HI was heated in a sealed tube at 400 degrees celsius till the equilibrium was reached .HI was found to be 22% decomposed. the equilibrium constant for dissociation is?

2HI(g)⇌H
2

(g)+I
2

(g)

Suppose, initially 2 moles of HI are present. 22% or 2×
100
22

=0.44 moles will dissociate. 2−0.44=1.56 moles of HI will be present.

0.44 moles of HI will form 0.44×
2
1

=0.22 moles of H
2

and 0.22 moles of I
2

.

K
C

=
[HI]
2

[H
2

][I
2

]


K
C

=
(1.56/V)
2

(0.22/V)×(0.22/V)


K
C

= 0.0199

A) the enthralling of formation for pure element in its standard state is always positivehope this : )

Bis the answerhope it !