± Enthalpy H rxn=products m H freactants n H f where the subscript \"rxn\" is fo

Question

± Enthalpy

Hrxn=productsmHfreactantsnHf

where the subscript "rxn" is for "enthalpy of reaction" and "f" is for "enthalpy of formation" and m and n represent the appropriate stoichiometric coefficients for each substance.

The following table lists some enthalpy of formation values for selected substances.

Part A

2NaOH(s)+CO2(g)Na2CO3(s)+H2O(l)

Express your answer in kilojoules per mole to one decimal place.

-169.9

SubmitHintsMy AnswersGive UpReview Part

Correct

Therefore

[1(1131.0kJ/mol) + 1(285.8kJ/mol)] [1(393.5kJ/mol) + 2(426.7kJ/mol)] = 169.9kJ/mol

The values 1, 1, 1, and 2 are the coefficients of the products and the reactants involved in the reaction.

Part B

Consider the reaction

Na2CO3(s)Na2O(s)+CO2(g)

Hrxn=321.5kJ/mol

Express your answer in kilojoules per mole to one decimal place.

(need help with the answer to part B, I already answered part A)

± Enthalpy

Enthalpy H is a measure of the energy content of a system at constant pressure. Chemical reactions involve changes in enthalpy, H, which can be measured and calculated:

Hrxn=productsmHfreactantsnHf

where the subscript "rxn" is for "enthalpy of reaction" and "f" is for "enthalpy of formation" and m and n represent the appropriate stoichiometric coefficients for each substance.

The following table lists some enthalpy of formation values for selected substances.

Substance Hf (kJ/mol) CO2(g) 393.5 NaOH(s) 426.7 H2O(l) 285.8 Na2CO3(s) 1131.0 H2O(g) 241.8

Part A

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.

Hrxn =

-169.9

  kJ/mol  

SubmitHintsMy AnswersGive UpReview Part

Correct

Therefore

[1(1131.0kJ/mol) + 1(285.8kJ/mol)] [1(393.5kJ/mol) + 2(426.7kJ/mol)] = 169.9kJ/mol

The values 1, 1, 1, and 2 are the coefficients of the products and the reactants involved in the reaction.

Part B

Consider the reaction

Na2CO3(s)Na2O(s)+CO2(g)

with enthalpy of reaction

Hrxn=321.5kJ/mol

What is the enthalpy of formation of Na2O(s)?

Express your answer in kilojoules per mole to one decimal place.

(need help with the answer to part B, I already answered part A)

Explanation / Answer

B) Na2CO3(s) Na2O(s)+CO2(g) Hrxn = 321.5 kJ/mol

Hrxn = Hfo(products) - Hfo( reactants)

  Hrxn = Hfo [Na2O(s)] + Hfo [CO2(g)] - Hfo [Na2CO3(s)]

321.5 kJ/mol = Hfo [Na2O(s)] - 393.5 kJ/mol - [-1131 kJ/mol ]

Hfo [Na2O(s)] = - 416 kJ/mol

Therefore,

  Enthalpy of formation of Na2O(s) = - 416 kJ/mol

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