Chapter 6 Principles of Reactivity: Energy and Chemical Reactions Thermochemistry

1
Chapter 6Principles of ReactivityEnergy and
Chemical ReactionsThermochemistry
2
Goals of Chapter

• Assess heat transfer associated with changes in
  temperature and changes of state.
• Apply the First Law of Thermodynamics.
• Define and understand the state functions
  enthalpy (H) and internal energy (E).
• Calculate the energy changes in chemical
  reactions and learn how these changes are
  measured.

3
Thermochemistry

• study of the relationships between energy changes
  and chemical processes

4
Energy

• The capacity to do work or to transfer heat
• Kinetic Energy
• energy of motion KE ½ mv2
• Potential Energy
• stored energy fuel of motor-cars, trains, jets.
• It is converted into heat and then to work.
• due to relative position water at the top of a
• water wheel. It is converted to mechanical E
• electrostatic lightning converts it to light and
• heat

5
Joule

• SI unit for energy
• the energy of a 2 kg mass moving at 1 m/s
• KE ½ mv2 ½(2 kg)(1 m/s)2 1 kg?m2/s2 1 J
• 1 cal is the amount of energy required to raise
  the temperature of 1 g water 1C
• 1 cal 4.184 J 1 cal 1 calorie
• 1 Cal 1000 cal 1 kcal
• 1 Cal dietary Calorie (nutritional calorie)

6
System

• the part of the universe under study
• the substances involved in the chemical and
  physical changes under investigation
• in chemistry lab, the system may be the chemicals
  inside a beaker

7
Surroundings

• the rest of the universe
• in chemistry lab, the surroundings are outside
  the beaker where the chemicals are
• The system plus the surroundings is the universe.

8
System and Surroundings

• SYSTEM
• The object under study
• SURROUNDINGS
• Everything outside the system

9
Thermodynamic State

• The set of conditions that specify all of the
  properties of the system is called the
  thermodynamic state of a system.
• For example the thermodynamic state could
  include
• The number of moles and identity of each
  substance.
• The physical states of each substance.
• The temperature of the system.
• The pressure of the system.
• The volume of the system.
• The height of a body relative to the ground.

10
First Law of Thermodynamics

• law of conservation of energy
• during any process, energy is neither created nor
  destroyed, it is merely converted from one form
  to another
• the mass of a substance is a form of energy
• E mc2 (Albert Einstein)
• e.g. in nuclear reactions mass is not conserved,
• it is transformed into heat (E)
• The combined amount of energy in the universe
  is constant.

11
Internal Energy (E)

• the total energy of a system S of kinetic and
• potential E of all atoms, molecules, or ions in
  the
• system
• E cannot be measured exactly
• E is a state function change in E does not
  depend on how change of state happens
• ?E change in E. ?E can be measured
• ?E Efinal Einitial (of final and initial
  states)
• ?E gt 0 () indicates system gains energy during
  process (E increases, ?)
• ?E lt 0 (-) indicates system loses energy during
  process (E decreases, ?)

12
?E q w

• first law of thermodynamics
• q heat
• w work done on the system
• w gt 0 () ? work done on system by surroundings
  (eg. compressing gas) E of system increases
• w lt 0 () ? work done by system on surroundings
  (expanding gas) E decreases
• q gt 0 () ? heat flows into system E ?
• q lt 0 () ? heat flows out of system E ?
• q and w are not state functions

13
Exothermic reactions give off energy in the form
of heat (they give off heat).Endothermic
reactions absorb heat.

• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) 890 kJ
• exothermic

In this case, heat is given off. It is released
by the system. It is a product of the reaction.
14
Directionality of Heat Transfer

• Heat always transfers from hotter object to
  cooler one.
• EXOthermic heat transfers from SYSTEM to
  SURROUNDINGS.

15
Directionality of Heat Transfer

• Heat always transfers from hotter object to
  cooler one.
• ENDOthermic heat transfers from SURROUNDINGS to
  the SYSTEM.

