Determining initial rate of decamp of N2O5

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Madison Davis 3F
Posts: 44
Joined: Fri Sep 26, 2014 2:02 pm

Determining initial rate of decamp of N2O5

Postby Madison Davis 3F » Thu Feb 19, 2015 10:49 pm

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

Postby Neil DSilva 1L » Thu Feb 19, 2015 11:08 pm

The value plugged into the rate law is the concentration of 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 in the reaction, so we don't need to multiply it by two.


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