model 3: standard molar enthalpy of formation. 35 - addendum model 3: standard molar enthalpy of...

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ChemActivity 35 - Addendum Model 3: Standard Molar Enthalpy of Formation. Enthalpies of atom combination may be used to estimate enthalpy changes for chemical reactions, but they are not commonly used because they are not widely available. Instead, tables of the standard molar enthalpy of formation Δ ! ° are used since they are available from many sources. The standard molar enthalpy of formation is for the formation of one mole of a compound from its elements in their standard states. The following are examples: ½ H 2 (g) + ½ I 2 (g) HI (g); Δ ! ° = 26.48 / H 2 (g) + ½ O 2 (g) H 2 O (l); Δ ! ° = 285.830 / Hg (l) + 1/8 S 8 (s) HgS (s); Δ ! ° = 58.2 / The standard enthalpy of formation for elements in their standard states are zero. Standard state means 298 K and 1 atm. Critical Thinking Questions 12. What are the standard states of the following elements? a) Hydrogen b) Oxygen c) Mercury d) Sulfur 13. The reaction for the formation of liquid bromine from the element in its standard state is Br 2 (l) Br 2 (l). What is Δ ! ° for this reaction? 14. Which of the following reactions correspond to the Δ ! ° for the product? Explain your choices. a) 2 H 2 (g) + O 2 (g) 2 H 2 O (g) b) MnS (s) Mn (s) + 1/8 S 8 (s) c) Pb (s) + 2 Cl 2 (g) PbCl 4 (s)

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Page 1: Model 3: Standard Molar Enthalpy of Formation. 35 - Addendum Model 3: Standard Molar Enthalpy of Formation. Enthalpies of atom combination may be used to estimate enthalpy changes

ChemActivity 35 - Addendum

Model 3: Standard Molar Enthalpy of Formation. Enthalpies of atom combination may be used to estimate enthalpy changes for chemical reactions, but they are not commonly used because they are not widely available. Instead, tables of the standard molar enthalpy of formation Δ𝐻!° are used since they are available from many sources. The standard molar enthalpy of formation is for the formation of one mole of a compound from its elements in their standard states. The following are examples:

½ H2 (g) + ½ I2 (g) à HI (g); Δ𝐻!° = 26.48  𝑘𝐽/𝑚𝑜𝑙 H2 (g) + ½ O2 (g) à H2O (l); Δ𝐻!° =  −285.830  𝑘𝐽/𝑚𝑜𝑙 Hg (l) + 1/8 S8 (s) à HgS (s); Δ𝐻!° =  −58.2  𝑘𝐽/𝑚𝑜𝑙

The standard enthalpy of formation for elements in their standard states are zero. Standard state means 298 K and 1 atm. Critical Thinking Questions 12. What are the standard states of the following elements?

a) Hydrogen b) Oxygen c) Mercury d) Sulfur

13. The reaction for the formation of liquid bromine from the element in its standard state is Br2 (l) à Br2 (l). What is Δ𝐻!° for this reaction? 14. Which of the following reactions correspond to the Δ𝐻!° for the product? Explain your choices.

a) 2 H2 (g) + O2 (g) à 2 H2O (g) b) MnS (s) à Mn (s) + 1/8 S8 (s) c) Pb (s) + 2 Cl2 (g) à PbCl4 (s)

Page 2: Model 3: Standard Molar Enthalpy of Formation. 35 - Addendum Model 3: Standard Molar Enthalpy of Formation. Enthalpies of atom combination may be used to estimate enthalpy changes

ChemActivity 35 - Addendum

Model 4: Enthalpy Change for Chemical Reactions Δ𝐻° for a reaction can be calculated by taking the differences between the total enthalpies of the products from the total enthalpies of the reactants:

Δ𝐻° =  Δ𝐻!"#$%&'(° − Δ𝐻!"#$%#&%'° The calculation of Δ𝐻° for the two reactions from Model 1 are shown below. Reaction 1: N2O4 (g) à 2 NO2 (g) Using Δ𝐻!"°

Δ𝐻° =  −1932.93𝑘𝐽𝑚𝑜𝑙 − 2 −937.86  𝑘𝐽/𝑚𝑜𝑙

Δ𝐻° = −57.21𝑘𝐽𝑚𝑜𝑙

Using Δ𝐻!°

Δ𝐻° =  9.16𝑘𝐽𝑚𝑜𝑙 − 2 33.18  𝑘𝐽/𝑚𝑜𝑙

Δ𝐻° = −57.20𝑘𝐽𝑚𝑜𝑙

Reaction 2: N2 (g) + 3 H2 (g) à 2 NH3 (g) Using Δ𝐻!"°

Δ𝐻° =  2 −1171.76𝑘𝐽𝑚𝑜𝑙 — 945.418  𝑘𝐽/𝑚𝑜𝑙 + 3 −435.30

𝑘𝐽𝑚𝑜𝑙

Δ𝐻° = −92.21𝑘𝐽𝑚𝑜𝑙

Using Δ𝐻!°

Δ𝐻° =  2 −46.11 − 0𝑘𝐽𝑚𝑜𝑙 + 3 0  𝑘𝐽/𝑚𝑜𝑙

Δ𝐻° = −92.22𝑘𝐽𝑚𝑜𝑙

Critical Thinking Questions 15. Circle the Δ𝐻!"#$%&'(° in each of the calculations shown in the Model. 16. Verify by calculation that the Δ𝐻° using Δ𝐻!° for each reaction shown above are correct. 17. Are the results using Δ𝐻!"° vastly different from those using Δ𝐻!°? Why might this be the case?