(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)
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
1 answer
1 answer