Calculate the value of $\Delta G$ for the following reaction at $300 \ K$.
$H_2O_{(s)} \longrightarrow H_2O_{(l)}$
$(\Delta H = 7 \ kJ, \Delta S = 24.8 \ J \ K^{-1})$

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:

The occurrence of a reaction is impossible if

Gibbs free energy $(G)$ is defined as

The value of ${\left[ {\frac{dG}{dT}} \right]_P}$ is ....

For reaction $Ag_2O_{(s)} \to 2Ag_{(s)} + 1/2 O_{2(g)}$,the value of $\Delta H = 30.56 \ kJ \ mol^{-1}$ and $\Delta S = 0.066 \ kJ \ K^{-1} \ mol^{-1}$. Temperature at which free energy change for reaction will be zero,is ............. $K$.