Energy!
Energy: the capacity to do work or supplying heat
Energy is detected only by its effects
Energy can be stored within molecules etc as chemical potential
We will most often deal with energy changes with heat (q)
For systems at constant pressure (no P V work) then
= H (enthalpy)
energy
q
Measuring Heat (q):
units: Joule, calorie, Calorie (1 calorie = 4.184 J
1000 calories)
Use these equivalencies to convert between heat units:
An 80 Cal apple has how many J? kJ?
80 Cal
1
X
1000 cal
1 Cal
334720 J
1
X
X
4.184 J
1 cal
1 kJ
1000 J
= 334720 J
= 334.720 kJ
1 Calorie =
Heat is very often involved in chemical reactions
There are two possibilities of how heat is involved:
1) heat is absorbed by the reaction ENDOTHERMIC
2) heat is given off by the reaction
EXOTHERMIC
How much heat is given off or absorbed is measured with enthalpy (H)
We are more interested in the change in enthalpy therefore H
H = Hproducts - Hreactants
Exothermic reactions:
combustion reactions are obvious examples of exothermic reactions
C6H12O6 + 6O2
6CO2 + 6H2O + 2803 kJ
Heat is given off
Heat is a product in a chemical reaction
H of the products must be less than the H of the reactants
Therefore H must be negative
For this reaction H = - 2803 kJ
We can also make a Enthalpy diagram for this reaction
Reactants
H = -2803 kJ
Products
Reaction progress
Endothermic reactions:
- Br2 + Cl2 + 29.4 kJ
2 BrCl
Heat is absorbed by the reaction
Heat is a reactant in a chemical reaction
H of the products is more than the H of the reactants
Therefore H must be positive
For this reaction H = 29.4 kJ
This reactions enthalpy diagram looks like:
Products
H H = 29.4 kJ reactants
reaction progress
Products
H = 29.4 kJ
Reactants
Heat with no change of state:
- Specific heat (C) the amount of heat needed to raise 1 g of a substance 1 degree
q=mC
T ( T = Tf - Ti)
see chart pg. 296, therefore the C of water = 4.184 J
- Calorimetry
measurement of heat change
o EX: How much heat is absorbed by 875 mL of water as it is heated from 25o to 75o C
q=mC
T
q = (875)(4.184)(50)
q = 183050 J
o EX: 10.0 g of an unknown metal is heated to 100.0o C. The hot metal is added to 50 mL of water
at 25.0o C. The final temperature of the solution is 35.6o C. What is the specific heat capacity of
the unknown metal?
heat lost by metal = heat gained by water
-(m C
T) = m C
T
-(10)(x)(-64.4) = (50)(4.184)(10.6)
x = 3.44 J/ g(C)
q=
Hfus (mol)
q=
Hvap (mol)
Things to know: (where might one write them?)
Hfus = 6.01 kJ/mol
Hvap = 40.7 kJ/mol
Cice = 2.1 J/g . oC
Cwater = 4.184 J/g . oC
Csteam = 1.7 J/g . oC
q=mC
q=
Hfus (mol)
T
q=
Hvap (mol)
EX: How much heat is required to heat 10.0 g of ice at -25.0 oC to steam at 130.0 oC?
From -25 to 0:
q=mC T
q= (10)(2.1)(25)
q= 525 J
Boil the water:
q= Hvap (mol)
q= (40.7)(.556)
q= 22.63 kJ = 22630 J
Melt the ice:
q= Hfus (mol)
q= (6.01)(.556)
q= 3.34 kJ = 3340 J
Heat the steam to 130:
q=mC T
q= (10)(1.7)(30)
q= 510 J
Heat the water to boiling:
q=mC T
q= (10)(4.184)(100)
q= 4184 J
Add up the total heat:
Total q=
525+3340+ 4184+22630+510 =
31189 J = 31.2 kJ
EX: How much heat is required to heat 10.0 g of ice at -25.0 oC to steam at 130.0 oC?
From -25 to 0:
q=mC T
q= (10)(2.1)(25)
q= 525 J
Boil the water:
q= Hvap (mol)
q= (40.7)(.556)
q= 22.63 kJ = 22630 J
Melt the ice:
q= Hfus (mol)
q= (6.01)(.556)
q= 3.34 kJ = 3340 J
Heat the steam to 130:
q=mC T
q= (10)(1.7)(30)
q= 510 J
Heat the water to boiling:
q=mC T
q= (10)(4.184)(100)
q= 4184 J
from ice at -25 to water at 0o
525 + 3340 = 3865 J
Add up the total heat:
Total q=
525+3340+ 4184+22630+510 =
31189 J =from
31.2icekJ
at -25 to water at 100o
525 + 3340 + 4184 = 8049 J
from ice at -25 to steam at 100o
525 + 3340 + 4184 + 22630 = 30679 J
REVIEW
Exothermic reactions:
Heat is given off by the reaction
Heat is a product in a chemical reaction
Heat term is on the right
H must be negative
Endothermic reactions:
Heat is absorbed by the reaction
Heat is a reactant in a chemical reaction
Heat term is on the left
H must be positive
Now we can use this information in stoichiometry type problems (aren’t you happy!)
The heat term can be treated just like balanced coefficient on a substance in a mole ratio.
Ex: When 12.5 g of glucose (C6H12O6) is burned how many kJ of energy are released?
C6H12O6 + 6O2
6CO2 + 6H2O + 2803 kJ