For the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$,$A$ is $33 \%$ dissociated at a total pressure $P$. The correct relation between $P$ and $K_{p}$ 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 $700 \, K$?

Attainment of the equilibrium $A_{(g)} \rightleftharpoons 3C_{(g)} + 2B_{(g)}$ gave the following graph. Find the correct option. $(\text{Percentage dissociation} = \text{fraction dissociated} \times 100)$

The equilibrium constants $K_{p1}$ and $K_{p2}$ for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ respectively are in the ratio of $1 : 9$. If the degree of dissociation of $X$ and $Z$ be equal,then the ratio of total pressures at these equilibria is:

$2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2_{(g)}}$. If $NOBr$ is $40\%$ dissociated at a certain temperature and a total pressure of $0.30 \text{ atm}$,the $K_p$ for the reaction $2NO_{(g)} + Br_{2_{(g)}} \rightleftharpoons 2NOBr_{(g)}$ is:

At a certain temperature and total pressure of $10^{5} \ Pa$,iodine vapour contains $40 \%$ by volume of $I$ atoms.
$I_{2(g)} \longleftrightarrow 2I_{(g)}$
Calculate $K_{p}$ for the equilibrium.