The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298 \text{ K}$? $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$

Hydrolysis of sucrose gives,
$Sucrose + H_{2}O \rightleftharpoons Glucose + Fructose$
Equilibrium constant $K_{c}$ for the reaction is $2 \times 10^{13}$ at $300 \ K$. Calculate $\Delta G^{\ominus}$ at $300 \ K$.

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,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

For a reaction at $298 \ K$,the equilibrium constant is ${K_p} = 0.17 \times {10^{12}}$. Find the standard Gibbs free energy change $\Delta {G^\Theta }$. (Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

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 $298 \, K$.