What will be the value of $\Delta G$ for the melting of ice at $283 \ K$?

$A$ reaction at $1 \ bar$ is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following:

The standard entropy change for the reaction $4 Fe ( s ) + 3 O_{2} ( g ) \rightarrow 2 Fe_{2} O_{3} ( s )$ is $-550 \ J K^{-1}$ at $298 \ K$. Given: The standard enthalpy change for the reaction is $-165 \ kJ \ mol^{-1}$. The temperature in $K$ at which the reaction attains equilibrium is ... . (Nearest Integer)

The values of $\Delta H$ and $\Delta S$ for the reaction,$C_{(graphite)} + CO_{2(g)} \rightarrow 2CO_{(g)}$ are $170 \ kJ$ and $170 \ J \ K^{-1}$ respectively. This reaction will be spontaneous at temperatures greater than ............ $K$.

For a certain reaction, $\Delta H^{\circ} = 40 \text{ kJ}$ and $\Delta S^{\circ} = 80 \text{ JK}^{-1}$. Find the temperature at which $\Delta G^{\circ} = 0$. (in $\text{ K}$)

The densities of graphite and diamond at $298 \, K$ are $2.25 \, g \, cm^{-3}$ and $3.31 \, g \, cm^{-3}$ respectively. If the standard free energy difference $(\Delta G^o)$ is $1895 \, J \, mol^{-1}$, the pressure at which graphite will be transformed into diamond at $298 \, K$ is: