Transcript General Chemistry - Valdosta State University

Principles of Reactivity:
Energy and Chemical
Reactions
Chapter 6
Chapter 6
1
Energy: Some Basics
From Physics:
Force – a kind of push or pull on an object.
Energy – the capacity to do work.
Work – force applied over a distance
w=Fd
Heat – energy transferred from a warmer object to a
cooler object.
Chapter 6
2
Energy: Some Basics
Kinetic and Potential Energy
Kinetic Energy (Thermal Energy) – energy due to motion.
Ek  1 2 mv2
Chapter 6
3
Energy: Some Basics
Kinetic and Potential Energy
Potential Energy (Stored Energy) – the energy an object
possesses due to its position.
- Potential energy can be converted into kinetic energy.
Example: a ball of clay dropping off a building.
Chapter 6
4
Energy: Some Basics
First Law of Thermodynamics
“The total amount of energy in the universe is fixed.”
Also referred to as the “Law of Conservation of Energy”
Chapter 6
5
Energy: Some Basics
Temperature and Heat
Temperature is a measure of heat energy
• Heat is not the same as temperature.
• The more thermal energy a substance has the greater
its molecular motion (kinetic energy).
• The total thermal energy in an object is the sum of the
energies of all the “bodies” in the object.
Chapter 6
6
Energy: Some Basics
Systems and Surroundings
System – portion of the universe we wish to study.
Surroundings – everything else.
Universe = System + Surroundings
Chapter 6
7
Energy: Some Basics
Directionality of Heat
Heat energy always flows from the hot object to the cold
object.
- this flow continues until the two objects are at the same
temperature (thermal equilibrium).
Chapter 6
8
Energy: Some Basics
Directionality of Heat
Exothermic – Heat is transferred from the system to the
surroundings (object will feel “hot”).
Endothermic – Heat is transferred to the system from
the surroundings (object will fell “cold”).
Chapter 6
9
Energy: Some Basics
Energy Units
SI Unit for energy is the joule, J:
1 J  1 kg m / s
2
2
A more traditional unit is the Calorie
Calorie (cal) – amount of energy required to raise 1.0 g
of water 1oC.
1cal = 4.184J
Chapter 6
10
Specific Heat Capacity
The amount of heat transferred is dependant on three
quantities:
– Quantity of material
– Size of temperature change
– Identity of the material
Chapter 6
11
Specific Heat Capacity
q  c  mT 
q = energy
c = specific heat capacity
T = temperature change
T  Tfinal  Tinitial
Chapter 6
12
Specific Heat Capacity
q  c  mT 
exothermic
endothermic
-T
+T
Chapter 6
-q
+q
13
Specific Heat Capacity
q  c  mT 
• Specific heat capacity can be either per gram
(J/g(oC) or per mole (J/mol(oC).
• The smaller a substances specific heat capacity, the
better a thermal conductor it is.
Chapter 6
14
Energy and Changes of State
Chapter 6
15
Energy and Changes of State
• In the previous slide there is a continuous, steady
application of energy.
• The sections that show increasing temperature are the
result of the particular phase being warmed.
q = cm(T)
• The “flat” sections occur when all the applied energy is
used to change the phase of the substance.
• Fusion – solid  liquid
• Vaporization – liquid  gas
Chapter 6
16
Energy and Changes of State
• The energy required to change the phase of a
substance is unique and is described in a physical
constant.
• Solid  Liquid
• Heat of Fusion (water, 333J/g)
• Liquid  Gas
• Heat of Vaporization (water, 2256J/g)
• These constants can be used to determine the energy
used in melting or vaporizing a substance.
q = (Heat of Fusion)(mass of sample)
q = (Heat of Vapor.)(mass of sample)
Chapter 6
17
Energy and Changes of State
q = cm(T)
q = (Heat of Vapor.)(mass)
Chapter 6
18
First Law of Thermodynamics
Internal Energy
Internal Energy – sum of all kinetic and potential energy
in an object.
• It is very hard to determine an objects internal energy,
but it is possible to determine the change in energy
(E).
• Change in internal energy, E = Efinal - Einitial
– A positive E means Efinal > Einitial
or the system gained energy from the surroundings
(endothermic)
– A negative E means Efinal < Einitial
or the system lost energy to the surroundings (exothermic)
Chapter 6
19
First Law of Thermodynamics
Relating E to Heat and Work
E = q + w
q = heat
w = work
• Both heat energy and work can change a systems internal
energy.
Chapter 6
20
First Law of Thermodynamics
State Functions
State function – a process that is determined by its
initial and final conditions.
Chapter 6
21
First Law of Thermodynamics
State Functions
State function – a process that is determined by its
initial and final conditions.
• “A process that is not path dependant.”
• Work (w) and heat (q) are not state functions.
• Energy change (E) is a state function.
Chapter 6
22
First Law of Thermodynamics
Enthalpy (H) - Heat transferred between the system
and surroundings carried out under constant pressure.
E = q + w
Most reactions occur under constant pressure, so
E = q + (-P(V))
If volume is also constant, V = 0
E = qp
So, Energy change is due to heat transfer,
E = H = qp
Chapter 6
23
Enthalpy
Enthalpy Change (H) – The heat evolved or absorbed
in a reaction at constant pressure
H = Hfinal - Hinitial = qP
Chapter 6
24
Enthalpy
Enthalpy Change (H) – The heat evolved or absorbed
in a reaction at constant pressure
• H and H are state functions, depending only on
the initial and final states.
Chapter 6
25
Enthalpies of Reaction
H reaction   H (products)   H (reactants)
2 H2(g) + O2(g)  2 H2O(g)
Chapter 6
H = -483.6 J
26
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) H = -1604 kJ
Chapter 6
27
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H = +802 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
Chapter 6
28
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
3. Change in enthalpy depends on state:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H = -802 kJ
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H = -890 kJ
Chapter 6
29
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
a) Is this reaction endothermic or exothermic?
Chapter 6
30
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
a) Is this reaction endothermic or exothermic?
Exothermic, this is indicated by the negative H.
Chapter 6
31
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
b) Calculate the amount of heat transferred when 2.4g
of Mg reacts at constant pressure.
2
.
4
g
molesMg 
24.3g / mol
Chapter 6
 0.10 mol Mg
32
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
b) Calculate the amount of heat transferred when 2.4g
of Mg reacts at constant pressure.
m olesMg  2.4 g
 0.10 m ol Mg
24.3g / m ol
t hereis a rat iobet ween t hemolesof Mg used
and t heheat produced
Chapter 6
33
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
b) Calculate the amount of heat transferred when 2.4g
of Mg reacts at constant pressure.
m olesMg  2.4 g
24.3g / m ol
 0.10 m ol Mg
- 1205kJ
x

