If the equilibrium constant for the formation of $NH_3$ is $K_c$,then the dissociation constant of $NH_3$ at the same temperature is:

If the volume of the container is $1 \ L$ and at equilibrium the amounts are $SO_3 = 48 \ g$,$SO_2 = 12.8 \ g$,and $O_2 = 9.6 \ g$,find the value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

If for ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$ and ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$ the equilibrium constants are $K_1$ and $K_2$ respectively,the reaction ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$ would have equilibrium constant

In a closed vessel,$PCl_{5(g)}$ is obtained by the chemical reaction between $PCl_{3(g)}$ and $Cl_{2(g)}$. If the equilibrium concentrations in this vessel of $PCl_3$,$Cl_2$,and $PCl_5$ at $500 \ K$ are $1.59 \ M$,$1.59 \ M$,and $1.41 \ M$ respectively,then find the equilibrium constant $K_c$ for the reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

$2$ moles of $PCl_5$ were heated in a closed vessel of $2 \ L$ capacity. At equilibrium,$40\%$ of $PCl_5$ is dissociated into $PCl_3$ and $Cl_2$. The value of equilibrium constant is

For the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be