`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))^(@)`

`K_(1)` and `K_(2)` are equilibrium constants for reaction (i) and (ii

1.	`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))^(@)`
Answer» Correct Answer - A `N_(2)+O_(2) hArr 2NO` …(i) `NO hArr 1/2 N_(2)+1/2O_(2)` …(ii) Equation (ii) is obtained by reversing equation (i) and dividing by `2`. `:. K_(2)=1/((K_(1))^(1//2))` `rArr (K_(2))^(2)=1/K_(1)` `rArr K_(1)=1/((K_(2))^(2))=(1/K_(2))^(2)`

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`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))^(@)`

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