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:

For a reaction to be spontaneous at all temperatures, the values of enthalpy change $\Delta H$ and entropy change $\Delta S$ should be:

For a reaction at $25^o C$,$\Delta H > 0$ and $\Delta S > 0$. Under what conditions will this reaction become spontaneous?

For a reaction at $25 \, ^oC$,the enthalpy change and entropy change are $-11.7 \times 10^3 \, J \, mol^{-1}$ and $-105 \, J \, mol^{-1} \, K^{-1}$ respectively. What is the Gibbs free energy in $kJ$?

The standard state Gibbs free energies of formation of $C$ (graphite) and $C$ (diamond) at $T = 298 \ K$ are:
$\Delta_f G^0[C(\text{graphite})] = 0 \ kJ \ mol^{-1}$
$\Delta_f G^0[C(\text{diamond})] = 2.9 \ kJ \ mol^{-1}$
The standard state means that the pressure should be $1 \ bar$,and the substance should be pure at a given temperature. The conversion of graphite [$C$ (graphite)] to diamond [$C$ (diamond)] reduces its volume by $2 \times 10^{-6} \ m^3 \ mol^{-1}$. If $C$ (graphite) is converted to $C$ (diamond) isothermally at $T = 298 \ K$,the pressure at which $C$ (graphite) is in equilibrium with $C$ (diamond) is:
[Useful information: $1 \ J = 1 \ kg \ m^2 \ s^{-2} ; 1 \ Pa = 1 \ kg \ m^{-1} \ s^{-2} ; 1 \ bar = 10^5 \ Pa$ ] (in $bar$)

Compare the following criteria for spontaneity of a reaction based on the values of $\Delta_{\text{r}}H^0$,$\Delta_{\text{r}}S^0$,and $\Delta_{\text{r}}G^0$:

$\Delta_{\text{r}}H^0$	$\Delta_{\text{r}}S^0$	$\Delta_{\text{r}}G^0$	Description
$(a) (+)$	$(-)$	$(+)$	$(i) \text{ Non-spontaneous at all temperatures}$
$(b) (-)$	$(-)$	$(-)$	$(ii) \text{ Spontaneous at low temperatures}$
$(c) (-)$	$(+)$	$(-)$	$(iii) \text{ Spontaneous at all temperatures}$