Question

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?

Answer 1

rate of NO2 formation = rate N2O5 x (4 moles NO2/2 moles N2O5) = ?
rate of O2 formation = rate of N2O5 x (1 mol O2/2 moles N2O5) = ?

Answer 2

Sorry, im still confused, any way you could elaborate a little more?

Answer 3

Just like stoichiometry.
If you had 2 mols N2O5 initially it can produce 4 moles NO2. That's basic stoichiometry. Rate works the same way. So if the rate of consumption of N2O5 is 0.008 M/s the rate of formation of NO2 is 4/2 x that or 0.008 x 4/2 = 0.008 x 2 = ?
From the equation, if you use up N2O5 at the rate of 1 mol N2O5 each second then it follows that you will form NO2 at the rate of 2 moles each second.
For the reaction A ==> B, if A is consumed at the rate of 0.1 M/s then B MUST be formed at the rate of 0.1 M/s. You KNOW B can't be formed any faster than A is consumed (where would B get anything over 0.1?) and you KNOW B can't be formed any slower (without having an excess of A going somewhere---if we could do that we could produce tons of A over and over and we're getting something for nothing).
Likewise, for the reaction A ==> 3C, if the rate of consumption of A is 0.1 M/s then the rate of formation C must be 3 times that.
Kinetics, in my opinion, is the most confusing part of chemistry.