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Author: Subject: can some one help me with this problem?
jaisachdeva
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[*] posted on 4-6-2010 at 10:58
can some one help me with this problem?


Please help I have my question in the attachment file.

Attachment: ques.doc (19kB)
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kclo4
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[*] posted on 4-6-2010 at 22:04


Is it not possible to post the contents of this file in a post?



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Magpie
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[*] posted on 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]




The single most important condition for a successful synthesis is good mixing - Nicodem
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