PCl5 dissociates into PCl3 and Cl2. In a first vessel at equilibrium with total pressure 3 atm, all three gases are present in equal moles. In a second vessel at the same temperature, an equimolar mixture of PCl5, PCl3 and Cl2 is taken at an initial total pressure of 9 atm. Find the partia

Question

PCl5 dissociates into PCl3 and Cl2. In a first vessel at equilibrium with total pressure 3 atm, all three gases are present in equal moles. In a second vessel at the same temperature, an equimolar mixture of PCl5, PCl3 and Cl2 is taken at an initial total pressure of 9 atm. Find the partial pressure of Cl2 at equilibrium in the second vessel (in atm).

Accepted Answer

2. Vessel 1: equal moles => each partial pressure = 3/3 = 1 atm. Kp = 1*1/1 = 1 atm. Vessel 2: equimolar at 9 atm => each initial partial pressure = 3 atm. Let reaction shift by x (x<0 means reverse direction). P(PCl5)=3-x, P(PCl3)=3+x, P(Cl2)=3+x. Kp=1: (3+x)²/(3-x)=1 => (3+x)²=3-x => x²+7x+6=0 => (x+1)(x+6)=0. x=-1 (x=-6 rejected). P(Cl2)=3+(-1)=2 atm.

Solution

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