Download Chapter 8 Thermochemistry: Thermochemistry: Chemical Energy

Energy Changes in Chemical Rxns
Chapter 8
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Thermochemistry:
Chemical Energy
Most reactions give off or absorb energy
Energy is the capacity to do work or supply
heat.
 Heat: transfer of thermal (kinetic) energy between
two systems at different temperatures (from hot to
cold)
Metal bar in water
Metal bar drilled
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Types of Energy
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Work (w): energy transfer when forces are
applied to a system
Heat (q): energy transferred from a hot object
to a cold one
 Radiant energyÆ heat from the sun
 Thermal energy Æ associated with motion of
particles
 Potential energy Æ energy associated with object’s
position or substance’s chemical bonds
 Kinetic energy Æ energy associated with object’s
motion
Energy and Energy Conservation
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Heat versus Temperature
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Describe the difference between the two.
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SI unit of energy: J 1 J = 1 kg ⋅ m
s2
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Molecular heat transfer
1 watt = 1 J/s,, so a 100 Watt bulb uses 100 J each
second
We often use the unit of kJ to refer to chemical heat
exchanges in a reaction. 1 kJ = 1000 J
Energy is also reported in calories:
 Amount of energy needed to raise 1 gram of water by 1oC
 1 cal = 4.184 J; 1 Cal = 4184 J
 Cal (or kcal) is used on food labels
Conservation of Energy
Heat: form of energy transferred from object at
higher temperature to one at lower temperature
(from hot object to cold object)
Thermochemistry: study of heat changes in
chemical reactions, in part to predict whether
or not a reaction will occur
Thermodynamics: study of heat and its
transformations
First Law of Thermodynamics: Energy can
be converted from one form to another but
cannot be created or destroyed
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System and Surroundings
Endothermic vs Exothermic
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Endothermic reaction: q is positive (q > 0)
 Reaction (system) absorbs heat
 Surroundings feel cooler
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Exothermic reaction: q is negative (q < 0)
 Reaction releases heat
 Surroundings feel warmer
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System loses heat (negative); gains heat
(positive)
Applications of heat
emission/absorption
Specific Heat and Heat Capacity
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Specific heat (sp. ht.): amount of heat
required to raise 1 gram of substance by 1oC
Use mass, specific heat, and ΔT to calculate
the amount of heat gained or lost:
q = msΔT
Æ
ms = C
Æ
q = CΔT
 Heat capacity (C): amount of heat required to raise
the temperature of a given quantity of a substance
by 1oC; C = q / ΔT = J / oC
 Molar heat capacity (Cm): amount of heat that can
be absorbed by 1 mole of material when
temperature increases 1oC; q = (Cm) x (moles of
substance) x (ΔT) = J / mol • oC
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Specific Heat and Heat Capacity
Practice Problem
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Calculate the amount of heat transferred when 250 g of
H2O (with a specific heat of 4.184 J/g·oC) is heated
from 22oC to 98oC.
q = msΔT
Is heat being put into the system or given off by the
system?
If a piece of hot metal is placed in cold water, what
gains heat and what loses heat? Which one will have
a positive q value and which will have a negative q
value?
Worked Ex. 8.4, 8.5; Problems 8.10 – 8.12
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Practice Problem
Calorimetry and Heat Capacity
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34.8 g of an unknown metal at 25.2oC is mixed
with 60.1 g of H2O at 96.2 oC (sp. ht. = 4.184
J/g·oC). The final temperature of the system
comes
co
es to
o 88
88.4oC
C. Identify
de y the
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unknown
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metal.
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Specific heats of metals:
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 Al
 Fe
 Cu
 Sn
0.897 J/g·oC
0.449 J/g·oC
0.386 J/g·oC
0.228J/g·oC
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Heat changes in a reaction
can be determined by
measuring the heat flow at
constant
co
sa p
pressure
essu e
Apparatus to do this is
called a calorimeter.
Heat evolved by a reaction
is absorbed by water; heat
capacity of calorimeter is
the heat capacity of water.
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Example
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Enthalpies of Physical/Chemical Changes
A 28.2 gram sample of nickel is heated to
99.8oC and placed in a coffee cup calorimeter
containing 150.0 grams of water at 23.5oC.
After the metal cools, the final temperature of
th metal
the
t l and
d water
t iis 25
25.0
0oC.
C
qabsorbed + qreleased = 0
Which substance absorbed heat?
Which substance released heat?
Calculate the heat absorbed by the substance
you indicated above.
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Why does T become
constant during
melting and
evaporating?
M lti
Melting,
vaporization,
i ti
and sublimation are
endothermic
We can calculate
total heat needed to
convert a 15 gram
piece of ice at -20oC
to steam at 120oC.
Enthalpy (H) describes heat flow into and out
of a system under constant pressure
Enthalpy (a measure of energy) is heat
transferred per mole of substance
substance.
At constant pressure,
 qp = ΔH = Hproducts – Hreactants
 ΔH > 0 Æ endothermic (net absorption of energy
from environment; products have more internal
energy)
 ΔH < 0 Æ exothermic (net loss of energy to
environment; reactants have more internal energy)
Heating a Pure Substance (Water)
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Enthalpies of Phase Changes
2.080 J/goC
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2250 J/g
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4.184 J/goC
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334 J/g
2.09 J/g
J/goC
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Heat of fusion (ΔHfus): Amount of heat required
to melt (solid Æ liquid)
Heat of vaporization (ΔHvap): Amount of heat
required to evaporate (liquid Æ gas)
Heat of sublimation (ΔHsub): Amount of heat
required to sublime (solid Æ gas)
Why are there no values for ΔHfreezing,
ΔHcondendsation, or ΔHdeposition?
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Enthalpies of Reaction
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Thermochemical Equations
Determine if the following processes are
endothermic or exothermic…
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 Combustion of methane
 Reacting Ba(OH)2 with NH4Cl
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 Neutralization of HCl
 Melting
Shows both mass and enthalpy relationships
2Al (s) + Fe2O3 (s) Æ 2Fe (s) + Al2O3 (s)
ΔHo = -852 kJ
Amount of heat given off depends on amount
of material:
 852 kJ of heat are released for every 2 mol Al, 1 mol
 CaCO3 (s) Æ CaO (s) + CO2 (g)
Fe2O3, 2 mol Fe, and 1 mol Al2O3
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Thermochemical Equations
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Hess’s Law
2Al (s) + Fe2O3 (s) Æ 2Fe (s) + Al2O3 (s)
ΔHo = 852 kJ
How much heat is released if 10.0 grams of Fe2O3
reacts with excess Al?
Worked Ex. 8.3; Problems 8.8, 8.9
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What if we reversed the reaction?
Heat would have to be put in to make the reaction
proceed:
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 2Fe (s) + Al2O3 (s) Æ 2Al (s) + Fe2O3 (s) ΔHo = +852 kJ
Hess’s Law
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If a compound cannot be directly synthesized
from its elements, we can add the enthalpies of
multiple reactions to calculate the enthalpy of
reaction
eac o in ques
question.
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Hess’s Law: change in enthalpy is the same
whether the reaction occurs in one step or in a
series of steps
Look at direction of reaction and amount of
reactants/products
Hess’s Law
Value changes sign with direction
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Values of enthalpy change
 For a reaction in the reverse direction, enthalpy is
numerically equal but opposite in sign
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Reverse direction, heat flow changes; endothermic
becomes exothermic (and vice versa); sign of ΔH changes
 Proportional to the amount of reactant consumed
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Twice as many moles = twice as much heat; half as many
moles = half as much heat
ΔHT = ΔH1 + ΔH2 + ΔH3 + ….
Figure 8.5
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Enthalpy of Chemical Reaction
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Thermochemical equation:
 H2(g) + I2(s) → 2HI(g)
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ΔH = +53.00 kJ
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Two possible changes:
Reverse the equation:
 2HI(g) → H2(g) + I2(s)
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Hess’s Law
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ΔH = -53.00 kJ
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Calculate ΔHo for
2NO (g) + O2 (g) Æ N2O4 (g)
ΔHo = ?
N2O4 (g)
( ) Æ 2NO2 (g)
( )
NO (g) + ½ O2 (g) Æ NO2 (g)
ΔH
Ho = 57.2
2 kJ
ΔHo = -57.0 kJ
Double the amount of material:
 2H2(g) + 2I2(s) → 4HI(g) ΔH = +106.00 kJ
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Hess’s Law
Hess’s Law
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We can use known values of ΔHo to calculate
unknown values for other reactions
P4 (s) + 3 O2 (g) Æ P4O6 (s) ΔH = -1640.1 kJ
P4 (s) + 5 O2 (g) Æ P4O10 (s) ΔH = -2940.1
2940 1 kJ
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What is ΔHo for the following reaction?
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P4O6 (s) + 2 O2 (g) Æ P4O10 (s)
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ΔH = ?
Worked Ex. 8.6, 8.7; Problem 8.13, 8.15
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Hess’s Law
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Given:
2NH3(g) Æ N2H4(l) + H2(g)
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N (g) + 32 H2(g) Æ NH3(g)
2 2
CH4O(l)
( ) Æ CH2O(g)
(g) + H2(g)
Standard Heats of Formation
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ΔH = 54 kJ
ΔH = -69 kJ
ΔH = -195 kJ
Find the enthalpy for the following reaction:
N2H4(l) + CH4O(l) Æ CH2O (g) + N2(g) + 3H2(g)
ΔH = ? kJ
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Standard heat of formation (ΔHof): heat
needed to make 1 mole of a substance from its
stable elements in their standard states
ΔHof = 0 for a stable (naturally occurring)
element
Which of these have ΔHof = 0?
 CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s)
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Do the following equations represent standard
enthalpies of formation? Why or why not?
 2Ag (l) + Cl2 (g) Æ 2AgCl (s)
 Ca (s) + F2 (g) Æ CaF2 (s)
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Standard Enthalpies of Formation
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Standard Enthalpies of Formation
Can use measured enthalpies of formation to
determine the enthalpy of a reaction (use
Appendix B in back of book)
ΔHorxn = ΣnΔHof (products) – ΣnΔHof (reactants)
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Table 8.2: Some Common Substances (25oC)
 Σ = sum; n = number of moles (coefficients)
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Direct calculation of enthalpy of reaction if the
reactants are all in elemental form
 Sr (s) + Cl2 (g) Æ SrCl2 (g)
 ΔHorxn = [ΔHof (SrCl2)] – [ΔHof (Sr) + ΔHof (Cl2)]
= -828.4 kJ/mol
Heats of Formation
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Bond Dissociation Energies
ΔHorxn = Σ ΔHof,products - Σ ΔHof,reactants
Calculate values of ΔHo for the following rxns:
1) CaCO3 (s) Æ CaO (s) + CO2 (g)
2)) 2C6H6 ((l)) + 15O2 (g) Æ 12CO2 (g) + 6H2O ((l))
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ΔHof values:
CaCO3: -1207.1 kJ/mol; CaO: -635.5 kJ/mol; CO2: 393.5 kJ/mol; C6H6: 49.0 kJ/mol; H2O(l): -285.8
kJ/mol
Worked Ex. 8.8, 8.9; Problems 8.16, 8.17
Bond Dissociation Energy (or Bond Energy,
BE): energy required to break a bond in 1 mole
of a gaseous molecule
Reactions generally proceed to form H Bond Energy
compounds with more stable (stronger) bonds
(greater bond energy)
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Bond Dissociation Energies
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Bond Dissociation Energies
Bond energies vary somewhat from one molecule to another so we use average bond
dissociation energy (D)
H-OH
502 kJ/mol
Avg O-H = 453
H-O
427 kJ/mol
kJ/mol
H-OOH
431 kJ/mol
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Bond Dissociation Energies
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ΔHorxn = ΣBE (reactants) + - ΣBE (products)
endothermic
energy input
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Heats of Reaction
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exothermic
energy released
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ΣBE(react) > ΣBE(prod) Æ endothermic
ΣBE(react) < ΣBE(prod) Æ exothermic
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Use only when heats of formation are not
available, since bond energies are average
values for gaseous molecules
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Use bond energies to calculate the enthalpy
change for the following reaction:
N2(g) + 3H2(g) → 2NH3(g)
ΔHrxn = [BEN ≡ N + 3BEH-H] + [[-6BE
6BEN-H]
ΔHrxn = [945 + 3(436)] – [6(390)] = -87 kJ
measured value = -92.2 kJ
Why are the calculated and measured values
different?
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Heats of Reaction
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Thermochemistry Calculation Summary
Use bond energies to calculate the enthalpy
change for the decomposition of nitrogen
trichloride:
NCl
C 3 (g) Æ N2 (g) +
C
Cl2 (g)
How many distinct bond types are there in
each molecule?
How many of each bond type do we need to
calculate ΔHrxn?
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 BE(N-Cl) = 200 kJ/mol
Use q = msΔT (s = J/g·oC)
If given mass of reactant, convert to moles and
multiply by enthalpy to find total heat
transferred
If given multiple equations with enthalpies, use
Hess’s Law
If given ΔHof values: products – reactants
If given bond energy (BE) values: +reactants +
-products
 BE(N≡N) = 945 kJ/mol
 BE(Cl-Cl) = 243 kJ/mol
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Practice Problems
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Practice Problems
Identify how to set up the following problems:
Calculate the ΔHo of reaction for:
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C3H8(g): -103.95
103 95 kJ/mol;
393 5
f CO2(g): -393.5
kJ/mol; ΔHof H2O(l): -285.8 kJ/mol
8750 J of heat are applied to a 170 g sample of metal,
causing a 56oC increase in its temperature. What is
the specific heat of the metal? Which metal is it?

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ΔHo
f
ΔHo = ?
 H2 (g) + F2 (g) Æ 2HF (g) ΔHo = -537 kJ
 C (s) + 2F2 (g) Æ CF4 (g) ΔHo = -680 kJ
 C3H8 (g) + 5O2 (g) Æ 3CO2 (g) + 4H2O (l)
ΔHo
C2H4(g ) + 6F2(g) Æ 2CF4(g) + 4HF(g)
 2C (s) + 2H2 (g) Æ C2H4 (g) ΔHo = 52.3 kJ
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Use average bond energies to determine the
enthalpy of the following reaction (from Table
7.1).
 CH4 (g) + Cl2 (g) Æ CH3Cl (g) + HCl (g)
(BEC-Cl = 328 kJ/mol)
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