NH4+ + NO2- = N2 + 2H2O
A reaction is run, and the liberated N2 gas is collected in a previously evacuated 500 mL container. After the reaction has gone on for 750 seconds, the pressure of N2 in the 500 mL container is 2.77 x 10^-2 atm , and the temperature of the N2 is 250 degrees celsius. Calculate the number of moles of N2 liberated and then calculate the average rate of the reaction
Please show steps! Thanks!
You should be able to ue PV = nRT and solve for n. Essy enough.
Then rate is= delta(N2)/delta Time)
Remember for (N2), that is mols/L. You will have mols and L.
That's the equation I used! I found moles to be
3.22 x 10^-4 mol and concentration to be 6.45 x 10^-4 M.
I found the rate to be 8.60 x 10^-7 M/s, but the answer in my textbook is 7.55 x 10^-7 M/s
So I am not sure where things are going wrong?
Except for some rounding differences I get the same thing. I get 3.23E-4 mols
Was the concentration of 6.45 x 10^-4 M correct? Just so I know I understand the concept?
I obtained that number also.