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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
`N_2 + O_2 " " 2NO " " ....(i)`
`NO " " 1/2N_2 + 1/2 O_2 " " ....(ii)`
Equation (ii) is obtained by reversing equation (i) and dividing by 2
` therefore K_2 = (1)/((K_1)^(1//2) ) rArr (K_2)^2 = (1)/(K_1) rArr K_1 = ( (1)/(K_2) )^2 `
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