Ch 6 - Oglethorpe University

```Chapter 6
Thermochemistry:
Energy Flow and
Chemical Change
6-1
Dr. Wolf’s CHM 101
Thermochemistry: Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy: Heats of Reaction and Chemical Change
6.3 Calorimetry: Laboratory Measurement of Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hess’s Law of Heat Summation
6.6 Standard Heats of Reaction (DH0rxn)
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Dr. Wolf’s CHM 101
Thermochemistry:
Energy Flow and Chemical Change
In the observation and measurement of a change
in energy, we talk about the System and the
Surroundings.
The System is what we are studying, e.g. a
reaction taking place in a flask.
The Surroundings are the rest of the universe, but
generally only the nearby portion that is relevant
to the system.
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A change in energy in the System is always
accompanied by an opposite change in energy of
the Surroundings.
Change in energy, DE = Efinal - Einitial = Eproducts - Ereactants
Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings.
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Dr. Wolf’s CHM 101
Heat (thermal energy) and Work, Two Forms of D E
The symbol for heat is q. The symbol for work is w.
They can be positive or negative values.
Energy flowing into a system is defined as positive.
Energy flowing out of a system is negative.
DE = q + w
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In a system transferring energy as heat only.
Dr. Wolf’s CHM 101
DE = q + 0
A system losing energy as work only.
DE = 0 + w
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Work is done on the surroundings,
so work on the system is negative.
The Sign Conventions* for q, w and DE
q
+
w
=
DE
+
+
+
+
-
depends on sizes of q and w
-
+
depends on sizes of q and w
-
-
-
* For q: + means system gains heat; - means system loses heat.
* For w: + means word done on system; - means work done by system.
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Dr. Wolf’s CHM 101
Law of Conservation of Energy
(First Law of Thermodynamics)
The total energy of the universe is constant. Energy transfers can
occur between the system and surroundings in the form of heat
and/or work. But the energy of the system and the energy of the
surroundings remains constant; energy is conserved.
DEuniverse = DEsystem + DEsurroundings = 0
Units of Energy
Joule (J)
Calorie (cal)
British Thermal Unit
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1 J = 1 kg*m2/s2
1 cal = 4.18J
1 Btu = 1055 J
(mass x accel x dist)
Sample Problem 6.1
PROBLEM:
PLAN:
System
When gasoline burns in a car engine, the heat released causes
the products CO2 and H2O to expand, which pushes the pistons
outward. Excess heat is removed by the car’s cooling system.
If the expanding gases do 451J of work on the pistons and the
system loses 325J to the surroundings as heat, calculate the
change in energy (DE) in J, kJ, and kcal.
Define system and surrounds, assign signs to q and w and calculate
DE. The answer should be converted from J to kJ and then to kcal.
SOLUTION:
q = - 325J
DE = q + w =
-776J
6-9
Determining the Change in Internal Energy of a
kJ
103J
w = - 451J
-325J + (-451J) = -776J
= 0.776kJ
Dr. Wolf’s CHM 101
0.776kJ
kcal
4.18kJ
= 0.186kcal
State functions such as internal energy of a system are
dependent only upon the initial state and the final state.
It is independent upon the pathway to get between the
two states.
Two different paths for the energy change of a system.
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Enthalpy, H
In chemistry the work is
usually electrical (Ch
21) or pressure-volume
(PV) work.....work done
by expanding gas
(change in volume, DV)
DV= Vfinal - Vinitial
So this work done on
surroundings is
negative.
DE = q + w
DE = q - PDV
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Dr. Wolf’s CHM 101
Pressure-volume
work.
The Meaning of Enthalpy, H
DE = q - PDV
qp = DE + PDV
(constant pressure)
DH = DE + PDV
In many reactions little or no
work done so DH ~ DE in
1. Reactions that do not involve gases.
2. Reactions in which the number of
moles of gas does not change.
3. Reactions in which the number of moles
of gas does change but q is >>> PDV.
DH is the change in enthalpy - change in internal
energy plus PDV work
qp = DE + PDV = DH
DH = qp , equals the heat gained or lost at constant pressure.
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Change of Enthalpy, D H
DH= Hfinal - Hinitial
Exothermic / Endothermic
If the products have less energy than the reactants,
i.e. Hfinal < Hinitial , then DH is negative. Heat has been
released. Exothermic processes have a negative DH.
Enthalpy diagram for exothermic process.
CH4(g) + 2O2(g)
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CO2(g) + 2H2O(g)
Dr. Wolf’s CHM 101
Change of Enthalpy, D H
DH= Hfinal - Hinitial
Exothermic / Endothermic
If the products have more energy than the reactants,
i.e. Hfinal > Hinitial , then DH is positive. Heat has been
absorbed. Endothermic processes have a positive DH.
Enthalpy diagram for endothermic process.
H2O(s)
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H2O(l)
Sample Problem 6.2
PROBLEM:
PLAN:
Drawing Enthalpy Diagrams and Determining
the Sign of DH
In each of the following cases, determine the sign of DH, state
whether the reaction is exothermic or endothermic, and draw
and enthalpy diagram.
(a) H2(g) + 1/2O2(g)
H2O(l) + 285.8kJ
(b) 40.7kJ + H2O(l)
H2O(g)
Determine whether heat is a reactant or a product. As a reactant,
the products are at a higher energy and the reaction is
endothermic. The opposite is true for an exothermic reaction
SOLUTION: (a) The reaction is exothermic.
H2(g) + 1/2O2(g) (reactants)
EXOTHERMIC
H2O(l)
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DH = -285.8kJ
(products)
(b) The reaction is endothermic.
H2O(g)
ENDOTHERMIC
H2O(l)
(products)
DH = +40.7kJ
(reactants)
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
C4H10(l) + 13/2O2(g)
4CO2(g) + 5H2O(g)
heat of formation (DHf)
K(s) + 1/2Br2(l)
KBr(s)
heat of fusion (DHfus)
NaCl(s)
NaCl(l)
heat of vaporization (DHvap)
C6H6(l)
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C6H6(g)
Table 6.3 Heats of Combustion of Some Fats and Carbohydrates
Substance
DHcomb(kJ/g)
Fats
vegetable oil
-37.0 (x 1/4.18) = -8.85 Cal / g
margarine
-30.1
butter
-30.0
Carbohydrates
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table sugar (sucrose)
-16.2 (x 1/4.18) = -3.88 Cal / g
brown rice
-14.9
maple syrup
-10.4
Dr. Wolf’s CHM 101
Components of Internal Energy
Contributions to the kinetic energy:
• The molecule moving through space, Ek(translation)
• The molecule rotating, Ek(rotation)
• The bound atoms vibrating, Ek(vibration)
• The electrons moving within each atom, Ek(electron)
Contributions to the potential energy:
• Forces between the bound atoms vibrating, Ep(vibration)
• Forces between nucleus and electrons and between electrons in each
atom, Ep(atom)
• Forces between the protons and neutrons in each nucleus, Ep(nuclei)
• Forces between nuclei and shared electron pair in each bond, Ep(bond)
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Components of internal
energy (E)
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Breaking and Forming Chemical Bonds
During a chemical reaction, chemical bonds are
broken and new chemical bonds are formed.
Breaking each chemical bond requires energy, Endothermic
Forming each chemical bond releases energy, Exothermic
The DH for a reaction is the sum of the energy
required and released during the bond breaking and
forming process.
The enthalpy for any reverse process has the same
numerical value but a change in sign.
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Specific Heat Capacity
When an object absorbs heat, it gets hotter.
The more heat it absorbs the hotter it gets,
i.e. q ~ DT
or q = constant x DT or q /DT = constant
This capacity of an object to absorb heat is called its:
heat capacity = q / DT (units of J/K)
The amount of heat to change the temperature by 1 K
A related property is Specific Heat Capacity (c)
This is the amount of heat to change the temperature of 1 gram of
substance by 1 K
c = q / mass x DT (units of J/g-K)
Another property is Molar Heat Capacity (C)
This is the amount of heat to change the temperature of 1 mole of
substance by 1 K
C = q / moles x DT (units of J/mol-K)
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Sample Problem 6.3
PROBLEM:
Calculating the Quantity of Heat from the
Specific Heat Capacity
A layer of copper welded to the bottom of a skillet weighs 125g.
How much heat is needed to raise the temperature of the
copper layer from 250C to 300.0C? The specific heat capacity
(c) of Cu is 0.387J/g-K.
PLAN: Given the mass, specific heat capacity and change in temperature, we
can use q = c x mass x DT to find the answer. DT in 0C is the same as for
K.
SOLUTION:
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q=
0.387J
g*K
Dr. Wolf’s CHM 101
x 125g x (300-25)0C = 1.33x104J
Calorimetry
Calorimetry is a way to measure heat from a process by
directing the heat into “surroundings” where the increase of
heat can be measured. The surroundings will be a container
called a calorimeter.
The heat lost by the system equals the heat gained by
the calorimeter.
- q sample = qcalorimeter (minus sign for heat lost)
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Coffe-cup calorimeter
- q sample = qcalorimeter
- (mass x heat capacity x DT)samp =
(mass x heat capacity x DT)cal.
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Sample Problem 6.4
PROBLEM:
PLAN:
Determining the Specific Heat Capacity of a
Solid
A 25.64g sample of a solid was heated in a test tube to 100.00oC in
boiling water and carefully added to a coffee-cup calorimeter
containing 50.00g of water. The water temperature increased from
25.10oC to 28.49oC. What is the specific heat capacity of the solid?
(Assume all the heat is gained by the water).
It is helpful to use a table to summarize the data given. Then work the
problem realizing that heat lost by the system must be equal to that
gained by the surroundings.
SOLUTION:
csolid =
Mass(g)
c(J/g-K)
solid
25.64
c
H2O
50.00
4.184
25.64g x
c x
- 50.00g x
4.18J/g-K x 3.39 K
Dr. Wolf’s CHM 101
Tfinal
100.00 28.49
25.10
-71.51K = - 50.00g x
25.64g x -71.51K
6-25
Tinitial
=
28.49
DT
-71.51
3.39
4.18J/g-K x 3.39 K
0.387J/g-K
A bomb calorimeter
- q sample = qcalorimeter
qcalorimeter = heat capacity x DT
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Dr. Wolf’s CHM 101
Sample Problem 6.5
A manufacturer claims that its new dietetic dessert has “fewer
than 10 Calories per serving.” To test the claim, a chemist at the
Department of Consummer Affairs places one serving in a bomb
calorimater and burns it in O2(the heat capacity of the
calorimeter = 8.15kJ/K). The temperature increases 4.9370C.
Is the manufacturer’s claim correct?
PROBLEM:
PLAN:
Calculating the Heat of Combustion
- q sample = qcalorimeter
SOLUTION:
qcalorimeter
= heat capacity x DT
= 8.151kJ/K x 4.937K
= 40.24kJ
40.24kJ
kcal
= 9.62kcal or Calories
4.18kJ
The manufacturer’s claim is true.
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AMOUNT (mol)
of compound A
Summary of the relationship between
amount (mol) of substance and the heat
(kJ) transferred during a reaction.
AMOUNT (mol)
of compound B
molar ratio from
balanced equation
HEAT (kJ)
DHrxn (kJ/mol)
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gained or lost
Sample Problem 6.6
PROBLEM:
Using the Heat of Reaction (DHrxn) to Find
Amounts
The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented by
Al2O3(s)
2Al(s) + 3/2O2(g)
DHrxn = 1676kJ
If aluminum is produced this way, how many grams of aluminum can
form when 1.000x103kJ of heat is transferred?
PLAN:
SOLUTION:
heat(kJ)
1676kJ=2mol Al
mol of Al
xM
g of Al
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Dr. Wolf’s CHM 101
1.000x103kJ x
2mol Al
26.98g Al
1676kJ
1mol Al
= 32.20g Al
Hess’s Law
Since enthalpy is a state function, the enthalpy for an overall
process is equal to the sum of the enthalpies of all individual
steps.
Procedure: Identify the target equation whose DH is
unknown.
Manipulate contributing equations by multiplying
coefficients and/or reversing to get the correct amounts
of compounds as is target equation. Remembering to
change the sign of DH when reversing equations and
multiplying DH when multiplying coefficients.
The sum of all equations must be the same as the target
equation. The sum of the individual DH values will then
equal the target DH .
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Sample Problem 6.7
PROBLEM:
Using Hess’s Law to Calculate an Unknown DH
Two gaseous pollutants that form auto exhaust are CO and NO.
An environmental chemist is studying ways to convert them to
less harmful gases through the following equation:
CO(g) + NO(g)
CO2(g) + 1/2N2(g)
DH = ?
Given the following information, calculate the unknown DH:
Equation A: CO(g) + 1/2O2(g)
Equation B: N2(g) + O2(g)
PLAN:
2NO(g) DHB = 180.6kJ
Equations A and B have to be manipulated by reversal and/or
multiplication by factors in order to sum to the first, or target, equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g)
CO(g) + NO(g)
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CO2(g) DHA = -283.0kJ
Dr. Wolf’s CHM 101
CO2(g) DHA = -283.0kJ
1/2N2(g) + 1/2O2(g)
DHB = -90.6kJ
CO2(g) + 1/2N2(g) DHrxn = -373.6kJ
Formation Equations
The chemical equation showing the formation of 1
mole of a compound from the elements in their
standard state. The enthalpy change for the formation
is called the Standard Heat of Formation, DHof .
Element in their standard state have a DHof = 0
Most compounds have a negative DHof , i.e. The
reaction is exothermic........the compound is lower in
energy than the elements from which it was formed.
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Table 6.5 Selected Standard Heats of Formation at 250C(298K)
Formula
calcium
Ca(s)
CaO(s)
CaCO3(s)
DH0f(kJ/mol)
0
-635.1
-1206.9
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
chlorine
Cl(g)
6-33
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
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Formula DH0f(kJ/mol)
Formula DH0f(kJ/mol)
0
-92.3
silver
Ag(s)
AgCl(s)
hydrogen
H(g)
H2(g)
218
0
sodium
nitrogen
N2(g)
NH3(g)
NO(g)
0
-45.9
90.3
oxygen
O2(g)
O3(g)
H2O(g)
0
143
-241.8
H2O(l)
-285.8
Cl2(g)
HCl(g)
Na(s)
Na(g)
NaCl(s)
0
-127.0
0
107.8
-411.1
sulfur
S8(rhombic) 0
S8(monoclinic) 2
SO2(g)
-296.8
SO3(g)
-396.0
Sample Problem 6.8
PROBLEM:
Writing Formation Equations
Write balanced equations for the formation of 1 mol of the following
compounds from their elements in their standard states and include
DH0f.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
PLAN:
Use the table of heats of formation for values.
SOLUTION:
(a) Ag(s) + 1/2Cl2(g)
(b) Ca(s) + C(graphite) + 3/2O2(g)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
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DH0f = -127.0kJ
AgCl(s)
CaCO3(s)
HCN(g)
DH0f = -1206.9kJ
DH0f = 135kJ
The general process for determining DH0rxn from DH0f values
Using Hess’s Law to Calculate DH0rxn
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Sample Problem 6.9
Calculating the Heat of Reaction from Heats of
Formation
Nitric acid, whose worldwide annual production is about 8 billion kg,
is used to make many products, including fertilizer, dyes, and
explosives. The first step in the industrial production process is the
oxidation of ammonia:
PROBLEM:
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate DH0rxn from DH0f values.
PLAN:
Look up the DH0f values and use Hess’s Law to find DHrxn.
SOLUTION:
DHrxn = S mDH0f (products) - S nDH0f (reactants)
DHrxn = [4(DH0f NO(g) + 6(DH0f H2O(g)] - [4(DH0f NH3(g) + 5(DH0f O2(g)]
= (4mol)(90.3kJ/mol) + (6mol)(-241.8kJ/mol) [(4mol)(-45.9kJ/mol) + (5mol)(0kJ/mol)]
DHrxn = -906kJ
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End of Chapter 6
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