Given three reactions and their equilibrium constants:
$N_2 + 3H_2 \rightleftharpoons 2NH_3 ; k_1$
$N_2 + O_2 \rightleftharpoons 2NO ; k_2$
$H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O ; k_3$
The equilibrium constant for the reaction $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $k_1, k_2,$ and $k_3$ is:

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is:

At $T \ K$,$K_{c}$ for the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$ is $16$. If initially one mole each of all the four gases are taken in a $1 \ L$ vessel,the equilibrium concentrations of $SO_{3(g)}$ and $SO_{2(g)}$ in $mol \ L^{-1}$ respectively are:

For the reversible reaction $2NO_2 \leftrightarrow[K_2]{K_1} N_2O_4$,the rate of disappearance of $NO_2$ is given by:

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 $(2)$

For the reaction $Br_2 \rightleftharpoons 2Br$,the equilibrium constants at $327\,^oC$ and $527\,^oC$ are $6.1 \times 10^{-12}$ and $1.0 \times 10^{-7}$ respectively. The reaction is ...........