May 11
Unit 3:
Thermochemistry
Chemistry 3202
1
May 11
Unit Outline
 Temperature
and Kinetic Energy
 Heat/Enthalpy Calculation
Temperature changes (q = mc∆T)
Phase changes (q = n∆H)
Heating and Cooling Curves
Calorimetry (q = C∆T & above
formulas)
2
May 11
Unit Outline
 Chemical
Reactions
PE
Diagrams
Thermochemical Equations
Hess’s Law
Bond Energy
 STSE:
What Fuels You?
3
May 11
Temperature and Kinetic Energy
Thermochemistry is the study of energy
changes in chemical and physical changes
eg. dissolving
burning
phase changes
4
May 11
Temperature , T, measures the average
kinetic energy of particles in a substance
- a change in temperature means
particles are moving at different speeds
- measured in either Celsius degrees or
degrees Kelvin
Kelvin = Celsius + 273.15
5
May 11
The Celsius scale is based on the
freezing and boiling point of water
The Kelvin scale is based on absolute
zero - the temperature at which
particles in a substance have zero
kinetic energy.
6
May 11
p. 628
7
May 11
K
°C
50.15
48
450.15
-200
8
May 11
300 K
# of particles
500 K
Kinetic Energy
9
May 11
Heat/Enthalpy Calculations
system - the part of the universe being studied
and observed
surroundings - everything else in the universe
open system - a system that can exchange
matter and energy with the surroundings
eg. an open beaker of water
a candle burning
closed system - allows energy transfer but is
closed to the flow of matter.
10
May 11
isolated system – a system completely closed
to the flow of matter and energy
heat - refers to the transfer of kinetic energy
from a system of higher temperature to a
system of lower temperature.
- the symbol for heat is q
WorkSheet: Thermochemistry #1
11
May 12
Part A: Thought Lab (p. 631)
12
May 12
Part B: Thought Lab (p. 631)
13
May 12
Heat/Enthalpy Calculations
specific heat capacity – the energy , in
Joules (J), needed to change the
temperature of one gram (g) of a
substance by one degree Celsius (°C).

The symbol for specific heat capacity is
a lowercase c
14
May 12

A substance with a large value of c can
absorb or release more energy than a
substance with a small value of c.
ie. For two substances, the substance
with the larger c will undergo a smaller
temperature change with the same loss
or gain of heat.
15
May 12
FORMULA
q = mc∆T
q = heat (J)
m = mass (g)
c = specific heat capacity
∆T = temperature change
= T2 – T1
= Tf – Ti
16
May 12
eg. How much heat is needed to raise the
temperature of 500.0 g of water from
20.0 °C to 45.0 °C?
q = m c ∆T
for c, m, ∆T, T2 & T1
 p. 634 #’s 1 – 4
 p. 636 #’s 5 – 8
WorkSheet: Thermochemistry #2
Solve
17
May 13
heat capacity - the quantity of energy , in
Joules (J), needed to change the
temperature of a substance by one
degree Celsius (°C)
The symbol for heat capacity is
uppercase C
 The unit is J/ °C or kJ/ °C

18
May 13
FORMULA
C = mc
q = C ∆T
Your Turn
C = heat capacity
c = specific heat
capacity
m = mass
∆T = T2 – T1
p.637 #’s 11-14
WorkSheet: Thermochemistry #3
19
May 18
Enthalpy Changes
enthalpy change - the difference between the
potential energy of the reactants and the
products during a physical or chemical
change
AKA: Heat of Reaction or ∆H
20
May 18
Endothermic Reaction
Products
PE
∆H
Reactants
Reaction Progress
21
May 18
Endothermic Reaction
Products
PE
Enthalpy
∆H
∆H
Reactants
Reaction Progress
22
May 18
Products
Enthalpy
∆H is +
Reactants
Endothermic
23
May 18
reactants
Enthalpy
∆H is products
Exothermic
24
May 18
Enthalpy Changes in Reactions
 All
chemical reactions require bond
breaking in reactants followed by
bond making to form products
 Bond breaking requires energy
(endothermic) while bond formation
releases energy (exothermic)
see p. 639
25
May 18
26
May 18
Enthalpy Changes in Reactions
endothermic reaction - the energy
required to break bonds is greater than
the energy released when bonds form.
ie. energy is absorbed
exothermic reaction - the energy
required to break bonds is less than the
energy released when bonds form.
ie. energy is produced
27
May 18
Enthalpy Changes in Reactions
1.
