 1200\ \mathrm{s} \)
\( \displaystyle \text{Decrease in } [\ce{N2O5}] = 2.0 \times 10^{-4} \times 1200 = 0.24\ \mathrm{mol\,L^{-1}} \)
\( \displaystyle [\ce{N2O5}] = 1.24 - 0.24 = 1.00\ \mathrm{mol\,L^{-1}} \)
NCERT (2026-27 edition):
3.1 Rate of a Chemical Reaction – “For expressing the rate of such a reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients.”