At $444 \, ^\circ C$,the equilibrium constant for the reaction $HI \rightleftharpoons 1/2 H_2 + 1/2 I_2$ is $64$. What will be the equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

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 reaction $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2_{(g)}}$ at $1060 \ K$,the equilibrium constant $K_p$ is $0.033 \ atm$. Find the value of $K_c$. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

For the reaction $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1$ at $298 \, K$. For the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ at the same temperature,the equilibrium constant is $K_2$. Then,which of the following is correct?

$(1) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \ ; \ K_1$
$(2) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \ ; \ K_2$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_2O_{(g)} \ ; \ K_3$
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

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