a) Determine the enthalpy change for this reaction: 2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)b_ image attached Consider the reactionNa2 CO3 (s)→Na2 0(s) + CO2 (g)with enthalpy of reaction321.5kJ/molrxnWhat is the enthalpy of formation of Na2 0(s)?Express your answer in kilojoules per mole to one decimal place. Enthalpy H is a measure of the energy content of asystem at constant pressure. Chemical reactionsinvolve changes in enthalpy, AH, which can bemeasured and calculated:ΔΗ Σ μιoducts mΔΗ -Σctants nΔΗ; οwhere the subscript "rxn" is for "enthalpy of reaction"and "f" is for "enthalpy of formation" and m and nrepresent the appropriate stoichiometric coefficients foreach substance.The following table lists some enthalpy of formationvalues for selected substances.SubstanceAH;° (kJ/mol)CO2 (g)-393.5NaOH(s)-426.7H2O(1)-285.8NażCO3 (s)-1131.0H2O(g)-241.8

Given : We have predict the enthalpy change for the given reaction: 2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)Calculations: a) 2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)∆H = ∆HProducts -∆Hreactants ∆H =∆HNa2CO3 + ∆HH2O-2∆HNaOH-∆HCO2∆H = -1131KJ/mol -285.8KJ/mol - 2 x (-426.7KJ/mol) -(-393.5KJ/mol)∆H = -169.9KJ/molb)Na2CO3(s)-->CO2(g) + Na2O(s)∆H = ∆HProducts -∆Hreactants ∆H =∆HCO2 + ∆HNa2O-∆HNa2CO3321.5KJ/mol = -393.5KJ/mol + ∆HNa2O +1131KJ/mol∆HNa2O = -416KJ/mol