For the reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Later during the reaction, the rate of consumption of N2O5 was determined to be 0.0080 M/s. What was the rate of formation of NO2 and of O2 at this point in time?
To determine the rate of formation of NO2 and O2 at a given time during the reaction, we need to use the stoichiometry of the balanced chemical equation.
The stoichiometric coefficient tells us the ratio of moles of reactants and products involved in the reaction. In this case, the balanced equation shows that 2 moles of N2O5 react to form 4 moles of NO2 and 1 mole of O2.
Given that the rate of consumption of N2O5 is 0.0080 M/s, we can use this information to find the rate of formation of NO2 and O2.
Rate of formation of NO2:
From the balanced equation, we know that for every 2 moles of N2O5 consumed, 4 moles of NO2 are formed. Hence, the rate of formation of NO2 is (0.0080 M/s) × (4/2) = 0.0160 M/s.
Rate of formation of O2:
From the balanced equation, we know that for every 2 moles of N2O5 consumed, 1 mole of O2 is formed. Hence, the rate of formation of O2 is (0.0080 M/s) × (1/2) = 0.0040 M/s.
Therefore, the rate of formation of NO2 at this point in time is 0.0160 M/s and the rate of formation of O2 is 0.0040 M/s.