The equilibrium constants for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ are $K_{1}$ and $K_{2}$,respectively. If the initial concentrations and the degree of dissociation of $X$ and $Z$ are the same,the ratio $K_{1} / K_{2}$ is

For the reaction $2SO_3 \rightleftharpoons 2SO_2 + O_2$,if $K_c = 100$ and the degree of dissociation $\alpha = 1$,determine the concentration of $O_2$ when the concentration of $SO_3$ is equal to the concentration of $SO_2$.

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:

$A$ reaction mixture containing $H_2, N_2$ and $NH_3$ has partial pressures of $2 \ atm, 1 \ atm$ and $3 \ atm$ respectively at $725 \ K.$ If the value of $K_P$ for the reaction,$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is $4.28 \times 10^{-5} \ atm^{-2}$ at $725 \ K,$ in which direction will the net reaction proceed?

When the reaction $A + 2B \rightleftharpoons 2C + D$ was studied,it was observed that the initial concentration of $B$ was $1.5$ times that of $A$,and the equilibrium concentrations of $A$ and $C$ were equal. Then $K_C$ for the given equilibrium is equal to

In a closed vessel of $1 \ L$ capacity,$2 \ mol$ of $N_2$ and $6 \ mol$ of $H_2$ are mixed. If at equilibrium $50\% \ N_2$ is converted into $NH_3$,then the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ will be: