.............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
a 1.00-mol sample of phosphorus pentachloride placed in 10.0-L reaction flask and allowed to come to equilibrium at 250 degrees celsius; PCL5(g) = PCL3(g) Cl2(g) if the amount of chlorine in the equilibrium mixture is0.470 mol, calculate (a) the equilibrium concentration of each gas, (b) Kc and (c) the fraction of PCL5 dissociated
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