PCl 5 dissociates as PCl 5( g )leftharpoons PCl 3( g )+ Cl 2( g ) 5 moles of PCl 5 are placed in a 200 litre vessel which contains 2 moles of N 2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K p for the dissociation of PCl 5 is × 10-3. (nearest integer) (Given: R =0.082 L atm K -1 mol -1: Assume ideal gas behaviour)

Question

PCl 5 dissociates as PCl 5( g )leftharpoons PCl 3( g )+ Cl 2( g ) 5 moles of PCl 5 are placed in a 200 litre vessel which contains 2 moles of N 2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K p for the dissociation of PCl 5 is × 10-3. (nearest integer) (Given: R =0.082 L atm K -1 mol -1: Assume ideal gas behaviour) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given: 2 mole of N2 gas was present as inert gas. Equilibrium pressure = 2.46 atm <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4456d973939e1ff1818dc10ee1a5ef2c-.png" /> from ideal gas equation PV = nRT 2.46 × 200=(5- x + x + x +2) × 0.082 × 600 x =3 K P =( n PCl 3 × n Cl 2/ n PCl 5) ×[( P text total / n text total )] (3 × 3/2) × (2.46/10)=1.107=1107 × 10-3

Given : 2 mole of N2​ gas was present as inert gas. Equilibrium pressure =2.46atm from ideal gas equation PV=nRT 2.46×200=(5−x+x+x+2)×0.082×600 x=3 KP​=nPCl5​​nPCl3​​×nCl2​​​×[ntotal ​Ptotal ​​] 23×3​×102.46​=1.107=1107×10−3