Download Chapter 6 Thermochem 110 F11 IP

Chapter 6
Thermochemistry:
Energy Flow and Chemical Change
1
Slide 6-1
Chapter 6: First Law
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 (ΔH0rxn)
2
Slide 6-2
Chapter 6: First Law
Thermodynamics: First Law
Thermodynamics: Study and measure of the nature of heat and energy and its
transformations
Thermochemistry is a branch of thermodynamics that deals with the heat
involved with chemical and physical changes.
✔Energy flow is key in science; understanding it will help us to understand WHY
things occur. When energy is transferred from one object to another, it appears as
work and/or as heat
Thermodynamics was developed independent of theory through observation of bulk
matter
To deal with large groups of molecules and atoms we need to define THREE
parameters: System, Surroundings and Thermodynamic Universe
3
Slide 6-3
Chapter 6: First Law
Thermodynamics: First Law
System:
The part of the universe of immediate interest;
composed of particles with their own internal
energies (E or U). The system has an internal
energy. When a change occurs, the internal
energy changes. Can be closed, open or isolated;
Sign is from the point of view
of the System
Surroundings:
The portion of the universe that can exchange
energy and matter with the system;
Thermodynamic Universe:
System + Surroundings
Open System: Exchanges both E and matter with Surroundings;
Closed System: Exchanges only E with Surroundings;
Isolated System: Exchanges NEITHER with Surroundings
Slide 6-4
Chapter 6: First Law
6-4
Figure 6.1
A chemical system and its surroundings.
The system in this case is the contents of the reaction flask. The
surroundings comprise everything else, including the flask itself.
5
Slide 6-5
Chapter 6: First Law
An Auto
Slide 6-6
A Battery
Chapter 6: First Law
A perfect Thermos
6
Energy diagrams for the transfer of internal energy (E) between a system
and its surroundings.
A. E of system decreasses
B. E of system increases
ΔE = Efinal - Einitial = Eproducts - Ereactants
7
Slide 6-7
Chapter 6: First Law
A system transferring energy as heat only.
A E lost as heat
B E gained as heat
8
Slide 6-8
Chapter 6: First Law
A system losing energy as work only.
9
Slide 6-9
Chapter 6: First Law
Work (w)
Motion against an opposing force; a directed motion thru a distance; w = F•Δx
Capacity of a system to do work is called
Internal Energy (U)
Most important: ΔU = Ufinal – Uinitial
If E transfer is only due to work then ΔU = w
Work is either expansion or nonexpansion
(ΔV≠ 0 vs. ΔV= 0)
w = –PextΔV (since w = – when system does work)
10
Slide 6-10
Chapter 6: First Law
Work
w = – PextΔV (Pext must be constant)
Work is in L-atm: 1 L-atm = 101.325 J
If is Pext = 0, then free expansion occurs
If Pext does change in infinitesimally small amts, then process is REVERSIBLE: a process
in which an infinitesimally small change in the opposite direction will cause the process
to reverse direction
Chapter 6: First Law
11
Slide 6-11
Figure 6.5
Pressure-volume work.
12
Slide 6-12
Chapter 6: First Law
Figure 6.7
Pressure-volume work.
An expanding gas pushing back the atmosphere does PV work (w =
-PΔV).
13
Slide 6-13
Chapter 6: First Law
Heat (q)
Energy transfer due to a temperature difference
 More familiar way to transfer E
 Sys and Surr are at different T s
 If no work is done, then ΔU = q
 Temperature is not heat; it is directly
proportional to it (q ∝ ΔT)!
 Temperature: a measure of the relative hotness
or coldness of an object
 q = –qsurr
 Exothermic vs Endothermic
Chapter 6: First Law
Slide 6-14
6-14
Calorimetry
q = c x m x ΔT
q = heat lost or gained
c = specific heat capacity
m = mass in g
ΔT = Tfinal - Tinitial
The specific heat capacity (c) of a substance is the
quantity of heat required to change the temperature of
1 gram of the substance by 1 K.
15
Slide 6-15
Chapter 6: First Law
Measuring Heat Using a Calorimeter
Constant Pressure
Slide 6-16
Constant Volume
Chapter 6: First Law
6-16
Figure 6.7
Coffee-cup calorimeter.
17
Slide 6-17
Chapter 6: First Law
Figure 6.8
A bomb calorimeter.
18
Slide 6-18
Chapter 6: First Law
19
Slide 6-19
Chapter 6: First Law
Selected Heat Capacities (Cs and Cm)
 
 
Slide 6-20
Amt of heat to raise 10.0 g of water 5°C: 209.2 J
Temp increase if same amount of heat is added to 10.0 g of gold (Cs = 0.106
J/g-C): 197.4 °C!
Chapter 6: First Law
6-20
21
Slide 6-21
Chapter 6: First Law
Cm,V vs. Cm,P
q
q
Cm =
=
n!T !T
Since q = nCmΔT then
(n = 1)
However, q = ΔU at constant volume or q = ΔH at constant pressure, So:
Cm,V
!U
=
n!T
OR
Cm, P
!H
=
n!T
and
Cm,P = Cm,V + R
For Ar: Cm,V =12.8 J/mol-K and Cm,P = 21.1 J/mol-K
Chapter 6: First Law
Slide 6-22
6-22
Sample Problem 6.3
Finding the Quantity of Heat from Specific Heat Capacity
PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How
much heat is needed to raise the temperature of the copper layer from
25oC to 300.oC? The specific heat capacity (c) of Cu is 0.387 J/g•K.
PLAN: Given the mass, specific heat capacity and change in temperature, we can use q
= c x mass x ΔT to find the answer. ΔT in oC is the same as for K.
SOLUTION:
q = 0.387 J/g•K
x 125 g x (300oC - 25oC)
= 1.33 x 104 J
23
Slide 6-23
Chapter 6: First Law
Figure 6.4
Two different paths for the energy change of a system.
24
Slide 6-24
Chapter 6: First Law
First Law of Thermodynamics
The internal energy of an isolated system is constant
•  ΔU = q + w
•  Isolated system, ΔU = 0
•  Container walls can be
either adiabatic (q = 0) or
diathermic (w = 0)
•  Both q and w are path
dependent
Internal Energy (U) is a State Function
To determine the change in a state function, choose any convenient path
(even if the real path is too complicated to calculate)
Chapter 6: First Law
Slide 6-25
6-25
ΔEuniverse = ΔEsystem + ΔEsurroundings
Units of Energy
Joule (J)
calorie (cal)
1 J = 1 kg•m2/s2
1 cal = 4.184 J
British thermal unit (Btu)
1 Btu = 1055 J
26
Slide 6-26
Chapter 6: First Law
Sample Problem 6.1
PROBLEM:
PLAN:
Determining the Change in Internal Energy of a
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 451 J of work on the pistons and the system loses 325 J to the
surroundings as heat, calculate the change in energy (ΔE) in J, kJ, and
kcal.
Define system and surroundings, assign signs to q and w and calculate ΔE. The
answer should be converted from J to kJ and then to kcal.
SOLUTION:
q = - 325 J
w = - 451 J
ΔE = q + w = -325 J + (-451 J) =
-776 J x
kJ
103J
= -0.776 kJ
-0.776 kJ x
-776 J
kcal
4.184 kJ
= -0.185 kcal
27
Slide 6-27
Chapter 6: First Law
Enthalpy (H)
The heat released or absorbed by a system at constant pressure
•  For Constant Volume systems ΔU = q
•  Most Processes WE study are Constant Pressure systems
•  Enthalpy is a State Function that allows us to keep track of
q at constant P; By definition: H = U + PV
•  ΔH = ΔU + PΔV = q – PextΔV + PΔV = q – PextΔV + PextΔV
•  So ΔH = q
•  Implication 1: At constant volume you measure ΔU; at
constant pressure you measure ΔH
•  Implication 2: The molar heat capacity Cm = q/(nΔT)
Therefore it will be different for a constant volume process
than for a constant pressure process
Chapter 6: First Law
Slide 6-28
6-28
The Meaning of Enthalpy
w = - PΔV
H = E + PV
ΔH ≈ ΔE in
1. Reactions that do not involve gases.
where H is enthalpy
ΔH = ΔE + PΔV
qp = ΔE + PΔV = ΔH
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 >>> PΔV.
29
Slide 6-29
Chapter 6: First Law
Mathematical Aspects of State Functions
Given X is a state function, then:
•  The change in X is equal to the sum of the X values for
the products minus the reactants:
ΔX = ΣniXpdts – ΣnjXrxtnts
•  The value for the reverse process is equal in magnitude
BUT opposite in sign:
ΔXrev = –ΔXfwd
•  Multiplication of a process by number multiplies the X
value by that number:
If A → 2B ΔX1 = 7 then 3A → 6B ΔX2 = 3ΔX1 = 21
•  The sum of all of the X values making up a process is
equal to the X overall for that process:
ΔXtotal = ΔXstep 1 + ΔXstep 2 + ΔXstep 3 + …
Chapter 6: First Law
Slide 6-30
6-30
Figure 6.6
Enthalpy diagrams for exothermic and endothermic processes.
31
Slide 6-31
Chapter 6: First Law
ΔH of Physical Processes
Chapter 6: First Law
Slide 6-32
6-32
Heating Curves
o
ΔHvap
ΔHfus
q
q
q
o
T1 (–5 C))
T2 (+120 C)
Chapter 6: First Law
Slide 6-33
6-33
Heating Curve Calculations
To take 0.5 mole (9.0 g) of ice from –5ºC to vapor at 120ºC:
Total heat = (heat to take 0.5 mol of ice from –5ºC to 0ºC)
+ (heat to melt 0.5 mol of ice)
+ (heat to take 0.5 mol of water from 0ºC to 100ºC)
+ (heat to vaporize 0.5 mol of ice)
+ (heat to take 0.5 mol of vapor from 100ºC to 120ºC)
Chapter 6: First Law
Slide 6-34
6-34
ΔH of Chemical Change
• Enthalpy change during chemical reactions important to
understanding chemistry.
• Reaction Enthalpies: ΔH or ΔHrxn
• ΔH associated with Thermochemical Equations:
2Al (s) + 6HCl (aq) → 2AlCl3 (aq) + H2 (g) ΔH = –1049 kJ
• ΔHrxn associated with certain types of reactions; corrected
to a per mole basis; used in tabular collections of data:
ΔHrxn = –524.5 kJ/mol for the HCl oxidation of Al
ΔHcombustion = –1560 kJ/mol for ethane (C2H6)
• Relation between ΔH and ΔU:
For reactions in which NO gas is generated: ΔH = ΔU
For reactions in which gas is generated: ΔH = ΔU + ΔngasRT
Chapter 6: First Law
Slide 6-35
6-35
Standard Reaction Enthalpies ΔH°
ΔH depends on conditions and phases; to allow
standardization for reporting the idea of Standard State was
developed.
The standard state of a property X (designated as X°) is
defined as such:
  Substances and elements: most stable form of that material
at the T of interest and 1 bar of pressure;
  Solutions: 1.0 M concentration;
Allows the tabulation and reporting of data from
various sources to be accomplished!
Chapter 6: First Law
Slide 6-36
6-36
Common ΔHx° s
Enthalpy of Formation (ΔHf°)
  Defined as: the standard reaction enthalpy for the formation of one mole
of a substance in its most stable form from its elements in their most stable
form at the temperature of interest:
2 Na(s) + C(s,gr) + 3/2 O2(g)  Na2CO3 (s)
  By Definition ΔHf°≡ 0 of an element in its most stable state;
  Allows ΔHrxn° to be calculated using ΔHf° of pdts - rxtnts:
aA + bB  dD
ΔHf° = [dΔH°f,D] – [aΔH°f,A – bΔH°f,B]
Enthalpy of Combustion (ΔHc° or ΔH°comb)
  Defined as: the standard reaction enthalpy for the combustion of one mole
of a substance in its most stable form at T of interest:
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O (l)
Chapter 6: First Law
Slide 6-37
6-37
Sample Problem 6.2
PROBLEM:
Drawing Enthalpy Diagrams and Determining the
Sign of ΔH
In each of the following cases, determine the sign of ΔH, state whether
the reaction is exothermic or endothermic, and draw an enthalpy
diagram.
1
(a) H2(g) +
O (g)
H2O(l) + 285.8 kJ
2 2
(b) 40.7 kJ + H2O(l)
PLAN:
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.
H2O(g)
SOLUTION:
38
Slide 6-38
Chapter 6: First Law
Sample Problem 6.4
Determining the Heat of a Reaction
PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00oC
and carefully add 25.0 mL of 0.500 M HCl, also at 25.00oC. After stirring, the
final temperature is 27.21oC. Calculate qsoln (in J) and ΔHrxn (in kJ/mol).
(Assume the total volume is the sum of the individual volumes and that the
final solution has the same density and specific heat capacity as water: d =
1.00 g/mL and c = 4.184 J/g•K)
PLAN:
We need to determine the limiting reactant from the net ionic equation. The
moles of NaOH and HCl as well as the total volume can be calculated. From the
volume we use density to find the mass of the water formed. At this point qsoln
can be calculated using the mass, c, and ΔT. The heat divided by the M of water
will give us the heat per mole of water formed.
39
Slide 6-39
Chapter 6: First Law
Sample Problem 6.4
Determining the Heat of a Reaction
continued
SOLUTION:
HCl(aq) + NaOH(aq)
H+(aq) + OH-(aq)
NaCl(aq) + H2O(l)
H2O(l)
For NaOH
0.500 M x 0.0500 L = 0.0250 mol OH-
For HCl
0.500 M x 0.0250 L = 0.0125 mol H+
HCl is the limiting reactant.
0.0125 mol of H2O will form during the reaction.
total volume after mixing = 0.0750 L
0.0750 L x 103 mL/L x 1.00 g/mL = 75.0 g of water
q = mass x specific heat x ΔT
= 75.0 g x 4.184 J/g•oC x (27.21oC - 25.00oC)
= 693 J
(693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed
40
Slide 6-40
Chapter 6: First Law
Sample Problem 6.5
PROBLEM:
PLAN:
Calculating the Heat of a Combustion Reaction
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 Consumer Affairs places one serving in a bomb calorimeter and
burns it in O2 (the heat capacity of the calorimeter = 8.151 kJ/K). The
temperature increases 4.937oC. Is the manufacturer s claim correct?
- q sample = qcalorimeter
SOLUTION:
qcalorimeter
(First Law of Thermodynamics)
= heat capacity x ΔT
= 8.151 kJ/K x 4.937 K
= 40.24 kJ
1 Calorie = 1 kcal = 4.184 kJ
10 Calorie = 41.84 kJ
The manufacturer’s claim is true.
41
Slide 6-41
Chapter 6: First Law
Figure 6.9
Summary of the relationship between amount
(mol) of substance and the heat (kJ)
transferred during a reaction.
42
Slide 6-42
Chapter 6: First Law
Sample Problem 6.6
PROBLEM:
PLAN:
Using the Heat of Reaction (ΔHrxn) to Find Amounts
The major source of aluminum in the world is bauxite (mostly aluminum
oxide). Its thermal decomposition can be represented by
3
Al2O3(s)
2Al(s) + O2(g) ΔHrxn = 1676 kJ
2
If aluminum is produced this way, how many grams of aluminum can form
when 1.000 x 103 kJ of heat is transferred?
SOLUTION:
1.000 x 103 kJ x
2 mol Al
1676 kJ
x
26.98 g Al
1 mol Al
= 32.20 g Al
43
Slide 6-43
Chapter 6: First Law
Sample Problem 6.8
PROBLEM:
Using Hess’s Law to Calculate an Unknown
ΔH
Two gaseous pollutants that form in auto exhausts are CO and
NO. An environmental chemist is studying ways to convert them
to less harmful gases through the following reaction:
CO(g) + NO(g)
→
CO2(g) + ½N2(g) ΔH = ?
Given the following information, calculate the unknown ΔH:
Equation A: CO(g) + ½ O2(g) → CO2(g) ΔHA = -283.0 kJ
Equation B: N2(g) + O2(g) → 2NO(g) ΔHB = 180.6 kJ
PLAN:
Manipulate Equations A and/or B and their ΔH values to get to the target
equation and its ΔH. All substances except those in the target equation
must cancel.
44
Slide 6-44
Chapter 6: First Law
Sample Problem 6.8
SOLUTION:
Multiply Equation B by ½ and reverse it:
NO(g) → ½ N2(g) + ½ O2(g); ΔH = - 90.3 kJ
Add the manipulated equations together:
Equation A: CO(g) + ½ O2(g) → CO2(g)
½ Equation B:
(reversed)
NO(g)
ΔH = -283.0 kJ
→ ½ N2(g) + ½ O2(g)
CO(g) + NO(g)
→
ΔH = - 90.3 kJ
CO2(g) + ½ N2(g)
ΔHrxn = -373.3 kJ
45
Slide 6-45
Chapter 6: First Law
46
Slide 6-46
Chapter 6: First Law
Table 6.3 Selected Standard Enthalpies of Formation at 25°C (298K)
Formula
ΔH°f (kJ/mol) Formula
Calcium
Ca(s)
CaO(s)
CaCO3(s)
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)
Slide 6-47
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
ΔH°f (kJ/mol) Formula
ΔH°f (kJ/mol)
0
-92.3
Hydrogen
H(g)
H 2 (g )
Silver
Ag(s)
AgCl(s)
218
0
Sodium
Nitrogen
N 2 (g )
NH3(g)
NO(g)
0
-45.9
90.3
Cl2(g)
HCl(g)
Oxygen
O2(g)
O3(g)
H2O(g)
H2O(l)
Chapter 6: First Law
0
143
-241.8
-285.8
Na(s)
Na(g)
NaCl(s)
0
-127.0
0
107.8
-411.1
Sulfur
S8(rhombic) 0
S8(monoclinic) 0.3
SO2(g)
-296.8
SO3(g)
-396.0
47
Sample Problem 6.9
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
ΔH°f.
(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:
Write the elements as reactants and 1 mol of the compound as the
product formed. Make sure all substances are in their standard
states. Balance the equations and find the value of ΔH°f in Table 6.3
or Appendix B.
48
Slide 6-48
Chapter 6: First Law
Sample Problem 6.9
SOLUTION:
(a) Silver chloride, AgCl, a solid at standard conditions.
Ag(s) + ½Cl2(g) → AgCl(s)
ΔH°f = -127.0 kJ
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
Ca(s) + C(graphite) +
3
O22(g) → CaCO3(s) ΔH°f = -1206.9 kJ
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
½H2(g) + C(graphite) + ½N2(g) → HCN(g) ΔH°f = 135 kJ
49
Slide 6-49
Chapter 6: First Law
Slide 6-50
Chapter 6: First Law
6-50
Slide 6-51
Chapter 6: First Law
6-51
Hess Law of Heat Summation
Let's say we want to determine the reaction enthalpy for the following reaction:
H2 (g) + Cl2 (g) ! 2HCl (g)
H°rxn = ??
given the following individual reactions:
NH3 (g) + HCl (g) ! NH4Cl (s)
H1° = -176.0 kJ
N2 (g) + 3H2 (g) ! 2NH3 (g)
H2° = -92.22 kJ
N2 (g) + 4H2 (g) + Cl2 (g) ! 2 NH4Cl (s)
H3° = -628.66 kJ
Start by adding two reactions that give us the reactants we want on the correct side:
Step 1
N2 (g) + 4H2 (g) + Cl2 (g) ! 2 NH4Cl (s)
H3°
2NH3 (g) ! N2 (g) + 3H2 (g)
-( H2°)
2NH3 (g) + H2 (g) + Cl2 (g) ! 2 NH4Cl (s)
H°(step 1) = H3° + -( H2°)
Step 2
2NH3 (g) + H2 (g) + Cl2 (g) ! 2 NH4Cl (s)
2[NH4Cl (s) ! NH3 (g) + HCl (g)]
H2 (g) + Cl2 (g) ! 2HCl (g)
So H° rxn = H°(step 1) + 2(- H1°) =
H° rxn = -184.64 kJ
Slide 6-52
H°(step 1)
2(- H1°)
H3° + -( H2°)+ 2(- H1°)
Chapter 6: First Law
6-52
Figure 6.10
The general process for determining ΔHorxn from ΔHof values.
53
Slide 6-53
Chapter 6: First Law
Sample Problem 6.9
Calculating the Heat of Reaction from Heats of
Formation
PROBLEM: Nitric acid, whose worldwide annual production is about 10 billion
kilograms, is used to make many products, including fertilizer, dyes, and
explosives. The first step in the industrial production process is the
oxidation of ammonia:
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate ΔHorxn from ΔHof values.
PLAN:
Look up the ΔHof values and use Hess s law to find ΔHrxn.
SOLUTION:
ΔHrxn = Σ mΔHof (products) - Σ nΔHof (reactants)
ΔHrxn = {4[ΔHof NO(g)] + 6[ΔHof H2O(g)]} - {4[ΔHof NH3(g)] + 5[ΔHof O2(g)]}
= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
ΔHrxn = -906 kJ
Slide 6-54
Chapter 6: First Law
54
Figure 6.11
The trapping of heat by the atmosphere.
55
Slide 6-55
Chapter 6: First Law
End of Chapter 6
Silberberg HW: 10, 14, 15, 19, 35, 37, 39, 41, 48,
52, 54, 57, 59, 64, 65, 67, 71, 77, 79, 81, 82, 92,
97, 98, 100
56
Slide 6-56
Chapter 6: First Law