Chemistry

A sample of gaseous PCl5 was introduced into an evacuated flask so that the pressure of pure PCl5 would be 0.54 atm at 425 K. However, PCl5 decomposes to gaseous PCl3 and Cl2, and the actual pressure in the flask was found to be 0.85 atm. Calculate Kp for the decomposition reaction below at 425 K. Also calculate K at this temperature.

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  1. ..........PCl5 ==> PCl3 + Cl2
    I.........0.54......0......0
    C..........-p.......p......p
    E.......0.54-p

    Ptotal = 0.85 = 0.54-p+p+p
    p = 0.31 atm.
    pPCl5 = 0.23
    pPCl3 = 0.31
    pCl2 = 0.31
    Substitute these values into Kp expression and solve for Kp.
    Kp = Kc*RTdeltan . Substitute Kp and solve for Kc.

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