In the reaction,$H_2 + I_2 \rightleftharpoons 2HI$. In a $2 \ L$ flask,$0.4 \ mol$ of each $H_2$ and $I_2$ are taken. At equilibrium,$0.5 \ mol$ of $HI$ are formed. What will be the value of equilibrium constant,$K_c$?

In the synthesis of $HI$,the amounts of $H_{2(g)}$,$I_{2(g)}$,and $HI_{(g)}$ at equilibrium were found to be $0.8 \ mol$,$0.8 \ mol$,and $2.4 \ mol$ respectively in a $10 \ L$ vessel. Calculate the equilibrium constant $(K_c)$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ and the equilibrium constant for the reverse reaction.

$2 \ mol$ of $PCl_5$ is heated in a closed vessel of $2 \ L$ capacity. When equilibrium is attained,$PCl_5$ is $40\%$ dissociated. The equilibrium constant $K_c$ for the reaction will be ........... $mol/L$.

The values of $K_p/K_c$ for the following reactions at $300 \ K$ are respectively (At $300 \ K, RT = 24.62 \ dm^3 \ atm \ mol^{-1}$):
$(i) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$
$(ii) \ N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
$(iii) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

The reaction for which $K_{c} = 2.3 \times 10^3 \ mol \ L^{-1}$ is $-$

$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$
In an equilibrium mixture,the partial pressures are
$P_{SO_{3}} = 43 \ kPa$,$P_{O_{2}} = 530 \ Pa = 0.53 \ kPa$,and
$P_{SO_{2}} = 45 \ kPa$. The equilibrium constant $K_{p} = ...... \times 10^{-2} \ kPa^{-1}$. (Nearest integer)