2 moles of PCl-5 were heated in a closed vessel of 2 litre capacity- At equilibrium- 40- of PCl-5 is dissociated into PCl-3 and Cl-2- The value of equilibrium constant is-0-530-2662-665-3

Question

Toppr Admin · Accepted Answer

PCl5PCl3Cl2Initial200Change-0-80-80-8Equilibrium1-20-80-8Molar concentration0-60-40-4The expression for the equilibrium constant is Kc-PCl3-Cl2-PCl5-0-4-xD7-0-40-6-0-266-

	$P C l_{5}$	$P C l_{3}$	$C l_{2}$
Initial	2	0	0
Change	-0.8	0.8	0.8
Equilibrium	1.2	0.8	0.8
Molar concentration	0.6	0.4	0.4

$2$ moles of $P C l_{5}$ were heated in a closed vessel of 2 litre capacity. At equilibrium, $40 %$ of $P C l_{5}$ is dissociated into $P C l_{3}$ and $C l_{2}$ . The value of equilibrium constant is-
$0.53$
$0.266$
$2.66$
$5.3$

$P C l_{5}$
$P C l_{3}$
$C l_{2}$
Initial
2
0
0
Change
-0.8
0.8
0.8
Equilibrium
1.2
0.8
0.8
Molar concentration
0.6
0.4
0.4

The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.4 \times 0.4}{0.6} = 0.266.$

2 moles of PCl5 were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is-0.530.2662.665.3

PCl5PCl3Cl2Initial200Change-0.80.80.8Equilibrium1.20.80.8Molar concentration0.60.40.4The expression for the equilibrium constant is Kc=[PCl3][Cl2][PCl5]=0.4×0.40.6=0.266.

$2$ moles of $P C l_{5}$ were heated in a closed vessel of 2 litre capacity. At equilibrium, $40 %$ of $P C l_{5}$ is dissociated into $P C l_{3}$ and $C l_{2}$ . The value of equilibrium constant is-
$0.53$
$0.266$
$2.66$
$5.3$

$P C l_{5}$
$P C l_{3}$
$C l_{2}$
Initial
2
0
0
Change
-0.8
0.8
0.8
Equilibrium
1.2
0.8
0.8
Molar concentration
0.6
0.4
0.4

The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.4 \times 0.4}{0.6} = 0.266.$