CHEM

A 0.229 mol sample of PCL5 is injected into an empty 3.20 L reaction vessel held at 250 degrees celsius. Calculate the concentration of PCl5 and PCl3 at equilibrium.

Kc=1.80 M

PCl5 -------> PCl3 + Cl2

asked by Mer
  1. (PCl5) = 0.229/3.20 = approx 0.072 but you should start over and calculate more accurately.
    ......PCl5 -------> PCl3 + Cl2
    I.....0.072..........0......0
    C.......-x...........x......x
    E....0.072-x.........x......x

    Substitute the E line into Kc expression and solve for x = (PCl3) = (Cl2) and 0.072-x = (PCl5)

    posted by DrBob222

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