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 $CO_2$ ?
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)} + \text{Heat}$

For the reaction $XCO_{3(s)} \rightleftharpoons XO_{(s)} + CO_{2(g)},$ $K_p = 1.642 \text{ atm}$ at $727^{\circ}C.$ If $4 \text{ moles}$ of $XCO_{3(s)}$ were placed into a $50 \text{ L}$ container and heated to $727^{\circ}C,$ what mole percent of the $XCO_3$ remains unreacted at equilibrium?

For a reaction $A_{(s)} \rightleftharpoons B_{(s)} + C_{(g)}$,the set of all correct statements are:
$(a) \ K$ is independent of $[A]$.
$(b) \ K$ is dependent on partial pressure of $C$ at a given temperature.
$(c) \ \Delta H$ will be independent of temperature.
$(d) \ \Delta H$ is independent of the catalyst addition.

The equilibrium constants $(K_C)$ of two reactions $H_2 + I_2 \rightleftharpoons 2 HI$ and $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ are $50$ and $1000$,respectively. The equilibrium constant of the reaction $N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$ is closest to

The equilibrium constants $K_{p1}$ and $K_{p2}$ for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ respectively are in the ratio of $1 : 9$. If the degree of dissociation of $X$ and $Z$ be equal,then the ratio of total pressures at these equilibria is:

$PCl_{5}$ dissociates as $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. $5 \, \text{moles}$ of $PCl_{5}$ are placed in a $200 \, L$ vessel which contains $2 \, \text{moles}$ of $N_{2}$ and is maintained at $600 \, K$. The equilibrium pressure is $2.46 \, atm$. The equilibrium constant $K_{p}$ for the dissociation of $PCl_{5}$ is $...... \times 10^{-3}$. (nearest integer) (Given: $R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$: Assume ideal gas behaviour)