The vapour density of partially dissociated iodine vapour at a temperature was found to be 80. What is its degree of dissociation?
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Vapour density of any substance is defined as fraction of molecules dissoaciating at a given time.
Given:
Molecular weight of Idoine=254
Let "α " be the degree of dissociation under equilibrium.
equation:
I₂₍g₎ ⇄ 2I₂₍g₎
1-α 2α
VApour density of undissociated vapour
d₀ =254/2 =127
Number of moles at equilibrium=1-α +2α⇒1+α
Vapour density of undissociated I₂
______________________________ = d/d₀ =1/1+α
vapour density of equilibrium molecule
degree of dissociation =α=d₀-d/d =127=80/80 =0.587
Given:
Molecular weight of Idoine=254
Let "α " be the degree of dissociation under equilibrium.
equation:
I₂₍g₎ ⇄ 2I₂₍g₎
1-α 2α
VApour density of undissociated vapour
d₀ =254/2 =127
Number of moles at equilibrium=1-α +2α⇒1+α
Vapour density of undissociated I₂
______________________________ = d/d₀ =1/1+α
vapour density of equilibrium molecule
degree of dissociation =α=d₀-d/d =127=80/80 =0.587
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