For the reaction at $298 \ K$,$2 \ A^{+}B \rightarrow C$. $\Delta H = 400 \ kJ \ mol^{-1}$ and $\Delta S = 0.2 \ kJ \ mol^{-1} \ K^{-1}$. The reaction will become spontaneous above $...... \ K$.

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

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 certain reaction,the enthalpy change and entropy change are $40.63 \ kJ \ mol^{-1}$ and $100 \ J \ K^{-1} \ mol^{-1}$ respectively. What will be the value of $\Delta G$ at $27 \ ^oC$?

Which of the following relations is correct?

An ideal gas expands from an initial volume of $5 \text{ dm}^3$ to $15 \text{ dm}^3$ against a constant external pressure of $2 \text{ atm}$. What is the work done by the gas (in $\text{ J}$)?