chem

PCL5=PCl3+Cl2

a. Initially 0.72 mole PCl5 are placed in 1.50 L flask. At equilibrium, there is 0.16 mole PCl3 in the flask. What is the equilibrium concentration of the PCl5?
b. What is the equilibrium concentration of the Cl2?
c.What is the equilibrium constant for the reaction?

asked by chemdummy79
  1. 0.72/1.50L = 0.48 M = (PCl5)
    0.16/1.50L = 0.107 M = (PCl3)
    .........PCl5 ==> PCl3 + Cl2
    I........0.48......0......0
    C........-x........x.......x
    E.......0.48-x.....x......x

    x = 0.107M
    (PCl3) = 0.107M
    (Cl2) = 0.107M
    (PCl5) = 0.48-0.107 = ?
    Plug these values into the Kc expression and solve for Kc.

    posted by DrBob222

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