Decomposition of N₂O₅ is expressed by the equation N₂O₅ → 2NO₂ + \(\frac{1}{2}\)O₂. If rate of decomposition of N₂O₅ is 1.8 x 10^-3 mol L^-1 min^-1, then what will be the rate of formation of NO₂ during the same time interval?

3.6 x 10^-3 mol L^-1 min^-1

Rate expression = -\(\frac{d[N_2O_5]}{dt}\) = \(\frac{1}{2}\)\(\frac{d[NO_2]}{dt}\) = 2\(\frac{d[O_2]}{dt}\)

\(\frac{d[NO_2]}{dt}\) = -2 x (-1.8 x 10^-3) = 3.6 x 10^-3 mol L^-1 min^-1