Question

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?

​

Answer 1

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