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Chapter 8
Thermochemistry:
Chemical Energy
Energy is the capacity to do work
•
Thermal energy is the energy associated with
the random motion of atoms and molecules
•
Chemical energy is the energy stored within the
bonds of chemical substances
•
Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
•
Electrical energy is the energy associated with
the flow of electrons
•
Potential energy is the energy available by virtue
of an object’s position, or stored energy
•
Kinetic energy is moving energy.
•E = qrxn + w
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
SURROUNDINGS
SYSTEM
open
Exchange: mass & energy
closed
isolated
energy
nothing
Calorimetry and Heat Capacity
Measure the heat flow at constant pressure (H).
Calorimetry and Heat Capacity
Measure the heat flow at constant volume (E).
First Law of Thermodynamics
• Energy is neither created nor destroyed.
• 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.
Use Fig. 5.5
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
H > 0
Enthalpy Changes 01
• Enthalpies of Physical
Change:
Enthalpies of Physical and
Chemical Change
Enthalpy of Fusion (Hfusion): The amount of heat
necessary to melt a substance without changing its
temperature.
Enthalpy of Vaporization (Hvap): The amount of heat
required to vaporize a substance without changing its
temperature.
Enthalpy of Sublimation (Hsubl): The amount of heat
required to convert a substance from a solid to a gas
without going through a liquid phase.
Thermochemical Equations
Is H negative or positive?
System absorbs heat
Endothermic
H > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
H = 6.01 kJ
Thermochemical Equations
Is H negative or positive?
System gives off heat
Exothermic
H < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) H = -890.4 kJ
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
H = 6.01 kJ
If you reverse a reaction, the sign of H changes
H2O (l)
•
H2O (l)
H2O (s)
H = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then H must change by the same factor n.
2H2O (s)
2H2O (l)
H = 2 x 6.01 = 12.0 kJ
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
H = 6.01 kJ
H2O (l)
H2O (g)
H = 44.0 kJ
How much heat is transfered when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
H = -3013 kJ
-3013 kJ
= -6470 kJ
1 mol P4
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity (m)
of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = mst
q = Ct
t = tfinal - tinitial
Calorimetry and Heat Capacity
Assuming that a can of soda has the same specific
heat as water, calculate the amount of heat (in
kilojoules) transferred when one can (about 350 g) is
cooled from 25 °C to 3 °C.
q = (specific heat) x (mass of substance) x T
J
Specific heat (Water) = 4.18
g °C
Mass = 350 g
Temperature change = 3 C-25 °C = - 22 °C
Calorimetry and Heat Capacity
Calculate the amount of heat transferred.
4.18 J x 350 g x -22 °C
Heat evolved =
= -32 000 J
g °C
-32 000 J x
1 kJ
1000 J
= -32 kJ
Calorimetry and Heat Capacity
Constant-Volume Calorimetry
Bomb Calorimetry
• Because the volume in the
bomb calorimeter is
constant, what is measured
is really the change in
internal energy, E, not H.
 E = qrxn + w
• W = VΔP
• For most reactions, the
difference is very small.
• ΔE almost = qrxn
Constant-Volume Calorimetry
qsys = qwater + qCal + qrxn
qsys = 0
qrxn = - (qwater + qCal)
qwater = mst
qCal = CCalt
No heat enters or leaves!
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mst
qcal = Ccalt
Reaction at Constant P
H = qrxn
No heat enters or leaves!
Hess’s Law
Hess’s Law: The overall enthalpy change for a reaction
is equal to the sum of the enthalpy changes for the
individual steps in the reaction.
Haber Process: 3H2(g) + N2(g)
2NH3(g) H° = -92.2 kJ
Multiple-Step Process
2H2(g) + N2(g)
N2H4(g)
H°1 = ?
N2H4(g) + H2(g)
2NH3(g)
H°2 = -187.6 kJ
3H2(g) + N2(g)
2NH3(g)
H°1+2 = -92.2 kJ
Hess’s Law
H°1 + H°2 = H°1+2
H°1 = H°1+2 - H°2
= -92.2 kJ - (-187.6 kJ) = 95.4 kJ
Hess’s Law 01
• Hess’s Law: The overall enthalpy change
for a reaction is equal to the sum of the
enthalpy changes for the individual steps in
the reaction.
