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General Chemistry Pressure and Rate laws

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data
    The kinetics of the decomposition of phosphine at 950 K
    4PH3 (g) -> P4 (g) + 6H2 (g)
    was studied by injecting PH3(g) into a reaction vessel and measuring the total pressure at constant volume.

    P total (Torr) Time (s)
    100 0
    150 40
    167 80
    172 120

    What is the rate constant of this reaction?

    2. Relevant equations

    PV = nRT
    rate = k[PH3] ^n

    3. The attempt at a solution

    Do I need to find the reaction order with respect to PH3 before finding the rate constant? I am lost.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 25, 2007 #2


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    Yes, you need to find the reaction order. But to do that, you need to relate the change in total pressure to the change in the number of mol/L of PH3. If at some time, t, there are some n moles of PH3 consumed, then what is the net change in the total number of moles of gas in the container?

    Also, you've written the differential form of the n'th order rate law. Can you collect the terms, integrate it and write the expression for the concentration of PH3 as a function of time (and the initial concentration)?
  4. Feb 25, 2007 #3
    Thanks for your help. This is what I have done so far, but it is still wrong.

    ln[(PPH3)t / (PPH3)o] = -kt

    Ptotal = PPH3 + PP4 + PH2

    PPH3 = Po - PP4
    PH2 = 6PP4

    Ptotal = (Po - PP4) + 6PP4 + PP4
    Ptotal= Po + 6PP4
    PP4 = (Ptotal - Po) / 6

    PPH3 = Po -PP4 = Po - [(Ptotal-Po) / 6]

    Po = 100 torr

    Pt= 100 - [(150-100) / 6] = 91.666 torr

    ln0.91666 = -k * 40s
    k = 0.0022

    The correct answer is k=0.027 s-1
  5. Feb 25, 2007 #4


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    This is only true for n=1, and is not the correct equation for n != 1. You need to figure out the order of the reaction rate; you can't assume it is 1.

    For n not 1, you have:

    [tex]dA/dt = -kA^n \implies A^{-n}dA = -kdt \implies A^{1-n} - A_0^{1-n} = -k(t - t_0)[/tex]

    Not sure I follow what you're doing here. For every 4 moles of PH3 consumed there are 7 moles of products produced, resulting in a net change of +3 moles (for every 4 moles of PH3 consumed). So the decrease in partial pressure of PH3 is 4/3 of the increase in total pressure over any interval of time.

    So, for instance, during the first 40s, since the p(tot) increases by 50 torr, p(PH3) must decrease by 50*4/3 torr. Using this method, you can translate the table for p(tot) into a similar table for p(PH3).

    Write down the values in this new table and plug them into the general equation for the n'th order rate law (above) to find k and n.

    Note that the above rate equation only applies if n is not 1. It is actually prudent to first check if n=1 works.

    PS: I just checked. It is first order; so you can indeed use the rate equation you used in post #3, only you need to use it on the new table of numbers.
    Last edited: Feb 25, 2007
  6. Feb 25, 2007 #5
    I think this problem can be done by substituting the pressure and time variables in the concentration-rate constant equations for various ordered reaction The equation are founsd as :

    r = -d[A]/dt
    r = k[A]^n n - order of reaction.


    kdt = d[A]/[A]^n. Integrate (with limits c0 - c initial and final concentration, and 0 - t) to find equation for 1st, 2nd, 3rd... order reactions.

    Substitute the given data in equations 1 by 1 till constant value of k for all pair of pressure values is obtained. This constant value gives the real k and n the order.

    Im not surwe of this method but i have seen similar problems that can be done this way.
  7. Feb 26, 2007 #6


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    This last bit is incorrect.

    The given data does not refer to the concentration (or partial pressure) of one species - it is the total pressure in the vessel.
  8. Feb 27, 2007 #7
    yeah u r right the data have to be converted into partial pressure of PH3 that would make my suggestion same as post #4. Sorry for the repetition.
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