## Determining initial rate of decamp of N2O5

Posts: 44
Joined: Fri Sep 26, 2014 2:02 pm

### Determining initial rate of decamp of N2O5

14.11

2N2O5--> 4NO2 + O2

3.45 g N2O5 , 0.750 L

When solving for the concentration of N2O5, I understand that you multiply the molar mass of N2O5 and then divide by the liters given, but why wouldn't you multiply by 2 from the balanced EQ?

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Determining initial rate of decamp of N2O5

The value plugged into the rate law is the concentration of $N_{2}O_{5}$ and we get that by converting the grams to moles and dividing by the total volume. The grams given already tells us the quantity of $N_{2}O_{5}$ in the reaction, so we don't need to multiply it by two.