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:

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.

Exo vs. Endo

Exothermic reactions get

HOT

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.

Endothermic reactions get

COLD

Measuring Energy Changes

• Units – Calorie and Joules

– 1 calorie = 4.184 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=mCDT q = heat (J, Joule)m = mass (g, grams)C = Specific heat (J/g°C)DT = change in temperature (°C)

Rearranging the heat equation

Solve q = mCDT for each of the other variables:

m =

C =

DT = qmC

qCDT

qmDT

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 = mCDTq = m = C = DT =

q = (100g)(0.45 J/g°C)(-55°C)q = - 2475 J

?100g0.45 J/g°CTf – Ti = 25°C - 80°C = -55°C

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 = mCDTq = m = C = DT =

q = (75g)(0.45 J/g°C)(85°C)q = 2868.8 J

?75g0.45 J/g°CTf – Ti = 105°C - 20°C = 85°C

Thermodynamics

• Study of matter and energy interactions

H: enthalpy – heat content of a substanceS: entropy – disorder of a substanceG: 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 = ∑∆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”

• 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

Practice problemsCalculate 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/mol0

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

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/meal Food item Amount Calories

Breakfast Honey bunches of oats

1 cup 350

1% milk ½ cup 90