in the reaction, N2O4-->2NO2, if the rate of formation of NO2 were 0.010 mol/L*s,, what would be the rate of decomposition for N2O4
Plz show steps thx
Well, let's break it down step by step, just like N2O4 breaks down into 2NO2. But remember, I'll be using a touch of humor to keep things interesting!
1. The balanced equation tells us that for every 1 mole of N2O4 that decomposes, 2 moles of NO2 are formed. So the rate of decomposition of N2O4 will be half of the rate of formation of NO2. It's like the N2O4 is saying, "Hey NO2, you form at 0.010 mol/L*s, but I'm going to do it better and at half the rate!"
2. Therefore, the rate of decomposition of N2O4 is simply half of the rate of formation of NO2. So, 0.010 mol/L*s divided by 2 gives us a rate of decomposition of 0.005 mol/L*s. It's like N2O4 is saying, "I may not be as fast as NO2, but I can still hold my own!"
So, the rate of decomposition of N2O4 would be 0.005 mol/L*s. It's like N2O4 is performing its own little comedy routine, trying to keep up with NO2's speedy formation!
To determine the rate of decomposition of N2O4, we can use the stoichiometry of the balanced equation:
N2O4 → 2NO2
The stoichiometry tells us that for every 1 mole of N2O4 reacting, 2 moles of NO2 are formed. This means that the rate of decomposition of N2O4 is half of the rate of formation of NO2.
Given that the rate of formation of NO2 is 0.010 mol/L*s, the rate of decomposition of N2O4 can be calculated as follows:
Rate of decomposition of N2O4 = Rate of formation of NO2 / 2
Rate of decomposition of N2O4 = 0.010 mol/L*s / 2
Rate of decomposition of N2O4 = 0.005 mol/L*s
Therefore, the rate of decomposition of N2O4 is 0.005 mol/L*s.