16
Calculate ?E of a system that absorbs 35 J of
heat and does 44 J of work on the surroundings.

• q 35 J (absorbed)
• w 44 J (the system did it)
• ?E q w
• ?E 35 J (44 J)
• ?E 9 J (internal E decreases)
• Note that Efinal and Einitial are not calculate,
  just ?E

17
Work symbol w

• w force x distance
• Expansion/compression work at constant P,
• w P?V ?V Vfinal Vinitial
• ?E q P?V
• under conditions of constant volume, P?V 0,
  w0, because ?V 0 (no work done on or by the
  system)
• ?, ?E q 0
• ?E q
• ?E qV This provides a way of measuring
  ?E in a reactor at constant V.

18
Heat Capacity (C)

• the amount of heat energy required to raise the
  temperature of an object 1 K (1C),
• units J/K, cal/C, ...
• q C ? ?T ?T Tfinal Tinitial
• The amount of heat can be calculated from ?T
• Specific Heat (c)
• the amount of heat energy required to raise the
  temperature of 1 g of something 1 K (1C)
• units J/gK, J/gC, cal/gC, ...
• q c ? m ? ?T m mass (grams)

19
Molar Heat Capacity

• the specific heat of water is
• 1 cal/gC 4.184 J/gC
• KNOW THIS!!!
• The molar heat capacity is the amount of heat
  energy required to raise the temperature of 1 mol
  of a substance 1 K (1C).
• units J/K mol, cal/C mol
• For water, it is
• cal 18.0 g
  cal
• C 1 ? 18.0
• g C 1 mol mol C
  or K

20
Heat/Energy TransferNo Change in State

• q transferred (sp. ht.)?(mass)?(?T)

21
Law Zero of Thermodynamics

• When two bodies, liquids, solutions,
  solid-liquid, etc.,() initially at different
  temperatures, are put in contact or mixed, the
  amount of heat absorbed and given off by the two
  samples have the same absolute value, but one is
  gt0 and the other is lt0
• q1 q2 0 q1 -q2
• If more than two components()
• q1 q2 q3 0

22
Example Calculate the amount of heat energy
given off when 45.3 g water cools from 77.9 C to
14.3 C

• ?T 14.3 C 77.9 C 63.6 C
• q c ? m ? ?T (know this formula)
• 1 cal
• q ? 45.3 g ? (63.6 C) 2.88 ? 103 cal
• g C
• ?T is negative because T lowers,
• hence q is negative (it is given off)

23
After absorbing 1.850 kJ of heat, the temperature
of a 0.500-kg block of copper is 37 C. What was
its initial temperature?

• J
• Cu specific heat c 0.385 ---
• g K
• q m ? c ? ?T m ? c ? (Tfinal - Tinitial)
• 1000 J 0.385 J
• 1.850 kJ?----- 500. g ? -----?(37C - Ti)
• 1 kJ g C
• 1.850 ? 1000
• 37C - Ti --------- 9.6 C Ti 27.4C
• 500. ? 0.385

24
A 182-g sample of Au at some temperature is added
to 22.1 g of water. The initial water T is 25.0
C, and the final T of the whole is 27.5 C. If
the specific heat of gold is 0.128 J/g.K, what is
the initial T of the gold?

• T of H2O increased, hence it absorbed heat. Then,
  
• Au gave off heat, i.e., its temperature
  decreased.
• qwater(absorbed) qAu(given off) 0 q m
  ? c ? ?T
• gt0 lt0
  ?T Tf Ti
• 22.1 g ? 4.184 J/g.C ? (27.5 25.0)C
• 182 g ? 0.128 J/g.C
  ? (27.5C Ti) 0
• 231 C 23.3 ?(27.5 C Ti) 231 C 641C
  23.3Ti 0
• 231 641
• Ti(Au) 37.4 C
• 23.3

25
One beaker contains 156 g of water at 22 C and a
second contains 85.2 g of water at 95 C. If the
water in the two beakers is mixed, what is the
final temperature?

• Water in beaker 1 (w1) will absorb heat, its T
  will ?.
• Water in beaker 2 will give off heat, its T will
  ?.
• q1(absorbed) q2(given off) 0 q m ? c
  ? ?T
• gt0 lt0
  ?T Tf Ti
• 156 g ? 4.184 J/g.C ? (Tf 22C)
• 85.2 g ?4.184 J/g.C
  ?(Tf 95 C) 0
• 156 Tf 3432 85.2 Tf 8094 0
• 8094 3432
• Tf 47.8 48 C
• 156 85.2

26
Bomb Calorimeter constant V
Thermometer
Ignition Filament
Stirrer
H2O
Insulated Box
Bomb
27
Example A 1.50g sample of methane was burned in
excess oxygen in a bomb calorimeter with a heat
capacity of 11.3kJ/C. The temperature of the
calorimeter increased from 25.0 to 32.3C.
Calculate the ?E in kJ per gram of methane for
this reaction.

• ?T (32.3 25.0) C 7.3 C
• CH4(g) 2O2(g) ? CO2(g) 2 H2O(l)
• In a bomb calorimeter V is constant ? ?E q
• qcalorim C ? ?T (11.3 kJ/C) ? 7.3C 83
  kJ
• q qcalorim 0 Then, ?E q -qcalorim, ?E
  83 kJ
• ?E is negative because the rxn gives off heat
  that the
• calorimeter absorbs
• 83 kJ kJ
  kJ
• ?E 55.3 55 (two SF)
• 1.50 g g
  g CH4

28
Example A bomb calorimeter was heated with a
heater that supplied a total of 8520 J of heat.
The temperature of the calorimeter increased
2.00C. A 0.455g sample of sucrose, C12H22O11,
was then burned in excess oxygen in that
calorimeter causing the temperature to increase
from 24.49C to 26.25C. Calculate the ?E for
this reaction in kJ/mol sucrose and the dietary
calories per gram of sucrose.

• 2C12H22O11 35 O2 ???24CO2 22H2O
• We will need MW of sucrose 342.3 g/mol

29
Calorimeter heat capacity (C)

• heat supplied q 8520 J
  J
• C 4260
• ?T ?T 2.00
  C C
• For the reaction, ?T (26.25 24.49)C
• ?V 0, then, ?E q qcalor (reaction gives
  off heat)
• qcalorim C ? ?T (4260 J/C) ? 1.76C 7.50
  x103 J
• 7.50 x103 J 342.3 g sucrose
  1 kJ
• ?E ? ?
• 0.455 g sucrose 1 mol sucrose 103
  J
• kJ
• ?E 5.64x103
• mol sucrose

30
Nutritional (dietary) Calories (Cal)

• Strategy divide J by grams of sucrose.
• Convert J to cal and cal to Cal.
• 1 Cal 1000 cal 1 kcal (see slide 5)
• 7.50 x103 J 1 cal 1 Cal
  3.94 Cal
• ? ?
• 0.455 g sucro 4.184 J 1000 cal g
  sucrose

31
Heat Transfer with Change of State

• Changes of state involve energy (at const T)
• Ice 333 J/g (heat of fusion) ? Liquid water
• q (heat of fusion)?(mass)

32
Heat Transfer and Changes of State
Liquid ? Vapor

• Requires energy (heat).
• This is the reason
• a) you cool down after swimming
• b) you use water to put out a fire

energy
33
Heating/Cooling Curve for Water
Note that T is constant as ice melts
34
Heat Changes of State

• What quantity of heat is required to melt
• 500. g of ice and heat the water to steam
• at 100 oC?

Heat of fusion of ice 333 J/g Specific heat of
water 4.2 J/gK Heat of vaporization 2260 J/g
35
Heat Changes of State

• What quantity of heat is required to melt 500.
• g of ice and heat the water to steam at 100 oC?
• 1. To melt ice
• q1 (500. g)(333 J/g) 1.67 x 105 J
• 2. To raise water from 0 oC to 100 oC
• q2 (500. g)(4.2 J/gK)(100 - 0)K 2.1 x 105
  J
• 3. To evaporate water at 100 oC
• q3 (500. g)(2260 J/g) 1.13 x 106 J
• 4. Total heat energy q1 q2 q3
• 1.51 x 106 J 1510 kJ

36
23. The freezing point of Hg is -38.8C. What
quantity of heat (J) is released to the
surroundings if 1.00 mL of Hg is cooled from 23.0
C to -38.8C and then frozen to a solid? d
13.6 g/mL c 0.140 J/g K qfus 11.4 J/g
AW 200.6 g/mol

• 
  13.6 g
• q q1 q2 1mL ?
  ---- 13.6 g
• 
  1 mL
• q m ? c ? ?T m ?(-qfus) negative for
  freezing
• 0.140 J
  (-11.4 J)
• q 13.6 g ?-----?(-38.8 -23)C 13.6 g ?-----
• g C
  g
• q -117 J -155 -272 J (given off)
• Exothermic process

37
Enthalpy, H

• Chemistry is commonly done in open beakers or
  flasks on a desk top at atmospheric pressure.
• Because atmospheric pressure only changes by
  small amounts, this is almost at constant
  pressure.
• Because heat at constant pressure is so frequent
  in chemistry and biology, it is helpful to have a
  measure of heat transfer under these conditions.
  That is the enthalpy change.

38
Enthalpy, H

• heat content
• state function change in H does not depend on
  how change of state happens
• H E PV (We do not measure
• ?H Hfinal Hinitial or calculate Hf,i
  but ?H)
• at constant pressure, ?H qP
• ?H gt 0 () ? heat flows into system H ?
• endothermic
• ?H lt 0 () ? heat flows out of system H ?
• exothermic

39
Calorimetry

• A coffee-cup
• calorimeter is used to
• measure the amount of
• heat produced (or
• absorbed) in a reaction
• at constant P. That is qp or
• ?H. The cup is under
• constant atmospheric
• pressure (P) because it is
• not completely sealed. It is
• isolated no heat transfer
• between system and
• surroundings.

40
Product- or Reactant-Favored Reactions

• nature favors processes that decrease energy of
  the system
• ?, nature favors exothermic processes
• 4 Fe(s) 3 O2(g)????Fe2O3(s) ?H 1651 kJ
• Exothermic. The formation of product is favored.
• CaCO3(s)?????CaO(s) CO2(g) ?H 179.0 kJ
• Heat is required for the reaction to occur. The
  
• reaction is endothermic. Reagent is favored.
• Is it always like that? Is ?H the only factor
  that
• matters? No, entropy change counts too

41
Enthalpy of Reaction, Heat of Reaction

• enthalpy of reaction, heat of reaction
• ?Hreaction when a reaction takes place
• ?Hfusion when a solid is melted
• ?Hcrystallization when a compound is
  crystallized
• from a solution

42
Hess Law

• if a reaction is the sum of two or more other
  reactions, the ?H for the overall reaction is
  equal to the sum of the ?H values of those
  reactions.
• also applies to ?E. E and H are state functions
• The ?H of some reactions can not be measured in a
  
• calorimeter, because some other reactions take
• place at the same time in the reactor.
• C(s) 2H2(g) ? CH4(g) C2H2, C2H4, C2H6, C3H6,
  etc.,
• are also produced. In a case like this, the ?H of
  other
• related reactions can be employed to calculate ?H

43
C(s) 2H2(g) ? CH4(g) ?Hrxn ?

• C(s) O2(g) ? CO2(g) ?H 393.5
  kJ
• 2H2(g) O2(g) ? 2H2O(l) ?H
  571.6 kJ
• CO2(g) 2H2O(l) ? CH4(g) 2O2(g) ?H 890.4 kJ
• _______________________________________
• C(s) 2H2(g) ? CH4(g) ?Hrxn 74.7 kJ
• ?Hr 393.5 kJ (571.6 kJ) 890.4 kJ
  74.7 kJ

44
Calculate ?H of the fourth equation out of ?Hs of
the first three C(s) O2(g) ? CO2(g)
?H 393.5 kJ Eq. (1) H2(g)
1/2O2(g) ? H2O(l) ?H 285.8
kJ Eq. (2) 2C2H2(g) 5O2(g) ? 4CO2(g)
2H2O ?H 2598.8 kJ Eq. (3) 2C(s)
H2(g) ? C2H2(g) ?Hrxn ? Eq. (4)

• We need to multiply Eq. (1) by 2 leave Eq.
  (2) as it is multiply Eq. (3) by -½, that is
  to reverse it and times ½.
• 2 ? C(s) O2(g) ? CO2(g) ?H
  393.5 kJ H2(g) 1/2O2(g) ? H2O(l)
  ?H 285.8 kJ -½ ?
  2C2H2(g) 5O2(g) ? 4CO2(g) 2H2O ?H 2598.8
  kJ
• 2C(s) 2O2(g) ? 2CO2(g) ?H
  2(393.5 kJ)
• H2(g) 1/2O2(g) ? H2O(l)
  ?H 285.8 kJ
• 2CO2(g) H2O???C2H2(g) 5/2O2(g) ?H
  1/2(2598.8 kJ)
• 2C(s) H2(g) ? C2H2(g) ?Hrxn 226.6 kJ
• ?Hrxn 2(393.5 kJ) 285.8 kJ 1/2(2598.8 kJ)

45
2C(s) H2(g) ? C2H2(g) ?Hrxn ? C(s) O2(g) ?
CO2(g) ?H 393.5 kJ H2(g) 1/2O2(g) ?
H2O(l) ?H 285.8 kJ 2C2H2(g) 5O2(g) ?
4CO2(g) 2H2O
?H 2598.8 kJ

• 2C(s) 2O2(g) ? 2CO2(g) ?H 2(393.5 kJ)
• H2(g) 1/2O2(g) ? H2O(l) ?H 285.8 kJ
• 2CO2(g) H2O???C2H2(g) 5/2O2(g)
• ?H
  1/2(2598.8 kJ)
• 2C(s) H2(g) ? C2H2(g) ?Hrxn 226.6 kJ
• ?Hrxn 2(393.5 kJ) 285.8 kJ 1/2(2598.8 kJ)

46
P4(s) 6 Cl2(g) ? 4 PCl3(l) ?Hrxn
?P4(s) 10 Cl2(g) ? 4 PCl5(s) ?H
1774.0kJPCl3(l) Cl2(g) ? PCl5(s)
?H 123.8 kJ

• P4(s) 10 Cl2(g) ? 4 PCl5(s) ?H
  1774.0kJ
• 4x(PCl3(l) ? PCl5(s) Cl2(g) ?H 123.8
  kJ)
• P4(s) 10 Cl2(g) ? 4 PCl5(s) ?H
  1774.0kJ
• 4PCl5(s) ? 4PCl3(l) 4Cl2(g) ?H 4(123.8
  kJ)
• P4(s) 6 Cl2(g) ? 4 PCl3(l) ?Hrxn
  1278.8 kJ
• ?Hrxn 1774.0 kJ 495.2 kJ 1278.8 kJ

47
Formation Reaction

• Reaction in which 1 mol of a substance is
• formed from its elements in their most stable,
• natural states.
• eg. formation reaction for C2H6SO(l)
• C(s) H2(g) S8(s) O2(g)???C2H6SO(l)
• 2C(s) 3H2(g) 1/8S8(s) 1/2O2(g)???C2H6SO(l)

48
Heat (Enthalpy) of Formation

• ?Hf
• enthalpy change associated with a formation
  reaction

49
Standard Conditions

• T 25C 298 K
• P 1 atm

50
Standard Heat of Formation orStandard Molar
Enthalpy of Formation

• ?Hf is the enthalpy change for the formation of
  1 mol of a compound directly from its elements in
  their standard states
• ?Hf may be gt0 or lt0 for a compound (tables)
• ?Hf 0 for any element in its most stable form,
• e.g. Na(s), Hg(l), Cl2(g), Br2(l), H2(g), P4(s),
  C(s)

51
Enthalpy Change for a Reaction

• a A b B ? c C d D
• ?Hrxn S(?Hf products) S(?Hf reagents)
• means take the sum
• ?Hrxn c??Hf(C) d??Hf(D)
• 
  a??Hf(A) b??Hf(B)
• ?Hf(element) 0

52
Example Calculate the ?H in kJ/mol B5H9 for
the following reaction.2B5H9 12O2 ? 5B2O3
9H2O

• Compound ?Hf (kJ/mol)
• B5H9 73.2
• B2O3 1272.8
• H2O 241.83
• ?HRxn (5(?Hf B2O3) 9(?Hf H2O)
• (2 (?Hf B5H9)
  12(?Hf O2)
• ?HRxn 5mol(1272.8 kJ/mol) 9mol(241.83
  kJ/mol)
• 2 mol(73.2 kJ/mol) 12 mol(0 kJ/mol) 8686.9
  kJ
• 8686.9 kJ/2 mol B5H9 4343.5 kJ/mol B5H9

53
Calculate the ?H in kJ/mol Mg for the following
reaction.3Mg SO2 ? MgS 2MgO

• Compound ?Hf (kJ/mol)
• MgO 601.7
• MgS 598.0
• SO2 296.8
• ?HRxn 1 mol(?Hf MgS) 2(?Hf MgO)
• 3(?Hf Mg) 1
  mol(?Hf SO2)
• ?HRxn 1mol(598.0 kJ/mol) 2 mol(601.7
  kJ/mol)
• 3 mol(0 kJ/mol) 1 mol(296.8 kJ/mol)
  1504.6 kJ
• 1504.6 kJ/3 mol Mg 501.5 kJ/mol Mg

54
The enthalpy change for the combustion of
styrene, C8H8, is measured by calorimetry
C8H8(l) 10 O2(g) ? 8 CO2(g) 4 H2O(l)
?Hrxn
4395.0 kJ

• Use this, along with the ?Hf of CO2 and H2O, to
• calculate the ?Hf C8H8, in kJ/mol
• ?Hf CO2(g) 393.51 kJ/mol
• H2O(l) 285.83 kJ/mol
• ?HRxn 8(?Hf CO2) 4(?Hf H2O(l) (?Hf
  C8H8)
• ?Hf C8H8 8(?Hf CO2) 4(?Hf H2O(l)
  ?Hrxn
• ?Hf C8H8 8(393.51) 4(285.83) (4395.0)
• ?Hf C8H8 103.6 kJ/mol

55
Nitroglycerin is a powerful explosive that forms
four different gases when detonated2C3H5(NO3)3(l
) ? 3N2(g) ½O2(g) 6CO2(g) 5H2O(g)

• Calculate the enthalpy change when 10.0 g of
• nitroglycerin is detonated. ?Hf(kJ/mol) are
• CO2(g) 393.5 H2O(g) 241.8
• NTG, C3H5(NO3)3(l) 364
• ?Hrxn 6(?Hf CO2) 5(?HfH2O(g)
  2(?HfNTG)
• ?Hrxn 6(393.5) 5(241.8) 2(364)
• 2842 kJ for 2 mol of NTG
• For 10.0 g of NTG we need its MW 227.1 g/mol

56
2842 kJ for 2 mol of NTGfor 10.0 g of NTG we
need its MW 227.1 g/mol

• 2842 kJ 1 mol NTG
• ? ?10.0 g NTG 62.6 kJ
• 2 mol NTG 227.1 g NTGJ
• 1421 kJ 1 mol NTG
• ? ?10.0 g NTG 62.6 kJ
• 1 mol NTG 227.1 g NTGJ

57
Calculate the amount of heat released in the
complete combustion of 8.17 grams of Al to form
Al2O3(s) at 25C and 1 atm. ?Hf for Al2O3(s)
1676 kJ/mol

• 4Al(s) 3O2(g) ? 2Al2O3(s)
• 1 mol Al
• 8.17 g Al? 0.303 mol Al
• 26.98 g Al
• 2 moles mol Al2O3 are produced out of 4 moles of
  Al
• -1676 kJ 2 mol Al2O3
• ? ? 0.303 mol Al -254 kJ
• 1 mol Al2O3 4 mol Al

58
How much heat energy is liberated when 11.0
grams of manganese is converted to Mn2O3 at
standard state conditions?

• 4Mn(s) 3O2(g) ? 2Mn2O3(s) ?H
  -1924.6 kJ
• 1 mol Mn
• 11.0 g Mn? 0.200 mol Mn
• 54.94 g Mn
• The given ?H corresponds to the reaction of 4
  moles Mn
• - 1924.6 kJ
• ? 0.200 mol Mn -96.2 kJ
• 4 mol Mn