Observe the following equilibrium in a $1 \text{ L}$ flask. $A_{(g)} \rightleftharpoons B_{(g)}$. At $T \text{ K}$, the equilibrium concentrations of $A$ and $B$ are $0.5 \text{ M}$ and $0.375 \text{ M}$ respectively. $0.1 \text{ moles}$ of $A$ is added into the flask and heated to $T \text{ K}$ to establish the equilibrium again. The new equilibrium concentrations (in $\text{M}$) of $A$ and $B$ are respectively.

Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

Write the expression for the equilibrium constant,$K_{c}$ for each of the following reactions:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$
$(ii)$ $2 Cu(NO_{3})_{2(s)} \longleftrightarrow 2 CuO_{(s)} + 4 NO_{2(g)} + O_{2(g)}$
$(iii)$ $CH_{3}COOC_{2}H_{5(aq)} + H_{2}O_{(l)} \longleftrightarrow CH_{3}COOH_{(aq)} + C_{2}H_{5}OH_{(aq)}$
$(iv)$ $Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} \longleftrightarrow Fe(OH)_{3(s)}$
$(v)$ $I_{2(s)} + 5 F_{2(g)} \longleftrightarrow 2 IF_{5(g)}$

For the gaseous reactions $(I)$ and $(II)$,the equilibrium constants are $X$ and $Y$,respectively.
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III. N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$