At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the partial pressures of $PCl_3$,$Cl_2$,and $PCl_5$ are $0.3 \ atm$,$0.2 \ atm$,and $0.6 \ atm$ respectively. If the partial pressures of $PCl_3$ and $Cl_2$ are doubled,what will be the partial pressure of $PCl_5$ in $atm$?

The value of $K_p$ for the following reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ is $1.2 \times 10^{-2}$ at $106.5 ^oC$. The value of $K_c$ for this reaction is

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: $A(g) \rightleftharpoons B(g) + C(g)$.

$1/T \text{ (K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)

$2 \ mol$ of $PCl_5$ is heated in a closed vessel of $2 \ L$ capacity. When equilibrium is attained,$PCl_5$ is $40\%$ dissociated. The equilibrium constant $K_c$ for the reaction will be ........... $mol/L$.

For the following three equilibria $X, Y$,and $Z$ in the gaseous state at $300 \ K$,the increasing order of the ratio of $K_p$ to $K_c$ is:
$X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$
$Y: PCl_5 \rightleftharpoons PCl_3 + Cl_2$
$Z: 2HI \rightleftharpoons H_2 + I_2$