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Question

# In a gaseous reaction at equilibrium. ‘n’ mole of reactant ‘A’ decomposes to give 1 mole each of C and D. It has been found that degree of dissociation of A at equilibrium is independent of total pressure. Value of ‘n’ is.

A
2
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B
3
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C
0.0
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D
None of these
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Solution

## The correct option is A 2nA⇌C+D Initially only A will be present in the reaction mixture and its concentration is given by "c". At equilibrium, the concentration of A will be (c−cα)n the concenrations of C and D will be cα Kc = α2 c2(c−cα)n Since α << 1 Kc = c2−nα2 Kc is a constant α is a constant Therefore α2 = Kcc2−n For α to be constant, c2−n should be equal to 1 c2−n = 0 2-n = 0 n= 2

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