Water decomposes at $2300 \ K$. $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$. The percent of water decomposing at $2300 \ K$ and $1 \ bar$ is $...........$ (Nearest integer). Equilibrium constant for the reaction is $2 \times 10^{-3}$ at $2300 \ K$.

One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_p/K_c$ is .....

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:
$X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$
$Y: PCl_5 \rightleftharpoons PCl_3 + Cl_2$
$Z: 2HI \rightleftharpoons H_2 + I_2$

One mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ (by mass) of water reacted with $CO_{(g)}$ as follows: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2_{(g)}} + CO_{2_{(g)}}$. The value of $K_p$ is: