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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?

  • chem -

    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.

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