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
7 answers
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.
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
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.
6 answers