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Section 15.1 –pg 517 Energy

Section 15.1 –pg 517 Energy . Main idea in this chapter: Energy can change forms and flow, but its always conserved. Key concepts: Energy, Law of Conservation of energy, chemical potential energy, measuring heat, specific heat, and calculating heat absorbed . Energy Types.

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Section 15.1 –pg 517 Energy

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  1. Section 15.1 –pg 517Energy Main idea in this chapter: Energy can change forms and flow, but its always conserved. Key concepts: Energy, Law of Conservation of energy, chemical potential energy, measuring heat, specific heat, and calculating heat absorbed

  2. Energy Types • Energy- is the ability to do work or produce heat. - Energy is expressed in Joules unit is J. 2 Basic forms of Energy: a) Potential Energy- the capacity and ability of energy to be in motion but is at rest . • Symbol PE expressed in J b) Kinetic energy- the energy of motion. • Symbol KE expressed in J

  3. A Rule followed by Energy • Law of conservation- in any chemical reaction or physical process, energy can be converted from one form to another, BUT it is neither created nor destroyed. ALSO called First law of Thermodynamics

  4. How do we measure energy? • One way to measure it is in the form of heat. • Heat- energy that is in the process of flowing from a warmer object to a cooler object. • We measure heat in calorie, units cal • Key concept: the amount of energy required to raise the temperature of one gram of pure water by one degree Celsius is defined as a calorie. • Calculations of how to Convert Energy Units- pg. 518

  5. Calculation Problem 1: Convert Calories to Joules • Question: A breakfast of cereal, orange juice, and milk might contain 230 nutritional Calories. Express it in joules. • What are we given? • What are we looking for? • What conversion factors or relationships are you going to use to find your answer? • Next setup a Stoichiometry setup

  6. Answer to Problem • Question: A breakfast of cereal, orange juice, and milk might contain 230 nutritional Calories. Express it in joules. • What are we given? • What are we looking for? • What conversion factors or relationship are you going to use to find your answer? • Given: 230 Calories • Looking for: Joules • Conversion factor: • 1000 cal / 1 Calories • 4.184 J/ 1cal Stoichiometry Setup: Use for Calories to Joules

  7. Heat versus Specific heat • The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celsius. • heat- which is energy that is in the process of flowing from a warmer object to a cooler object. • Some objects require more heat than others to raise their temperature.

  8. Calculating the Heat absorbed or specific heat. • The quantity of heat absorbed or released by a substance is equal to the product of its specific heat, the mass of the substance, and the change in its temperature. • Equation for calculating heat • q = c × m × ΔT • q = heat absorbed or released • c = specific heat of substance • m = mass of substance in grams • ΔT = change in temperature in Celsius

  9. Calculation Problem 2: Calculate Specific Heat • Question: The temperature of a sample of iron with a mass of 10 g changed from 50 degrees Celsius to 25 Celsius with the release of 114 J. What is the specific heat of iron? • What is given? • 1O g, 50 degrees Celsius and 25 degrees Celsius, 114 J • Use equation : q = c × m × ΔT Approach: • 1st calculate ΔT , do this by subtracting initial time to final time • 2nd solve for “c” by using c = q/ m x ΔT

  10. Answer to Calculation Problem 2:Calculate Specific Heat • What is given? • 1O g Fe, 50 degrees Celsius and 25 degrees Celsius, 114 J • Use equation : q = c × m × ΔT Approach: • 1st calculate ΔT , do this by subtracting initial temperature to final temperature • 2nd solve for “c” by using c = q/ m x ΔT

  11. Practice problem continued • 1st: 50 degrees Celsius minus 25 degrees equals 25 degrees Celsius • 2nd: c = q/ m x ΔT Remember: q = c × m × ΔT • q = heat, • c = specific heat, J/ (g x degrees Celsius ) • m = mass of substance ΔT = change in temperature • Solve: c = 114 J / (10g x 25 degrees Celsius) • c = .456 J/ (g x degrees Celsius)

  12. Recap vocabulary The heat required to raise one gram of a substance by one degree Celsius is called ____. A.joule B.calorie C.specific heat D.energy Which of the following is an example of chemical potential energy? A.the moon orbiting Earth B.a car battery C.a compressed spring D.a roller coaster at the top of a hill

  13. Recap vocabulary The heat required to raise one gram of a substance by one degree Celsius is called ____. A.joule B.calorie C.specific heat D.energy Which of the following is an example of chemical potential energy? A.the moon orbiting Earth B.a car battery C.a compressed spring D.a roller coaster at the top of a hill

  14. Section 15.2 –pg 523Heat Main idea in this chapter: The enthalpy change for a reaction is the enthalpy of the products minus of the reactants. Key concepts: calorimeter, thermochemistry, system, surroundings, universe, enthalpy, enthalpy (heat) of a reaction

  15. Energy can be released or absorbed • Energy that is released is Exothermic -Exothermic- the products in a balanced equation have less energy than reactants • Energy that is absorbed is Endothermic -Endothermic products in a balanced equation have more energy than reactants Exothermic Reaction example Endothermic Reaction example

  16. How do we measure heat? • A calorimeteris an insulated device used for measuring the amount of heat absorbed or released in a chemical reaction or physical process. How does this work? A sample is positioned in a steal inner chamber called the bomb, which is filled with oxygen at high pressure. Surrounding the bomb is a measured mass of water stirred by a low-friction stirrer to ensure uniform temperature. The reaction is initiated by a spark, and the temperature is recorded until it reaches its maximum.

  17. Who studies heat? • Thermochemistry- the study of heat changes that accompany chemical reactions and phase changes. • Three parts to Thermochemistry: • System, Surrounding, and Universe. • The system is the specific part of the universe that contains the reaction or process you wish to study. • The surroundingsare everything else other than the system in the universe. • The universeis defined as the system plus the surroundings.

  18. What is a system, universe, and surroundings? Closed system The pot, the steam, and the ingredients inside are the system. The Surrounding includes everything expect the system. The Universe- is the system plus the surroundings Open System

  19. Measuring the Heat of a Reaction • Energy can be either lost or gained in a reaction. • Enthalpy change- used to measure the energy absorbed or released as heat during a process • Symbol of Enthalpy: ΔHrxn and the units are kJ • How to calculate enthalpy ? • ΔHrxn = Hfinal– Hinitial ΔHrxn = Hproducts– Hreactants Thus, ΔHrxn is the difference between the enthalpy of the substance that exist at the end of the reaction and the enthalpy of the substance present at the start.

  20. How to read a Enthalpy change? How do you know if a reaction is endothermic or exothermic looking at its enthalpy number? • Example1: 2S + 2O2 2SO2ΔH or ΔH rxn = -594 kJ • Enthalpy change for example 1 is Negative = exothermic rxn Example 2: 2H2 + 02 2H20 ΔH or ΔH rxn = +483 kJ • Enthalpy change for example 2 is Positive = endothermic rxn.

  21. Recap Questions: 1. In thermochemistry, the specific part of the universe you are studying is called ____. A.system B.area C.enthalpy D.surroundings 2. What is the heat content of a system at constant pressure called? A.heat of reaction B.heat of the system C.enthalpy D.entropy

  22. Section 15.3 –pg 529Thermochemical Equations Main idea in this chapter: Thermochemcial equations express the amount of heat released or absorbed by chemical reactions. Key concepts: thermochemcial equation, enthalpy of (heat) of combustion, molar enthalpy (heat) of vaporization, and molar enthalpy (heat) of fusion.

  23. What are Thermochemcial equations? • A thermochemical equationis a balanced chemical equation that includes the physical states of all reactants and products, and energy change. • Example of thermochemcial equations: • C(graphite, s) + O2(g) → CO2(g) ΔH rxn = - 283. 0 kJ • Is this a exothermic or endothermic rxn? exothermic • How do we know this? The ΔH rxn is negative

  24. We have Standard Enthalpies of combustion for some substances • The enthalpy (heat) of combustion of a substance is the enthalpy change for the complete burning of one mole of the substance. • Symbol: ΔHcomb0 or ΔHc0 • Combustion is the reaction of a fuel with oxygen.

  25. Vaporization and Fusion enthalpies • Molar enthalpy (heat) of vaporizationrefers to the heat required to vaporize one mole of a liquid substance. • Symbol: ΔHvapo • Or ΔHvo • Molar enthalpy (heat) of fusionis the amount of heat required to melt one mole of a solid substance. • Symbol: ΔHfus0 or ΔHf0

  26. Phase change and enthalpy values • Look on the chart • Enthalpy of water gone through condensation is noted by ΔHcond • Enthalpy of water in a solid state is ΔHsolid

  27. How to calculate Energy release in a reaction? • Question: • A bomb calorimeter is useful for measuring the energy released in combustion reactions. The reaction is carried out in a constant-volume bomb with a high pressure of oxygen. How much heat is evolved when 54.0 g glucose (C6H12O6) is burned according to this equation? Equation: C6H12O6 (s) + 6O2(g)  6C02(g) + 6H20 (l) ΔHcomb = -2808 kJ

  28. Mass to Heat First Step: calculate moles • What is given? 54.0 g glucose C6H12O6 (s) + 6O2(g)  6C02(g) + 6H20 (l) ΔHcomb = -2808 kJ • What are we trying to find? • Heat, q in kJ. Second Step: Calculate Heat

  29. Mass to Heat: Answer • First Step: calculate moles • Second Step: Calculate Heat Your answer is 842 kJ

  30. Section 15.4 –pg 534Calculating Enthalpy Change Main idea in this chapter: The enthalpy change for a reaction can be calculated using Hess’s law Key concepts: Hess’s law, standard enthalpy (heat) of formation.

  31. Elements in their standard states have a When calorimeters don’t work, we use Hess’s law to measure the Enthalpy of a reaction. • Hess’s lawstates that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction. The formation The formation of compounds are placed above or below elements in their standard states.

  32. Calculating Enthalpy change of a reaction using Hess’s law • Given: 2S (s) + 3O2 (g) 2SO3(g) ΔH =? 2S + 2O2 2SO2ΔH = -594 kJ 2SO2 + O2 2SO3ΔH = -194 kJ _________________________________________________ • Just add the two enthalpies since the products and reactants are on the correct side for BOTH the equations: • ΔH = -788kJ

  33. Work this Hess’s law problem when the reactants/products order is rearranged. • Use equations (a) and (b) to determine ΔH for the following reaction: The red and blue are the reactants and the green is the product. 3CO(g) +2NO(g)  2CO2 (g) + N2 (g) ΔH = ? • (a) 2CO(g) + O2 (g) 2CO2(g) ΔH = -566.0 kJ • (b) N2 (g) + O2 (g)  2NO (g) ΔH = -180. 6 kJ I CAN”T ADD THE NUMBERS IN THIS ORDER! • I have to flip equation (b) to get 2NO on the reactant side! • 2NO  N2 (g) + O2 (g) ΔH = 180. 6 kJ (positive value) • 2CO(g) + O2 (g) 2CO2(g) ΔH = -566.0 kJ • Now I can add them to get ΔH = -385.4 kJ

  34. TRY IT ON YOUR OWN! USE HESS’S LAW TO FIND ΔH

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