jaisachdeva - 4-6-2010 at 10:58
Please help I have my question in the attachment file.
Attachment: ques.doc (19kB)
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kclo4 - 4-6-2010 at 22:04
Is it not possible to post the contents of this file in a post?
Magpie - 6-6-2010 at 10:32
2N2O5 ---> 4NO2 + O2
Assuming a 1st-order decomposition:
-d[N2O5]/dt = k[N2O5]
integrating:
[N2O5]/[N2O5]o = e^(-kt)
[N2O5]/2 = e^[-3.38*10^(-5) sec^(-1)](10)(60sec) =0.9799
[N2O5] = 1.960
at t = 0 there are 2 moles
at t= 10 min there are 1.960 + 5/2(2-1.960) = 2.06 moles
P10 = (moles10)/(moles0)Po = (2.06/2)500atm = 515 atm
This assumes there is no heat of reaction effect.
This is the final pressure in the constant volume reaction vessel. The partial pressure of just the N2O5 would be:
[(1.96)/2.06](515) = 490 atm
[Edited on 6-6-2010 by Magpie]