For the given reaction: N_(2)(g) + 3H_2(g) hArr 2NH_3(g) Equilibrium constant K_c=[NH_3]^2/([N_2][H_2])^3 Some reactions are written below in Column-I and their equilibrium constants in terms of K_c are written in Column-II. Match the following reactions with the corresponding equilibrium constant

For the given reaction: N_(2)(g) + 3H_2(g) hArr 2NH_3(g) Equilibrium

1.	For the given reaction: N_(2)(g) + 3H_2(g) hArr 2NH_3(g) Equilibrium constant K_c=[NH_3]^2/([N_2][H_2])^3 Some reactions are written below in Column-I and their equilibrium constants in terms of K_c are written in Column-II. Match the following reactions with the corresponding equilibrium constant
Answer» Solution :For the reaction, `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))` Equilibrium `K_c=[NH_3]^2/([N_2][H_2]^3)` (A) The GIVEN reaction `[2N_(2(g)) + 6H_(2(g)) hArr 4NH_(3(g))]` is twice the above reaction. Hence `K=K_c^2` (B) The reaction `[2NH_(3(g)) hArr N_(2(g)) + 3H_(2(g))]` is REVERSE of the above reaction. hence, `K=1/K_c` (C) The reaction `[1/2N_(2(g)) +3/2 H_(2(g)) hArr NH_(3(g))]` is half of the above reaction Hence, `K=sqrtK_c=K_c^(1/2)`