Question 1[Maximum number: 2]Whether or not a chemical reaction is spontaneous (feasible) can be deduced by calculating the change in free energy, ΔG⊖\Delta G^{\ominus}ΔG⊖, at a given temperature.MgCO3( s)→MgO( s)+CO2( g)\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})MgCO3( s)→MgO( s)+CO2( g)ΔH⊖=+117 kJ mol−1ΔS⊖=+175JK−1 mol−1\begin{aligned} & \Delta H^{\ominus}=+117 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta S^{\ominus}=+175 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}ΔH⊖=+117 kJ mol−1ΔS⊖=+175JK−1 mol−1Calculate the value of ΔG⊖\Delta G^{\ominus}ΔG⊖ at 298 K for the above reaction.Mark as masteredShow AnswerΔG⊖=ΔH⊖−TΔS⊖=117−((298×175)/1000)=(+)64.85( kJ mol−1)\begin{aligned} \Delta G^{\ominus}=\Delta H^{\ominus}-\mathrm{T} \Delta S^{\ominus} =117-((298 \times 175) / 1000) =(+) 64.85\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right) \end{aligned}ΔG⊖=ΔH⊖−TΔS⊖=117−((298×175)/1000)=(+)64.85( kJ mol−1)Add to Test