ENERGY

1. ENERGY

2. Energy Review • Temperature – measurement of the random motion of the components of a substance • Heat – flow of energy due to temperature differences

3. In general, the universe is made up of two parts for thermodynamic purposes. • System – part of the universe in which you are interested • Surroundings – everything outside of that system

4. EXOTHERMIC REACTIONS • Energy is released. (Negative value) Examples: Combustion: 2C26H54 + 79 O2 52 CO2 + 54 H2O + Heat Precipitating: Na+(aq) + CH3COO-(aq)  NaCH3COO(s) + Heat Phase change: H2O(l) H2O(s) + Heat • Energy of reactants is greater than products. (See diagram on next slide.) • Energy flows out of the system into the surroundings.

6. Exo vs. Endo Exothermic reactions get HOT

7. Endothermic Reactions • Energy is absorbed. (Positive value) Examples: Phase changes: Heat + H2O(s) H2O(l) Dissolving: Heat + NH4Cl(s)  NH4+(aq) + Cl-(aq) • Energy of products is greater than energy of reactants. (See next slide.) • Energy flows INTO the system from the surroundings.

9. Endothermic reactions get COLD

10. Measuring Energy Changes • Units – Calorie and Joules • 1 calorie = 4.184 Joules • Example – Convert 60.1 calories of energy to joules

11. SPECIFIC HEAT CAPACITY • Amount of heat needed to raise 1 gram of a substance 1 Celsius. • Measures the ability of a substance to store heat energy. • When the temp of something, is changed heat is required. • The amount of heat depends on the amount (mass) and nature of the substance.

12. Heat equation q=mCDT q = heat (J, Joule) m = mass (g, grams) C = Specific heat (J/g°C) DT = change in temperature (°C)

13. q CDT q mDT q mC Rearranging the heat equation Solve q = mCDT for each of the other variables: m = C = DT =

14. Practice problems 1) How much heat is released when a 100g piece of iron (CFe =0.45 J/g°C) goes from 80°C to 25°C? q = mCDT q = m = C = DT = q = (100g)(0.45 J/g°C)(-55°C) q = - 2475 J ? 100g 0.45 J/g°C Tf – Ti = 25°C - 80°C = -55°C

15. Practice problems 2) How much heat is required to heat a 75g piece of iron (CFe = 0.45 J/g°C) from 20°C to 105°C? q = mCDT q = m = C = DT = q = (75g)(0.45 J/g°C)(85°C) q = 2868.8 J ? 75g 0.45 J/g°C Tf – Ti = 105°C - 20°C = 85°C

16. Thermodynamics • Study of matter and energy interactions H: enthalpy – heat content of a substance S: entropy – disorder of a substance G: Gibb’s free energy – chemical potential

17. Types of Thermodynamic Reactions • Exothermic • heat is given off • ∆H<0 - number • Endothermic • Heat is absorbed • ∆H>0 + number

18. ∆H˚rxn = ∑∆Hf˚products - ∑∆Hf˚reactants • ∆H˚rxn = enthalpy change for a rxn • ∆Hf˚ = heat of formation, how much E it takes to put substance together • ˚ = standard conditions (25˚C, 101.3kPa, 1.0M) • ∑ = “sum of”

19. Use tables to look up ∑∆Hf values. • Unit – KJ mol • All lone elements in a rxn: ∆Hf = 0 • ∆Hf Al = 0∆Hf O2 = 0 • Need Balanced equations • Must account for moles

20. Practice problems Calculate the ∆Hrxn for the following rxn: Cl2(g) + HBr(g) → 2HCl (g) + Br2 (g) ∆Hrxn = ∑Products - ∑Reactants Cl2= HBr= HCl= Br2 = ∆Hrxn= (2mol(-92.30KJ/mol) - (2mol(-36.23KJ/mol) ∆Hrxn= - 112.14KJ 0 -36.23 KJ/mol -92.30 KJ/mol 0

21. Heat of Vaporization – energy change from Liquid → gas Calculate the heat of vaporization for water: H2O(l) → H2O(g) ∆Hrxn = ∑Products - ∑Reactants H2O(g) = H2O(l) = ∆Hrxn = -241.8 KJ/mol – (-285.85KJ/mol) ∆Hrxn= - 44.05 KJ/mol Endothermic -241.8 KJ/mol -285.85 KJ/mol

23. Homework – Due 5/3 Food assignment: For one entire day keep track of what and how much you eat in a table. Use the food labels or the USDA website to determine how many calories each item contains. Due Tuesday.