Consider the following reaction:
$A_{(g)} + 3 B_{(g)} \longrightarrow 2 C_{(g)} ; \Delta H^{\ominus} = -24 \ kJ$.
At $25^{\circ} C$,if $\Delta G^{\ominus}$ of the reaction is $-9 \ kJ$,the standard entropy change (in $J \ K^{-1}$) of the same reaction at the same temperature is:

For a reaction $A + B \rightarrow$ products,$\Delta H = -84.2 \ kJ$ and $\Delta S = -200 \ J \ K^{-1}$. Calculate the highest value of temperature (in $K$) so that the reaction will proceed in the forward direction.

Which of the following set of thermodynamic conditions is true for the reaction to be spontaneous at high temperature?

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 $bar$)

What is the nature of the reaction at $298 \ K$,if the entropy change and enthalpy change for a chemical reaction are $7.4 \ cal \ K^{-1}$ and $-2.5 \times 10^3 \ cal$,respectively?

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: