(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:

Question

(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: (A) (K1.K2/K3) (B) (K1.K23/K2) (C) K1K2K3 (D) (K2.K33/K1)

Tardigrade Admin · Accepted Answer

To calculate the value of K4 in the given equation we should apply: eqn. (2) + eqn.(3) x 3 - eqn. (1) hence K4=(K2K33/K1)

$\frac{K _{1} . K _{2}}{K _{3}}$

$\frac{K _{1} . K _{3}^{2}}{K _{2}}$

$K_{1} K_{2} K_{3}$

$\frac{K _{2} . K _{3}^{3}}{K _{1}}$

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

Q. (1) N2​(g)+3H2​(g)⇌2NH3​(g),K1​ (2) N2​(g)+O2​(g)⇌2NHO(g),K2​ (3) H2​(g)+21​O2​(g)⇌H2​O(g),K3​ The equation for the equilibrium constant of the reaction 2NH3​(g)+25​O2​(g)⇌2NO(g),3H2​O(g),(K4​) in terms of K1​ and K3​ is:

K3​K1​.K2​​

K2​K1​.K32​​

K1​K2​K3​

K1​K2​.K33​​

To calculate the value of K4​ in the given equation we should apply: eqn.(2)+eqn.(3)x3−eqn.(1) hence K4​=K1​K2​K33​​

$\frac{K _{1} . K _{2}}{K _{3}}$

$\frac{K _{1} . K _{3}^{2}}{K _{2}}$

$K_{1} K_{2} K_{3}$

$\frac{K _{2} . K _{3}^{3}}{K _{1}}$

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