An equimolar mixture of PCl 3 ( g ) and Cl 2 ( g ) were placed in an evacuated closed vessel of fixed volume at 25 0 ∘ C and the initial total pressure of the gaseous mixture was 'R' atm

Question

An equimolar mixture of PCl 3 ​ ( g ) and Cl 2 ​ ( g ) were placed in an evacuated closed vessel of fixed volume at 25 0 ∘ C and the initial total pressure of the gaseous mixture was 'R' atm. After some time, the following equilibrium was established in which 75% of PCl 3 ​ ( g ) were converted into PCl 5 ​ ( g ) PCl 3 ​ ( g ) + C l 2 ​ ( g ) ⇌ PCl 5 ​ ( g ) , K P ​ = 3 1 ​ a t m − 1 Choose the correct option(s) :

TestHub · Accepted Answer

0 a mole a mole PCl 5 ​ ⇌ PCl 3 ​ + Cl 2 ​ 3a/4 a/4 a/4 K P ​ = P PCl 5 ​ ​ P PCl 3 ​ ​ × P Cl 2 ​ ​ ​ = 4 3 ​ a 4 a ​ × 4 a ​ ​ × ( 4 5 ​ a P T ​ ​ ) Given K P ​ = 3 , we have: 3 = 15 P T ​ ​ ⇒ P T ​ = 45 atm Since volume is constant, P ∝ m o l e Therefore, 45 ∝ 4 5 ​ a , a ∝ 36 atm .

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