Calculate the enthalpy of desorption of CO2 from sodium hydroxide (NaOH) pellets given the following information: - The heat of absorption of CO2 by NaOH pellets is -20.9 kJ/mol.- The standard enthalpy of formation of Na2CO3 is -1130 kJ/mol.- The standard enthalpy of formation of water is -286 kJ/mol.- The standard enthalpy of formation of NaOH is -425 kJ/mol. Assuming that the desorption process is the reverse of the absorption process, determine the enthalpy change for the reaction: Na2CO3(s) + H2O(l) → 2NaOH(s) + CO2(g).

Question

Calculate the enthalpy of desorption of CO2 from sodium hydroxide (NaOH) pellets given the following information: - The heat of absorption of CO2 by NaOH pellets is -20.9 kJ/mol.- The standard enthalpy of formation of Na2CO3 is -1130 kJ/mol.- The standard enthalpy of formation of water is -286 kJ/mol.- The standard enthalpy of formation of NaOH is -425 kJ/mol. Assuming that the desorption process is the reverse

Answer 1

To determine the enthalpy change for the reaction, we can use the following equation:H_reaction =  H_f products  -  H_f reactants For the given reaction: Na2CO3 s  + H2O l   2NaOH s  + CO2 g H_f Na2CO3  = -1130 kJ/molH_f H2O  = -286 kJ/molH_f NaOH  = -425 kJ/molThe enthalpy of formation for CO2 is not given, but we can calculate it using the heat of absorption of CO2 by NaOH pellets:H_absorption = -20.9 kJ/molSince the desorption process is the reverse of the absorption process, the enthalpy of desorption of CO2 is the opposite of the heat of absorption:H_desorption CO2  = 20.9 kJ/molNow we can calculate the enthalpy change for the reaction:H_reaction = [2  H_f NaOH  + H_desorption CO2 ] - [H_f Na2CO3  + H_f H2O ]H_reaction = [2   -425 kJ/mol  + 20.9 kJ/mol] - [-1130 kJ/mol - 286 kJ/mol]H_reaction = [-850 kJ/mol + 20.9 kJ/mol] - [-1416 kJ/mol]H_reaction = -829.1 kJ/mol + 1416 kJ/molH_reaction = 586.9 kJ/molThe enthalpy change for the reaction Na2CO3 s  + H2O l   2NaOH s  + CO2 g  is 586.9 kJ/mol.