Values of $\Delta H$ and $\Delta S$ for five different reactions are given below. On the basis of these values,predict which one of these will be spontaneous at all temperatures.

Reaction	$\Delta H \ (kJ \ mol^{-1}) / \Delta S \ (J \ K^{-1} \ mol^{-1})$
$I$	$+98.0, +14.8$
$II$	$-55.5, -84.6$
$III$	$+28.3, -17.0$
$IV$	$-40.5, +24.6$
$V$	$+34.7, 0.0$

For the reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$,$\Delta H = -285.8 \ kJ \ mol^{-1}$,$\Delta S = -0.163 \ kJ \ mol^{-1} K^{-1}$. What is the value of free energy change at $27 \ ^\circ C$ for the reaction in $kJ \ mol^{-1}$?

Which of the following reactions is expected to never be spontaneous?

Calculate the work done if $1 \ mole$ of a certain gas is compressed isothermally and reversibly at $300 \ K$ from an initial pressure $x \ bar$ to a final pressure $2x \ bar$ $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$. (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 $bar$)

Under isothermal and reversible conditions,the term "free energy" in thermodynamics signifies