$2 \ mol$ of $N_2O_{4(g)}$ is kept in a closed container at $298 \ K$ and under $1 \ atm$ pressure. It is heated to $596 \ K$ when $20 \%$ by mass of $N_2O_{4(g)}$ decomposes to $NO_2$. The resulting pressure is (in $atm$)

The equilibrium constants for the following reactions are $K_1$ and $K_2$,respectively.
$2 P_{(g)} + 3 Cl_{2(g)} \rightleftharpoons 2 PCl_{3(g)}$
$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$
Then,the equilibrium constant for the reaction,$2 P_{(g)} + 5 Cl_{2(g)} \rightleftharpoons 2 PCl_{5(g)}$ is

Solid carbon, $CaO$ and $CaCO_3$ are mixed and allowed to attain equilibrium at $T \text{ K}$. $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ $K_{p1} = 0.08 \text{ atm}$. $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$ $K_{p2} = 2 \text{ atm}$. The partial pressure of $CO$ is . . . . . . $\times 10^{-1} \text{ atm}$.

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,at equilibrium,the mole fraction of $PCl_5$ is $0.4$ and the mole fraction of $Cl_2$ is $0.3$. What will be the mole fraction of $PCl_3$?

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 gas phase reaction; $C_2H_{4(g)} + H_{2(g)} \rightleftharpoons C_2H_{6(g)}$,$\Delta H = -32.7 \ kcal$ carried out in a vessel,the equilibrium concentration of $C_2H_4$ can be increased by: