ENERGY

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

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

EXOTHERMIC REACTIONS Energy is released. (Negative value) Examples: Combustion: 2C 26 H O 2  52 CO H 2 O + Heat Precipitating: Na + (aq) + CH 3 COO - (aq)  NaCH 3 COO (s) + Heat Phase change: H 2 O (l)  H 2 O (s) + Heat Energy of reactants is greater than products. (See diagram on next slide.) Energy flows out of the system into the surroundings.

Exo vs. Endo Exothermic reactions get HOT

Endothermic Reactions Energy is absorbed. (Positive value) Examples: Phase changes: Heat + H 2 O (s)  H 2 O (l) Dissolving: Heat + NH 4 Cl (s)  NH 4 + (aq) + Cl - (aq) Energy of products is greater than energy of reactants. (See next slide.) Energy flows INTO the system from the surroundings.

Endothermic reactions get COLD

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

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.

Heat equation q=mC  T q = heat (J, Joule) m = mass (g, grams) C = Specific heat (J/g°C)  T = change in temperature (°C)

Rearranging the heat equation Solve q = mC  T for each of the other variables: m = C =  T = q mC qCTqCT qmTqmT

Practice problems 1) How much heat is released when a 100g piece of iron (C Fe =0.45 J/g°C) goes from 80°C to 25°C? q = mC  T q = m = C =  T = q = (100g)(0.45 J/g°C)(-55°C) q = J ? 100g 0.45 J/g°C Tf – Ti = 25°C - 80°C= -55°C

Practice problems 2) How much heat is required to heat a 75g piece of iron (C Fe = 0.45 J/g°C) from 20°C to 105°C? q = mC  T q = m = C =  T = q = (75g)(0.45 J/g°C)(85°C) q = J ? 75g 0.45 J/g°C Tf – Ti = 105°C - 20°C= 85°C

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

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

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

Use tables to look up ∑∆H f values. Unit – KJ mol All lone elements in a rxn: ∆H f = 0 ∆H f Al = 0 ∆H f O 2 = 0 Need Balanced equations Must account for moles

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

Heat of Vaporization – energy change from Liquid → gas Calculate the heat of vaporization for water: H 2 O (l) → H 2 O (g) ∆H rxn = ∑Products - ∑Reactants H 2 O (g) = H 2 O (l) = ∆H rxn = KJ/mol – ( KJ/mol) ∆H rxn = KJ/mol Endothermic KJ/mol KJ/mol

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. Time/mealFood itemAmountCalories BreakfastHoney bunches of oats 1 cup350 1% milk½ cup90