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Unit 12 (Chapter 11) Thermochemistry. The Flow of Energy--Heat  Thermochemistry—heat changes that occur during chemical reactions  Energy—ability to.

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Presentation on theme: "Unit 12 (Chapter 11) Thermochemistry. The Flow of Energy--Heat  Thermochemistry—heat changes that occur during chemical reactions  Energy—ability to."— Presentation transcript:

1 Unit 12 (Chapter 11) Thermochemistry

2 The Flow of Energy--Heat  Thermochemistry—heat changes that occur during chemical reactions  Energy—ability to do work or supply heat  Chemical potential energy—energy stored within structural units of a chemical substance  Heat (q)—energy that flows from warmer objects to cooler objects –Heat cannot be detected, only changes caused by heat

3  All reactions and changes in state involve the release or absorption of heat  System—chemicals and their reaction  Surroundings—everything else (beaker, air, etc)  Combined = universe  Direction of heat flow is always given from the point of view of the system  + q = endothermic—heat flows into the system from the surroundings—heat is absorbed  System gains heat; surroundings cool down  -q = exothermic—system loses heat  System loses heat; surroundings heat up Look at Page 294

4 Heat Capacity and Specific Heat  calorie—quantity of heat needed to raise temperature of 1 g of pure H 2 O 1°C 1 Calorie = 1kc = 1000 calories  Calorie—dietary use only, refers to the energy in food

5  joule—SI unit of heat and energy 1 J =.2390 calories 4.184 J = 1 cal

6  heat capacity—amount of heat needed to raise the temperature of a whole sample of substance exactly 1° C –Depends on mass and composition mass = heat capacity mass = heat capacity

7  specific heat capacity—(C) amt of heat needed to raise temperature of 1g of a substance 1° C q = mC∆T q = heat ∆T = T f – T i C = specific heat (from table) m = mass (g) C = q__ m∆T m ∆T See table 11.2 (p. 296) – Use values that contain J as a unit

8 Sample problems When 435 J of heat is added to 3.4 g olive oil at 21ºC, the temperature increases to 85ºC.What is the specific heat of olive oil? q = 435 J m = 3.4 g ∆T = 85 °C – 21 °C = 64 °C C = 435 J_____ 3.4 g x 64 °C = 1.999 J/g °C = 2. 0 J/g °C

9 Measuring and Expressing Heat Changes  Calorimetry—the accurate and precise measurement of heat change for chemical and physical processes. –The heat released by the system = heat gained by the surroundings  Heat lost = Heat gained  Calorimeter—device used to measure heat changes  styrofoam cup—simple calorimeter, run at a constant pressure  See page 300  bomb calorimeter –run at constant volume; measures heat released from burning a compound

10 Enthalpy (H)  for systems at constant pressure, the heat content (q) is the same as a property called enthalpy (H) of the system Enthalpy – the heat content of a system at constant pressure.  heat changes symbolized as ∆H  so…q = ∆H = mC∆T  Exothermic Reaction-- ∆H is -, (∆H < 0)  Endothermic Reaction-- ∆H is +, (∆H > 0)

11 A student mixed 50.0 mL of water containing 0.50 mol HCl at 22.5ºC with 50.0 mL of water containing 0.50 mol NaOH at 22.5 ºC in a foam cup calorimeter. The temperature of the resulting solution increased to 26.0ºC. How much heat in joules (J) was released by this reaction?  50.0 mL HCl + 50.0 mL NaOH = 100.0 mL solution  100.0 mL x 1 g/mL = 100.0 g of solution  T f = 26.0 °C  T i = 22.5 °C  C of water = 4.18 J/g°C  ∆H = ?  ∆H = mc∆T  ∆H = 100.0 g ∙ 4.18 J/g°C (26.0 °C – 22.5 °C)  ∆H = 100.0 g ∙ 4.18 J/g°C ∙ 3.5 °C  ∆H = 1463 J = 1500 J

12 Thermochemical Equations  Thermochemical equations are equations that include the heat change  Heat of Reaction—heat change for the equation exactly as written  usually reported as ∆H  physical states must also be given  heat as reactant = endothermic  heat as product = exothermic Exothermic CaO(s) + H 2 O(l) → Ca(OH) 2 + 65.2 kJ Endothermic 2NaHCO 3 (s) + 129 kJ → Na 2 CO 3 (s) + H 2 0(l) + CO 2 (g)

13 How much heat is required to raise the temperature of 250.0 g of mercury 52ºC? c Hg = 0.14 J/g °C - (chart value) q = ? m = 250.0 g Hg ∆T = 52 °C q = m c ∆T q = 250.0 g ∙ 0.14 J ∙ 52 °C g °C q = 1820 J q = 1.8 kJ


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