Asked by jack
For the reaction:
PCl3(g) + Cl2(g) --> PCl5(g) at 85 degreeC, Kp = 1.19
If one starts with 2.00 atm pressure of PCl3, 1.00 atm pressure of Cl2 and no PCl5, what is the partial pressure of PCl5(g) at equilibrium?
PCl3(g) + Cl2(g) --> PCl5(g) at 85 degreeC, Kp = 1.19
If one starts with 2.00 atm pressure of PCl3, 1.00 atm pressure of Cl2 and no PCl5, what is the partial pressure of PCl5(g) at equilibrium?
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