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Chemistry - Half-life

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Phosphene decomposes to phosphorus and hydrogen in a first order mechanism. 4PH3(g) --> P4(g) + 6H2(g)
This reaction's half-life is 35.0 s at 680 degrees C. If you had 520 mmHg of phosphene in an 8.00-L 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 - Half-life - ,

    4PH3(g)==> P4(g) + 6H2(g)
    For a 1st order rxn k = 0.693/t1/2 = 0.693/35 = 0.0198

    pPH3 decreases, pP4 increases, pH2 increases; we want the total to be 760 mm.
    Therefore 520-4p+p+6p = 760 and p = 80 mm
    So pPH3 must decrease from 520 to 520-4(80) = 200.
    Then ln(No/N) = kt and
    ln(520/200) = 0.0198t
    Solve for t in seconds.

  • Chemistry - Half-life - ,

    I can follow this until we get to the

    520-4p+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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