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Thermochemistry

Thermochemistry. Thermochemistry is the study of _________________ during chemical reactions. Energy. Capacity to do ______________ or supply ______________ 2 Types of Energy ______________ - stored energy ______________ - energy of motion Energy is weightless, odorless, and tasteless.

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Thermochemistry

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  1. Thermochemistry

  2. Thermochemistry is the study of _________________ during chemical reactions.

  3. Energy • Capacity to do ______________ or supply ______________ • 2 Types of Energy • ______________ - stored energy • ______________ - energy of motion • Energy is weightless, odorless, and tasteless

  4. Energy • Kinds of ______________ and their arrangements in a substance determines the amount of ______________ stored in that substance.

  5. Heat Energy • Energy ______________ from one object to another causing a temperature change. • It cannot be detected - only the ______________ by heat can be measured.

  6. Heat Energy • Thermometers measure the average ______________, not heat. • Heat flows from a ______________ object to a ______________ object.

  7. Heat Energy • All chemical reactions and changes in physical states have either an ______________ of heat or a ______________ of heat.

  8. Heat Energy • ______________ - What we focus on. • ______________ - everything else. • Heat flow is given in relation to the ______________

  9. System and Surroundings • The ______________ includes the molecules we want to study (here, the hydrogen and oxygen molecules). • The ______________ are everything else (here, the cylinder and piston).

  10. Heat Energy • Law of Conservation of Energy - energy cannot be ______________ nor ______________, it just changes from one form to another.

  11. 2 Processes of Heat Energy • ______________ - a process in which the system ______________ heat. • Value is ______________. • Symbol - +H • ______________ - a process in which the system ______________ heat. • Value is ______________. • Symbol - -H

  12. Units of Heat • Calorie - food calorie • ______________ - quantity of heat required to raise the temperature of 1 gram of water 1°C • Joule - more commonly used (J) • 1 C = 1000 cal • 4.184 J = 1 cal

  13. Heat Capacity • The amount of energy needed to ______________ the temperature of an object 1°C

  14. Specific Heat • Amount of ______________ needed to raise 1 gram of a substance 1°C • q = m • C • T • q = heat change • m = mass • C = specific heat • T = change in temperature

  15. Calorimetry • Measurement of heat changes • ______________ - heat content of a system at constant pressure • Heat released = Heat absorbed • ______________ is the device used to measure the change

  16. Calorimeter

  17. qlost = qgained • If q = m • C • T, then: • (m • C • T)lost = (m • C • T)gained

  18. How much heat is needed to warm 250 g of water (about 1 cup) from 22°C (about room temperature) to near its boiling point, 98°C? The specific heat of water is 4.18 J/g•ºC.

  19. Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.082 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0°C.

  20. In the laboratory, a student heats a 100 g sample of metal to 101.5ºC. The student then adds the metal to a calorimeter that has 100 mL of water at 22.5C. Calculate the specific heat of the metal if the final temperature of the metal-water mixture is 38.7C.

  21. Thermochemical Equations • An equation that includes the ______________.

  22. Reactant • Heat as a reactant will give a ______________ q value. • Heat as a reactant will be an ______________ process.

  23. Product • Heat as a product will give a ______________ q value. • Heat as a product will be an ______________ process.

  24. 2H2O(l) 2H2(g) + O2(g) Hrxn = 572 kJ • 572 kJ + 2H2O(l) 2H2(g) + O2(g)

  25. Al2O3(s) 2Al(s) + 3/2 O2(g) Hrxn = -1676kJ • Al2O3(s) 2Al(s) + 3/2 O2(g) + 1676 kJ

  26. Heat of Reaction • Conditions are standard conditions. • 1 atm or 101.3 kPa and 25°C • Hrxn

  27. Heat of Combustion • Complete ______________ of 1 mole of a substance. • Hcomb

  28. Heat of Fusion • Heat ______________ by one mole of a solid to change it to a liquid. • Hfus

  29. Heat of Solidification • Heat ______________ by one mole of a liquid to change it to a solid. • Hsolid

  30. Hfus = -Hsolid • Same process just in opposite direction.

  31. Heat of Vaporization • Amount of heat needed to change 1 mole of a ______________ to a ______________. • Hvap

  32. Heat of Condensation • Amount of heat needed to change 1 mole of ______________ to a ______________. • Hcond

  33. Hvap =-Hcond • Same process just in opposite direction

  34. Heat of Solution • Change in heat when dissolving a ______________ in a solvent. • Hsoln

  35. Hess’s Law of Summation Hess’s law states that “If 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.”

  36. If we know the individual reactions that make up the final equation, we can determine the Hrxn for the final equation.

  37. Individual Steps C3H8 (g) 3C(graphite) + 4H2 (g) 3C(graphite) + 3O2 (g)  3CO2 (g) 4H2 (g) + 2O2 (g)  4H2O (l)

  38. C3H8 (g) 3C(graphite) + 4H2 (g) 3C(graphite) + 3O2 (g)  3CO2 (g) 4H2 (g) + 2O2 (g)  4H2O (l) _______________________________ C3H8(g) + 5O2(g) 3CO2(g) + 4 H2O(l)

  39. Standard Heats of Formation • An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

  40. Standard Heats of Formation • Standard enthalpies of formation, Hf, are measured under standard conditions (25°C and 1.00 atm pressure).

  41. When we do not know the individual steps of a reaction, we can use the standard heats of formation to calculate H. • H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients.

  42. C3H8(g) + 5O2(g) 3CO2(g) + 4 H2O(l) H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ

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