1 / 25

Thermochemistry

Thermochemistry. AP Chemistry Seneca Valley SHS. Chapter 6. 6.1 The Nature of Energy. Kinetic and Potential Energy Energy: the capacity to do work or produce heat. Kinetic energy: energy an object has due to its motion Potential energy: energy due to position or composition.

chet
Télécharger la présentation

Thermochemistry

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. Thermochemistry AP Chemistry Seneca Valley SHS Chapter 6

  2. 6.1 The Nature of Energy Kinetic and Potential Energy • Energy:the capacity to do work or produce heat. • Kinetic energy:energy an object has due to its motion • Potential energy:energy due to position or composition

  3. 6.1 The Nature of Energy Kinetic and Potential Energy Potential energy can be converted into kinetic energy. Example: a ball of clay dropping off a building.

  4. 6.1 The Nature of Energy Temperature:the average of energy of particles in motion. temp.  heat Heat:the transfer of energy from a hotter object to a colder one. State Function:property of a system that depends only on its present condition

  5. 6.1 The Nature of Energy Systems and Surroundings System:part of the universe we are interested in. Surroundings:the rest of the universe. Exothermic:reaction that involves the release of heat to the surroundings Endothermic:reaction that absorbs energy from the surroundings

  6. 6.1 The Nature of Energy Thermodynamics • Def: the study of energy and its interconversions First Law of Thermodynamics(Law of Conservation of Energy) Def:Energy cannot be created nor destroyed in ordinary chemical reactions

  7. 6.1 First Law of Thermodynamics Internal Energy • Internal Energy: total energy of a system. • Cannot measure absolute internal energy. • Change in internal energy, DE = Efinal - Einitial

  8. 6.1 First Law of Thermodynamics • Relating DE to Heat and Work • Energy cannot be created or destroyed. • Energy of (system + surroundings) is constant. • when a system undergoes a physical or chemical change, the change in internal energy is given by: the heat added to or absorbed by the system plus the work done on or by the system: • DE = q + w

  9. 6.1 First Law of Thermodynamics Relating DE to Heat and Work Thermodynamic quantities are measured from the perspective of the system Example: in an endothermic process where heat is flowing into the system, q would be positive (+)

  10. 6.2 Enthalpy Enthalpy: Heat transferred between the system and surroundings carried out under constant pressure. One can only measure the change(D in enthalpy (H): DH = Hfinal - Hinitial = qP

  11. 6.2 Enthalpy For any reaction: DHrxn = H(products) - H (reactants) Enthalpy is an extensive property (magnitude DH is directly proportional to amount): CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) DH = -1604 kJ When we reverse a reaction, we change the sign of DH: CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) DH = +802 kJ Change in enthalpy depends on state: H2O(g)  H2O(l) DH = -88 kJ

  12. 6.2 Calorimetry Heat Capacity and Specific Heat Calorimetry = science of measuring of heat flow. Calorimeter = device that measures heat flow. Heat capacity = the amount of energy required to raise the temperature of an object. Molar heat capacity = heat capacity of 1 mol of a substance. Specific heat capacity = heat capacity of 1 g of a subst. q = (specific heat)  (grams of substance) T.

  13. 6.2 Calorimetry Constant-Pressure Calorimetry Atmospheric pressure is constant! DH = qP qrxn = -qsoln = -(specific heat of solution)  (grams) of solution) DT.

  14. 6.2 Calorimetry Constant-Pressure Calorimetry Problem 1: When a student mixes 50.0 mL of 12.0 M HCl and 50.0 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0C to 27.5C. Calculate the enthalpy change per mole of water formed, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, the density of the solution is 1.00 g/mL, and its specific heat is 4.18 J/gC.

  15. 6.2 Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry) The reaction is carried out under constant vol. Usually used to study combustion. qrxn = -CcalorimeterT.

  16. 6.3 Hess’s Law • Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is equal to the sum of H’s from each individual step. • For example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ 2H2O(g)  2H2O(l) H = -88 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ

  17. 6.3 Hess’s Law Notes on Hess’s Law • If a reaction is reversed in order to solve a Hess’s Law problem, then the sign of the Hof the reaction is also reversed. • The magnitude of H is directly proportional to the quantities of reactants and products. If the coefficients in an equation are multiplied by an integer, the H is also multiplied by that integer • Hess’s Law can be used to calculate theH in a reaction; however, the component equations do not necessarily tell us how the overall reaction occurred.

  18. 6.3 Hess’s Law In the above enthalpy diagram note that: H1 = H2 + H3

  19. 6.3 Hess’s Law Example (ex. 6.8 pg. 258): Diborane (B2H6) is a highly reactive boron hydride. Calculate H for the synthesis of diborane from its elements, according to the following equation:2 B(s) + 3 H2 (g) --> B2H6 (g) Use the following data to help you: 2 B(s) + 3/2 O2(g) --> B2O3(s) H = -1273 kJ B2H6(g) + 3 O2(g) --> B2O3(s) + 3 H2O(g)H = -2035 kJ H2 (g) + 1/2 O2(g) --> H2O (l) H = -286 kJ H2O(l) --> H2O(g)H = 44 kJ

  20. 6.4 Enthalpies of Formation • If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hof . • Degree symbol in Hof indicates standard conditions… • Standard conditions 1 atm and 25 oC (298 K). • If there is more than one state for a substance under standard conditions, the more stable one is used. • Learn the distinction between standard state and standard conditions---they are not the same!

  21. 6.4 Enthalpies of Formation • Standard enthalpy of formation of the most stable form of an element is zero.

  22. 6.4 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction:

  23. 6.4 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction Example 1: (Ex. 6.9)Use the standard enthalpies of formation listed in Table 6.2 to calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid.

  24. 6.4 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction Example 2: (Ex. 6.10) Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction, which occurs when a mixture of powdered aluminum and iron (III) oxide is ignited with a magnesium fuse: 2 Al(s) + Fe2O3(s) --> Al2O3 (s) + 2 Fe(s)

  25. To save time we will not formally cover 6.5-6.6 in class (thus you won’t be tested on it!) Thermochemistry End of Chapter 6

More Related