Question

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)

Answer 1

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