For the reversible reaction $2NO_2 \leftrightarrow[K_2]{K_1} N_2O_4$,the rate of disappearance of $NO_2$ is given by:

At $1 \ bar$ pressure and $310 \ K$ temperature,$25\%$ of $N_2O_4$ decomposes. The reaction is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
$(i)$ Find $K_p$.
$(ii)$ At $0.1 \ bar$ pressure and $310 \ K$,what is the percentage of $N_2O_4$ decomposed?

In which of the following equilibria is $K_{eq} > 1$?

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.]

$A$ reaction mixture containing $H_2, N_2$ and $NH_3$ has partial pressures of $2 \ atm, 1 \ atm$ and $3 \ atm$ respectively at $725 \ K.$ If the value of $K_P$ for the reaction,$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is $4.28 \times 10^{-5} \ atm^{-2}$ at $725 \ K,$ in which direction will the net reaction proceed?

One mole each of $He$ and $A(g)$ are taken in a $10 \text{ L}$ closed flask and heated to $400 \text{ K}$ to establish the following equilibrium: $A(g) \rightleftharpoons B(g)$. $K_{c}$ for this reaction at $400 \text{ K}$ is $4.0$. The partial pressures (in $\text{atm}$) of $He$ and $B(g)$ are respectively (at equilibrium) (Assume $He$, $A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$)