Unit 13 Thermochemistry

1
Unit 13Thermochemistry
CHM 1046 General Chemistry and Qualitative
Analysis

• Dr. Jorge L. Alonso
• Miami-Dade College Kendall Campus
• Miami, FL

• Textbook Reference
• Chapter 15 (sec. 1-11)
• Module 3 (sec. I-VIII)

2
Energy

• Potential Energy an object possesses by virtue
  of its position or chemical composition (bonds).

2 C8H18 (l) 25 O2 (g)
16 CO2(g) 18 H2O(g)
Chemical bond energy
Energy (-?E) work heat
Kinetic Energy an object possesses by virtue of
its motion.
3
Energy
Chemical bond energy
?? gas25?34 9?
9? x 22.4L/? 202 L
2 C8H18 (l) 25 O2 (g)
16 CO2(g) 18 H2O(g)

?H - 5.5 x 106 kJ/?
Energy produced by chemical reactions heat or
work.

• Heat (q) spontaneous transfer (flow) of energy
  (energy in transit) from one object to another
  causes molecular motion vibrations (kinetic
  energy) measured as temperature.

• Work (w) Energy (force) used to cause an object
  that has mass to move (a distance) kinetic
  energy.

W F x d
q w
?E
Law of Conservation of Energy
4
Units of Energy Joule calorie

• The joule (J) SI unit of energy (work)
• The calorie (cal)
• heat required to raise 1 g of H2O 10C
• 1 cal 4.184 J
• Nutritional Calorie 1000 calories
  (1kcal)4,184 J.

Energy (Work) F x d
Joule (J) Nm
m
The Newton (N) is the amount of force that is
required to accelerate a kilogram of mass at a
rate of one meter per second squared.
5
System and Surroundings

• The system includes the molecules we want to
  study.
• The surroundings are everything else (here, the
  cylinder and piston).

Work
-?Ework
?Ew
Heat
System
-?Eq(heat)
?Eq
?E q w
Surroundings
6
First Law of Thermodynamics

• Energy is neither created nor destroyed.

• Also known as the Law of Conservation of Energy

Work

• In other words, the total energy of the universe
  is a constant if the system loses energy, it
  must be gained by the surroundings, and vice
  versa.

-?Ew
?Ew
UNIVERSE
Heat
System
-?Eq
?Esys ?Esurr 0
?Eq
?Esys -?Esurr
or ?E sys (q w)surr
Surroundings
7
Changes in Internal Energy (?E)
energy released/absorbed as either work or
heat, is looked at from the perspective of the
system (the chemicals)
?E, q, w, and Their Signs
q
- q
gt q
Syst looses heat
Syst gains heat
gt w
w
- w
Work done by Syst

• ?E q w

Work done on Syst
q
w
?E
- q
- w
- ?E
q
w
?E
8
Why consider Energy Change (?E) and not the total
Internal Energy (E)?
?Esys ?Esurr 0
Esys Esurr ETotal (constant)

• Total Internal Energy (E) the sum of all
  kinetic and potential energies of all components
  of the system. How can it be determined?

• Usually we have no way of knowing the total
  internal energy of a system finding that value
  is simply too complex a problem.

• But when a change occurs we can measure the
  change in internal energy
• ?E Efinal - Einitial

9
Energy as Work (w) of Gas Expansion
F P A
Work - (Force x distance)
w - (F x ?h)
w - (P x A x ?h)
w - P ?V (_at_ constant P)
w - ?nRT (_at_ constant V)
WorkGasExpansion
10
Exchange of Heat (?H) by chemical systems

• When heat is released by the system to the
  surroundings, the process is exothermic.

- ?H
q

• When heat is absorbed by the system from the
  surroundings, the process is endothermic.

?H
q
11
Enthalpies of Reaction (?H)
Reactants

• The change in enthalpy, ?H, is the enthalpy of
  the products minus the enthalpy of the reactants
  
• ?H Hproducts - Hreactants

This quantity, ?H, is called the enthalpy of
reaction, or the heat of reaction.
Products
Notice it is the same value, but of opposite sign
for the reverse reaction
12
State Functions

• Are ?E, q and w all state functions?

• A state function depends only on the present
  state of the system, not on the path by which the
  system arrived at that state.

?E qw

• However, q and w are not state functions.

• Other state functions are P, V and T.

13
Calorimetry

• The experimental methods of measuring of heats
  (?H) involved in chemical reactions

• Methods utilized
• Constant Pressure Calorimetry (open container in
  contact with atmosphere)
• Constant Volume Calorimetry closed container
  (bomb)

1 atm
14
Thermochemical Equations
2 C8H18 (l) 25 O2 (g)
16 CO2(g) 18 H2O(g)
?H - 5.5 x 106 kJ/? C8H18

1. ?H is per mole of reactant, at the indicated
   states (s, l, g).
2. Direct quantitative relationship between
   coefficients of balanced equation and ?H.
3. Reverse equation has ?H of opposite sign.

Problem Calculate ?H when 25g of C8H18 (l) (MM
114) are burned in excess oxygen?
15
(1) Calorimetry at Constant Pressure

• Most chemical reactions are carried out in
  beakers or flasks that are open to the
  atmosphere (i.e., at constant pressure,
  isobaric).

?E q w
1 atm
?E qp - P?V
SOLVE FOR
?E P?V qp
q m ? c ? ?T
qp Heat of Reaction DH or
Enthalpy of Reaction
?E P?V ?H
Big Problem calorimeter also absorbs heat
?H ?E P?V
?H ?Epv
? and when the volume is constant
16
(a) Determining the Heat Capacity of a Calorimeter
50 mL of H2O _at_ 650C are added.
50 mL H2O
50 mL H2O
..to 50 mL of H2O _at_ 250C inside the calorimeter
The final temperature of the 100 mL of H2O inside
the calorimeter is 420C
Heat lost by heat gained by heat gained
by hot water cold water
calorimeter
qhot qcold qcalorimeter
17
Determining the Heat Capacity of a Calorimeter
(Ccal)
qhot qcold qcalorimeter
(mc?T)H2O (mc?T)H2O (Ccal ?T)
Includes both the mass and heat capacity
(mc?T)H2O - (mc?T)H2O
(?T)cal
Ccal
18
(b) Calculating the Heats of Reactions
(?H)
HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
50 mL_at_25C 50 mL_at_25C ? 100
mL of NaCl(aq) _at_ 30C
Heat of heat gained by
heat gained by reaction NaCl solution
calorimeter
?H(qrxn) qNaCl soln qcal
qrxn (mc?T)NaClsoln (Ccal?T)
qrxn - ?Hrxn heat of
neutralization rxn
How do you determine the cNaCl soln?
19
How do you determine the cNaCl soln?
50 mL of H2O _at_ 650C are added.
50 mL H2O
50 mL NaCl soln
..to 50 mL of NaCl solution _at_ 250C inside the
calorimeter
The final temperature of the 100 mL of solution
inside the calorimeter is 410C
Heat lost by heat gained by heat gained
by hot water salt water soln
calorimeter
qh qNaClsoln qcal
(mc?T)H2O (mc?T)NaClsoln (Ccal?T)
20
(2) Calorimetry at Constant Volume (Bomb Cal.)
?E q w

• w P ?V
• ?V 0
• w 0

?E qv
q m ? c ? ?T
Bomb calorimeter
21
Bomb Calorimetry

• Because the volume in the bomb calorimeter is
  constant, what is measured is really the change
  in internal energy, ?E, not ?H.
• For most reactions, the difference is very small.

?E qv
?H ?E ?nRT
22
Review of Thermochemical Equations
?E q w
?E qp - P?V
?E qp - ?nRT
since ?H qp
?E ?H - P?V
1 atm
?H ?E P?V
?E q w
?E qv
23
Comparing ?H and ?E
?H ?E P?V
2 C8H18 (l) 25 O2 (g)
16 CO2(g) 18 H2O(g)
?H - 5.5 x 106 kJ/?
?E ?H - P?V (useful at const P)
?E ?H - ?nRT (useful at const V)
Problem calculate ?E _at_ 25C for above equation _at_
const. V R 8.314J/K?.
?? gas34-25 9? 4.5? C8H18
24
Constant-Volume Calorimetry

• Problem When one mole of CH4 (g) was combusted
  at 250C in a bomb calorimeter, 886 kJ of energy
  were released. What is the enthalpy change for
  this reaction?

CH4 (g) 2 O2 (g)
CO2(g) 2 H2O(l)
?E qv - 886 kJ R 8.31 J/ ?.K ?n 1 (12)
-2 ?
?H ?E P?V ?E ?nRT
25
Energy in Foods

• Most of the fuel in the food we eat comes from
  carbohydrates fats.

26
Methods of determining ?H

1. Calorimetry (experimental)
2. Hesss Law using Standard Enthalpy of Reaction
   (?Hrxn) of a series of reaction steps (indirect
   method).
3. Standard Enthalpy of Formation (?Hf ) used with
   Hesss Law (direct method)
4. Bond Energies used with Hesss Law

?
Experimental data combined with theoretical
concepts
?
27
(2) Determination of ?H using Hesss Law

• ?Hrxn is well known for many reactions, but it is
  inconvenient to measure ?Hrxn for every reaction.

• However, we can estimate ?Hrxn for a reaction of
  interest by using ?Hrxn values that are published
  for other more common reactions.

?

• The Standard Enthalpy of Reaction (?Hrxn) of a
  series of reaction steps are added to lead to
  reaction of interest (indirect method).

Standard conditions (25C and 1.00 atm
pressure). (STP for gases T 0C)
?
28
Hesss Law

• If a reaction is carried out in a series of
  steps, ?H for the overall reaction will be equal
  to the sum of the enthalpy changes for the
  individual steps.

- 1840, Germain Henri Hess (180250), Swiss
29
Calculation of ?H by Hesss Law
3 C(graphite) 4 H2 (g) ?? C3H8 (g) ?H
-104 3 C(graphite) 3 O2 (g) ?? 3 CO2 (g)
?H-1181 4 H2 (g) 2 O2 (g) ?? 4 H2O (l)
?H-1143
C3H8 (g) ?? 3 C(graphite) 4 H2 (g) ?H 104
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Appropriate set of Equations with their ?H values
  are obtained (or given), which containing
  chemicals in common with equation whose ?H is
  desired.
• These Equations are all added to give you the
  desired equation.
• These Equations may be reversed to give you the
  desired results (changing the sign of ?H).
• You may have to multiply the equations by a
  factor that makes them balanced in relation to
  each other.
• Elimination of common terms that appear on both
  sides of the equation .

30
Calculation of ?H by Hesss Law
3 C(graphite) 3 O2 (g) ?? 3 CO2 (g)
?H-1181 4 H2 (g) 2 O2 (g) ?? 4 H2O (l)
?H-1143
C3H8 (g) ?? 3 C(graphite) 4 H2 (g) ?H 104

• ?Hrxn
• 104 kJ
• 1181 kJ
• 1143 kJ
• 2220 kJ

C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)
31
Calculation of ?H by Hesss Law

• Calculate heat of reaction
• W C (graphite) ? WC (s) ?H ?
• Given data
• 2 W(s) 3 O2 (g) ? 2 WO3 (s) ?H -1680.6 kJ
• C (graphite) O2 (g) ? CO2 (g) ?H -393.5
  kJ
• 2 WC (s) 5 O2 (g) ? 2 WO3 (s) CO2 (g) ?H
  -2391.6 kJ

½(2 W(s) 3 O2 (g) ? 2 WO3 (s) ) ½(?H
-1680.6 kJ)
W(s) 3/2 O2 (g) ? WO3 (s) ) ?H -840.3 kJ
C (graphite) O2 (g) ? CO2 (g) ?H -393.5
kJ
½(2 WO3 (s) CO2 (g) ? 2 WC (s) 5 O2 (g))
½ (?H 2391.6 kJ)
WO3 (s) CO2 (g) ? WC (s) 5/2 O2 (g) ?H
1195.8 kJ)
W C (graphite) ? WC (s) ?H - 38.0
32
Hesss Law
Problem Chloroform, CHCl3, is formed by the
following reaction Desired ?Hrxn equation
CH4 (g) 3 Cl2 (g) ? 3 HCl (g) CHCl3 (g)
Determine the enthalpy change for this reaction
(?Hrxn), using the following 2 C (graphite)
H2 (g) 3Cl2 (g) ? 2CHCl3 (g)?Hf 103.1
kJ/mol CH4 (g) 2 O2 (g) ? 2 H2O (l) CO2
(g) ?Hrxn 890.4 kJ/mol 2 HCl (g) ? H2 (g)
Cl2 (g) ?Hrxn 184.6 kJ/mol C (graphite)
O2 (g) ? CO2(g) ?Hrxn 393.5 kJ/mol H2 (g)
½ O2 (g) ? H2O (l) ?Hrxn 285.8
kJ/mol answers a) 103.1 kJ b) 145.4
kJ c) 145.4 kJ d) 305.2 kJ
e) 305.2 kJ f) 103.1 kJ This is a
hard question. To make is easer give C
(graphite) ½ H2(g) 3/2 Cl2(g) ? CHCl3(g) ?Hf
103.1 kJ/mol
33
Methods of determining ?H

