Which of the following statements is/are true about equilibrium?
$(a)$ Equilibrium is possible only in a closed system at a given temperature.
$(b)$ All the measurable properties of the system remain constant at equilibrium.
$(c)$ Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

The surface of copper gets tarnished by the formation of copper oxide. $N_2$ gas was passed to prevent the oxide formation during heating of copper at $1250 \ K$. However,the $N_2$ gas contains $1 \ \text{mole}\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2 Cu_{(s)} + H_2O_{(g)} \longrightarrow Cu_2O_{(s)} + H_{2(g)}$
$p_{H_2}$ is the minimum partial pressure of $H_2$ (in $\text{bar}$) needed to prevent the oxidation at $1250 \ K$. The value of $\ln(p_{H_2})$ is . . . . .
(Given: total pressure $= 1 \ \text{bar}$,$R = 8 \ J \ K^{-1} \ mol^{-1}$,$\ln(10) = 2.3$. $Cu_{(s)}$ and $Cu_2O_{(s)}$ are mutually immiscible.
At $1250 \ K$: $2 Cu_{(s)} + 1/2 O_{2(g)} \longrightarrow Cu_2O_{(s)}; \Delta G^\theta = -78,000 \ J \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)}; \Delta G^\theta = -1,78,000 \ J \ mol^{-1}$)

Which of the following statements is correct?

For a reaction,$A \rightleftharpoons P$,the plots of $[A]$ and $[P]$ with time at temperatures $T_1$ and $T_2$ are given below. If $T_2 > T_1$,the correct statement$(s)$ is (are) (Assume $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ are independent of temperature and ratio of $\ln K$ at $T_1$ to $\ln K$ at $T_2$ is greater than $T_2 / T_1$. Here $H, S, G$ and $K$ are enthalpy,entropy,Gibbs energy and equilibrium constant,respectively.)
$(A)$ $\Delta H^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(B)$ $\Delta G^{\ominus} < 0, \Delta H^{\ominus} > 0$
$(C)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(D)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} > 0$

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:

$A$ reaction mixture containing $H_2, N_2$ and $NH_3$ has partial pressures of $2 \ atm, 1 \ atm$ and $3 \ atm$ respectively at $725 \ K.$ If the value of $K_P$ for the reaction,$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is $4.28 \times 10^{-5} \ atm^{-2}$ at $725 \ K,$ in which direction will the net reaction proceed?