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

Question

2 moles of PCl5 were heated in a closed vessel of 2 L capacity. At equilibrium, 40 % of PCl5 is dissociated into PCl3 and Cl2 . The value of equilibrium constant is (A)  0.266  (B)  0.366  (C)  2.66  (D)  3.66

Tardigrade Admin · Accepted Answer

PCl5 PCl3 Cl2 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 equillibrium constant is Kc = ([PCl3][Cl2]/[PCl2]) = (0.4 × 0.4/0.6) = 0.266.

Q. $2$ moles of $PC l_{5}$ were heated in a closed vessel of $2 L$ capacity. At equilibrium, $40%$ of $PC l_{5}$ is dissociated into $PC l_{3}$ and $C l_{2}$ . The value of equilibrium constant is

$0.266$

$0.366$

$2.66$

$3.66$

$PC l_{5}$ $PC l_{3}$ $C l_{2}$

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 equillibrium constant is
$K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{2} ]}$
$= \frac{0.4 \times 0.4}{0.6} = 0.266$ .

	$PC l_{5}$	$PC l_{3}$	$C l_{2}$
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

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

0.266

0.366

2.66

3.66