$2SO_2 + O_2 \to$ product. If the pressure of the reaction increases $3$ times, then the rate will be (in $\times$)?

$A$ mixture of $SO_2$ and $O_2$ at $5 \, atm$ pressure reacts $30\%$ until equilibrium is reached. Determine the total pressure of the equilibrium mixture in $atm$.
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$

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

For the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$,the equilibrium constants $K_p$ and $K_q$ are in the ratio $1:9$. If the degree of dissociation of $X$ and $Z$ is the same,then the ratio of their total pressures is:

At $700 \, K$,the equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \longleftrightarrow 2 HI_{(g)}$ is $54.8$. If $0.5 \, mol \, L^{-1}$ of $HI_{(g)}$ is present at equilibrium at $700 \, K$,what are the concentrations of $H_{2(g)}$ and $I_{2(g)}$,assuming that we initially started with $HI_{(g)}$ and allowed it to reach equilibrium at $700 \, K$?

Calculate the equilibrium constant for the reaction $H_{2(g)} + CO_{2(g)} \rightleftharpoons H_2O_{(g)} + CO_{(g)}$ at $1395 \ K$ by using the following data:
$2H_2O_{(g)} \rightleftharpoons 2H_{2(g)} + O_{2(g)}; K_1 = 2.1 \times 10^{-13}$
$2CO_{2(g)} \rightleftharpoons 2CO_{(g)} + O_{2(g)}; K_2 = 1.4 \times 10^{-12}$