Equivalent amounts of $H_2$ and $I_2$ are heated in a closed vessel till equilibrium is obtained. If $80\%$ of the hydrogen is converted to $HI$,the $K_c$ at this temperature is

The equilibrium constants $K_{p_1}$ and $K_{p_2}$ for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ are in the ratio of $1 : 4$. If the degree of dissociation of $X$ is $2$ times that of $Z$,then the ratio of total pressure $(P_1 : P_2)$ at these equilibria is: (Assume degree of dissociation for both reactions are very small)

In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,initially $1 \text{ mole}$ each of $PCl_5$ and $PCl_3$ are present. At equilibrium,$x \text{ moles}$ of $PCl_5$ remain. What is the total number of moles at equilibrium (in $- x$)?

$K_{c}$ for the following reaction is $99.0$: $A_{2(g)} \rightleftharpoons B_{2(g)}$. In a $1 \ L$ flask,$2 \ moles$ of $A_{2}$ were heated to $T(K)$ and equilibrium was reached. The concentrations at equilibrium of $A_{2}$ and $B_{2}$ are $C_{1}(A_{2})$ and $C_{2}(B_{2})$ respectively. Now,$1 \ mole$ of $A_{2}$ was added to the flask and heated to $T(K)$ to establish equilibrium again. The concentrations of $A_{2}$ and $B_{2}$ are $C_{3}(A_{2})$ and $C_{4}(B_{2})$ respectively. What is the value of $C_{3}(A_{2})$ in $mol \ L^{-1}$?

Consider the following reaction equilibrium:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initially,$1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$ are taken in a $2 \ L$ flask. At equilibrium,if the number of moles of $N_2$ is $0.6$,what is the total number of moles of all gases present in the flask?

The equilibrium constants of the following are
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is: