In a $1 \, L$ container,the reaction of $2 \, mol$ $N_2$ and $5 \, mol$ $H_2$ occurs. If the equilibrium concentration of $NH_3$ is half the equilibrium concentration of $N_2$,what is the equilibrium constant $(K_c)$?

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:

For the equilibrium:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$; $K_{p} = 1.64 \ atm$ at $1000 \ K$.
$50 \ g$ of $CaCO_{3}$ in a $10 \ L$ closed vessel is heated to $1000 \ K$. The percentage of $CaCO_{3}$ that remains unreacted at equilibrium is:
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

At $783 \, K$ in the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the molar concentrations $(mol \, L^{-1})$ of $H_2, I_2$ and $HI$ at some instant of time are $0.1, 0.2$ and $0.4$,respectively. If the equilibrium constant is $46$ at the same temperature,then as the reaction proceeds:

Calculate:
$(a)$ $\Delta G^{\circ}$ and
$(b)$ the equilibrium constant for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \, K$
$NO_{(g)} + 1/2 O_{2(g)} \longleftrightarrow NO_{2(g)}$
Given:
$\Delta G^{\circ}_f(NO_2) = 52.0 \, kJ/mol$
$\Delta G^{\circ}_f(NO) = 87.0 \, kJ/mol$
$\Delta G^{\circ}_f(O_2) = 0 \, kJ/mol$

Ammonia under a pressure of $15 \ atm$ at $27 \ ^{\circ}C$ is heated to $347 \ ^{\circ}C$ in a closed vessel in the presence of a catalyst. Under these conditions,$NH_3$ is partially decomposed according to the equation,$2NH_3 \rightleftharpoons N_2 + 3H_2$. The vessel is such that the volume remains constant,and the pressure increases to $50 \ atm$. Calculate the percentage of $NH_3$ actually decomposed.