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?
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) = ?
Sorry, im still confused, any way you could elaborate a little more?
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.
To determine the rate of formation of NO2 and O2, we need to use the stoichiometry of the balanced chemical equation. In this case, the reaction shows that 2 moles of N2O5 react to form 4 moles of NO2 and 1 mole of O2.
Since we know the rate of consumption of N2O5 (0.0080 M/s), we can use this information to calculate the rates of formation of NO2 and O2.
Given:
Rate of consumption of N2O5 = 0.0080 M/s
Using the stoichiometry of the balanced chemical equation, we can set up the following ratio:
(0.0080 M/s of N2O5 consumed) : (x M/s of NO2 formed) : (x M/s of O2 formed)
According to the balanced chemical equation, 2 moles of N2O5 react to form 4 moles of NO2 and 1 mole of O2. Therefore, the molar ratio of N2O5 to NO2 to O2 in terms of rates is 2:4:1.
To find the rate of formation of NO2, we can set up the following ratio:
(0.0080 M/s of N2O5 consumed) : (x M/s of NO2 formed) = 2:4
Simplifying the ratio gives us:
0.0080 M/s : x M/s = 2 : 4
Cross-multiplying and solving for x, we get:
x = (0.0080 M/s) * (4/2) = 0.016 M/s
So, the rate of formation of NO2 at this point in time is 0.016 M/s.
Similarly, to find the rate of formation of O2, we can set up the following ratio:
(0.0080 M/s of N2O5 consumed) : (x M/s of O2 formed) = 2:1
Simplifying the ratio gives us:
0.0080 M/s : x M/s = 2 : 1
Cross-multiplying and solving for x, we get:
x = (0.0080 M/s) * (1/2) = 0.004 M/s
So, the rate of formation of O2 at this point in time is 0.004 M/s.
Therefore, the rate of formation of NO2 is 0.016 M/s and the rate of formation of O2 is 0.004 M/s at this point in time.