Hi was heated in a sealed tube at 440

Asked by admin @ 16/08/2021 in Chemistry viewed by 257 People

Two moles of HI were heated in a sealed tube at 440 degree C till the equilibrium was reached. HI was found to be 22% decomposed. Find equilibrium constant for dissociation.

Answered by admin @ 16/08/2021

Hey dear,

● Answer -
Kc = 0.02

● Explaination -
Dissociation of HI occurs as follows -
Reaction :- 2HI ---> H2 + I2
Initial :- 2 0 0
Equilibrium :- 2-2x x x

Given that, HI is 22% reduced,
(2-2x)/78% = (2x)/22%
44 - 44x = 156x
44 = 200x
x = 0.22

Dissociation constant is calculated by -
Kc = x.x / (2-2x)^2
Kc = 0.22 × 0.22 / (2-0.44)^2
Kc = 0.02

Therefore, equilibrium constant for dissociation is 0.02 .

Hope this helps you ...

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