$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Hence,the resultant pressure is: (in $atm$)

In a chemical reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times of $A$ but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the aforesaid chemical reaction is

The equilibrium constant at $298 \ K$ for a reaction $A + B \rightleftharpoons C + D$ is $100$. If the initial concentration of all the four species were $1 \ M$ each,then the equilibrium concentration of $D$ (in $mol \ L^{-1}$) will be:

Consider the following reaction approaching equilibrium at $27^{\circ} C$ and $1 \ atm$ pressure. Given the rate constants for the forward and backward reactions are $K_{f} = 10^{3} \ s^{-1}$ and $K_{b} = 10^{2} \ s^{-1}$ respectively,calculate the standard Gibb's energy change $(\Delta_{r} G^{\circ})$ at $27^{\circ} C$ in $kJ \ mol^{-1}$ (Nearest integer). (Given: $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ and $\ln 10 = 2.3$)

In a $1 \, L$ container,the reaction of $2 \, mol$ $N_2$ and $5 \, mol$ $H_2$ occurs. If the equilibrium concentration of $NH_3$ is half the equilibrium concentration of $N_2$,what is the equilibrium constant $(K_c)$?

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$56 \ g$ of nitrogen and $8 \ g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \ g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen and ammonia are respectively: