Question

Calculate the standard enthalpy change for the phase transition reaction of solid calcium carbonate (CaCO3) to gaseous carbon dioxide (CO2) at 298 K, given the following formation enthalpies: ΔHf°(CaCO3) = -1206.9 kJ/mol ΔHf°(CO2) = -393.5 kJ/mol Also, assume that the specific heat capacity of CaCO3 is 0.82 J/g·K and that of CO2 is 0.85 J/g·K.

Answer 1

To calculate the standard enthalpy change for the phase transition reaction of solid calcium carbonate  CaCO3  to gaseous carbon dioxide  CO2  at 298 K, we can use the following equation:H reaction  =  Hf products  -  Hf reactants The reaction is: CaCO3 s   CaO s  + CO2 g First, we need to find the formation enthalpy of CaO. We can do this using the given formation enthalpies for CaCO3 and CO2:Hf CaCO3  = Hf CaO  + Hf CO2 -1206.9 kJ/mol = Hf CaO  - 393.5 kJ/molHf CaO  = -1206.9 kJ/mol + 393.5 kJ/mol = -813.4 kJ/molNow we can calculate the standard enthalpy change for the reaction:H reaction  = [Hf CaO  + Hf CO2 ] - Hf CaCO3 H reaction  = [ -813.4 kJ/mol  +  -393.5 kJ/mol ] -  -1206.9 kJ/mol H reaction  = -1206.9 kJ/mol + 1206.9 kJ/molH reaction  = 0 kJ/molSince the standard enthalpy change for the reaction is 0 kJ/mol, the phase transition reaction of solid calcium carbonate to gaseous carbon dioxide at 298 K is thermodynamically neutral. The specific heat capacities of CaCO3 and CO2 are not needed for this calculation, as they would be used to calculate the heat absorbed or released during a temperature change, not a phase transition.