Energy Changes in Chemical
Reactions
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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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Work (w): energy transfer when forces are
applied to a system
Heat (q): energy transferred from a hot object
to a cold one
◦ Radiant energy heat from the sun
◦ Thermal energy  associated with motion of
particles
◦ Potential energy  energy associated with object’s
position or substance’s chemical bonds
◦ Kinetic energy  energy associated with object’s
motion
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Describe the difference between the two.
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1 kg  m
SI unit of energy: J
1J 
s2
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
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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 loses heat (negative); gains heat
(positive)
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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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Determine if the following processes are
endothermic or exothermic…
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Combustion of methane
Reacting Ba(OH)2 with NH4Cl
Neutralization of HCl
Melting
CaCO3 (s)  CaO (s) + CO2 (g)
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Combustion of methane
Reacting Ba(OH)2 with NH4Cl
Neutralization of HCl
Melting
CaCO3 (s)  CaO (s) + CO2 (g)
exothermic
endothermic
exothermic
endothermic
endothermic
◦ Combustion, neutralization, and combination
reactions tend to be exothermic
◦ Decomposition reactions tend to be endothermic
◦ Melting, boiling, and sublimation are endothermic
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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 DT to calculate
the amount of heat gained or lost:
q = msDT
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ms = C
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q = CDT
◦ Heat capacity (C): amount of heat required to raise
the temperature of a given quantity of a substance
by 1oC; C = q / DT = 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 (DT) = J / mol • oC
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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 = msDT
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?
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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 to 88.4oC. Identify the
unknown metal.
Specific heats of metals:
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Al0.897 J/g·oC
Fe
0.449 J/g·oC
Cu
0.386 J/g·oC
Sn
0.228J/g·oC
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Heat changes in a reaction
can be determined by
measuring the heat flow
at constant pressure
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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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
the metal and water is 25.0oC.
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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A hot piece of copper (at 98.7oC, specific heat
= 0.385 J/g•oC) weighs 34.6486 g. When
placed in room temperature water, it is
calculated that 915.1 J of heat are released by
the metal.
What gains heat?
What loses heat?
What is the final temperature of the metal?
Watch signs!!!!
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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.
At constant pressure,
◦ qp  DH = Hproducts – Hreactants
◦ DH > 0  endothermic (net absorption of energy
from environment; products have more internal
energy)
◦ DH < 0  exothermic (net loss of energy to
environment; reactants have more internal energy)
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Why does T become
constant during
melting and
evaporating?
Melting,
vaporization, and
sublimation are
endothermic
We can calculate
total heat needed
to convert a 15
gram piece of ice at
-20oC to steam at
120oC.
2.080 J/goC
2250 J/g
4.184 J/goC
334 J/g
2.09 J/goC
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Heat of fusion (DHfus): Amount of heat
required to melt (solid  liquid)
Heat of vaporization (DHvap): Amount of heat
required to evaporate (liquid  gas)
Heat of sublimation (DHsub): Amount of heat
required to sublime (solid  gas)
Why are there no values for DHfreezing,
DHcondendsation, or DHdeposition?
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Shows both mass and enthalpy relationships
2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s)
DHo = -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 Fe2O3, 2 mol Fe, and 1 mol Al2O3
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2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s)
DHo = -852 kJ
How much heat is released if 10.0 grams of Fe2O3
reacts with excess Al?
What if we reversed the reaction?
Heat would have to be put in to make the reaction
proceed:
◦ 2Fe (s) + Al2O3 (s)  2Al (s) + Fe2O3 (s)
DHo = +852 kJ
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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 in question.
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
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Value changes sign with direction
Figure 8.5
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Values of enthalpy change
◦ For a reaction in the reverse direction, enthalpy is
numerically equal but opposite in sign
 Reverse direction, heat flow changes; endothermic
becomes exothermic (and vice versa); sign of DH
changes
◦ Proportional to the amount of reactant consumed
 Twice as many moles = twice as much heat; half as
many moles = half as much heat
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DHT = DH1 + DH2 + DH3 + ….
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Thermochemical equation:
◦ H2(g) + I2(s)  2HI(g)
kJ
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Two possible changes:
Reverse the equation:
◦ 2HI(g)  H2(g) + I2(s)
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DH = +53.00
DH = -53.00 kJ
Double the amount of material:
◦ 2H2(g) + 2I2(s)  4HI(g)
+106.00 kJ
DH =
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Calculate DHo for
2NO (g) + O2 (g)  N2O4 (g)
N2O4 (g)  2NO2 (g)
57.2 kJ
NO (g) + ½ O2 (g)  NO2 (g)
DHo = ?
DHo =
DHo = -57.0 kJ
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We can use known values of DHo to calculate
unknown values for other reactions
P4 (s) + 3 O2 (g)  P4O6 (s)
DH = -1640.1
kJ
P4 (s) + 5 O2 (g)  P4O10 (s) DH = -2940.1
kJ
What is DHo for the following reaction?
P4O6 (s) + 2 O2 (g)  P4O10 (s)
DH = ?
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Given:
2NH3(g)  N2H4(l) + H2(g)
1 N (g) + 3
H2(g)  NH3(g)
2
2
2
CH4O(l)  CH2O(g) + H2(g)
DH = 54 kJ
DH = -69 kJ
DH = -195 kJ
Find the enthalpy for the following reaction:
N2H4(l) + CH4O(l)  CH2O (g) + N2(g) + 3H2(g)
DH = ? kJ
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Given the following equations:
2CO2 (g)  O2 (g) + 2CO (g) DH = 566.0 kJ
½ N2 (g) + ½ O2 (g)  NO (g) DH = 90.3 kJ
Calculate the enthalpy change for:
2CO (g) + 2NO (g)  2CO2 (g) + N2 (g) DH = ?
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Standard heat of formation (DHof): heat
needed to make 1 mole of a substance from
its stable elements in their standard states
DHof = 0 for a stable (naturally occurring)
element
Which of these have DHof = 0?
◦ CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s)
Do the following equations represent
standard enthalpies of formation? Why or
why not?
◦ 2Ag (l) + Cl2 (g)  2AgCl (s)
◦ Ca (s) + F2 (g)  CaF2 (s)
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Can use measured enthalpies of formation to
determine the enthalpy of a reaction (use
Appendix B in back of book)
DHorxn = SnDHof (products) – SnDHof (reactants)
◦ S  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)  SrCl2 (g)
◦ DHorxn = [DHof (SrCl2)] – [DHof (Sr) + DHof (Cl2)]
= -828.4 kJ/mol
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Some Common Substances (25oC)
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DHorxn = S DHof,products - S DHof,reactants
Calculate values of DHo for the following rxns:
1) CaCO3 (s)  CaO (s) + CO2 (g)
2) 2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O (l)
DHof 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
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Use Standard Heat of Formation values to
calculate the enthalpy of reaction for:
C6H12O6(s)  C2H5OH(l) + CO2(g)
Hint: Is the equation balanced?
DHof (C6H12O6(s)) = -1260.0 kJ/mol
DHof (C2H5OH(l)) = -277.7 kJ/mol
DHof (CO2(g)) = -393.5 kJ/mol
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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
compounds with more stable (stronger)
H Bond Energy
bonds (greater bond energy)
2
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Bond energies vary somewhat from one
mole- cule to another so we use average
bond dissociation energy (D)
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H-OH
= 453
H-O
H-OOH
502 kJ/mol
Avg O-H
427 kJ/mol
431 kJ/mol
kJ/mol
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DHorxn = SBE (reactants) + - SBE (products)
released
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endothermic
energy input
exothermic
energy
SBE(react) > SBE(prod)  endothermic
SBE(react) < SBE(prod)  exothermic
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)
DHrxn = [BEN  N + 3BEH-H] + [-6BEN-H]
DHrxn = [945 + 3(436)] – [6(390)] = -87 kJ
measured value = -92.2 kJ
Why are the calculated and measured values
different?
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Use bond energies to calculate the enthalpy
change for the decomposition of nitrogen
trichloride:
NCl3 (g)  N2 (g) +
Cl2 (g)
How many distinct bond types are there in
each molecule?
How many of each bond type do we need to
calculate DHrxn?
◦ BE(N-Cl) = 200 kJ/mol
◦ BE(N≡N) = 945 kJ/mol
◦ BE(Cl-Cl) = 243 kJ/mol
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6(N-Cl) + -1(N N) + -3(Cl-Cl)
≡
6(200) + -(945) + -3(243) = -474 kJ
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Use q = msDT (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 DHof values: products – reactants
If given bond energy (BE) values:
+reactants + -products
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Identify how to set up the following
problems:
Calculate the DHo of reaction for:
◦ C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
◦ DHof C3H8(g): -103.95 kJ/mol; DHof CO2(g): -393.5
kJ/mol; DHof 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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C2H4(g ) + 6F2(g)  2CF4(g) + 4HF(g)
?
◦ H2 (g) + F2 (g)  2HF (g)
DHo = -537 kJ
◦ C (s) + 2F2 (g)  CF4 (g)
DHo = -680 kJ
◦ 2C (s) + 2H2 (g)  C2H4 (g) DHo = 52.3 kJ
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DHo =
Use average bond energies to determine the
enthalpy of the following reaction
◦ CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)
(BEC-Cl = 328 kJ/mol)