1. Calorimetry (experimental)
2. Hesss Law using Standard Enthalpy of Reaction
   (?Hrxn) of a series of reaction steps (indirect
   method).
3. Standard Enthalpy of Formation (?Hf ) used with
   Hesss Law (direct method)
4. Bond Energies used with Hesss Law

?
Experimental data combined with theoretical
concepts
?
34
(3) Determination of ?H using Standard
Enthalpies of Formation (?Hf )
?
Enthalpy of formation, ?Hf, is defined as the
enthalpy change for the reaction in which a
compound is made from its constituent elements in
their elemental forms.
?
C O2 ? CO2 ?Hf -393.5 kJ/ ?
?

• Standard Enthalpy of formation ?Hf are measured
  under standard conditions (25C and 1.00 atm
  pressure).

35
Calculation of ?H
CH4(g) O2(g) ? CO2(g) H2O(g)
C 2H2(g) ? CH4(g) ?Hf -74.8 kJ/?
C(g) O2(g) ? CO2(g) ?Hf -393.5
kJ/?
2H2(g) O2(g) ? 2H2O(g) ?Hf -241.8
kJ/?

• We can use Hesss law in this way
• ?H ??n??Hf(products) - ??m??Hf(reactants)
• where n and m are the stoichiometric
  coefficients.

?
?
-
n CH4(g) n O2(g)
n CO2(g) n H2O(g)
??????H 1(-393.5 kJ) 1(-241.8 kJ) -
1(-74.8 kJ) 1(-0 kJ) - 560.5 kJ
36
Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)
?H ??n??Hf(products) - ??m??Hf(reactants)

• ??????H 3(-393.5 kJ) 4(-285.8 kJ) -
  1(-103.85 kJ) 5(0 kJ)
• (-1180.5 kJ) (-1143.2 kJ) - (-103.85
  kJ) (0 kJ)
• (-2323.7 kJ) - (-103.85 kJ)
• -2219.9 kJ

Table of Standard Enthalpy of formation, ?Hf
37
(4) Determination of ?H using Bond Energies

• Most simply, the strength of a bond is measured
  by determining how much energy is required to
  break the bond.
• This is the bond enthalpy.
• The bond enthalpy for a ClCl bond,
• D(ClCl), is measured to be 242 kJ/mol.

38
Average Bond Enthalpies (H)

• Average bond enthalpies are positive, because
  bond breaking is an endothermic process.

• NOTE These are average bond enthalpies, not
  absolute bond enthalpies the CH bonds in
  methane, CH4, will be a bit different than the
• CH bond in chloroform, CHCl3.

39
Enthalpies of Reaction (?H )

• Yet another way to estimate ?H for a reaction is
  to compare the bond enthalpies of bonds broken to
  the bond enthalpies of the new bonds formed.

• In other words,
• ?Hrxn ?(bond enthalpies of bonds broken) ?
• ?(bond enthalpies of bonds formed)

40
Hesss Law ?Hrxn ?(bonds broken) ? ?(bonds
formed)
CH4(g) Cl2(g) ??? CH3Cl(g) HCl(g)

• ?Hrxn D(CH) D(ClCl) ? D(CCl) D(HCl)
• (413 kJ) (242 kJ) ? (328 kJ) (431 kJ)
• (655 kJ) ? (759 kJ)
• ?104 kJ

41
Bond Enthalpy and Bond Length

• We can also measure an average bond length for
  different bond types.
• As the number of bonds between two atoms
  increases, the bond length decreases.

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2003 B Q3
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2005 B
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2002
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2002 8
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2003 A