4.0 moles of argon and 5.0 moles of PCI 5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K p for the reaction is [Given: R =0.082 L atm K -1 mol -1 ]

Question

4.0 moles of argon and 5.0 moles of PCI 5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K p for the reaction is [Given: R =0.082 L atm K -1 mol -1 ] (A) 2.25 (B) 6.24 (C) 12.13 (D) 15.24

Tardigrade Admin · Accepted Answer

PCl 5=5 mole Ar =4 mole P text Total =(9 × 0.82 × 610/100)=4.5 atm P PCl 5=(5 × 4.5/9)=2.5 ; P Ar =(4 × 4.5/9)=2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/8651d0ccf1ec7910b86bcbf6cf43ff34-.png" /> P text total =2.5- P + P + P + P Ar =6 P =1.5 K p =(1.5 × 1.5/1)=2.25

$2.25$

$6.24$

$12.13$

$15.24$

$PC l_{5} = 5$ mole
$A r = 4$ mole
$P_{Total} = \frac{9 \times 0.82 \times 610}{100} = 4.5 a t m$
$P_{PC l_{5}} = \frac{5 \times 4.5}{9} = 2.5; P_{A r} = \frac{4 \times 4.5}{9} = 2$

$P_{total} = 2.5 - P + P + P + P_{A r} = 6$
$P = 1.5$
$K_{p} = \frac{1.5 \times 1.5}{1} = 2.25$

Q. 4.0 moles of argon and 5.0 moles of PCI5​ are introduced into an evacuated flask of 100 litre capacity at 610K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0atm. The Kp​ for the reaction is [Given : R=0.082L atm K−1mol−1 ]

2.25

6.24

12.13

15.24

PCl5​=5 mole Ar=4 mole PTotal ​=1009×0.82×610​=4.5atm PPCl5​​=95×4.5​=2.5;PAr​=94×4.5​=2 Ptotal ​=2.5−P+P+P+PAr​=6 P=1.5 Kp​=11.5×1.5​=2.25

$2.25$

$6.24$

$12.13$

$15.24$

$PC l_{5} = 5$ mole
$A r = 4$ mole
$P_{Total} = \frac{9 \times 0.82 \times 610}{100} = 4.5 a t m$
$P_{PC l_{5}} = \frac{5 \times 4.5}{9} = 2.5; P_{A r} = \frac{4 \times 4.5}{9} = 2$

$P_{total} = 2.5 - P + P + P + P_{A r} = 6$
$P = 1.5$
$K_{p} = \frac{1.5 \times 1.5}{1} = 2.25$