1. Unit VIII: Energy and Chemical Change

2. Introduction Energy is the essence of our very existence as individuals and as a society The food that we eat furnishes the energy to live, work, and play The dwindling supply (and rising cost) of fossil fuels have forced use to look for alternative energy sources

3. The Nature of Energy Energy is the ability to do work or produce heat It exists in 2 basic forms: 1. Potential energy Energy due to composition or position of an object Examples: Water behind a dam, calories in food, energy stored in bonds (chemical PE) 2. Kinetic energy The energy of motion (KE = ½ mv 2 ) Examples: A roller coaster “rolling” down a hill, Motion of particles in states of matter

4. One of the most important characteristics of energy is that it is conserved The Law of Conservation of Energy states that energy can be converted from one form to another, but can neither be created nor destroyed The energy in the universe is constant Also known as the First Law of Thermodynamics

5. Temperature & Heat Temperature is the measure of the average KE of particles Measured in Kelvin (K – SI unit), °C, and °F Note: Recall your conversion formulas Heat is how thermal energy is transferred The flow of energy (due to a temperature difference) Measured in Joules (SI unit for energy) Symbol → q

6. There are 2 common units of energy 1. Calorie (c) [little c - C represents kilocalories] Unit for the metric system The amount of energy (heat) required to raise the temp. of 1 g of water 1°C 1 kilocalorie (or Calorie) = 1000 calories 2. Joule Unit for the SI system 1 cal = 4.184 J (expressed as 4.184 J / g °C) Named after James Prescott Joule

7. Specific Heat Substances respond differently to being heated 4.184 J raises 1 g of H 2 O 1°C The same amount of energy applied to 1 g of Au would raise the temp. by  32°C This difference is known as Specific Heat Capacity The amount of heat required to raise the temp. of 1 g of a substance by 1°C

8. Formula: Examples → Next Page! Q = c m Δt Energy (heat) absorbed or released (J) Specific Heat Capacity (J / g °C) Mass (g) Change in temperature (°C) Δt = t f - t i

9. Some common specific heat values:

10. Example 1: If the temperature of 34.4 g of ethanol increases from 25.0°C to 78.8°C, how much heat has been absorbed? Q = c m Δt = 2.44 J/g°C 34.4 g (78.8°C - 25.0°C) = 4.52 x 10 3 J Specific Heat (J/g°C ) (looked up in table) Mass (g) Change in temperature (°C) Δt = t f - t i

11. Example 2: A.158 oz nugget of pure gold absorbed 66.0 cal of energy. If the initial temperature was 77.0°F, what was the final temperature? Δt = Q c m = 276 J (.129 J/g°C 4.51 g) 66.0 cal = 276 J = 475°C Δt = t f - t i t f = Δt + t i = 475°C + 25°C = 500°C.158 oz = 4.51 g Specific Heat for Gold

12. Example 3: A 5.50 g sample of an unknown metal is heated from 25.0°C to 95.0°C. During the process, 345 J are absorbed by the metal. A) What is the specific heat of the metal? B) What is the identity of the unknown metal sample? c = Q m Δt = 345 J (5.50 g 70.0°C) =.896 J/g°C = Aluminum

13. Heat Have you ever wondered how food chemists obtain the Calorie information that appears on packaged food? These packages record the results of combustion reactions carried out in calorimeters An insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process Bomb calorimeter

14. Satisfactory results can be obtained in calorimetry experiments using a much simpler foam-cup calorimeter

15. Thermochemistry Virtually every chemical reaction either releases or absorbs heat Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes The universe is divided into 2 systems: 1. System Specific part of the universe the reaction you want to study 2. Surroundings Everything else

16. Thermodynamic quantities consist of 2 parts: 1. Number - Gives the magnitude of the change 2. Sign - Indicates the direction A. Exothermic exo – a prefix meaning “out of” Energy flows out of the system Example: Match (warmer to the touch) B. Endothermic endo – a prefix meaning “from within” Process that absorbs energy from the surroundings Example: Boiling water forms steam (cooler to the touch)

17. Consider lighting a match… Is this an exothermic or endothermic process?

18. Chemists like to know how much energy is produced or absorbed by a given reaction This is known as enthalpy (H) You might also see this written as follows: Example: Heat pack ΔH rxn = H final - H initial 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) ΔH = -1625 KJ

19. Thermochemical Stoichiometry Consider the following reaction… How much heat is released when 1.00 g of Fe (s) is reacted with excess O 2 (g)? 1.00 g Fe → ΔH 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) ΔH = -1625 KJ 1.00g Fe mol Fe 55.85 1 4 -1625 KJ = - 7.27 KJ

20. Hess’ Law In going from a particular set of reactants to a particular set of products, the change in enthalpy (ΔH) is the same whether the reaction takes place in one step of in a series of steps

21. Hess’ Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ Step #1: CH 4 must appear on the reactant side, so we reverse reaction #1 and change the sign on  H. CH 4  C + 2H 2 +74.80 kJ

22. Hess’ Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ Step #2: Keep reaction #2 unchanged, because CO 2 belongs on the product side C + O 2  CO 2 -393.50 kJ

23. Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ C + O 2  CO 2 -393.50 kJ Hess’ Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Step #3: Multiply reaction #2 by 2 2H 2 + O 2  2 H 2 O -571.66 kJ

24. Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ C + O 2  CO 2 -393.50 kJ Hess’ Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O 2H 2 + O 2  2 H 2 O -571.66 kJ CH 4 + 2O 2  CO 2 + 2H 2 O-890.36 kJ Step #4: Sum up reaction and  H