For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

For the reaction $A_{(g)} \rightarrow B_{(g)},$ the value of the equilibrium constant at $300 \ K$ and $1 \ atm$ is equal to $100.0.$ The value of $\Delta_{r}G^{\circ}$ for the reaction at $300 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is $-xR,$ where $x$ is ........... (Rounded off to the nearest integer) ($R = 8.31 \ J \ mol^{-1} K^{-1}$ and $\ln 10 = 2.3$)

Which of the following correctly represents the relationship between $\Delta G$ and $\Delta G^{\circ}$? $[P = \text{products}, R = \text{reactants}]$

For a system in equilibrium,$\Delta G = 0$ under conditions of constant:

For the following reaction at $50^\circ$ $C$ and at $2 \text{ atm}$ pressure, $2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$. $N_2O_5$ is $50\%$ dissociated. The magnitude of standard free energy change at this temperature is $x$. $x = . . . . . . \text{ J mol}^{-1}$.

In a closed flask at $600 \text{ K}$, one mole of $X_2Y_4(g)$ attains equilibrium as given below: $X_2Y_4(g) \rightleftharpoons 2XY_2(g)$. At equilibrium, $75\%$ of $X_2Y_4(g)$ is dissociated and the total pressure is $1 \text{ atm}$. The magnitude of $\Delta_r G^\circ$ (in $\text{kJ mol}^{-1}$) at this temperature is . . . . . . .