$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

If $2 \ mol$ of $H_2$ and $I_2$ are taken initially in a $1 \ L$ vessel,and the equilibrium concentration of $HI$ is $2 \ mol/L$,find the $K_p$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.

The reaction between $N_2$ and $H_2$ to form ammonia has $K_c = 6 \times 10^{-2}$ at the temperature $500 \ ^oC$. The numerical value of $K_p$ for this reaction is

Equilibrium constants for the following reactions at $1200 \ K$ are given:
$2 \ H_2O_{(g)} \rightleftharpoons 2 \ H_{2(g)} + O_{2(g)}$
$K_1 = 6.4 \times 10^{-8}$
$2 \ CO_{2(g)} \rightleftharpoons 2 \ CO_{(g)} + O_{2(g)}$
$K_2 = 1.6 \times 10^{-6}$
The equilibrium constant for the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
at $1200 \ K$ will be

$K_p$ for the reaction $2SO_3 \rightleftharpoons 2SO_2 + O_2$ at $700 \ K$ is $1.3 \times 10^{-3} \ atm$. The $K_c$ at the same temperature for the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$ will be:

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to: