Question

suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 m/s. a) at what rate is n2o5 being form?

Answer 1

At a particular moment during the reaction, with molecular oxygen reacting at the rate of 0.024 M/s, the rate of N2O5 being formed is 0.048 M/s. This is due to the stoichiometry of the reaction, which is 4NO2 (g) + O2 (g) → 2N2O5 (g). This means that for every mole of molecular oxygen reacting, two moles of N2O5 will be formed. Therefore, the rate of N2O5 being formed is double the rate of molecular oxygen being consumed (0.024 M/s x 2 = 0.048 M/s).