If Kc is the equilibrium constant for the formation of NH3 . The dissociation constant of ammonia under the same temperature will be

Question

If Kc is the equilibrium constant for the formation of NH3 . The dissociation constant of ammonia under the same temperature will be (A)  Kc  (B)  √Kc  (C)  Kc2  (D)  (1/Kc)

Tardigrade Admin · Accepted Answer

NH 3 leftharpoons (1/2) N 2+(3/2) H 2 Kc=([ N 2]1 / 2[ H 2]3 / 2/[ NH 3]) ...(i) (1/2) N 2+(3/2) H 2 leftharpoons NH 3 Kc'=([ NH 3]/[ N 2]1 / 2[ H 2]3 / 2)...(ii) So, for dissociation K'=(1/Kc)

Q. If $K_{c}$ is the equilibrium constant for the formation of $N H_{3}$ . The dissociation constant of ammonia under the same temperature will be

$K_{c}$

$K_{c}$

$K_{c}^{2}$

$\frac{1}{K _{c}}$

Q. If Kc​ is the equilibrium constant for the formation of NH3​ . The dissociation constant of ammonia under the same temperature will be

Kc​

Kc​​

Kc2​

Kc​1​

NH3​⇌21​N2​+23​H2​ Kc​=[NH3​][N2​]1/2[H2​]3/2​ ...(i) 21​N2​+23​H2​⇌NH3​ Kc′​=[N2​]1/2[H2​]3/2[NH3​]​...(ii) So, for dissociation K′=Kc​1​

Q. If $K_{c}$ is the equilibrium constant for the formation of $N H_{3}$ . The dissociation constant of ammonia under the same temperature will be

$K_{c}$

$K_{c}$

$K_{c}^{2}$

$\frac{1}{K _{c}}$