For the reaction,$2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$
$\Delta H = -57.2 \ kJ \ mol^{-1}$ and $K_C = 1.7 \times 10^{16}$
Which of the following statement is $INCORRECT$?

One mole of $O_{2(g)}$ and two moles of $SO_{2(g)}$ were heated in a closed vessel of $1 \, L$ capacity at $1098 \, K$. At equilibrium,$1.6 \, moles$ of $SO_{3(g)}$ were found. The equilibrium constant $K_c$ of the reaction would be

For the reaction $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta G_r^o$ will be:

For the equilibrium reaction $A + B \rightleftharpoons C + D$,if we start with equal concentrations of $A$ and $B$,at equilibrium,the concentration of $C$ is $2$ times that of $A$. Find the value of $K_c$.

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:

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 $1250 \ K$: $2 Cu_{(s)} + 1/2 O_{2(g)} \longrightarrow Cu_2O_{(s)}; \Delta G^\theta = -78,000 \ J \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)}; \Delta G^\theta = -1,78,000 \ J \ mol^{-1}$)