The incorrect expression among the following is :

1. In isothermal process, ${\mathrm{W}}_{\text{reversible\hspace{0.17em}}}=-\mathrm{nRT}<mspace\; width="1em"></mspace>\ln <mspace\; width="1em"></mspace>\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}$

2. $\ln K=\frac{{\mathrm{\Delta H}}^{\circ }-{\mathrm{T\Delta S}}^{\circ }}{\mathrm{RT}}$

3. $\mathrm{K}={\mathrm{e}}^{-{\mathrm{\Delta G}}^{\circ }/\mathrm{RT}}$

4. $\frac{{\mathrm{\Delta G}}_{\text{system\hspace{0.17em}}}}{{\mathrm{\Delta S}}_{\text{total\hspace{0.17em}}}}=-\mathrm{T}$

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm³ to a volume of 100 dm³ at 27º C is :

1. 38.3 J mol⁻¹ K⁻¹

2. 35.8 J mol⁻¹ K⁻¹

3. 32.3 J mol⁻¹ K⁻¹

4. 42.3  J mol⁻¹ K⁻¹ 

Which of the following conditions is correct for a reversible reaction having both enthalpy change (ΔH, enthalpy change) and entropy change (ΔS, entropy change) positive, if Te is the equilibrium temperature?

1. T = Te (reaction is at equilibrium condition)

2.  Te > T (equilibrium temperature is greater than reaction temperature)

3.  T > Te (reaction temperature is greater than equilibrium temperature)

4.  Te = 5T (equilibrium temperature is five times reaction temperature)

Standard entropy of ${\mathrm{X}}_2,{\mathrm{Y}}_2$ and ${\mathrm{XY}}_3$ are $60,40$ and $50\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1},$ respectively. For the reaction, $\frac{1}{2}<mspace\; width="1em"></mspace>{\mathrm{X}}_2+\frac{3}{2}<mspace\; width="1em"></mspace>{\mathrm{Y}}_2<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>{\mathrm{XY}}_3∆\mathrm{H}=-30<mspace\; width="1em"></mspace>\mathrm{kJ},$ to be at equilibrium, the temperature will be : 

1. 500 K

2. 750 K

3. 1000 K

4. 1250 K

If a reaction is non-spontaneous at the freezing point of water but is spontaneous at the boiling point of water, then: