chemistry mcmurry fay

Transcription

Chapter 8
Thermochemistry: Chemical Energy
國防醫學院生化學科
王明芳老師
2011-11-8 & 2011-11-15
Chapter 8/1
Energy and Its Conservation
Conservation of Energy Law: Energy cannot be
created or destroyed; it can only be converted from one
form to another.
Energy: is the capacity to do work, or supply heat.
Energy = Work + Heat
Kinetic Energy: is the energy of motion.
(due to motion of the object)
Kinetic energy of a body in motion =
1 2 1
2




mv  mass  velocity
Chapter 8/2
2
2
2
kg

m
1 Joule = 1
s2
1 calorie = 4.184 J
Potential Energy: is stored energy. (due to position or
composition - can be converted to work)
Chapter 8/3
CONSERVATION OF ENERGY. The total amount of energy
contained by the water in the reservoir is constant.
Chapter 8/4
Thermal Energy: The kinetic energy of molecular
motion and is measured by finding the temperature of
an object. (including rotational and vibrational)
Thermal energy is proportional to the temperature in
degrees Kelvin.
Ethermal  T (oK)
Heat is the amount of thermal energy transferred
between two objects at different temperatures.
Heat involves a transfer of energy between 2 objects
due to a temperature difference
Chapter 8/5
Energy Units
calorie (cal):
One calorie defined as the amount of energy
necessary to raise the temperature of 1g of
water by 1 ℃ (specifically, from 14.5 ℃ to
15.5 ℃)
Kilocalorie = Kcal
SI unit： joule
If a 2.0 kg object is moving with a velocity of
1.0 m/s, the kinetic energy is:
1/2mv2 = 1/2(2.0 kg)(1.0 m/s)2
=1.0 kg*m2/s2=1.0 J (joule)
Chapter 8/6
Internal Energy and State
Functions
First Law of Thermodynamics: The total internal
energy E of an isolated system is constant
DE = Efinal - Einitial
Chapter 8/7
Functions
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + 802 kJ energy
DE = Efinal - Einitial = -802 kJ
802 kJ is released when 1 mole of methane, CH4,
reacts with 2 moles of oxygen to produce 1 mole of
carbon dioxide and two moles of water.
Chapter 8/8
Functions
State Function: A function or property whose
value depends only on the present state, or
condition, of the system, not on the path used
to arrive at that state
Chapter 8/9
Functions
State function (or property): refers to a property of the system
that depends only on its present state
does not depend on the history of the system
A change in this function (or
property) in going from one
state to another state is
independent of the particular
pathway taken between the
two states
Energy is a state function; work and heat are not state function
Chapter 8/10
Expansion Work
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
6 mol of gas
7 mol of gas
Expansion work in a chemical reaction.
Chapter 8/11
Expansion Work
The energy change in a system equals the work done on the
system + the heat added.
ΔE = Efinal – Einitial = E2 – E1 = q + w
q = heat, w = work
Work is a force (F) that produces an object’s movement, times
the distance moved (d):
Force F
Pressure =

Work = Force x Distance
Area A
Pressure is the force per unit area.
1 N/m2 = 1 Pa
(1 Newton per
 square meter)
1 atm = 101,325 Pa
(Pascal, Pa for short)
atmospheres (symbol = atm)
1 L • atm = 101 J
millimeters of mercury (symbol = mm Hg)
Chapter 8/12
Expansion Work
Expansion Work: Work done as the result of a volume
change in the system
Chapter 8/13
Expansion Work
The most common type of work encountered
in chemical systems is the expansion work
(also called pressure-volume, or PV, work)
done as the result of a volume change in the
system
Chapter 8/14
Example 8.1 Calculating The Amount Of PV Work
Calculate the work in kilojoules done during a reaction in which the volume expands from 12.0 L
to 14.5 L against an external pressure of 5.0 atm.
Solution
1. w = PDV, where P is the external pressure opposing the change in volume.
2. In this instance, P = 5.0 atm and DV = (14.5  12.0).
3. Remember that an expanding system loses work energy, which thus has a
negative sign.
Energy and Enthalpy
DE = q + w
q = heat transferred
w = work = -PDV
q = DE + PDV
Constant Volume (DV = 0):
qV = DE
Constant Pressure: qP = DE + PDV
Chapter 8/16
Energy and Enthalpy
qP = DE + PDV = DH
Enthalpy change
or
Heat of reaction (at constant pressure)
Enthalpy is a state function
whose value depends only on
the current state of the system,
not on the path taken to arrive
at that state.
DH = Hfinal - Hinitial
= Hproducts - Hreactants
The amount of heat exchanged between the system and the
surroundings is given the symbol q.
Chapter 8/17
The Thermodynamic Standard
State
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
DH = -2044 kJ
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
DH = -2220 kJ
The different of 176 kJ between the values of ΔH for the two reactions
arises because the conversion of liquid water to gaseous water
requires energy.
Chapter 8/18
The Thermodynamic Standard
State
Thermodynamic Standard State: Most stable form of a
substance at 1 atm pressure and at a specified
temperature, usually 25 oC; 1 M concentration for all
substances in solution
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
DHo = -2044 kJ
These are indicated by a superscript ° to the symbol
of the quantity reported. standard enthalpy change is
indicated by the symbol DH°
Chapter 8/19
Example 8.2 Calculating DE For A Reaction
The reaction of nitrogen with hydrogen to make ammonia has DH = 92.2 kJ. What is the value of DE in kilojoules
if the reaction is carried out at a constant pressure of 40.0 atm and the volume change is 1.12 L?
N2(g) + 3 H2(g)  2 NH3(g)
Solution
DH = 92.2 kJ
Enthalpies of Physical and
Chemical Change
Enthalpies of Physical Change
Enthalpy of Fusion or heat of fusion (DHfusion): The
amount of heat necessary to melt a substance without
changing its temperature
Enthalpy of Vaporization or heat of vaporization
(DHvap): The amount of heat required to vaporize a
substance without changing its temperature
Enthalpy of Sublimation or heat of sublimation
(DHsubl): The amount of heat required to convert a
substance from a solid to a gas without going through
Chapter 8/21
a liquid phase
Chemical Change
Enthalpy as a state function.
Because enthalpy is a state function, the value of the enthalpy
change from solid to vapor does not depend on the path taken
between the two states
Chapter 8/22
Chemical Change
2Al(s) + Fe2O3(s)
Thermite reaction of
aluminum with iron (III) oxide
2Fe(s) + Al2O3(s)
DHo = -852 kJ
exothermic
2Fe(s) + Al2O3(s)
2Al(s) + Fe2O3(s)
DHo = +852 kJ
endothermic
Enthalpies of Chemical Change
Chapter 8/23
Chemical Change
Endothermic: Heat flows into the system from the
surroundings and DH has a positive sign.
Exothermic: Heat flows out of the system into the
surroundings and DH has a negative sign.
Reversing a reaction changes the sign of DH for a reaction.
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O(l) DH = –2219 kJ
3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2 (g) DH = +2219 kJ
Multiplying a reaction increases DH by the same factor.
3C3H8 (g)+15O2 (g) 9CO2 (g)+12H2O(l) DH = –6657 kJ
Chapter 8/24
Example 8.3 Calculating The Amount Of Heat Released In A Reaction
How much heat in kilojoules is evolved when 5.00 g of aluminum reacts with a
stoichiometric amount of Fe2O3?
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
Solution
DH = 852 kJ
Calorimetry and Heat Capacity
Calorimetry is the science of measuring heat changes (q)
for chemical reactions.
There are two types of calorimeters:
1. Constant Pressure Calorimetry: A constant pressure
calorimeter measures the heat change at constant
pressure such that q = DH.
2. Bomb Calorimetry: A bomb calorimeter measures the
heat change at constant volume such that q = DE.
Chapter 8/26
Measure the heat flow at constant pressure (DH).
Chapter 8/27
Measure the heat flow at constant volume (DE).
A bomb calorimeter
for measuring the
heat evolved at
constant volume (DE)
in a combustion
reaction.
Chapter 8/28
Heat Capacity (C): The amount of heat necessary to
raise the temperature of an object or substance a given
amount
q = C x DT
Specific Heat (Capacity): The amount of heat required
to raise the temperature of 1 g of a substance by 1 oC
q = (Specific heat) x (Mass of substance) x DT
Chapter 8/29
Molar Heat Capacity (Cm): The amount of heat
necessary to raise the temperature of 1 mol of a
substance by 1 oC
q = Cm x Moles of substance x DT
Chapter 8/30
As indicated in Table 8.1,
the specific heat of liquid
water is considerably higher
than that of most other
substances, so a large
transfer of heat is
necessary to either cool or
warm a given amount of
water.
Chapter 8/31
Assuming that a can of soda has the same specific heat
as water, calculate the amount of heat (in kilojoules)
transferred when one can (about 350 g) is cooled from
25 oC to 3 oC.
q = (Specific heat) x (Mass of substance) x DT
J
Specific heat = 4.18 o
g C
Mass = 350 g
Temperature change = 3 oC - 25 oC = -22 oC
Chapter 8/32
Calculate the amount of heat transferred.
4.18 J
x 350 g x -22 oC = -32 000 J
Heat evolved =
o
g C
1 kJ
= -32 kJ
-32 000 J x
1000 J
Chapter 8/33
Example 8.4 Calculating A Specific Heat
What is the specific heat of silicon if it takes 192 J to raise the temperature of 45.0 g
of Si by 6.0 C?
Solution
Example 8.5 Calculating DH In A Calorimetry Experiment
Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate
of solid silver chloride:
Ag+(aq) + Cl(aq)  AgCl(s)
When 10.0 mL of 1.00 M AgNO3 solution is added to 10.0 mL of 1.00 M NaCl
solution at 25.0 C in a calorimeter, a white precipitate of AgCl forms and the
temperature of the aqueous mixture increases to 32.6 C. Assuming that the
specific heat of the aqueous mixture is 4.18 J/(g  C), that the density of the
mixture is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount
of heat, calculate DH in kilojoules for the reaction.
Example 8.5 Calculating DH In A Calorimetry Experiment
Continued
Solution
According to the balanced equation, the number of moles of AgCl produced equals the number of moles
of Ag+ (or Cl) reacted:
Therefore, DH = 64 kJ (negative because heat is released)
Hess’s Law
Hess’s Law: The overall enthalpy change for a reaction is
equal to the sum of the enthalpy changes for the individual
steps in the reaction.
Haber Process
3H2(g) + N2(g)
2NH3(g)
DHo = -92.2 kJ
Chapter 8/37
Hess’s Law
Hess’s Law: The overall enthalpy change for a reaction is
equal to the sum of the enthalpy changes for the individual
steps in the reaction.
Multiple-Step Process
2H2(g) + N2(g)
N2H4(g)
DHo1 = ?
N2H4(g) + H2(g)
2NH3(g)
DHo2 = -187.6 kJ
3H2(g) + N2(g)
2NH3(g)
DHo1+2 = -92.2 kJ
Chapter 8/38
Hess’s Law
DHo1 + DHo2 = DHo1+2
DHo1 = DHo1+2 - DHo2
= -92.2 kJ - (-187.6 kJ) = +95.4 kJ
Enthalpy changes for
steps in the synthesis
of ammonia from
nitrogen and
hydrogen. If DHo
values for step 2 and
for the overall reaction
are known, the DHo
for step 1 can be
calculated.
Example 8.6 Using Hess’s Law To Calculate DH
Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Use the following information to calculate in kilojoules for the combustion of methane:
CH4(g) + O2(g)  CH2O(g) + H2O(g)
DH = 275.6 kJ
CH2O(g) + O2(g)  CO2(g) + H2O(g)
DH = 526.7 kJ
H2O(l)  H2O(g)
DH = 44.0 kJ
Solution
+
Example 8.7 Using Hess’s Law To Calculate DH
Water gas is the name for the mixture of CO and H2 prepared by reaction of steam with carbon at 1000 C:
The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following
information to calculate in DH kilojoules for the water-gas reaction:
C(s) + O2(g)  CO2(g)
DH = 393.5 kJ
2 CO(g) + O2(g)  2 CO2(g)
DH = 566.0 kJ
2 H2(g) + O2(g)  2 H2O(g)
DH = 483.6 kJ
Solution
The water-gas reaction is endothermic by 131.3 kJ.
Standard Heats of Formation
Standard Heat of Formation (DHof ): The enthalpy
change for the formation of 1 mol of a substance in its
standard state from its constituent elements in their
standard states
Standard states
C(s) + 2H2(g)
CH4(g)
DHof = -74.8 kJ
1 mol of 1 substance
Chapter 8/42
The standard heat of formation for any element in its standard state
is defined as being ZERO.
DHof= 0 for an element in its standard state
Chapter 8/43
DHo = DHof (Products) - DHof (Reactants)
aA + bB
cC + dD
DHo = [c DHof (C) + d DHof (D)] - [a DHof (A) + b DHof (B)]
Products
Reactants
Chapter 8/44
Using standard heats of formation, calculate the
standard enthalpy of reaction for the photosynthesis of
glucose (C6H12O6) and O2 from CO2 and liquid H2O.
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g)
DHo = ?
Chapter 8/45
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g)
DHo = ?
DHo = [DHof (C6H12O6(s))] -
[6 DHof (CO2(g)) + 6 DHof (H2O(l))]
DHo = [(1 mol)(-1273.3 kJ/mol)] [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]
DHo = 2802.5 kJ
Chapter 8/46
Example 8.8 Using Standard Heats Of Formation To Calculate DH
Calculate DH in kilojoules for the synthesis of lime (CaO) from limestone (CaCO3),
the key step in the manufacture of cement.
CaCO3(s)  CaO(s) + CO2(g)
DHf [CaCO3(s)] = 1207.6 kJ/mol
DHf [CaO(s)] = 634.9 kJ/mol
DHf [CO2(g)] = 393.5 kJ/mol
Solution
DH = [DHf (CaO) + DHf (CO2)]  [DHf (CaCO3)]
= (1 mol)(634.9 kJ/mol) + (1 mol)(393.5 kJ/mol)  (1 mol)(1207.6 kJ/mol)
= 179.2 kJ
The reaction is endothermic by 179.2 kJ.
Example 8.9 Using Standard Heats Of Formation To Calculate DH
Oxyacetylene welding torches burn acetylene gas, C2H2(g). Use the information in Table 8.2 to
calculate DH in kilojoules for the combustion reaction of acetylene to yield CO2(g) and H2O(g).
Solution
The balanced equation is
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(g)
The necessary heats of formation are
DHf[C2H2(g)] = 227.4 kJ/mol
DHf[CO2(g)] = 393.5 kJ/mol
DHf [H2O(g)] = 241.8 kJ/mol
Remember also that DHf (O2) = 0 kJ/mol.
The standard enthalpy change for the reaction is
DH = [4 DHf (CO2)] + 2 DHf (H2O)]  [2 DHf (C2H2)]
= (4 mol)(393.5 kJ/mol) + (2 mol)(241.8 kJ/mol)  (2 mol)(227.4 kJ/mol)
=  2512.4 kJ
Bond Dissociation Energies
Bond dissociation energies are standard enthalpy
changes for the corresponding bond-breaking reactions.
Chapter 8/49
Bond Dissociation Energy: Can be used to
determine an approximate value for DH°
Apply Hess’s law, we can calculate an approximate
enthalpy change for any reaction by subtracting the
sum of the bond dissociation energies in the
products from the sum of the bond dissociation
energies in the reactants
DHo = D(Reactant bonds) – D(Product bonds)
Chapter 8/50
H2(g) + Cl2(g)
2HCl(g)
DHo = D(Reactant bonds) - D(Product bonds)
DHo = (DH-H + DCl-Cl) - (2DH-Cl)
DHo = [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] (2mol)(432 kJ/mol)
DHo = -185 kJ
Chapter 8/51
Example 8.10 Using Bond Dissociation Energies To Calculate DH
Use the data in Table to find an approximate DH in kilojoules for the industrial synthesis
of chloroform by reaction of methane with Cl2.
CH4(g) + 3 Cl2(g)  CHCl3(g) + 3 HCl(g)
Example 8.10 Using Bond Dissociation Energies To Calculate DH
Continued
Solution
The reactants have four C–H bonds and three Cl–Cl bonds; the products have one C–H
bond, three C–Cl bonds, and three H–Cl bonds. The bond dissociation energies are:
C–H
D = 410 kJ/mol
Cl–Cl
D = 243 kJ/mol
C–Cl
D = 330 kJ/mol
H–Cl
D = 432 kJ/mol
Subtracting the product bond dissociation energies from the reactant bond dissociation
energies gives the enthalpy change for the reaction:
DH = [3 DCl–Cl + 4 DC–H]  [DC–H + 3 DH–Cl + 3 DC–Cl]
= [(3 mol)(243 kJ/mol) + (4 mol)(410 kJ/mol)]  [(1 mol)(410 kJ/mol)
+ (3 mol)(432 kJ/mol) + (3 mol)(330 kJ/mol)]
=  327 kJ
The reaction is exothermic by approximately 330 kJ.
Fossil Fuels, Fuel Efficiency,
and Heats of Combustion
The objective of the exercise is to calculate the energy
density of several hydrocarbon fuels.
Heats of Combustion is also called combustion enthalpy,
DHc.
Chapter 8/54
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
DHo = [DHof (CO2(g)) + 2 DHof (H2O(l))] [DHof (CH4(g))]
DHo = [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] [(1 mol)(-74.8 kJ/mol)]
DHo = -890.3 kJ
Note that the H2O product in giving heats of
combustion is H2O(l) rather than H2O(g)
Chapter 8/55
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
It is more useful to calculate combustion enthalpy per gram, or
per milliliter of substance rather than per mole.
Chapter 8/56
Coal Gasification
Chapter 8/57
coal gasification
exothermic:
endothermic:
energy balance
from a solid to a gas
coal gasification
from a solid to a gas
can be directly converted to methanol
synthetic fibers
methanol
plastics
fuel – South Africa
The products of petroleum
refining
An Introduction to Entropy
Spontaneous Process: A process that, once started,
proceeds on its own without a continuous external
influence
Sledding downhill is a
spontaneous process that,
once started, continues on
its own. Dragging the sled
back uphill is a
nonspontaneous process
that requires c continuous
input of energy.
Chapter 8/61
Entropy (S): The amount of molecular randomness in a
system
Entropy is a measure of
molecular randomness
Chapter 8/62
Second Law of Thermodynamics: Reactions proceed in the direction that
increases the entropy of the system plus surroundings.
A spontaneous process is one that proceeds on its own without any
continuous external influence.
A nonspontaneous process takes place only in the presence of a continuous
external influence.
The measure of molecular disorder in a system is called the system’s
entropy; this is denoted S.
Entropy has units of J/K (Joules per Kelvin).
DS = Sfinal – Sinitial
Positive value of DS indicates increased disorder.
Negative value of DS indicates decreased disorder.
Chapter 8/63
Spontaneous processes are
• favored by a decrease in H (negative DH).
• favored by an increase in S (positive DS).
Nonspontaneous processes are
• favored by an increase in H (positive DH).
• favored by a decrease in S (negative DS).
Chapter 8/64
Example 8.11 Predicting The Sign Of DS For A Reaction
Predict whether DS is likely to be positive or negative for each of the following reactions:
(a) H2C=CH2(g) + Br2(g)  BrCH2CH2Br(l)
(b) 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
Solution
Reactions that increase the number of gaseous molecules generally have
a positive DS, while reactions that decrease the number of gaseous
molecules have a negative DS.
(a). The amount of molecular randomness in the system decreases when 2
mol of gaseous reactants combine to give 1 mol of liquid product, so the
reaction has a negative DS.
(b). The amount of molecular randomness in the system increases when 9
mol of gaseous reactants give 10 mol of gaseous products, so the reaction
has a positive DS.
An Introduction to Free Energy
Gibbs Free Energy Change (DG)
DG = DH - T DS
Enthalpy of
reaction
Temperature
(Kelvin)
Entropy
change
Chapter 8/66
Gibbs Free Energy Change (DG)
DG = DH - T DS
DG < 0
Process is spontaneous
DG = 0
Process is at equilibrium
(neither spontaneous nor nonspontaneous)
DG > 0
Process is nonspontaneous
Chapter 8/67
DGo = DHo - TDSo
Melting and freezing. The melting of ice disfavored by enthalpy (DH >
0) but favored by entropy (DS > 0). The freezing of water is favored by
enthalpy (DH < 0) but disfavored by entropy (DS < 0).
Chapter 8/68
Example 8.12 Using The Free-Energy Equation To Calculate Equilibrium
Temperature
Lime (CaO) is produced by heating limestone (CaCO3) to drive off CO2 gas, a reaction
used to make Portland cement. Is the reaction spontaneous under standard conditions at
25 C? Calculate the temperature at which the reaction becomes spontaneous.
CaCO3(s)  CaO(s) + CO2(g)
DH = 179.2 kJ;
DS = 160.0 J/K
Solution
At 25 C (298 K), we have
Because DG is positive at this temperature, the reaction is nonspontaneous.
The changeover point between spontaneous and nonspontaneous is
approximately
The reaction becomes spontaneous above approximately 1120 K (847 C).
Example 8.13 Predicting The Signs Of DH, DS, And DG For a Reaction
What are the signs of DH, DS, And DG for the following nonspontaneous transformation?
DG=DH-TDS
Solution
1. The drawing shows ordered particles in a solid subliming to give a gas.
2. Formation of a gas from a solid increases molecular randomness, DS so is positive.
3. Because we’re told that the process is nonspontaneous, DG is also positive.
Because the process is favored by DS (positive) yet still nonspontaneous, DH must
be unfavorable (positive).
4. This makes sense, because conversion of a solid to a liquid or gas requires energy
and is always endothermic.

chemistry mcmurry fay

Transcription

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