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If equilibrium constant for reaction 2AB...

If equilibrium constant for reaction `2AB = A_(2) + B_(2)`. Is 49, then the equilibrium constant for reaction `AB ltimplies 1/2_(A_(2)) + 1/2_(B_(2))`, will be:

A

7

B

20

C

49

D

21

Text Solution

Verified by Experts

The correct Answer is:
A
`2A,B ltimplies A_(2) + B_(2)`
`K_( c) = ([A_(2)][B_(2)])/([AB]^(2))`
For reaction `AB ltimplies 1/2A_(2) + 1/2B_(2)`
`K_(c )^(') ([A_(2)]^(1//2)[B_(2)]^(1//2)]/([AB]), K_(c )^(') = sqrt(K_(c )) = sqrt(49) = 7`
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