For the reaction $PCl_5 \to PCl_3 + Cl_2$,$\Delta H = 75 \ kJ \ mol^{-1}$ and $\Delta S = 120 \ J \ K^{-1} \ mol^{-1}$. At which temperature will the reaction be spontaneous? (in $K$)

Calculate the work done when $1 \text{ mole}$ of an ideal gas is expanded reversibly and isothermally from an initial pressure $P_1 = 10 \text{ bar}$ to a final pressure $P_2 = 1 \text{ bar}$ at a constant temperature $T = 300 \text{ K}$. $[R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}]$ (in $\text{ kJ}$)

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

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 $bar$)

The occurrence of a reaction is impossible if

$A$ reaction,$A + B \rightarrow C + D + q$,is found to have a positive entropy change. The reaction will be: