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=
q
CDT
C=
q
mDT
q
DT = mC
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 = 100g
C = 0.45 J/g°C
DT = Tf – Ti = 25°C - 80°C = -55°C
q = (100g)(0.45 J/g°C)(-55°C)
q = - 2475 J
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 = 75g
C = 0.45 J/g°C
DT = Tf – Ti = 105°C - 20°C = 85°C
q = (75g)(0.45 J/g°C)(85°C)
q = 2868.8 J
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 = ∑∆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 problems
Calculate the ∆Hrxn for the following rxn:
Cl2 (g) + HBr (g) → 2HCl (g) + Br2 (g)
∆Hrxn = ∑Products - ∑Reactants
Cl2= 0
HBr = -36.23 KJ/mol
HCl = -92.30 KJ/mol
Br2 = 0
∆Hrxn = (2mol(-92.30KJ/mol) - (2mol(-36.23KJ/mol)
∆Hrxn = - 112.14KJ
Heat of Vaporization – energy change from Liquid → gas
Calculate the heat of vaporization for water:
H2O(l) → H2O(g)
∆Hrxn = ∑Products - ∑Reactants
H2O(g) = -241.8 KJ/mol
H2O(l) = -285.85 KJ/mol
∆Hrxn = -241.8 KJ/mol – (-285.85KJ/mol)
∆Hrxn = - 44.05 KJ/mol
Endothermic
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
Breakfast
Food item
Honey bunches
of oats
1% milk
Amount
Calories
1 cup
350
½ cup
90