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Q.

Ammonia dissociates into N2 and H2 such that degree of dissociation α is very less than 1 and equilibrium pressure is P0 then the value of α is[If KP for 2NH3(g)  ⇌  N2(g)  +  3H2(g) is  27 × 10-8  P02]

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a

10-4

b

4 × 10-4

c

0.02

d

can't be calculated

answer is C.

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Detailed Solution

2NH3(g)  ⇌  N2(g)  +3 H2(g)  1-α                α2            3α2 KP=  α  PO21+α 3α   PO21+α31-α1+α PO21-α ≃  1 and 1 + α ≃  1    27 ×10-8   PO2 = 2716PO2  × α4α  = 2 × 10-2
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