In the conversion of limestone to lime,$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$,the values of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are $+179.1 \ kJ \ mol^{-1}$ and $160.2 \ J \ K^{-1} \ mol^{-1}$ respectively at $298 \ K$ and $1 \ bar$. Assuming that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not change with temperature,the temperature above which the conversion of limestone to lime will be spontaneous is ........... $K$.

For a chemical reaction,if $\Delta H > 0$ and $\Delta S > 0$,under what condition will the reaction not be spontaneous?

The criterion for a spontaneous process is

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 $bar$)

For a reaction $\Delta H = 9.08 \ kJ \ mol^{-1}$ and $\Delta S = 35.7 \ J \ K^{-1} \ mol^{-1}$. Which of the following statements is correct for the reaction?

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 ............ $K$.