What will happen to the following equilibrium reaction if pressure is applied on it?

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The value of ${\mathrm{K}}_{\mathrm{c}} \mathrm{us} 64 \mathrm{at} 800 \mathrm{K}$ for the reaction

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2 {\mathrm{NH}}_3 \left(\mathrm{g}\right)$

The value of ${\mathrm{K}}_{\mathrm{c}}$ for the following reaction is :

${\mathrm{NH}}_3 \left(\mathrm{g}\right)\to \frac{1}{2} {\mathrm{N}}_2 \left(\mathrm{g}\right)+\frac{3}{2} {\mathrm{H}}_2 \left(\mathrm{g}\right)$

The reaction

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

is in equilibrium in a closed vessel at $298 \mathrm{K}.$ The partial pressure (in atm) of ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ in the reaction vessel is closest to:

[Given: The change in Gibbs energies of formation at $298 \mathrm{K}$ and $1$ bar for

$\mathrm{CaO}\left(\mathrm{s}\right)=-603.501 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{CO}}_2\left(\mathrm{g}\right)=-394.389 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)=-1128.79 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Gas constant $\mathrm{R}=8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$]

Using the data provided, find the value of equilibrium constant for the following reaction at $298 \mathrm{K}$ and $1$ atm pressure. ${\mathrm{NO}}_{\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons {\mathrm{NO}}_2( \mathrm{g})$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{NO}}_{\left(\mathrm{g}\right)}\right)=90.4\mathrm{kJ}.{\mathrm{mol}}^{-1}$

${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{NO}}_2\left(\mathrm{g}\right)\right)=32.48\mathrm{kJ}\cdot {\mathrm{mol}}^{-1}$

${\mathrm{\Delta S}}^{\circ }\mathrm{at} 298\mathrm{K}=-70.8\mathrm{J}\cdot {\mathrm{K}}^{-1}\cdot {\mathrm{mol}}^{-1}$

$\mathrm{antilog}(0.50)=3162$

Find the total pressure of cases over a mixture of X and Y, ​​​​​​​​when two solid compounds 'X' and 'Y' dissociates at a certain temperature is as follows:
$\mathrm{X}\left(\mathrm{s}\right)\rightleftharpoons \mathrm{A}\left(\mathrm{g}\right)+2\mathrm{B}\left(\mathrm{g}\right){\mathrm{Kp}}_1=9\times {10}^{-3}{\mathrm{atm}}^3$

$\mathrm{Y}\left(\mathrm{s}\right)\leftrightharpoons 2\mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right) {\mathrm{Kp}}_2=4.5\times {10}^{-3} {\mathrm{atm}}^3$

$\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+2\mathrm{D}.....\left(\mathrm{i}\right)$

$\mathrm{A}+\mathrm{E}\rightleftharpoons \mathrm{C}+\mathrm{F}\dots \dots \dots .\left(\text{ii}\right)$

If ${\mathrm{K}}_1$ and ${\mathrm{K}}_2$ are equilibrium constants for the reactions $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, respectively, then find out the equilibrium constant for the reaction $\mathrm{E}+2\mathrm{D}\rightleftharpoons \mathrm{B}+\mathrm{F}$.

In which direction does the reaction tend to shift to reach equilibrium given at a particular time, the analysis shows that composition of reaction mixture is 3.0 mol L^-1 N₂, 2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃.

Equilibrium constant, K_c for the below reaction is

${\text{N}}_2\left(\text{g}\right)+{\text{3H}}_2\left(\text{g}\right)\rightleftharpoons {\text{2NH}}_3\left(\text{g}\right)\text{at 500 K is 0.061}$