$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of $CO_2$ is converted into $CO$ on addition of graphite. If total pressure at equilibrium is $0.8 \ atm$,then $K_P$ is : (in $atm$)

In a closed vessel of $1 \ L$ capacity,$2 \ mol$ of $N_2$ and $6 \ mol$ of $H_2$ are mixed. If at equilibrium $50\% \ N_2$ is converted into $NH_3$,then the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ will be:

For the reaction $N_2 + O_2 \rightleftharpoons 2NO$ at $300 \, ^\circ C$,the value of $K_c$ is $9 \times 10^{-4}$. If equivalent amounts of $N_2$ and $O_2$ are used,what is the concentration of $NO$ at equilibrium (in terms of $a$) (in $, a$)?

For the reaction $P_{(g)} + 3Q_{(g)} \rightleftharpoons 4R_{(g)}$,the initial concentrations of $P$ and $Q$ are equal. If the equilibrium concentrations of $P$ and $R$ are equal,then the equilibrium constant $K_c$ for the reaction will be .....

For the reaction $X(s) \rightleftharpoons Y(s) + Z(g)$,the plot of $\ln \frac{p_z}{p^\ominus}$ versus $\frac{10^4}{T}$ is given below,where $p_z$ is the pressure (in bar) of the gas $Z$ at temperature $T$ and $p^\ominus = 1 \ bar$.
(Given,$\frac{d(\ln K)}{d(\frac{1}{T})} = -\frac{\Delta H^\ominus}{R}$,where the equilibrium constant,$K = \frac{p_z}{p^\ominus}$ and the gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
$(1)$ The value of standard enthalpy,$\Delta H^\ominus$ (in $kJ \ mol^{-1}$) for the reaction is. . . . . . .
$(2)$ The value of $\Delta S^\ominus$ (in $J \ K^{-1} \ mol^{-1}$) for the given reaction,at $1000 \ K$ is. . . . . .
Give the answer for $(1)$ and $(2)$

At $67\,^{\circ}C$ and $1\ bar$ pressure,dinitrogen tetraoxide is $50\%$ dissociated into nitrogen dioxide. $\Delta G^{\circ}$ for the process $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is $(R = \frac{25}{3} \ J \ K^{-1} \ mol^{-1}, \ln 2 = 0.7, \ln 3 = 1.1)$.