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K(1) and K(2) are equilibrium constants ...

`K_(1)` and `K_(2)` are equilibrium constants for reaction (i) and (ii)
`N_(2)(g)+O_(2)(g) hArr 2NO(g)` …(i)
`NO(g) hArr 1//2 N_(2)(g)+1//2O_(2)(g)` …(ii)
then,

A

`K_1=[(1)/(K_2)]^2`

B

`K_1=K_2^2`

C

`K_1=(1)/(K_2)`

D

`K_1=(K_2)^(0)`

Text Solution

Verified by Experts

The correct Answer is:
A
Constant reaction (i).
`N_2(g) +O_2(g) hArr2NO(g)` ...(i)
`K_1=([NO]^2)/([N_2][O_2])`
Now, consider reaction (ii),
`NO(g) hArr(1)/(2)N_2(g) +(1)/(2)O_2(g)`.....(ii)
`K_2=([N_2]^(1//2)[O_2]^(1//2))/([NO])`
`(1)/(K_2)=(1)/(([N_2]^(1//2)[O_2]^(1//2))/([NO])`
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