$S_1$: In case of endothermic reactions,the equilibrium shifts in the forward direction on increasing temperature.
$S_2$: The value of $K_{eq}$ depends only on temperature and is independent of pressure.
$S_3$: For the reaction,$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,the equilibrium constant,$K_{eq}$ is dimensionless because the number of moles of gaseous products equals the number of moles of gaseous reactants.

$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Hence,the resultant pressure is: (in $atm$)

Using the data provided,find the value of the equilibrium constant for the following reaction at $298 \ K$ and $1 \ atm$ pressure: $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
$\Delta_{f} H^0(NO_{(g)}) = 90.4 \ kJ \cdot mol^{-1}$
$\Delta_{f} H^0(NO_{2(g)}) = 32.48 \ kJ \cdot mol^{-1}$
$\Delta S^{\circ} = -70.8 \ J \cdot K^{-1} \cdot mol^{-1}$
$\text{antilog}(6.4) = 2.51 \times 10^6$ (Note: Calculation based on standard thermodynamic relations)

Consider the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. The equilibrium constant of the above reaction is $K_p$. If pure ammonia is left to dissociate,the partial pressure of ammonia at equilibrium is given by (Assume that $P_{NH_3} \ll P_{total}$ at equilibrium and $P$ is the total pressure).

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

For the reaction $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$,if $25\%$ of $CO_2$ is converted to $CO$ at equilibrium and the total equilibrium pressure is $12 \ atm$,then the partial pressure of $CO_2$ at equilibrium will be............$atm$.