The reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2_{(g)}} + S_{2_{(g)}}$ is in equilibrium. If $0.5 \ mol$ of $H_2S$,$0.10 \ mol$ of $H_2$,and $0.4 \ mol$ of $S_2$ are taken in a $1 \ L$ vessel,the value of the equilibrium constant $(K)$ is .... $mol \ L^{-1}$.

The value of $K_{c} = 4.24$ at $800 \, K$ for the reaction,
$CO_{(g)} + H_{2}O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$
Calculate equilibrium concentrations of $CO_{2}$,$H_{2}$,$CO$ and $H_{2}O$ at $800 \, K$,if only $CO$ and $H_{2}O$ are present initially at concentrations of $0.10 \, M$ each.

At $298 \ K$,
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), K_1 = 4 \times 10^5$
$N_2(g) + O_2(g) \rightleftharpoons 2NO(g), K_2 = 1.6 \times 10^{12}$
$H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g), K_3 = 1.0 \times 10^{-13}$
Based on the above equilibria,the equilibrium constant of the reaction,
$2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$
is $.......... \times 10^{-33}$ (Nearest integer).

In a reaction,$A + B \rightleftharpoons C + D$,$40 \%$ of $B$ has reacted at equilibrium,when $1 \ mol$ of $A$ was heated with $1 \ mol$ of $B$ in a $10 \ L$ closed vessel. The value of $K_C$ is

In a one-litre flask,$6$ moles of $A$ undergoes the reaction $A_{(g)} \rightleftharpoons P_{(g)}$. The progress of product formation at two temperatures (in Kelvin),$T_1$ and $T_2$,is shown in the figure:
If $T_1=2 T_2$ and $(\Delta G_2^{\Theta}-\Delta G_1^{\Theta})=R T_2 \ln x$,then the value of $x$ is. . . . .
$[\Delta G_1^{\Theta}$ and $\Delta G_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$,respectively.]

$28 \, g$ of $N_2$ and $6 \, g$ of $H_2$ were kept at $400 \, ^oC$ in a $1 \, L$ vessel. The equilibrium mixture contained $27.54 \, g$ of $NH_3$. The approximate value of $K_c$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is (in $L^2 \, mol^{-2}$):