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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)

Samantha Barber

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Amanda Johnson on April 26, 2018

We use the equation for the standard enthalpy change of formation: ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)for the calculation of 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 products and 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, taking into account the reaction of Na2CO3(s)→Na2O(s)+CO2(g) with the enthalpy of the reaction ΔHoreaction=321.5 kJ/molwe also the use of 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)]Given that the coefficients are 1, ΔHfo[Na2O(s)] = 321.5 - (-393.509) + (-1130.68) ΔHfo[Na2O(s)] = -415.671 kJ/mol ≈ -415.7 kJ/mol


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