CGPET 2018 Solved Paper

For the given reaction. $\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s)$ At low temperature, the value of $\Delta n = 0 - 1 = (-) 1$ $\because \Delta n =$ moles of gaseous products $-$ moles of gaseous reactants For the reaction contain gaseous species. $\Delta H = \Delta E + \Delta n R T$ where, $\Delta H =$ enthalpy change $\Delta E =$ internal energy change $\Delta n =$ change in number of moles $R =$ gas constant $T =$ temperature in Kelvin. ∵ $\Delta n = (-) 1$ $\Delta H = \Delta E - R T$ means $\Delta H < \Delta E$ (negative) and of exothermic nature. For the reaction to occur spontaneously, $\Delta G = \Delta H - T \Delta S$, value of $\Delta G$ must be negative. At low temperature for $\Delta G$ and $\Delta H$ to be negative. $\because \Delta n = (-) 1$ So, $\Delta S$ must be negative, as number of gaseous moles in product side are less than of reactant side. Hence, the correct set is $\Delta H =$ negative $\Delta S =$ negative $\Delta G =$ negative $\therefore (b)$ is the correct option.