If $0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ are reacted in a $10 \ L$ vessel at $444 \ ^\circ C$,and the equilibrium constant $K_c$ at the same temperature is $49$,then the ratio of $[HI]$ to $[I_2]$ is .......

For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$,the equilibrium constants are $K_1$ and $K_2$ respectively. What is the relationship between $K_1$ and $K_2$?

For a reversible reaction,the rate constants for the forward and backward reactions are $2.38 \times 10^{-4}$ and $8.15 \times 10^{-5}$ respectively. The equilibrium constant for the reaction is

The standard equilibrium constant,$K_p$ at $298 \, K$ for the reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is $5.8 \times 10^5$. The value of standard equilibrium constant,if the concentration of gases is expressed in terms of $mol/L$,will be:
[Given : $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$]

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

$(1) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \ ; \ K_1$
$(2) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \ ; \ K_2$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_2O_{(g)} \ ; \ K_3$
The equation for the equilibrium constant of the reaction
$2NH_{3(g)} + \frac{5}{2}O_{2(g)} \rightleftharpoons 2NO_{(g)} + 3H_2O_{(g)}$
$(K_4)$ in terms of $K_1$,$K_2$,and $K_3$ is