The equilibrium constant of a reaction is related to

For a reaction,$\Delta G^{\circ} = -115 \, kJ$. What is the value of $\log \, K_p$ at $298 \, K$?

Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

At $320 \ K,$ a gas $A_2$ is $20 \%$ dissociated to $A_{(g)}.$ The standard free energy change at $320 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is approximately $(R = 8.314 \ J \ K^{-1} \ mol^{-1}; \ ln \ 2 = 0.693; \ ln \ 3 = 1.098).$

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 $300 \ K$.

$STATEMENT-1$: For every chemical reaction at equilibrium,standard Gibbs energy of reaction is zero. $STATEMENT-2$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.