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Hess’s Law

Hess’s Law. Ch. 17.4. The First Law of Thermodynamics:. Energy can’t be created or destroyed, but can change form. Amount of energy in a system remains constant. A. Enthalpy (H). Heat content of a system at constant pressure

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Hess’s Law

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  1. Hess’s Law Ch. 17.4

  2. The First Law of Thermodynamics: • Energy can’t be created or destroyed, but can change form. • Amount of energy in a system remains constant

  3. A. Enthalpy (H) • Heat content of a system at constant pressure • The amount of heat that a system can potentially give to other systems (heat content of a system).

  4. 1. Why do we use ΔH vs H? • This internal energy (the sum of all possible forms of energy in the system) cannot be measured. • THEREFORE, instead of talking about how much enthalpy something has, we instead talk about how much the enthalpy of a system changes (hence the triangle) when heat is taken away from it or added to it. • This term is given the symbol ∆H, where ∆ represents the change in enthalpy that occurs during a process.

  5. 2. Adding Enthalpy to what you already have learned • ENDOTHERMIC: System to Surroundings • [bonds breaking] • Feels cold and WORK is done BY the SYSTEM • When energy of reactants is less than energy of products : + ΔH • Energy is absorbed • EXOTHERMIC: Surroundings to System • [bonds forming] • Feels hot and WORK is done ON the SYSTEM • When energy of reactants is greater than energy of products : - ΔH • Energy is released

  6. 3. Types of Enthalpy • A. Heat of reaction (∆Hrxn) • describes the enthalpy change that occurs as the result of a chemical reaction. • If ∆Hrxn is positive the reaction is endothermic (it has absorbed energy) and if ∆Hrxn is negative, it is exothermic (and has given off energy). • Heat of combustion (∆Hcomb) describes the change in enthalpy when something undergoes a combustion reaction. • B. Enthalpy changes having to do with phase changes: • Molar heat of vaporization (∆Hvap) is the change in enthalpy when one mole of a compound boils. • Molar heat of fusion (∆Hfus) is the change in enthalpy when one mole of a compound melts. • C. Standard heat of formation (∆H°f) • is the change in enthalpy when a compound is formed from its elements. Think of this as a standard or accepted value just as molecules always have the same molar mass. An important point to be made about the standard enthalpy of formation is that when a pure element is in its reference form it is zero.

  7. 4. Enthalpy is equal to… • The amount of energy + the change in pressure(P) and volume(V) ΔH = Δ E + P V • Since we are talking about solids and liquids, P and V are negligible. • A side note Δ E = q (heat) + W (work done on the system) • Δ H = Hfinal– Hinitial = Hproducts– Hreactants

  8. A. Heat of reaction (∆Hrxn) and Heat of solution (∆Hsoln)

  9. Example

  10. Sample Problem 17.3 (p. 515 -516) Use the balanced equation to determine the conversion factor to convert to the needed information/

  11. B. Standard heat of formation (∆H°f) ∆H0rxn = Σ∆H0f(products) – Σ∆H0f(reactants) • What the terms in this equation mean: • ∆H0rxn­ is the “standard heat of reaction” – The enthalpy change for the process under standard conditions. • Standard conditions: 1 atm, 250C • ∆H0f = “standard heat of formation” - the change in enthalpy that accompanies the formation of one mole of the compound in standard state. • Standard state is the form of the compound that exists at 250 C and 1 atm. • ∆H0f for all elements in their standard state is defined as zero. • Σ= “sum of”; just add them all together. • Let’s see how this works with some sample problems…

  12. Example: Standard heat of formation (∆H°f) Example: Determine the heat of combustion of methane: CH4(g) + 2 O2(g)­CO2(g) + 2 H­2O(g) Given the following standard heats of formation: carbon dioxide: -394 kJ/mol methane: -75 kJ/mol water: -242 kJ/mol Solution: Add the heats of formation of the products together and subtract the heats of formation of the reactants: Heats of formation of the products: CO2: -394 kJ H2O: 2(-242 kJ/mol) = -484 kJ Total: -878 kJ Heats of formation of the reactants: methane: -75 kJ oxygen: 2(0 kJ/mol) = 0 kJ Total: -75 kJ Subtracting, we get ∆H0rxn = (-878 kJ) – (-75 kJ) = - 803 kJ

  13. C. Hess’ Law of Heat of Summation • Allows you to determine the heat of reaction indirectly • Using two or more equations, you can add the enthalpy of each thermo-equation to find the heat of summation (∆Hrxn) • Hess’s Law: You can find the enthalpy of reaction for any process if you can add up two or more known reactions to describe the process. • Watch this BrightStorm for an example: • http://www.youtube.com/watch?v=u7aTBxA7sL8

  14. Example • Example: Given the following equations: S(s) + O2(g) SO2(g) ∆H = -297 kJ 2 SO3(g) 2 SO2(g) + O2(g) ∆H = 198 kJ • What is the heat of reaction for 2 S(s) + 3 O2(g) 2 SO3(g)?

  15. Example: Hints • Solution: Add the given equations together to get the equation we’re looking for. Because simply adding them up doesn’t always give us the answer we want, we can do the following tricks: • If you want to switch the products and reactants with one another, you need to also change the sign of the ∆Hrxn. • For example: If S(s) + O2(g) SO2(g) ∆H = -297 kJ, • then SO2(g) S(s) + O2(g)_ ∆H = +297 kJ. • If you need to multiply the number of times you use an equation, multiply the ∆H value by the same amount. • For example, If S(s) + O2(g) SO2(g) ∆H = -297 kJ, • then 2 S(s) + 2 O2(g) 2 SO2(g) ∆H = (2)(+297) = 594 kJ

  16. Let’s Apply and Solve • Given these rules, let’s see how we can add the equations together to give us the one we want: • 2 S(s) + 2 O2(g)2 SO2(g) ∆H = 2(-297 kJ) = -594 kJ • 2 SO2(g) + O2(g) 2 SO3(g) ∆H = -198 kJ • 2 S(s) + 3 O2(g) 2 SO3(g)∆H = -792 kJ

  17. D. Enthalpy changes having to do with phase changes:

  18. Δ H and Freezing/Melting • The amount of energy that’s added/removed from a substance during the freezing or melting process is described by the equation: ∆H = n ∆Hfus • ∆H = the enthalpy change for this process. • n = the number of moles of the compound melting or freezing (if ∆H is given in grams, then n is replaced with m (mass in g)) • ∆Hfus = the molar heat of fusion (which is a constant) • Example: How much energy is required to melt 56 grams of frozen water? • ∆H = (3.11 moles)(6.01 kJ/mol) = 18.7 kJ = 18,700 J

  19. Δ H and Freezing/Melting • Important: When undergoing the phase transition from liquid to solid (or vice versa), all of the energy goes into breaking intermolecular forces. As a result, the temperature of the material doesn’t change as it undergoes the transition! • This is why ice cubes keep a cold drink at 00 C until they have completely disappeared.

  20. Δ H and Boiling/Condensing • The amount of energy that’s added/removed from a substance during the boiling or condensing process is described by the equation: ∆H = n ∆Hvap • ∆H = the enthalpy change for this process. • n = the number of moles of the compound boiling or condensing(if ∆H is given in grams, then n is replaced with m (mass in g)) • ∆Hfus = the molar heat of vaporization (which is a constant) • Example: How much energy is required to boil 56 grams of frozen water? • ∆H = (3.11 moles)(40.7 kJ/mol) = 126.6 kJ = 126,600 J

  21. Δ H and Boiling/Condensing • Important: When undergoing the phase transition from liquid to gas (or vice versa), all of the energy goes into breaking intermolecular forces. As a result, the temperature of the material doesn’t change as it undergoes the transition! • This is why boiling water remains at 1000 C until all of the water has been boiled away rather than increasing in temperature!

  22. Proportional ΔHfusion = - ΔHsolidification ΔHvaporization = - ΔHsolidification SOLID to LIQUID ΔH of fusion LIQUID to SOLID ΔH of solidification LIQUID TO GAS ΔH of vaporization GAS TO LIQUID ΔH of condensation

  23. Thermochemical Equations • Thermochemical equations are balanced chemical equations that include the physical states of all reactants and products and the energy change. • If energy is a reactant, the reaction is endothermic • If energy is a product, the reaction is exothermic.

  24. More resources • http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html (Hess’s Law overview with perspective of the principle of conservation of energy and some practice problems) • http://www.youtube.com/watch?v=iETCSFit-zA (A video working examples)

  25. How are heat of _______ related to enthalpy? • http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Enthalpy_Of_Vaporization#Definitions

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