∆H can represent the enthalpy change for
a number of processes
Chemical reactions
∆Hrxn – enthalpy of reaction
∆Hcomb – enthalpy of combustion
(see p. 643)
28
May 18
2.
Formation of compounds from elements
∆Hof – standard enthalpy of formation
The standard molar enthalpy of formation is
the energy released or absorbed when one
mole of a compound is formed directly from
the elements in their standard states. (see
p. 642)
eg.
C(s) + ½ O2(g) → CO(g)
ΔHfo = -110.5 kJ/mol
29
May 18
Use the equation below to determine the ΔHfo
for CH3OH(l)
2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ
1 C(s) + 2 H2(g) + ½ O2(g) → 1 CH3OH(l) + 238.6 kJ
∆H = -238.6 kJ/mol
30
May 18
Use the equation below to determine the ΔHfo
for CaCO3(s)
2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)
2 Ca(s) + 2 C(s) + 3 O2(g) → 2 CaCO3(s) + 2413.8kJ
1 Ca(s) + 1 C(s) + 1.5 O2(g) → 1 CaCO3(s) + 1206.9 kJ
∆H = -1206.9 kJ/mol
31
May 18
Use the equation below to determine
the ΔHfo for PH3(g)
4 PH3(g) → P4(s) + 6 H2(g) + 21.6 kJ
a) +21.6 kJ/mol
b) -21.6 kJ/mol
c) +5.4 kJ/mol
d) -5.4 kJ/mol
32
May 18
Phase Changes (p.647)
∆Hvap – enthalpy of vaporization (l → g)
3.
∆Hfus – enthalpy of melting (fusion: s → l)
∆Hcond – enthalpy of condensation (g → l)
∆Hfre – enthalpy of freezing (l → s)
eg. H2O(l)  H2O(g)
Hg(l)  Hg(s)
ΔHvap = +40.7 kJ/mol
ΔHfre = -23.4 kJ/mol
33
May 18
4.
Solution Formation (p.647, 648)
∆Hsoln – enthalpy of solution
eg.
ΔHsoln, of ammonium nitrate is +25.7 kJ/mol.
NH4NO3(s) + 25.7 kJ → NH4NO3(aq)
ΔHsoln, of calcium chloride is −82.8 kJ/mol.
CaCl2(s) → CaCl2(aq) + 82.8 kJ
34
May 18
Three ways to represent an enthalpy change:
1. thermochemical equation - the energy
term written into the equation.
2. enthalpy term is written as a separate
expression beside the equation.
3. enthalpy diagram.
35
May 18
eg. the formation of water from the elements
produces 285.8 kJ of energy.
1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ
thermochemical
equation
2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol
36
enthalpy
diagram
H2(g) + ½ O2(g)
3.
Enthalpy
(H)
examples:
questions
∆Hf = -285.8 kJ/mol
H2O(l)
May 18
pp. 641-643
p. 643 #’s 15-18
WorkSheet: Thermochemistry #4
37
May 24
Calculating Enthalpy Changes
FORMULA:
q = n∆H
q = heat (kJ)
n = # of moles
m
n
M
∆H = molar enthalpy
(kJ/mol)
38
May 24
eg. How much heat is released when
50.0 g of CH4 forms from C and H ?
(p. 642)
n 

50.0 g
16.05 g / mol
3.115 mol
q = nΔH
= (3.115 mol)(-74.6 kJ/mol)
= -232 kJ
39
May 24
eg. How much heat is released when
50.00 g of CH4 undergoes complete
combustion?
(p. 643)
50.0 g
n 
16.05 g / mol
 3.115 mol
q = nΔH
= (3.115 mol)(-965.1 kJ/mol)
= -3006 kJ
40
May 24
eg. How much energy is needed to change
20.0 g of H2O(l) at 100 °C to steam at 100 °C ?
Mwater = 18.02 g/mol
ΔHvap = +40.7 kJ/mol
20.0 g
n 
18.02 g / mol
 1.110 mol
q = nΔH
= (1.110 mol)(+40.7 kJ/mol)
= +45.2 kJ
41
May 24
∆Hfre and ∆Hcond have the opposite sign
of the above values.
