Given the equilibrium constants for the following three reactions:
$(1) N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1$
$(2) N_2 + O_2 \rightleftharpoons 2NO; K_2$
$(3) H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K_3$
The equilibrium constant for the reaction of $NH_3$ with oxygen to form $NO$ and $H_2O$ is:

In a one-litre flask,$6$ moles of $A$ undergoes the reaction $A_{(g)} \rightleftharpoons P_{(g)}$. The progress of product formation at two temperatures (in Kelvin),$T_1$ and $T_2$,is shown in the figure:
If $T_1=2 T_2$ and $(\Delta G_2^{\Theta}-\Delta G_1^{\Theta})=R T_2 \ln x$,then the value of $x$ is. . . . .
$[\Delta G_1^{\Theta}$ and $\Delta G_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$,respectively.]

If the pressure in a reaction vessel for the following reaction is increased by decreasing the volume,what will happen to the concentrations of $CO$ and $CO_2$ ?
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)} + \text{Heat}$

At a certain temperature,$2HI \rightleftharpoons H_2 + I_2$. If only $50\%$ of $HI$ is dissociated at equilibrium,the equilibrium constant $(K_c)$ is:

Thermal decomposition of gaseous $X_2$ to gaseous $X$ at $298 \ K$ takes place according to the following equation :
$X_{2(g)} \rightleftharpoons 2 X_{(g)}$
The standard reaction Gibbs energy,$\Delta_r G^{\circ}$,of this reaction is positive. At the start of the reaction,there is one mole of $X_2$ and no $X$. As the reaction proceeds,the number of moles of $X$ formed is given by $\beta$. Thus,$\beta_{\text{equilibrium}}$ is the number of moles of $X$ formed at equilibrium. The reaction is carried out at a constant total pressure of $2 \ bar$. Consider the gases to behave ideally. (Given : $R=0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)
$(1)$ The equilibrium constant $K_P$ for this reaction at $298 \ K$,in terms of $\beta_{\text{equilibrium}}$,is
$(A)$ $\frac{8 \beta_{\text{equilibrium}}^2}{2-\beta_{\text{equilibrium}}}$ $(B)$ $\frac{8 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$ $(C)$ $\frac{4 \beta_{\text{equilibrium}}^2}{2-\beta_{\text{equilibrium}}}$ $(D)$ $\frac{4 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$
$(2)$ The $INCORRECT$ statement among the following,for this reaction,is
$(A)$ Decrease in the total pressure will result in formation of more moles of gaseous $X$
$(B)$ At the start of the reaction,dissociation of gaseous $X_2$ takes place spontaneously
$(C)$ $\beta_{\text{equilibrium}}=0.7$
$(D)$ $K_c < 1$

At $1000 \ K$ in a $0.654 \ L$ vessel,$CaCO_{3(s)}$ is taken. For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the equilibrium constant $K_p$ is $3.9 \times 10^{-2} \ bar$. Find the weight of $CaO$ produced at equilibrium. $(Ca=40, C=12, O=16)$