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Honors Chemistry Hess’s Law, Heats of Formation

Honors Chemistry Hess’s Law, Heats of Formation. Hess’s Law.  H is well known for many reactions, and it is inconvenient to measure  H for every reaction in which we are interested. However, we can estimate  H using published  H values and the properties of enthalpy. Hess’s Law.

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Honors Chemistry Hess’s Law, Heats of Formation

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  1. Honors ChemistryHess’s Law, Heats of Formation

  2. Hess’s Law • H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. • However, we can estimate H using published H values and the properties of enthalpy.

  3. Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

  4. Hess’s Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

  5. Hess’s Law What this means is that if you know the enthalpies of some reactions, you can find the enthalpies of others through mathematical calculation.

  6. Hess’s Law Using the following data, calculate ΔH3. (1) C(s) + O2(g)  CO2 (g) ΔH1 = -393.5 kJ (2) CO(g) + ½ O2(g)  CO2(g) ΔH2 = -283.0 kJ (3) C(s) + ½ O2(g)  CO(g) ΔH3 = ? We can rearrange the equations and add them, cancelling out reactants and products just like if it were an algebra equation. (Yes, in thermodynamic equations, it’s common to have fractional coefficients, because ΔH is measured in kJ/mol of the leading reactant.)

  7. Hess’s Law Using the following data, calculate ΔH3. (1) C(s) + O2(g)  CO2 (g) ΔH1 = -393.5 kJ (2) CO(g) + ½ O2(g)  CO2(g) ΔH2 = -283.0 kJ (3) C(s) + ½ O2(g)  CO(g) ΔH3 = -110.5 kJ

  8. Hess’s Law Calculate ΔH for the reaction 2 C(s) + H2(g)  C2H2(g) given the following chemical equations and their respective enthalpy changes: C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l) ΔH = -1299.6 kJ C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ H2(g) + ½ O2(g)  H2O(l) ΔH = -285.8 kJ

  9. Enthalpies of Formation An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which one moleof a compound is made from its constituent elements in their elemental forms. Example: Formation of HCl from H2 and Cl2 ½ H2 + ½ Cl2 HCl

  10. Standard Enthalpies of Formation Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). See Appendix of textbook for tables of Hf°.

  11. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g)

  12. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g)

  13. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

  14. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

  15. Calculation of H We can use Hess’s law in this way: H = nHf°products – mHf°reactants where n and m are the stoichiometric coefficients. Take the sum of the products’ heats of formation and subtract the sum of the reactants’ heats of formation.

  16. Practice • For which of the following reactions at 25 ºC would the enthalpy change represent a standard enthalpy of formation? • 2 Na(s) + ½ O2(g)  Na2O (s) • 2 K(l) + Cl2(g)  2 KCl(s) • C6H12O6(s)  6 C(diamond) + 6 H2(g) + 3O2(g)

  17. Practice • Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l). • ΔHfº: CO2(g) = -393.5 kJ/mol H2O (l) = -285.8 kJ/mol C6H6(l) = 49.0 kJ/mol O2 (g) = 0 kJ/mol

  18. Practice • Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l). • ΔHfº: CO2(g) = -393.5 kJ/mol H2O (l) = -285.8 kJ/mol C6H6(l) = 49.0 kJ/mol O2 (g) = 0 kJ/mol • -3267 kJ/mol

  19. Practice • The standard enthalpy change for the reaction CaCO3(s)  CaO(s) + CO2(g) is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) (-635.5 kJ) and CO2(g) (-393.5 kJ), calculate the standard enthalpy of formation of CaCO3(s).

  20. Practice • The standard enthalpy change for the reaction CaCO3(s)  CaO(s) + CO2(g) is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) (-635.5 kJ) and CO2(g) (-393.5 kJ), calculate the standard enthalpy of formation of CaCO3(s). • -1207.1 kJ/mol

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