2 Mg
0.10 m ol
Chapter 6
34
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
b) Calculate the amount of heat transferred when 2.4g
of Mg reacts at constant pressure.
m olesMg  2.4 g
24.3g / m ol
 0.10 m ol Mg
- 1205kJ
x

2 Mg
0.10 m ol
x   60kJ
Chapter 6
35
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
c) How many grams of MgO are produced during an
enthalpy change of 96.0 kJ?
2 MgO
x

 1205 kJ  96.0kJ
Chapter 6
36
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
c) How many grams of MgO are produced during an
enthalpy change of 96.0 kJ?
2 MgO
x

 1205kJ  96.0kJ
x  0.16 m ol MgO
Chapter 6
37
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
c) How many grams of MgO are produced during an
enthalpy change of 96.0 kJ?
2 MgO
x

 1205kJ  96.0kJ
x  0.16 m ol MgO
g MgO  0.16m ol(40.3g / m ol)
Chapter 6
38
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
c) How many grams of MgO are produced during an
enthalpy change of 96.0 kJ?
2 MgO
x

 1205kJ  96.0kJ
x  0.16 m ol MgO
g MgO  0.16m ol( 40.3g / m ol)
 6.42g
Chapter 6
39
Enthalpies of Reaction
2 Mg(s) + O2(g)  2 MgO(s)
H = -1205 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
Chapter 6
40
Enthalpies of Reaction
2 MgO(s)  2 Mg(s) + O2(g)
H = 1205 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
molesMgO  7.50g
40.3g / mol
 0.186mol
Chapter 6
41
Enthalpies of Reaction
2 MgO(s)  2 Mg(s) + O2(g)
H = 1205 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
m olesMgO  7.50g
40.3g / m ol
 0.186m ol
1205kJ
x

2 MgO 0.186m ol
Chapter 6
42
Enthalpies of Reaction
2 MgO(s)  2 Mg(s) + O2(g)
H = 1205 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
m olesMgO  7.50g
40.3g / m ol
 0.186m ol
1205kJ
x

2 MgO 0.186m ol
x  112kJ
Chapter 6
43
Calorimetry
Constant-Pressure Calorimetry
Chapter 6
44
Calorimetry
Constant-Pressure Calorimetry
Atmospheric pressure is constant!
H = qP
qsystem = -qsurroundings
- The surroundings are composed of the water in the
calorimeter and the calorimeter.
qsystem = -(qwater + qcalorimeter)
Chapter 6
45
Calorimetry
Constant-Pressure Calorimetry
Atmospheric pressure is constant!
H = qP
qsystem = -qsurroundings
- The surroundings are composed of the water in the
calorimeter and the calorimeter.
- For most calculations, the qcalorimeter can be ignored.
qsystem = - qwater
csystemmsystem Tsystem = - cwatermwater Twater
Chapter 6
46
Calorimetry
Bomb Calorimetry
(Constant-Volume Calorimetry)
Chapter 6
47
Calorimetry
Bomb Calorimetry
(Constant-Volume Calorimetry)
- Special calorimetry for combustion reactions
- Substance of interest is placed in a “bomb” and filled
to a high pressure of oxygen
- The sealed bomb is ignited and the heat from the
reaction is transferred to the water
- This calculation must take into account the heat
capacity of the calorimeter (this is grouped together
with the heat capacity of water).
qrxn = -Ccalorimeter(T)
Chapter 6
48
Calorimetry
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Twater = 16.9oC – 22.0oC = -5.1oC
mwater = 60.0g
cwater = 4.184J/goC
msample = 4.25g
qsample = -qwater
qsample = -cwatermwater Twater
qsample = -(4.184J/goC)(60.0g)(-5.1oC)
qsample = 1280.3J
- Now calculate H in kJ/mol
Chapter 6
49
Calorimetry
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Twater = 16.9oC – 22.0oC = -5.1oC
mwater = 60.0g
cwater = 4.184J/goC
msample = 4.25g
qsample = 1280.3J
moles NH4NO3 = 4.25g/80.032g/mol = 0.0529 mol
H = qsample/moles
H = 1280.3J/0.0529mol
H = 24.2 kJ/mol
Chapter 6
50
Calorimetry
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -Ccal (Twater)
qrxn = -11.66kJ/oC(7.42oC)
qrxn = -86.52kJ
Chapter 6
51
Calorimetry
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -86.52kJ
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g =
Chapter 6
52
Calorimetry
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -86.52kJ
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g = -48.1 kJ/g
Hcombustion(in kJ/mol)
Hcombustion = -86.52kJ/0.01577mol =
Chapter 6
53
Calorimetry
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -86.52kJ
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g = -48.1 kJ/g
Hcombustion(in kJ/mol)
Hcombustion = -86.52kJ/0.01577mol = -5485 kJ/mol
Chapter 6
54
Hess’s Law
Hess’s law - if a reaction is carried out in a series of steps,
H for the overall reaction is the sum of H’s for each
individual step.
For example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
2H2O(g)  2H2O(l)
H = -88 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -890 kJ
Chapter 6
55
Enthalpies of Formation
(Heat of Formation)
- There are many type of H, depending on what you
want to know
Hvapor – enthalpy of vaporization (liquid  gas)
Hfusion – enthalpy of fusion (solid  liquid)
Hcombustion – enthalpy of combustion
(energy from burning a substance)
Chapter 6
56
Enthalpies of Formation
(Heat of Formation)
- A fundamental H is the Standard Enthalpy of
o
Formation ( H f )
Standard Enthalpy of Formation ( H of ) – The enthalpy
change that accompanies the formation of one mole of
a substance from the most stable forms of its
component elements at 298 Kelvin and 1 atmosphere
pressure.
“The standard enthalpy of formation of the most stable
form on any element is zero”
Chapter 6
57
Enthalpies of Formation
Chapter 6
58
Enthalpies of Formation
Using Enthalpies of Formation to Calculate
Enthalpies of Reaction
For a reaction:
o
Hrxn
 nH of products  mH of reactants
Chapter 6
59
Homework Problems
4, 14, 20, 24, 28, 36, 40, 44, 46, 52, 54, 56a
Chapter 6
60