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At $127 ∘ C$ , for given reaction taking place in closed rigid vessel of volume VL
$2 SO 3 (g) ⇌ 2 SO 2 (g) + O 2 (g); K c = 2 × 10 − 5 mol / L$
If degree of dissociation is $α$ and total pressure is P at equilibrium, then the correct option is:

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α ∝ (V) 13

α ∝ P

If pressure is doubled then

K p

will be halved of initial at constant temperature

α ∝ P 3

answer is A.

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Detailed Solution

2 SO 3 (g) ⇌ 2 SO 2 (g) + O 2 (g)

t = 0 mola

t eq mol a (1 − α) a α a α 2

Since

K C

is very small it means that

α

is very small wrt 1

K C = a α V 2 a α 2 V a V 2 = a α 3 2 V α 3 = 2 K C V a α = 2 K C V a 1 / 3

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