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Thermochemistry The study of heat energy changes in chemical reactions. Importance: 1. Heat energy used for work. Driving car to school. 2. Safety considerations.

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Presentation on theme: "Thermochemistry The study of heat energy changes in chemical reactions. Importance: 1. Heat energy used for work. Driving car to school. 2. Safety considerations."— Presentation transcript:

1 Thermochemistry The study of heat energy changes in chemical reactions. Importance: 1. Heat energy used for work. Driving car to school. 2. Safety considerations. KABOOM!

2 What is Energy? Energy = capacity to do work Work = Force x distance “working” defn of energy = capacity to make objects move

3 Energy forms:

4 Link to video

5

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7 How badly do you want the “Free Stuff?” What could POTENTIALLY happen?

8 Potential Energy (PE) Stored energy due to position OR chemical composition. “Position” examples:

9 Potential Energy (PE) “Chemical composition” examples:

10 Kinetic Energy (KE) Energy of Motion

11 1 st Law of Thermodynamics Energy can neither be created nor destroyed! Energy can only be transformed from one form to another. Also known as the Conservation of Energy

12 The Ethanol Cannon: Example of First Law in Action.. Converting Stored Potential Chemical Energy into Kinetic Energy!

13 C 2 H 5 OH + O 2 → What type of reaction pattern is this? ANS: Combustion of a hydrocarbon What are the products? ANS: CO 2 + H 2 O

14 The reaction between ethanol and oxygen is thermodynamically favorable, yet nothing is happening, why? The collisions between molecules don’t have enough energy to overcome the activation barrier. What is happening What we want to happen

15 Meet Mr. Tesla Coil! As electricity travels across the gap between the two nails it will heat the gas molecules past the Ea

16 How does the potential energy (PE) and kinetic energy (KE) of the reactant ethanol and oxygen molecules compare to the PE and KE of the product CO 2 and H 2 O molecules? Ans: Reactants have higher PE but lower KE. Products are lower in PE (stronger bonds, more stable) but at a higher temp (higher KE). Extra KE pushes cork across room!

17 Since final product molecules CO 2 and H 2 O have lower potential energy but higher kinetic energy the difference in energy has been released as heat and light. Link to gases phet

18 Hot gases have higher kinetic energy push outwards against the cork, sending it across the room.

19 Reaction Profile Diagram for Ethanol Cannon

20 2 nd Law of Thermodynamics: Heat always flows from hot to cold Heat ≠ temperature Heat = energy (units: Joules or calories) Temperature is proportional to KE of molecules (units: ºC or K)

21 Heat ≠ Temperature Temperature : proportional to average kinetic energy of particles Heat – energy that flows from a hotter object (higher temperature) to a colder object (lower temperature)

22 Is the amount of material (mass) important in heat transfer? The cup will have the higher final temperature; Each individual molecule absorbs more energy

23 When same total amount of energy is distributed amongst a smaller number of molecules, each individual molecule receives more energy. 10 Energy Units split 2 ways 10 energy units split 5 ways Link to video

24 Do all objects absorb and store heat in the same way? If both cars are left in the sun in the senior lot for the same amount of time will the interiors be at the same temperature at lunch time? Black Porsche White Porsche

25 If both are left in the sun for one hours which will be at a higher temperature? WoodMetal Different substances absorb and store heat differently – called different specific heat capacities, symbol = s

26 “s” is Specific Heat Capacity s is the amount of energy needed to raise the temp of 1 g of a substance by 1ºC. H 2 O: s = 4.18 J/gºC Gold:s = 0.13 J/gºC If 1 g of water and 1 g of Au raise 1ºC, which substance has stored more heat?

27 Q = s×m×ΔT Q = heat absorbed or lost +Q if absorbed, -Q if lost m = mass in grams ΔT = temperature change: ΔT = T final – T initial s = specific heat capacity units of s are J/gºC

28 Link to heat transfer video

29 11-6 Self-Check #1 Calculate J of energy required to heat 454 g of water from 5.4 o C to 98.6 o C. s of H 2 O = 4.18 J/g o C. Q = ? or x m = 454 g T i = 5.4 o C ; T f = 98.6 o C ; ∆T = T f - T i = 98.6 -5.4 =93.2 o C Q = sm ∆T = (4.18 J/ g o C)(454 g)(93.2 o C) Q = 1.77 x 10 5 J


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