Posted by MIndy on Wednesday, October 19, 2011 at 3:06pm.
.............PCl5 ==> PCl3 + Cl2
initial......1.0 mol....0.....0
change.......-x.........x.....x
equil......................0.470
Therefore, at equilibrium PCl3 must be the same or 0.470 mol and PCl5 must be 1.0-0.470.
Cl2 = 0.470/10L = ?
PCl3 = 0.470/10L = ?
PCl5 = (1-0.470)/10L = ?
Kc = (PCl3)(Cl2)/(PCl5).
Substitute and solve for Kc.
fraction = moles Cl2/initial moles PCl5
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