8/6/2019 Chemistry Chapter 01
1/56
Chapter 1Thermochemistry
CHEMISTRYThe Central Science
David P. White
1
8/6/2019 Chemistry Chapter 01
2/56
Kinetic Energy and Potential Energy Kinetic energy is the energy of motion:
Potential energy is the energy an object possesses by
virtue of its position.
Potential energy can be converted into kinetic energy.Example: a bicyclist at the top of a hill.
The Nature of Energy
2
8/6/2019 Chemistry Chapter 01
3/56
Kinetic Energy and Potential Energy
Electrostatic potential energy,Ed, is the attraction
between two oppositely charged particles, Q1 and Q2, a
distance dapart:
The constant = 8.99 109 J-m/C2. (a constant ofproportionality)
If the two particles are of opposite charge, thenEd is the
electrostatic repulsion between them.
The Nature of Energy
3
8/6/2019 Chemistry Chapter 01
4/56
Kinetic Energy and Potential Energy
When Q1 and Q2 have the same sign (Q1=+ve, Q2=+ve),
its repels each other, pushing them apart, then Ed is
positive.
For opposite sign (Q1=+ve, Q2=-ve), they are attract
each other pulling them each other, then Ed is negative.
The Nature of Energy
4
8/6/2019 Chemistry Chapter 01
5/56
Units of Energy
SI Unit for energy is the joule, J:
We sometimes use the calorie instead of the joule:
1 cal = 4.184 J (exactly)
A nutritional Calorie:
1 Cal = 1000 cal = 1 kcal
The Nature of Energy
5
8/6/2019 Chemistry Chapter 01
6/56
Systems and Surroundings System: part of the universe we are interested in.
Surroundings: the rest of the universe.
The Nature of Energy
6
8/6/2019 Chemistry Chapter 01
7/56
2009 Prentice-Hall Inc
Definitions:
System and Surroundings
The system includes themolecules we want to
study (here, the hydrogenand oxygen molecules).
The surroundings areeverything else (here, the
cylinder and piston).
8/6/2019 Chemistry Chapter 01
8/56
Transferring Energy: Work and Heat
Force is a push or pull on an object. Work is the product of force applied to an object over a
distance:
Energy is the work done to move an object against aforce.
Heat is the transfer of energy between two objects.
Therefore, Energy is the capacity to do work or transferheat.
The Nature of Energy
8
8/6/2019 Chemistry Chapter 01
9/56
2009 Prentice-Hall Inc
Heat
Energy can also betransferred as heat.
Heat flows fromwarmer objects tocooler objects.
8/6/2019 Chemistry Chapter 01
10/56
Internal Energy
Internal Energy: total energy of a system.
Cannot measure absolute internal energy.
Change in internal energy,
The First Law of
Thermodynamics
10
8/6/2019 Chemistry Chapter 01
11/56
Relating E to Heat and Work Energy cannot be created or destroyed.
Energy of (system + surroundings) is constant.
Any energy transferred from a system must be transferred
to the surroundings (and vice versa).
From the first law of thermodynamics:
when a system undergoes a physical or chemical change, thechange in internal energy is given by the heat added to or
absorbed by the system plus the work done on or by the
system:
The First Law of
Thermodynamics
11
8/6/2019 Chemistry Chapter 01
12/56
The First Law of
Thermodynamics
12
8/6/2019 Chemistry Chapter 01
13/56
13
8/6/2019 Chemistry Chapter 01
14/56
Exothermic and EndothermicProcesses
Endothermic: absorbs heat from the surroundings.
Exothermic: transfers heat to the surroundings. An endothermic reaction feels cold.
An exothermic reaction feels hot.
The First Law of
Thermodynamics
14
http://d/Media_Portfolio/ThermiteReaction/Thermite.htmlhttp://d/Media_Portfolio/FormationofWater/FormationofWater.html8/6/2019 Chemistry Chapter 01
15/562009 Prentice-Hall Inc
Exchange of Heat between
System and Surroundings
When heat is absorbed by the system fromthe surroundings, the process is endothermic.
8/6/2019 Chemistry Chapter 01
16/562009 Prentice-Hall Inc
Exchange of Heat between
System and Surroundings
When heat is absorbed by the system fromthe surroundings, the process is endothermic.
When heat is released by the system into thesurroundings, the process is exothermic.
8/6/2019 Chemistry Chapter 01
17/56
State Functions
State function: a property of a system that depends only
on the initial and final states of system, not on how the
internal energy is used. (depends only on the physical state (P,T,etc) of the system and not on the route used to achieve the current state).
The First Law of
Thermodynamics
17
8/6/2019 Chemistry Chapter 01
18/56
State Functions
18
However, q and w are not state functions.
Whether the battery is shorted out or is discharged by
running the fan, its Eis the same.
8/6/2019 Chemistry Chapter 01
19/56
Chemical reactions can absorb or release heat. However, they also have the ability to do work.
For example, when a gas is produced, then the gas
produced can be used to push a piston, thus doing work.
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
The work performed by the above reaction is called
pressure-volume work.
When the pressure is constant,
(when work is done on the surrounding,
w0 and positive.)
Enthalpy
19
http://d/Media_Portfolio/WorkofGasExpansion/WorkofGasExpansion.htmlhttp://d/Media_Portfolio/WorkofGasExpansion/WorkofGasExpansion.html8/6/2019 Chemistry Chapter 01
20/56
8/6/2019 Chemistry Chapter 01
21/56
Enthalpy,H: Heat transferred between the system andsurroundings carried out under constant pressure.
(internal energy + the product of the pressure and
volume of the system)
Enthalpy is a state function becauseE, P & Vare all state
functions.
If the process occurs at constant pressure,
Enthalpy
21
http://d/Media_Portfolio/ChangesofState/ChangesofState.html8/6/2019 Chemistry Chapter 01
22/56
Since we know that
We can write
When H, is positive, the system gains heat from the
surroundings. (endothermic)
When H, is negative, the surroundings gain heat from
the system. (exothermic)
Enthalpy
VPw
22
8/6/2019 Chemistry Chapter 01
23/56
Enthalpy
23
8/6/2019 Chemistry Chapter 01
24/56
For a reaction:
Enthalpy is an extensive property (magnitude His
directly proportional to amount):
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H= -802 kJ2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H= 1604 kJ
Enthalpies of Reaction
24
http://d/Media_Portfolio/WorkofGasExpansion/WorkofGasExpansion.html8/6/2019 Chemistry Chapter 01
25/56
When we reverse a reaction, we change the sign ofH:
CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H= +802 kJ
Change in enthalpy depends on state:
2H2O(g) 2H2O(l) H= -88 kJ
H2O(l) H2O(g) H= +44 kJ
H2O(s) H2O(l) H= +6.01 kJ
Enthalpies of Reaction
25
8/6/2019 Chemistry Chapter 01
26/56
Heat Capacity and Specific Heat
Calorimetry = measurement of heat flow by expt
Calorimeter = apparatus that measures heat flow.
Heat capacity = the amount of energy required to raisethe temperature of an object (by one degree).
Molar heat capacity = heat capacity of 1 mol of a
substance. Specific heat = specific heat capacity = heat capacity of 1
g of a substance.
Calorimetry
26
8/6/2019 Chemistry Chapter 01
27/56
8/6/2019 Chemistry Chapter 01
28/56
Calorimetry
Constant Pressure Calorimetry
28
8/6/2019 Chemistry Chapter 01
29/56
Calorimetry
Reaction carried out under
constant volume.
Use a bomb calorimeter.
Usually study combustion.
Ccal is the heat capacity of
the calorimeter.
Bomb Calorimetry (Constant Volume
Calorimetry)
29
8/6/2019 Chemistry Chapter 01
30/56
Hesss law: if a reaction is carried out in a number of
steps, Hfor the overall reaction is the sum ofHfor
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
Hesss Law
30
8/6/2019 Chemistry Chapter 01
31/56
Note that:
H1 = H2 + H3
Hesss Law
31
http://d/Media_Portfolio/ThermiteReaction/Thermite.html8/6/2019 Chemistry Chapter 01
32/56
32
SAMPLE EXERCISE 5.8 Using Hesss Law to Calculate H
The enthalpy of reaction for the combustion of C to CO2 is 393.5 kJ/mol C, and the enthalpy for thecombustion of CO to CO
2is 283.0 kJ/mol CO:
Using these data, calculate the enthalpy for the combustion of C to CO:
Solution
In order to use equations (1) and (2), we arrange them so that C(s) is on the reactant side and CO(g)is on the product side of the arrow, as in the target reaction, equation (3). Because equation (1) hasC(s) as a reactant, we can use that equation just as it is. We need to turn equation (2) around,however, so that CO(g) is a product. Remember that when reactions are turned around, the sign of His reversed. We arrange the two equations so that they can be added to give the desired equation:
8/6/2019 Chemistry Chapter 01
33/56
33
SAMPLE EXERCISE 5.8 continued
When we add the two equations, CO2(g) appears on both sides of the arrow and therefore cancels out.Likewise, is eliminated from each side.
8/6/2019 Chemistry Chapter 01
34/56
34
SAMPLE EXERCISE 5.9 Using Three Equations with Hesss Law to Calculate H
Calculate Hfor the reaction
given the following chemical equations and their respective enthalpy changes:
Solution
Analyze: We are given a chemical equation and asked to calculate its Husing three chemicalequations and their associated enthalpy changes.
Plan: We will use Hesss law, summing the three equations or their reverses and multiplying each byan appropriate coefficient so that they add to give the net equation for the reaction of interest. At thesame time, we keep track of the Hvalues, reversing their signs if the reactions are reversed andmultiplying them by whatever coefficient is employed in the equation.
Solve:Because the target equation has C
2H
2as a product, we turn the first equation around; the
sign of H is therefore changed. The desired equation has 2 C(s) as a reactant, so we multiply thesecond equation and its Hby 2. Because the target equation has as a reactant, we keep the thirdequation as it is. We then add the three equations and their enthalpy changes in accordance withHesss law:
8/6/2019 Chemistry Chapter 01
35/56
35
SAMPLE EXERCISE 5.9 continued
When the equations are added, there are on both sides of the arrow. Theseare canceled in writing the net equation.
8/6/2019 Chemistry Chapter 01
36/56
If 1 mol of compound is formed from its constituent
elements, then the enthalpy change for the reaction is
called the enthalpy of formation, Hof .
Standard conditions (standard state): 1 atm and 25oC
(298 K).
Standard enthalpy,
H
o
, is the enthalpy measured wheneverything is in its standard state.
Standard enthalpy of formation: 1 mol of compound is
formed from substances in their standard states.
Enthalpies of Formation
36
8/6/2019 Chemistry Chapter 01
37/56
If there is more than one state for a substance under
standard conditions, the more stable one is used.
Standard enthalpy of formation of the most stable form ofan element is zero.
Enthalpies of Formation
37
8/6/2019 Chemistry Chapter 01
38/56
Enthalpies of Formation
38
8/6/2019 Chemistry Chapter 01
39/56
Using Enthalpies of Formation tocalculate Enthalpies of Reaction
We use Hess Law to calculate enthalpies of a reaction
from enthalpies of formation.
Enthalpies of Formation
39
8/6/2019 Chemistry Chapter 01
40/56
40
8/6/2019 Chemistry Chapter 01
41/56
Using Enthalpies of Formation tocalculate Enthalpies of Reaction
For a reaction
Enthalpies of Formation
41
8/6/2019 Chemistry Chapter 01
42/56
42
SAMPLE EXERCISE 5.10 Identifying Equations associated with Enthalpies of Formation
For which of the following reactions at 25C would the enthalpy change represent a standard enthalpy
of formation? For those where it does not, what changes would need to be made in the reactionconditions?
Solution
Analyze:The standard enthalpy of formation is represented by a reaction in which each reactant is
an element in its standard state and the product is one mole of the compound.
Plan: To solve these problems, we need to examine each equation to determine, first of all, whetherthe reaction is one in which a substance is formed from the elements. Next, we need to determinewhether the reactant elements are in their standard states.
Solve: In (a) Na2O is formed from the elements sodium and oxygen in their proper states, a solidand O2 gas, respectively. Therefore, the enthalpy change for reaction (a) corresponds to a standard
enthalpy of formation.In (b) potassium is given as a liquid. It must be changed to the solid form, its standard state atroom temperature. Furthermore, two moles of product are formed, so the enthalpy change for thereaction as written is twice the standard enthalpy of formation of KCl(s). The proper equation for theformation reaction is
8/6/2019 Chemistry Chapter 01
43/56
43
SAMPLE EXERCISE 5.10 continued
Reaction (c) does not form a substance from its elements. Instead, a substance decomposes to itselements, so this reaction must be reversed. Next, the element carbon is given as diamond, whereasgraphite is the lowest-energy solid form of carbon at room temperature and 1 atm pressure. Theequation that correctly represents the enthalpy of formation of glucose from its elements is
8/6/2019 Chemistry Chapter 01
44/56
44
SAMPLE EXERCISE 5.11 Calculating an Enthalpy of Reaction fromEnthalpies of Formation
(a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g)
and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane to thatproduced by 1.00 g benzene.
SolutionAnalyze: (a) We are given a reaction [combustion of C6H6(l) to form CO2(g) and H2O(l)] and askedto calculate its standard enthalpy change, H(b) We then need to compare the quantity of heatproduced by combustion of 1.00 g C6H6 with that produced by 1.00 g of C3H8, whose combustion was
treated above in the text.Plan: (a) We need to write the balanced equation for the combustion of C6H6. We then look upvalues in Appendix C or inTable 5.3 and apply Equation 5.31 to calculate the enthalpy change for thereaction. (b) We use the molar mass of C6H6 to change the enthalpy change per mole to that pergram. We similarly use the molar mass of C3H8 and the enthalpy change per mole calculated in thetext above to calculate the enthalpy change per gram of that substance.
Solve: (a) We know that a combustion reaction involves O2(g) as a reactant. Thus, the balancedequation for the combustion reaction of 1 mol C6H6(l) is
SAMPLE EXERCISE 5 11 continued
8/6/2019 Chemistry Chapter 01
45/56
45
We can calculate Hfor this reaction by using Equation 5.31 and data in Table 5.3. Remember tomultiply the
value for each substance in the reaction by that substances stoichiometric coefficient. Recall
also that
for any element in its most stable form under standard conditions, so
Comment: Both propane and benzene are hydrocarbons. As a rule, the energy obtained from thecombustion of a gram of hydrocarbon is between 40 and 50 kJ.
SAMPLE EXERCISE 5.11 continued
(b) From the example worked in the text, H =2220 kJ for the combustion of 1 mol of propane. In
part (a) of this exercise we determined that H=3267 kJ for the combustion of 1 mol benzene. Todetermine the heat of combustion per gram of each substance, we use the molar masses to convertmoles to grams:
8/6/2019 Chemistry Chapter 01
46/56
Foods
Fuel value = energy released when 1 g of substance is
burned.
1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. Energy in our bodies comes from carbohydrates and fats
(mostly).
Intestines: carbohydrates converted into glucose which
react with O2 to produce energy.C6H12O6 + 6O2 6CO2 + 6H2O, H = -2816 kJ
Fats break down as follows:2C57H110O6 + 163O2 114CO2 + 110H2O, H
o = -75,520 kJ
Foods and Fuels
46
8/6/2019 Chemistry Chapter 01
47/56
Foods Fats: store excess energy; are insoluble in water, so are
good for energy storage.
Foods and Fuels
47
8/6/2019 Chemistry Chapter 01
48/56
Fuels
In 2000 the United States consumed 1.03 1017 kJ of
fuel.
Most fuel derived from petroleum and natural gas. Remainder from coal, nuclear, and hydroelectric.
Fossil fuels are not renewable.
Foods and Fuels
48
8/6/2019 Chemistry Chapter 01
49/56
Foods and
Fuels
49
8/6/2019 Chemistry Chapter 01
50/56
Fuels
Fuel value = energy released when 1 g of substance is
burned.
Hydrogen has great potential as a fuel with a fuel value of142 kJ/g.
Foods and Fuels
50
8/6/2019 Chemistry Chapter 01
51/56
51
Each Cl- ion is surrounded by six Na+ ions in NaCl
molecules
There is a regular arrangement of Na
+
and Cl
-
in 3D. Note that the ions are packed as closely as possible.
Note that it is not easy to find
a molecular formula to describe
the ionic lattice.
Lattice Energy in Ions
8/6/2019 Chemistry Chapter 01
52/56
52
Energetics of Ionic Bond
Formation
The formation of Na+(g) and Cl-(g) from Na(g) and
Cl(g) is endothermic.
Why is the formation of Na(s) exothermic? The reaction NaCl(s) Na+(g) + Cl-(g) is endothermic
(H= +788 kJ/mol).
The formation of a crystal lattice from the ions in the
gas phase is exothermic:
Na+(g) + Cl-(g) NaCl(s) H= -788 kJ/mol
8/6/2019 Chemistry Chapter 01
53/56
53
Lattice Energy
Lattice energy: the energy required to completely
separate a mole of an ionic solid into its gaseous ions.
Lattice energy depends on the charges on the ions and
the sizes of the ions:
is a constant (8.99 x 109 Jm/C2), Q1 and Q2 are the
charges on the ions, and dis the distance between ions.
d
QQEl
21
L tti i
8/6/2019 Chemistry Chapter 01
54/56
54
Lattice energy increases as
- the charges on the ions increase
- the distance between the ions decreases.
8/6/2019 Chemistry Chapter 01
55/56
8/6/2019 Chemistry Chapter 01
56/56