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,$K_c$?

The value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)

At $800 \ K$ in a closed vessel,the molar concentrations of $N_2, O_2$ and $NO$ at equilibrium are $3.2 \times 10^{-3} \ M, 4.2 \times 10^{-3} \ M$ and $2.8 \times 10^{-3} \ M$ respectively. The approximate values of $K_{c}$ and $\frac{1}{K_{c}}$ for the following reaction are respectively: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$

For a reversible reaction $A \rightleftharpoons B$,the equilibrium concentration of $B$,denoted as $[B]_e$,is given by which expression?

For the reaction $A + B \rightleftharpoons C + D$ at $298 \ K$,the equilibrium constant is $10.0$. If the initial concentration of all four species is $1 \ M$,what will be the equilibrium concentration of $D$ (in $mol \ L^{-1}$)?

In the equilibrium system $AB_{(s)} \rightleftharpoons A_{(g)} + B_{(g)}$,if the equilibrium concentration of $A$ is doubled,the equilibrium concentration of $B$ will be .....