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 $x$. $x = . . . . . . \text{ J mol}^{-1}$.

The value of $\Delta G^{\ominus}$ for the phosphorylation of glucose in glycolysis is $13.8 \, kJ \, mol^{-1}$. Find the value of $K_{c}$ at $298 \, K$.

The standard Gibbs free energy change $\Delta G^{\circ}$ at $25^{\circ} C$ for the dissociation of $N_2O_{4(g)}$ to $NO_{2(g)}$ is (given, equilibrium constant $K_{eq} = 0.15, R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)

At $60^{\circ} C$,dinitrogen tetroxide is $50 \%$ dissociated. Find its standard free energy change at this temperature and $1 \ atm$. [ Given: $\log 1.33 = 0.1239 ]$

Consider the reaction $X \rightleftharpoons Y$ at $300 \text{ K}$. If $\Delta H^\circ$ and $K$ are $28.40 \text{ kJ mol}^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^\circ$ for the reaction in $\text{J K}^{-1} \text{ mol}^{-1}$ is . . . . . . . (Nearest integer) (Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $\ln 10 = 2.3$, $\log 3 = 0.48$, $\log 2 = 0.30$)

If the equilibrium constant for a reaction is $10$,then the value of $\Delta G^o$ is ....... $(R = 8 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K)$