Determine the enthalpy for this reaction: 2naoh(s)+co2(g)→na2co3(s)+h2o(l) express your answer in kilojoules per mole to one decimal place. view available hint(s) δhrxn∘ = kj/mol part b consider the reaction na2co3(s)→na2o(s)+co2(g) with enthalpy of reaction δhrxn∘=321.5kj/mol what is the enthalpy of formation of na2o(s)? express your answer in kilojoules per mole to one decimal place. view available hint(s)

We use the equation for the standard enthalpy change of formation:
     ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)
to calculate for the enthalpy for the reaction 
     2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)
We now have
     ΔHoreaction = { ΔHfo[Na2CO3(s)] + ΔHfo[H2O(l)] } - { ΔHfo[NaOH(s)] +   
                             ΔHfo[CO2(g)] }
where we use the following Enthalpy of Formation (∆Hfo) values:
     Substance      ΔHf∘ (kJ/mol) 
     CO2(g)           −393.509 
     H2O(l)            −285.830 
     Na2CO3(s)    −1130.68
     NaOH(s)        −425.609 

and taking note of the coefficients of the products and the reactants,
     ΔHoreaction = [1*(−1130.68) + 1*(−285.830)] − [2*(−425.609) + 1*(−393.509)]
                          = -1416.51 - (-1244.727)
                          = -171.783 kJ/mol 
                          ≈ -171.8 kJ/mol as our enthalpy for the given reaction.

Now considering the reaction 
     Na2CO3(s)→Na2O(s)+CO2(g) with enthalpy of reaction ΔHoreaction=321.5kJ/mol
we also use the equation for the standard enthalpy change of formation:
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)ΔHoreaction
                          = { ΔHfo[Na2O(s)] +  ΔHfo[CO2(g)] } - { ΔHfo[Na2CO3(s)] }
to solve for the enthalpy of formation of Na2O(s): 
     ΔHfo[Na2O(s)] = ΔHoreaction -  ΔHfo[CO2(g)] + ΔHfo[Na2CO3(s)]
Since the coefficients are all 1,
     ΔHfo[Na2O(s)] = 321.5 - (-393.509) + (-1130.68) 
     ΔHfo[Na2O(s)] = -415.671 kJ/mol ≈ -415.7 kJ/mol

29.1-12=17.1the answer is this one