For the gas phase decomposotion of dinitrogen pentoxide at 335 k the average rate of disapperance of N2N2O5 over the time period from t= 0 s to t= 154 s is found to be 6.0 × 10-4 M s-1.
2N2O5 -->4 NO2 + O2
The average rate of formation of NO2 over the same time period is how many moles per liter per second
From the balanced equation, we can see that for every 2 moles of N2O5 that decompose, 4 moles of NO2 are formed. Therefore, the rate of formation of NO2 (in units of M/s) is twice the rate of disappearance of N2O5:
rate of formation of NO2 = 2(6.0 x 10^-4 M/s) = 1.2 x 10^-3 M/s
However, the question asks for the rate of formation of NO2 in moles per liter per second. We need to know the concentration of the reactants and products to convert between these units. Unfortunately, this information is not given in the question. Without the concentrations, we cannot provide an answer for this part of the question.