Transcript AP Notes Chapter 6

AP Notes Chapter 5
Principles of Reactivity:
Energy and Chemical Reactions
Thermodynamics
•study of energy transfer or heat flow
Energy
•Kinetic Energy
Thermal - Heat
Mechanical
Electrical
Sound
•Potential Energy
Chemical
Gravitational
Electrostatic
Energy is...



The ability to do work.
Conserved.
made of heat and work.



Work is a force acting over a distance.
W=F x d
Power is work done over time.
P=W
t
Heat is energy transferred between objects
because of temperature difference.
First Law of Thermodynamics

Law of conservation of energy
Total energy of the universe is constant

Temperature and heat

Heat is not temperature
 Thermal energy = particle motion
 Total thermal energy is sum of all a materials
individual energies

The Universe




is divided into two halves.
the system and the surroundings.
The system is the part you are concerned
with.
The surroundings are the rest.
q into system = -q from surroundings
System: region of space
where process occurs
Surroundings: region of
space around system
Thermodynamic Properties
1. State functions
-depend on nature of
process NOT method
-path independent
-denoted by capital letters
E → state function
Thermodynamic Properties
2. Path functions
-depends on how process
is carried out
-path dependent
-denoted by lower case
q & w → path functions
Work →w
w = -  (PV)
w = -PV - VP
Constant P
Constant V
wP = - P  V
wV = - V  P
Work →w
w=-PV
assume ideal piston
(i.e. no heat loss)
w=E
E=-PV
in isothermics
w = -q
PV = nRT
R=0.0821L atm/mol K
In isothermic conditions
-q = w = -nRT
R=8.314 J/mol K
U = q + w
In isothermics U = 0 so q=-w
U = q - PV
In constant V or constant P
w=0 so U = q
Calorie - Amount of heat needed to
raise the temperature of 1 gram H20, 1
degree centigrade
Nutritional calorie [Calorie]= 1 Kcal
kg m 

J

2

s 
2
SI unit → Joule
1 cal = 4.184 J
1000cal = 1Kcal=4.184KJ
Direction of energy flow
Every energy measurement has three
parts.
1. A unit ( Joules or calories).
2. A number how many.
3. and a sign to tell direction.




negative - exothermic
positive- endothermic
No change in energy - isothermic
Factors Determining
Amount of Heat

Amount of material

Temperature change

Heat capacity
Specific Heat
Amount of heat needed to raise the
temperature of 1 gram of material
1 degree centigrade
 J
Cp  
 g  oC

Heat = q




q = m Cp T
where Cp = Heat capacity
T = T2 – T1
.o
(J/mol C)
CP
H2O(s) = 37.11
H2O(l) = 75.35
H2O(g) = 33.60
What amount of heat is
needed to just boil away 100.
mL of water in a typical lab?
Surroundings
System
Energy
E <0
Exothermic reactions release energy to the surroundings.
q < 0 (-)
Exothermic
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
Surroundings
System
Energy
E >0
Endothermic reactions absorb energy from the
surroundings.
q > 0 (+)
Endothermic
N 2 + O 2 + heat  2NO
Potential energy
2NO
Heat
N2 + O2
1. How much heat is absorbed
when 10.0 g of H20 are heated
from 200C to 250C?
2. How much heat is needed
to completely vaporize 10.0 g
of H2O at its boiling point?
Heat of Vaporization
Hvap(H2O) = 40.66 kJ/mol
Heat of Fusion
Hfus(H2O) = 6.01 kJ/mol
Energy processes:
•Within a phase
q=mHv or mHf
•Between phases
q=mCpT
5
q total   q i
i 1
3. How much total heat is used
to raise the temperature of
100.0 g H2O from-15o to 125oC?
Some rules for heat and work





Heat given off is negative.
Heat absorbed is positive.
Work done by system on surroundings is
positive.
Work done on system by surroundings is
negative.
Thermodynamics- The study of energy
and the changes it undergoes.
First Law of Thermodynamics






The energy of the universe is constant.
Law of conservation of energy.
q = heat
w = work
E = q + w
Take the systems point of view to decide
signs.
What is work?







Work is a force acting over a distance.
w= F x d
P = F/ area
d = V/area
w= (P x area) x  (V/area)= PV
Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
units of liter - atm L-atm
Work needs a sign






If the volume of a gas increases, the
system has done work on the
surroundings.
work is negative
w = - PV
Expanding work is negative.
Contracting, surroundings do work on the
system w is positive.
1 L atm = 101.3 J
Examples



What amount of work is done when 15 L
of gas is expanded to 25 L at 2.4 atm
pressure?
If 2.36 J of heat are absorbed by the gas
above. what is the change in energy?
How much heat would it take to change
the gas without changing the internal
energy of the gas?
Calorimetry







Measuring heat.
Use a calorimeter.
Two kinds
Constant pressure calorimeter (called a
coffee cup calorimeter)
heat capacity for a material, C is
calculated
C= heat absorbed/ T = H/ T
specific heat capacity = C/mass
Calorimetry








molar heat capacity = C/moles
heat = specific heat x m x T
heat = molar heat x moles x T
Make the units work and you’ve done the
problem right.
A coffee cup calorimeter measures H.
An insulated cup, full of water.
The specific heat of water is 1 cal/gºC
Heat of reaction= H = sh x mass x T
Bomb Calorimeter
Calorimeter Constant
heat capacity that is constant over a
temperature range
where
q(cal) = CC x TC
(CC = calorimeter constant)
4. Determine the calorimeter
constant for a bomb calorimeter by
mixing 50.0 mL of water at 25oC
with 50.0 mL of water at 60.0 oC.
The final temp. attained is 40oC.
5. What is the molar heat of
combustion of sulfur, if 2.56 grams
of solid flowers of sulfur are
burned in excess oxygen in the
bomb calorimeter in #4?
6. Determine the molar heat of
solution of LiOH. HINT: Find the
calorimeter constant first.
7. The heat was absorbed by 815 g
of water in calorimeter which had
an initial temp of 21.25 C. After
equilibrium was reached, the final
temperature was 26.72 C.
Examples


The specific heat of graphite is 0.71 J/gºC.
Calculate the energy needed to raise the
temperature of 75 kg of graphite from 294
K to 348 K.
A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The
final temperature of both the water and
the copper is 21.8ºC. What is the specific
heat of copper?
Calorimetry




Constant volume calorimeter is called a
bomb calorimeter.
Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
The heat capacity of the calorimeter is
known and tested.
Since V = 0, PV = 0, E = q
Bomb Calorimeter

thermometer

stirrer

full of water

ignition wire

Steel bomb

sample
Properties




intensive properties not related to the
amount of substance.
density, specific heat, temperature.
Extensive property - does depend on the
amount of stuff.
Heat capacity, mass, heat from a reaction.
Enthalpy







abbreviated H
H = E + PV (that’s the definition) PV = w
at constant pressure.
H = E + PV
the heat at constant pressure qp can be
calculated from
E = qp + w = qp - PV
qp = E + P V = H
Hess’s Law




Enthalpy is a state function.
It is independent of the path.
We can add equations to come up with
the desired final product, and add the H
Two rules


If the reaction is reversed the sign of H is
changed
If the reaction is multiplied, so is H
Enthalpy and Hf
N2 + 2O2  2NO2
N2 + 2O2 + 68KJ/mol 2NO2
Which side is the heat written on? Why?
Look on your Heat Sheets!
Why is it only 34KJ/mol?
H (kJ)
2NO2
N2 + 2O2
68 kJ
H (kJ)
O2 + NO2
-112 kJ
180 kJ
N2 + 2O2
2NO2
H (kJ)
O2 + NO2
-112 kJ
180 kJ
N2 + 2O2
2NO2
68 kJ
Hess’ Law
Reactants → Products
The enthalpy change is
the same whether the
reaction occurs in one
step or in a series of
steps
Molar heat capacity
J
mol  deg
8. Calculate the heat
needed to decompose
one mole of calcium
carbonate.
Standard Enthalpy




The enthalpy change for a reaction at
standard conditions (25ºC, 1 atm , 1 M
solutions)
Symbol Hº
When using Hess’s Law, work by adding
the equations up to make it look like the
answer.
The other parts will cancel out.
Example
Given
5
C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l)
2
Hº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
Hº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
2
Hº= -286 kJ
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
5
C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l) +1300. kJ
2
C(s) + O 2 (g)  CO 2 (g) + 394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l) + 286 kJ
2
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
5
C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l) +1300. kJ
2
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g)
C(s) + O 2 (g)  CO 2 (g) + 394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l) + 286 kJ
2
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g)
2 C(s) +2 O 2 (g)  2 CO 2 (g) + 2(394 kJ)
1
H 2 (g) + O 2 (g)  H 2 O(l) + 286 kJ
2
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g)
2 C(s) +4 O 2 (g)  2 CO 2 (g) + 788 kJ
2
1
H 2 (g) + O 2 (g)  H 2 O(l) + 286 kJ
2
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g)
2 C(s) +4 O 2 (g)  2 CO 2 (g) + 788 kJ
2
1
H 2 (g) + O 2 (g)  H 2 O(l) + 286 kJ
2
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
1300.KJ  C2H2 (g)
2C(s)  788KJ
H2(g)  286KJ
1300.KJ  788kJ + 286KJ
1300.KJ  1074KJ
calculate Hº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Given
1300.KJ  1074KJ
226KJ 
Hf = 1300KJ – 1074KJ
Hf = 226KJ (+sign shows Endothermic)
calculate Hº for this reaction
226KJ +2C(s) + H 2 (g)  C 2 H 2 (g)
Example
Now, You try it!
Given
O 2 (g) + H 2 (g)  2OH(g) Hº= +77.9kJ
O 2 (g)  2O(g) Hº= +495 kJ
H 2 (g)  2H(g) Hº= +435.9kJ
Calculate Hº for this reaction
O(g) + H(g)  OH(g)
Standard Enthalpies of
Formation





Hess’s Law is much more useful if you
know lots of reactions.
Made a table of standard heats of
formation. The amount of heat needed to
for 1 mole of a compound from its
elements in their standard states.
Standard states are 1 atm, 1M and 25ºC
For an element it is 0
There is a table in Appendix 4 (pg A22)
Standard Enthalpies of
Formation





Need to be able to write the equations.
What is the equation for the formation of NO2 ?
½N2 (g) + O2 (g)  NO2 (g)
Have to make one mole to meet the definition.
Write the equation for the formation of
methanol CH3OH.
Since we can manipulate the
equations




We can use heats of formation to figure
out the heat of reaction.
Lets do it with this equation.
C2H5OH +3O2(g)  2CO2 + 3H2O
which leads us to this rule.
Since we can manipulate the
equations




We can use heats of formation to figure
out the heat of reaction.
Lets do it with this equation.
C2H5OH +3O2(g)  2CO2 + 3H2O
which leads us to this rule.
( H of products) - ( H of reactants) = H o
9. Calculate Hf of SO3(g)
from the data given.
Forms of Equations:
2SO2(g) + O2(g)  2SO3(g)
H = -197.8 kj
or
2SO2(g) + O2(g)  2SO3(g) +197.8 kj
10. If the metabolism of glucose is
combustion of C6H12O6(s), how much
heat is produced by the metabolism
of 1.00 g of glucose? H0  1268 kJ

f
mol

Additivity of Heats of Reaction
H = Enthalpy
H = E + PV
H = E + PV
at constant pressure:
E = qP + w
or E = qP - PV
now, at constant P:
VP = 0
thus, H = E + PV
& if E = qP - PV
then H = qP
H = Enthalpy
a “state” function
cannot be measured
measure H
H = Hproducts -
Standard Enthalpy
of Formation
Change in enthalpy that
accompanies the formation
of one mole of a compound
from its elements with all
substances in their
standard states
Thermodynamic Standard State
T = 25oC and P = 1 atm
Thermodynamic Data
Appendix L of text p. A31
reference pointfor ELEMENTS in standard state
H  0
0
f
11.
Combustion of Methane
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O(l)
HC = ?
Write equations and find ΔHf0 for methane,
carbon dioxide, & water
1. C ( s ) + 2 H 2 ( g )  CH 4 ( g )
2. C ( s ) + O2 ( g )  CO2 ( g )
3. H 2 ( g ) + O 2 ( g )  H 2O(l )
1
2
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O(l)
CH 4 ( g )  C ( s ) + 2 H 2 ( g )
- H1
C ( s ) + O2 ( g )  CO2 ( g )
H 2
2 H 2 ( g ) + O 2 ( g )  2 H 2O(l )
H 3
CH 4 ( g ) + 2 O2 ( g )  CO2 ( g ) + 2 H 2O(l )
H C  (H1 ) + H 2 + 2H 3
HC = ?