42
May 24
eg. The molar enthalpy of solution for
ammonium nitrate is +25.7 kJ/mol. How
much energy is absorbed when 40.0 g of
ammonium nitrate dissolves?
40.0 g
n
80.06g / mol
 0.4996 mol
q = nΔH
= (0.4996 mol)(+25.7 kJ/mol)
= +12.8 kJ
43
May 24
What mass of ethane, C2H6, must be burned
to produce 405 kJ of heat?
ΔH = -1250.9 kJ
- 405 kJ
n
q = - 405 kJ
 1250.9 kJ
m=?
q = nΔH
q
n
H
n = 0.3238 mol
m= nxM
= (0.3238 mol)(30.08 g/mol)
= 9.74 g
44
Complete:
p. 645; #’s 19 – 23
pp. 648 – 649; #’s 24 – 29
p. 638 #’ 4 – 8
pp. 649, 650 #’s 3 – 8
p. 657, 658 #’s 9 - 18
May 24
WorkSheet: Thermochemistry #5
45
19. (a) -8.468 kJ (b) -7.165 kJ 20. -1.37 x103 kJ
21. (a) -2.896 x 103 kJ (b) -6.81 x104 kJ
21. (c) -1.186 x 106 kJ
22. -0.230 kJ
23. 3.14 x103 g
24. 2.74 kJ
25.(a) 33.4 kJ (b) 33.4 kJ
26.(a) absorbed (b) 0.096 kJ
27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq)
(b) 1.69 kJ
(c) cool; heat absorbed from water
28. 819.2 g
29. 3.10 x 104 kJ
46
May 25
Heating and Cooling Curves
Demo: Cooling of p-dichlorobenzene
Time (s) Temperature (°C) Time (s)
Temperature (°C)
47
May 25
Cooling curve for p-dichlorobenzene
Temp. 80
(°C )
50
KE
liquid
PE
freezing
KE
solid
20
Time
48
May 25
Heating curve for p-dichlorobenzene
80
Temp.
(°C )
50
KE
PE
20
KE
Time
49
May 25
What did we learn from this demo??
 During
a phase change temperature
remains constant and PE changes
 Changes
in temperature during
heating or cooling means the KE of
particles is changing
50
May 25
p. 651
51
May 25
p. 652
q = mc∆T
q = n∆H
52
May 25
p. 656
q = n∆H
q = mc∆T
53
May 25
54
May 25
Heating Curve for H20(s) to H2O(g)
1.
2.
A 40.0 g sample of ice at -40 °C is
heated until it changes to steam and
is heated to 140 °C.
Sketch the heating curve for this
change.
Calculate the total energy required for
this transition.
55
May 25
q = mc∆T
140
100
q = n∆H
q = mc∆T
Temp.
(°C )
q = n∆H
0
q = mc∆T
-40
Time
56
May 25
Data:
cice = 2.01 J/g.°C
cwater = 4.184 J/g.°C
csteam = 2.01 J/g.°C
ΔHfus = +6.02 kJ/mol
ΔHvap = +40.7 kJ/mol
57
warming ice: (from -40 ºC to 0 ºC)
q = mc∆T
= (40.0)(2.01)(0 - -40)
= 3216 J
May 26
warming water: (from 0 ºC to 100 ºC)
q = mc∆T
= (40.0)(4.184)(100 – 0)
= 16736 J
58
May 26
warming steam: (from 100 ºC to 140
ºC)
q = mc∆T
= (40.0)(2.01)(140 -100)
= 3216 J
moles of water:
n = 40.0 g
18.02 g/mol
= 2.22 mol
59
melting ice: (fusion)
q = n∆H
= (2.22 mol)(6.02 kJ/mol)
= 13.364 kJ
May 26
boiling water: (vaporization)
q = n∆H
= (2.22 mol)(40.7 kJ/mol)
= 90.354 kJ
60
May 27
Total Energy
90.354 kJ
13.364 kJ
3216 J
3216 J
16736 J
127 kJ
61
May 27
Practice
p. 655: #’s 30 – 34
WorkSheet:
pp. 656: #’s 1 - 9
Thermochemistry #6
p. 657 #’s 2, 9
p. 658 #’s 10, 16 – 20
30.(b) 3.73 x103 kJ
31.(b) 279 kJ
32.(b) -1.84 x10-3 kJ
33.(b) -19.7 kJ -48.77 kJ
34. -606 kJ
62
May 30
Law of Conservation of Energy (p. 627)
The total energy of the universe is constant
∆Euniverse = 0
First Law of Thermodynamics
Universe = system + surroundings
∆Euniverse = ∆Esystem + ∆Esurroundings
∆Euniverse = ∆Esystem + ∆Esurroundings = 0
OR
OR
∆Esystem = -∆Esurroundings
qsystem = -qsurroundings
63
May 30
Calorimetry (p. 661)
calorimetry - the measurement of heat
changes during chemical or physical
processes
calorimeter - a device used to measure
changes in energy
2 types of calorimeters
1. constant pressure or simple
calorimeter (coffee-cup calorimeter)
2. constant volume or bomb calorimeter.
64
May 30
Simple
Calorimeter
p.661
65
May 30
a
simple calorimeter consists of an
insulated container, a thermometer, and
a known amount of water
 simple calorimeters are used to measure
heat changes associated with heating,
cooling, phase changes, solution
formation, and chemical reactions that
occur in aqueous solution
66
May 30
 to
calculate heat lost or gained by a
chemical or physical change we apply the
first law of thermodynamics:
qsystem = -qcalorimeter
Assumptions:
- the system is isolated
- c (specific heat capacity) for water is not
affected by solutes
- heat exchange with calorimeter can be
ignored
67
May 30
eg.
A simple calorimeter contains 150.0 g of
water. A 5.20 g piece of aluminum alloy at
525 °C is dropped into the calorimeter
causing the temperature of the calorimeter
water to increase from 19.30°C to
22.68°C.
Calculate the specific heat capacity of the
alloy.
68
May 31
aluminum alloy
m = 5.20 g
T1 = 525 ºC
T2 = 22.68 ºC
FIND c for Al
water
m = 150.0 g
T1 = 19.30 ºC
T2 = 22.68 ºC
c = 4.184 J/g.ºC
qsys = - qcal
mcΔT = - mc ΔT
(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)
-2612 c = -2121
c = 0.812 J/g.°C
69
May 31
eg.
The temperature in a simple calorimeter
with a heat capacity of 1.05 kJ/°C changes
from 25.0 °C to 23.94 °C when a very cold
12.8 g piece of copper was added to it.
Calculate the initial temperature of the
copper. (c for Cu = 0.385 J/g.°C)
70
copper
m = 12.8 g
T2 = 23.94 ºC
c = 0.385 J/g.°C
May 31
calorimeter
C = 1.05 kJ/°C
T1 = 25.00 ºC
T2 = 23.94 ºC
FIND T1 for Cu
qsys = - qcal
mcT = - CT
(12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0)
4.928 (23.94 – T1) = 1113
23.94 – T1= 1113/4.928
23.94 – T1= 225.9
T1= -202 ºC
71
May 31
Homework
p. 664, 665 #’s 1b), 2b), 3 & 4
 p. 667, #’s 5 - 7

72
p. 665 # 4.b)
(60.4)(0.444)(T2 – 98.0) = -(125.2)(4.184)(T2 – 22.3)
26.818(T2 – 98.0) = -523.84(T2 – 22.3)
26.818T2 - 2628.2 = -523.84T2 + 11681
550.66T2 = 14309.2
T2 = 26.0 °C
73
6. System (Mg)
m = 0.50 g
= 0.02057 mol
Find ΔH
Calorimeter
v = 100 ml
so m = 100 g
c = 4.184
T2 = 40.7
qMg = -qcal
T1 = 20.4
nΔH = -mcΔT
7. System
ΔH = -53.4 kJ/mol
n = CV
= (0.0550L)(1.30 mol/L)
= 0.0715 mol
Calorimeter
v = 110 ml
so m = 110 g
c = 4.184
T1 = 21.4
Find T2
74
June 1
Bomb
Calorimeter
75
Bomb Calorimeter
used to accurately measure enthalpy
changes in combustion reactions
 the inner metal chamber or bomb contains
the sample and pure oxygen
 an electric coil ignites the sample
 temperature changes in the water
surrounding the inner “bomb” are used to
calculate ΔH

76
 to
accurately measure ΔH you need to
know the heat capacity (kJ/°C) of the
calorimeter.
 must account for all parts of the
calorimeter that absorb heat
Ctotal = Cwater + Cthermom.+ Cstirrer + Ccontainer
NOTE: C is provided for all bomb
calorimetry calculations
77
eg. A technician burned 11.0 g of octane in a
steel bomb calorimeter. The heat capacity of
the calorimeter was calibrated at 28.0 kJ/°C.
During the experiment, the temperature of the
calorimeter rose from 20.0 °C to 39.6 °C.
What is the enthalpy of combustion for octane?
78
May 31
system (octane)
m = 11.0 g
Find ΔHcomb
n = 11.0 g
114.26 g/mol
= 0.09627 mol
calorimeter
C = 28.0 kJ/ºC
T2 = 39.6 ºC
T1 = 20.0 ºC
qsys = - qcal
n ΔH = -CΔT
(0.09627) ΔH = - (28.0)(39.6 – 20.0)
ΔH = -5700 kJ/mol
79
May 31
eg.
1.26 g of benzoic acid, C6H5COOH(s), is burned in
a bomb calorimeter. The temperature of the
calorimeter and contents increases from 23.62 °C
to 27.14 °C. Calculate the heat capacity of the
calorimeter. (∆Hcomb = -3225 kJ/mol)
benzoic acid
m = 1.26 g
ΔHcomb = -3225 kJ/mol
calorimeter
T1 = 23.62 ºC
T2 = 27.14 ºC
Find C
80
May 31
n = 1.26 g
122.13 g/mol
= 0.01032 mol
qsys = - qcal
n ΔH = -CΔT
(0.01032) ΔH = - (C)(27.14 – 23.62)
C = 9.45 kJ/ ºC
Homework
p. 675 #’s 8 – 10
WorkSheet: Thermochemistry #7
81
June 2
Hess’s Law of Heat Summation
 the
enthalpy change (∆H) of a physical or
chemical process depends only on the
beginning conditions (reactants) and the
end conditions (products)
 ∆H is independent of the pathway and/or
the number of steps in the process
 ∆H is the sum of the enthalpy changes of
all the steps in the process
82
June 2
eg. production of carbon dioxide
Pathway #1:
2-step mechanism
C(s) + ½ O2(g) → CO(g)
∆H = -110.5 kJ
CO(g) + ½ O2(g) → CO2(g)
∆H = -283.0 kJ
C(s) + O2(g) → CO2(g)
∆H = -393.5 kJ
83
June 2
eg. production of carbon dioxide
Pathway #2: formation from the elements
C(s) + O2(g) → CO2(g)
∆H = -393.5 kJ
84
Using Hess’s Law
June 2
 We
can manipulate equations with
known ΔH to determine an unknown
enthalpy change.
NOTE:
 Reversing an equation changes the sign
of ΔH.
 If we multiply the coefficients we must
also multiply the ΔH value.
85
June 2
multiply ?
reverse ?
eg.
Determine the ΔH value for:
H2O(g) + C(s) → CO(g) + H2(g)
using the equations below.
C(s) + ½ O2(g) → CO(g)
ΔH = -110.5 kJ
H2(g) + ½ O2(g) → H2O(g)
ΔH = -241.8 kJ
86
June 2
eg.
Determine the ΔH value for:
4 C(s) + 5 H2(g) → C4H10(g)
using the equations below.
Switch
ΔH (kJ)
C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)
-110.5
H2(g) + ½ O2(g) → H2O(g)
-241.8
C(s) + O2(g) → CO2(g)
Multiply
by 4
Multiply
by 5
-393.5
87
June 2
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g)
+110.5
5(H2(g) + ½ O2(g) → H2O(g)
4(C(s) + O2(g) → CO2(g)
-241.8)
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g)
5 H2(g) + 2½ O2(g) → 5 H2O(g)
4C(s) + 4 O2(g) → 4 CO2(g)
+110.5
-1209.0
-1574.0
-393.5)
Ans: -2672.5 kJ
88
June 2
Practice
pg. 681 #’s 11-14
WorkSheet:
Thermochemistry #8
89
Review
June 3
∆Hof (p. 642, 684, & 848)
The standard molar enthalpy of formation is
the energy released or absorbed when one
mole of a substance is formed directly from
the elements in their standard states.
∆Hof = 0 kJ/mol
for elements in the standard state
The more negative the ∆Hof , the more
stable the compound
90
Using Hess’s Law and ΔHf
June 3
Use the formation equations below to determine
the ΔH value for:
C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)
4 C(s) + 5 H2(g) → C4H10(g)
H2(g) + ½ O2(g) → H2O(g)
C(s) + O2(g) → CO2(g)
ΔHf (kJ/mol)
-2672.5
-241.8
-393.5
91
June 3
Using Hess’s Law and ΔHf
ΔHrxn = ∑ΔHf (products) - ∑ΔHf (reactants)
eg. Use ΔHf , to calculate the enthalpy of
reaction for the combustion of glucose.
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
92
June 3
ΔHrxn = ∑ΔHf (products) - ∑ΔHf (reactants)
CO2(g)
H2O(g)
C6H12O6(s)
ΔHf
-393.5 kJ/mol
-241.8 kJ/mol
-1274.5 kJ/mol
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
ΔHrxn = [6(-393.5) + 6(-241.8)] – [1(-1274.5)+ 6(0)]
= [-2361 + -1450.8] - [-1274.5 + 0]
= - 2537.3 kJ
93
June 3
Use the molar enthalpy of formation to calculate
ΔH for this reaction:
Fe2O3(s) + 3 CO(g) → 3 CO2(g) + 2 Fe(s)
−824.2 kJ/mol
−110.5 kJ/mol
−393.5 kJ/mol
ΔHrxn = [3(-393.5) + 2(0) ] – [3(-110.5)+ 1(-824.2)]
= [-1180.5 + 0] - [-331.5 + -824.20]
= - 24.8 kJ
94
June 7
Eg.
The combustion of phenol is represented by
the equation below:
−393.5 kJ/mol
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(g)
−241.8 kJ/mol
If ΔHcomb = -3059 kJ/mol, calculate the heat of
formation for phenol.
ΔHcomb = -27.4 kJ/mol
95
Bond Energy Calculations (p. 688)
 The
energy required to break a bond
is known as the bond energy.
 Each type of bond has a specific
bond energy (BE).
(table p. 847)
 Bond
Energies may be used to
estimate the enthalpy of a reaction.
96
Bond Energy Calculations (p. 688)
ΔHrxn = ∑BE(reactants) - ∑BE (products)
eg. Estimate the enthalpy of reaction for the
combustion of ethane using BE.
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
Hint: Drawing the structural formulas for all
reactants and products will be useful here.
97
C-C = 347
C-H = 338
O=O = 498
2
C
C
C=O = 745
H-O = 460
+ 7O=O
→ 4 O=C=O + 6 H-O-H
[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)]
8236 - 11480
= -3244 kJ
p. 690 #’s 23,24,& 26
p. 691 #’s 3, 4, 5, & 7
98
Energy Comparisons
 Phase
changes involve the least amount
of energy with vaporization usually
requiring more energy than melting.
 Chemical changes involve more energy
than phase changes but much less than
nuclear changes.
 Nuclear reactions produce the largest ΔH
eg.
nuclear power, reactions in the sun
99
STSE

What fuels you? (Handout)
100
aluminum alloy
m = 5.20 g
T1 = 525 ºC
T2 =
ºC
FIND c for Al
water
m = 150.0 g
T1 = 19.30 ºC
T2 = 22.68 ºC
c = 4.184 J/g.ºC
qsys = - qcal
mcT = - mc T
(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)
-2612 c = -2121
c = 0.812 J/g.°C
101
copper
m = 12.8 g
T2 =
ºC
c = 0.385 J/g.°C
calorimeter
C = 1.05 kJ/°C
T1 = 25.00 ºC
T2 = 23.94 ºC
FIND T1 for Cu
qsys = - qcal
mcT = - CT
(12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0)
4.928 (23.94 – T1) = 1113
23.94 – T1= 1113/4.928
23.94 – T1= 225.9
T1= -202 ºC
102
q
heat
J or kJ
c
Specific heat
capacity
Heat capacity
J/g.ºC
C
ΔH
Molar heat or
molar enthalpy
kJ/ ºC or
J/ ºC
kJ/mol
103