For the reaction \( 2N_2O_5→4NO_2 + O_2 \) at a given instant of time, the rate and rate constant are \( 1.02 × 10^{-4}\, mol\, L^{-1}\, s^{-1} \) and \( 3.4 × 10^{-5}\, s^{-1} \), respectively. The concentration of \( N_2O_5 \) at that time will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3.0\, mol\, L^{-1}\)
The unit of the rate constant tells the order of a reaction: a unit of s−1 (time−1 only) means a first order reaction. For the decomposition of \( \ce{N2O5} \) the rate law is therefore
\( \displaystyle \text{Rate} = k\,[\ce{N2O5}] \)
\( \displaystyle [\ce{N2O5}] = \frac{\text{Rate}}{k} = \frac{1.02 \times 10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}}{3.4 \times 10^{-5}\ \mathrm{s^{-1}}} \)
\( \displaystyle [\ce{N2O5}] = 3.0\ \mathrm{mol\,L^{-1}} \)
So the concentration of \( \ce{N2O5} \) at that instant is 3.0 mol L−1.
Shortcut: Options (3) and (4) are only the given rate and rate constant written again, and \( 1.02/0.34 = 3 \) gives the answer at once.