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Q = P PCL5 = 0.10 = 0.5 P PCL3 * P Cl2 0.30 * 0.60 Q (0.58) > K (0.0870) therefore reverse PCl 3(g) + Cl 2(g)  PCl 5(g).

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  1. Q = PPCL5 = 0.10 = 0.5 PPCL3 * PCl2 0.30 * 0.60Q (0.58) > K (0.0870) therefore reverse PCl3(g) + Cl2(g) PCl5(g)

  2. K = PPCL5 = (0.10-X) PPCL3 * PCl2 (0.30+X) (0.60+X)PCl3(g) + Cl2(g) PCl5(g)

  3. 0.0870= (0.10-X) (0.30+X) (0.60+X) (0.10-X) (0.18 + 0.90X + X2)

  4. 0.0870X2 + 0.0783X + 0.0157 = 0.10-X 0.0870X2 + 1.0783X - 0.0843 = 0 X= -1.0783 + (1.0783)2 -4(0.0870)(-0.0843) 2(0.0870)

  5. X= -1.0783 + (1.0783)2 -4(0.0870)(-0.0843) 2(0.0870)X= -1.0783 + 1.0918 = 0.078 0.174PCL3 = 0.30 + 0.078 = 0.378 atmCl2 = 0.60 + 0.078 = 0.678 atmPCl3 = 0.10 – 0.078 = 0.022 atm

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