Posted by Aria on Wednesday, April 25, 2012 at 5:17pm.
Phosphene decomposes to phosphorus and hydrogen in a first order mechanism. 4PH3(g) > P4(g) + 6H2(g)
This reaction's halflife is 35.0 s at 680 degrees C. If you had 520 mmHg of phosphene in an 8.00L flask, how long does it take for the pressure in the flask to rise to standard pressure?
The answer is 48.3 s, I am just not sure how to do it out.

Chemistry  Halflife  DrBob222, Wednesday, April 25, 2012 at 9:20pm
4PH3(g)==> P4(g) + 6H2(g)
For a 1st order rxn k = 0.693/t_{1/2} = 0.693/35 = 0.0198
pPH3 decreases, pP4 increases, pH2 increases; we want the total to be 760 mm.
Therefore 5204p+p+6p = 760 and p = 80 mm
So pPH3 must decrease from 520 to 5204(80) = 200.
Then ln(No/N) = kt and
ln(520/200) = 0.0198t
Solve for t in seconds.

Chemistry  Halflife  JD, Saturday, February 11, 2017 at 5:43pm
I can follow this until we get to the
5204p+p+6p=760 and p=80 mm
The 520 is obviously the starting pressure and the 760 the final, but I'm not sure how you are getting the amounts that you are subtracting since they don't seem to match the equation to me. Could you please explain a bit more where the specific numbers come from?
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