1. Tardigrade
  2. Question
  3. Chemistry
  4. (1) N 2( g )+3 H 2( g ) leftharpoons 2 NH 3( g ), K 1 (2) N 2( g )+ O 2( g ) leftharpoons 2 NHO ( g ), K 2 (3) H 2( g )+(1/2) O 2( g ) leftharpoons H 2 O ( g ), K 3 The equation for the equilibrium constant of the reaction 2 NH 3( g )+(5/2) O 2( g ) leftharpoons 2 NO ( g ), 3 H 2 O ( g ),(K4) in terms of K1 and K3 is:

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Q. (1) $N _2( g )+3 H _2( g ) \rightleftharpoons 2 NH _3( g ), K _1$
(2) $N _2( g )+ O _2( g ) \rightleftharpoons 2 NHO ( g ), K _2$
(3) $H _2( g )+\frac{1}{2} O _2( g ) \rightleftharpoons H _2 O ( g ), K _3$
The equation for the equilibrium constant of the reaction
$2 NH _3( g )+\frac{5}{2} O _2( g ) \rightleftharpoons 2 NO ( g ), 3 H _2 O ( g ),\left(K_4\right)$ in terms of $K_1$ and $K_3$ is:

JEE MainJEE Main 2013Equilibrium

Solution:

To calculate the value of $K_4$ in the given equation we should apply:
$eqn. (2) + eqn.(3) x 3 - eqn. (1)$
hence $K_{4}=\frac{K_{2}K^{3}_{3}}{K_{1}}$