Question

For the reaction $N _{2}( g )+ O _{2}( g ) \rightleftharpoons 2 NO ( g )$ the equilibrium constant is $K _{1}$. The equilibrium constant is $K _{2}$ for the reaction $2 NO ( g )+ O _{2}( g ) \rightleftharpoons 2 NO _{2}( g ) .$ What is $K$ for the reaction $NO _{2}( g ) \rightleftharpoons 1 / 2 N _{2}( g )+ O _{2}( g ) ?$

• A. $1 /\left( K _{1}\, K _{2}\right)$
• B. $1 /\left(2\, K _{1}\, K _{2}\right)$
• C. $1 /\left(4 \,K _{1}\, K _{2}\right)$
• D. $\left[1 / K _{1}\, K _{2}\right]^{1 / 2}$

Answer 1

Answer: (D)

$\left[1 / K _{1}\, K _{2}\right]^{1 / 2}$

Answer 2

$N _{2}( g )+ O _{2}( g ) \rightleftharpoons 2 NO ( g ) ; K _{1}$
$2 NO ( g )+ O _{2}( g ) \rightleftharpoons 2 NO _{2}( g ) ; K _{2}$

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$N _{2}( g )+2 O _{2}( g ) \rightleftharpoons 2 NO _{2}( g ) ; K = K _{1} \times K _{2}$
$\therefore For NO _{2}( g ) \rightleftharpoons \frac{1}{2} N _{2}( g )+ O _{2}( g )$
$K'=\left[\frac{1}{ K _{1} \cdot K _{2}}\right]^{1 / 2}$