1 / 30

7-7 Indirect Determination of  H : Hess’s Law

7-7 Indirect Determination of  H : Hess’s Law.  H is an extensive property. Enthalpy change is directly proportional to the amount of substance in a system. N 2 (g) + O 2 (g) → 2 NO(g)  H° = +180.50 kJ . ½ N 2 (g) + ½ O 2 (g) → NO(g)  H° = +90.25 kJ .

prince
Download Presentation

7-7 Indirect Determination of  H : Hess’s Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 7-7 Indirect Determination of H:Hess’s Law • H is an extensive property. • Enthalpy change is directly proportional to the amount of substance in a system. N2(g) + O2(g) → 2 NO(g) H° = +180.50 kJ ½N2(g) + ½O2(g) → NO(g) H° = +90.25 kJ • H changes sign when a process is reversed NO(g) → ½N2(g) + ½O2(g) H° = -90.25 kJ General Chemistry: Chapter 7

  2. Hess’s Law and Enthalpies • The ideas on the previous slide are familiar. ∆H is an intensive quantity. → The bigger the steak we want to barbecue the more moles (or grams) of propane we’ll have to burn. • H changes sign when a process is reversed. An analogy here would be climbing up or down a ladder. Climbing up our gravitational potential energy increases – climbing down our potential energy decreases (mgh rule!).

  3. All ∆H’s Need Not Be Measured • Elemental nitrogen and oxygen react to form a variety of oxides. We need not measure every ∆H experimentally – some can be calculated using data for a subset of all possible reactions. Reactions forming NXOY molecules can be exothermic or endothermic. • N2(g) + O2(g) → 2 NO(g) Endothermic • In the atmosphere lightning supplies the energy needed to form NO(g).

  4. Nature’s Synthesis of NO(g)! • The reaction of N2(g) and O2(g) to give NO(g) is an endothermic process. In nature, lightning storms supply the needed energy!

  5. All ∆H’s Need Not Be Measured • The reaction of NO(g) with additional oxygen is exothermic. • NO(g) + ½ O2(g) → NO2(g) Exothermic • Knowing the ∆H for this rxn and the one on the previous slide (by experiment) we can calculate ∆H for the rxn • ½ N2(g) + O2(g) → NO2(g) (red brown, LAX)

  6. Smog in Southern California • The formation of NO2(g) is one factor in smog formation.

  7. Hess’s Law Schematically General Chemistry: Chapter 7

  8. Hess’s Law of Constant Heat Summation ½N2(g) + O2(g) → NO(g) + ½ O2 (g) H° = +90.25 kJ NO(g) + ½O2(g) → NO2(g)H°= -57.07 kJ ½N2(g) + O2(g) → NO2(g)H° = +33.18 kJ • If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. General Chemistry: Chapter 7

  9. 7-8 Standard Enthalpies of Formation, Hof • The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. The standard enthalpy of formation of a pure element in its reference state is 0. General Chemistry: Chapter 7

  10. Br2(l) Br2(g) DHf° = 30.91 kJ Liquid bromine vaporizing General Chemistry: Chapter 7

  11. Diamond and graphite General Chemistry: Chapter 7

  12. Element 79! • Gold (Au), like most metals, is a shiny metal at room temperature (standard state!).

  13. You can Mix a Metal and a Nonmetal • If the “chemistry” is right! (One standard state?)

  14. Enthalpies of Formation • Heats of combustion of hydrocarbons are always –ve (exothermic reactions). Heats of formation of compounds can be +ve or –ve (endothermic and exothermic reactions). • ½ H2(g) + ½ Cl2(g) → HCl(g) ∆Hof = -92.3 kJ • ½ H2(g) + ½ I2(g) → HI(g) ∆Hof = +26.5 kJ • If we knew the ∆Hof for every substance we could calculate the ∆Ho value for every reaction.

  15. General Chemistry: Chapter 7

  16. number of carbons number of hydrogens Some standard enthalpies of formation at 298.15 K FIGURE 7-18 General Chemistry: Chapter 7

  17. Combining Heats of Formation • The process for calculating a general enthalpy change from heat of formation relies on the fact that enthalpy, H, is a state function. Thus, we can imagine proceeding from the reactants to the products by a one step (direct process) or by a two step process: • Step 1: Reactants → Constituent Elements • Step 2: Constituent Elements → Products

  18. PRODUCTS REACTANTS CONSTITUENT ELEMENTS

  19. Heats of Formation • The previous slide outlines a simple process for calculating Heats of Reactions (ΔHo’s) if the relevant Heats of Formation for all reactants and all products are known. • ΔHoRxn= ΣΔHof(Products) – ΣΔHof(Reactants) • In practice a particular temperature is specified for the Heats of Formation (usually 298K). Care must be taken to ensure that the correct phase for reactants and products has been specified.

  20. Standard Enthalpies of Reaction Hoverall = -2Hf°NaHCO3+ Hf°Na2CO3 +Hf°CO2 + Hf°H2O FIGURE 7-20 • Computing heats of reaction from standard enthalpies of formation General Chemistry: Chapter 7

  21. ∆H° = ∑np∆Hf°(products) - ∑nr∆Hf°(reactants) (7.21) Diagramatic representation of equation (7.21) FIGURE 7-21 General Chemistry: Chapter 7

  22. Examples: Use of Heats of Formation • Class examples: Show how to calculate the standard enthalpy change for each of the following reactions using Heats of Formation data. • (a) CaCO3(s) → CaO(s) + CO2(g) • (b) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) • (c) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

  23. Combustion Reactions – History • Combustion reactions for hydrocarbons are important sources of energy for many processes. One could easily put together a spreadsheet to calculate the enthalpies of combustion for all hydrocarbons – if the heats of formation of the hydrocarbons were known. Historically this is backwards as enthalpies of combustion are used in most cases to determine heats of formation for hydrocarbons. n-octane will be used as an example.

  24. Combustion of Liquid n-Octane – C8H18(l) “Elements” H (Enthalpy) (kJ∙mol-1) Reactants 8CO2(g) + 9H2O(l) Products

  25. Combustion of n-octane Class example: The standard enthalpy of combustion of n-octane was determined by experiment to be -5470.3 kJ∙mol-1 at 298K. Use heat of formation data for the products of combustion to determine the heat of formation of n-octane.

  26. Combining Thermochemical Equations • Unknown ΔH (and ΔU) values can be calculated by combining known thermochemical equations which are often more complex than those written for ΔHof’s. • It is often necessary to combine several thermochemical equations to obtain a ΔHo value not known from experiments. This requires that we compare the coefficients of substances appearing only once in the given and desired thermochemical equations.

  27. Combining ThemochemicalEqtns - Hydrazine • Example: Determine ∆Ho for the reaction • N2H4(l) + 2 H2O2(l) → N2(g) + 4 H2O(l) • Given: • N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ∆Ho = - 622.2 kJ • H2(g) + ½ O2(g) → H2O(l) ∆Ho = - 285.8 kJ • H2(g) + ½ O2(g) → H2O2(l) ∆Ho = - 187.8 kJ

  28. Combining Thermochemical Equations

  29. Results of Combining Equations

More Related