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 $\ln 10 = 2.3$)

At $298 \ K$,for the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the $K_p$ value is $0.98$. Predict whether the reaction is spontaneous or not.

The standard Gibbs free energy change $\Delta G^o$ is related to the equilibrium constant $K_p$ as:

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 . . . . . . .

The reaction,$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ is in equilibrium in a closed vessel at $298 \, K$. The partial pressure (in $atm$) of $CO_{2(g)}$ in the reaction vessel is closest to $....$ [Given : The change in Gibbs energies of formation at $298 \, K$ and $1 \, bar$ for $CaO_{(s)} = -603.501 \, kJ \, mol^{-1}$,$CO_{2(g)} = -394.389 \, kJ \, mol^{-1}$,$CaCO_{3(s)} = -1128.79 \, kJ \, mol^{-1}$,Gas constant,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

At $60^{\circ} C$,dinitrogen tetroxide is $50$ per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.