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