•
3 H2(g) + N2(g)  2 NH3(g) H° = –92.2 kJ
Hess’s Law
• The industrial degreasing solvent methylene
chloride (CH2Cl2, dichloromethane) is
prepared from methane by reaction with
chlorine:
• CH4(g) + 2 Cl2(g)
CH2Cl2(g) + 2 HCl(g)
• Use the following data to calculate H° (in
kilojoules) for the above reaction:
• CH4(g) + Cl2(g)
• CH3Cl(g) + Cl2(g)
CH3Cl(g) + HCl(g) H° = –98.3 kJ
CH2Cl2(g) + HCl(g) H° = –104 kJ
Standard Heats of Formation
H2(g) + 1/2 O2(g)  H2O(l) H°f = –286 kJ/mol
3/
2 H2(g)
+ 1/2 N2(g)  NH3(g) H°f = –46 kJ/mol
2 C(s) + H2(g)  C2H2(g) H°f = +227 kJ/mol
*Standard state (°):
P = 1atm, t = 25°C,
Concentration = 1 mol/Lit
Standard Heats of Formation
• Standard Heats of Formation (H°f): The enthalpy
change for the formation of 1 mole of substance in its
standard state from its constituent elements in their
standard states.
• The standard heat of formation for any element in its
standard state is defined as being ZERO.
H°f = 0 for an element in its standard state
Standard Heats of Formation
• Calculating H° for a reaction:
H° = H°f (Products) – H°f (Reactants)
• For a balanced equation, each heat of formation must be
multiplied by the stoichiometric coefficient.
aA + bB
cC + dD
H° = [cH°f (C) + dH°f (D)] – [aH°f (A) + bH°f (B)]
Standard Heats of Formation
Some Heats of Formation, Hf° (kJ/mol)
CO(g)
-111
C2H2(g)
227
Ag+(aq)
106
CO2(g)
-394
C2H4(g)
52
Na+(aq)
-240
H2O(l)
-286
C2H6(g)
-85
NO3-(aq)
-207
NH3(g)
-46
CH3OH(g)
-201
Cl-(aq)
-167
N2H4(g)
95.4
C2H5OH(g)
-235
AgCl(s)
-127
HCl(g)
-92
C6H6(l)
49
Na2CO3(s)
-1131
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
H0rxn = [ 12H0f (CO2) + 6H0f (H2O)] - [ 2H0f (C6H6)]
H0rxn = [ 12x(–393.5) + 6x( -286 ) – [ 2x(49.04) ] = -6542.0
kJ
-6542.0
= - 3271.0 kJ/mol C6H6
2 mol
See Page 292 for Standard Heat of Formation
Standard Heats of Formation
Using standard heats of formation, calculate the
standard enthalpy of reaction for the photosynthesis of
glucose (C6H12O6) and O2 from CO2 and liquid H2O.
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g)
H° = ?
H° = [H°f (C6H12O6(s))] - [6 H°f (CO2(g)) + 6 H°f (H2O(l))]
H° = [(1 mol)(-1260 kJ/mol)] [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]
= 2816 kJ
Bond Dissociation Energy
• Bond Dissociation Energy: Can be used to determine an
approximate value for H°f .
H°f = D (Bonds Broken) – D (Bonds Formed)
ΔHorxn = D (Reactant bonds) - D (Product bonds)
• For the reaction:
H-H (g) + Cl-Cl (g) 2H-Cl (g)
H°rxn = (D(H–H) + D(Cl-Cl))-(2 X D(H–Cl) )
H°rxn = -185 kJ
H°f = -92.5 KJ/mol
Bond Dissociation Energy 02
Calculate an approximate H° (in kilojoules) for the synthesis of
ethyl alcohol from ethylene:
C2H4(g) + H2O(g)  C2H5OH(g)
H
H
C
H
C
+ H-O-H
H
H
H
H
C
C
O-H H
ΔHorxn = D (Reactant bonds) - D (Product bonds)
ΔHorxn = (DC=C + 4 DCH + 2 DOH) - (DCC + DCO + 5 DCH + DOH)
H
ΔHorxn = (DC=C + 4 DCH + 2 DOH) - (DCC + DCO + 5 DCH + DOH)
ΔHorxn = [(1 mol)(611 kJ/mol) + (4 mol)(410 kJ/mol)
+(2 mol)(460 kJ/mol)] - [(1 mol)(350 kJ/mol)
+ (1 mol)( 350 kJ/mol) + (5 mol)(410 kJ/mol)
+ (1 mol)(460 kJ/mol)]
ΔHorxn = 39 kJ
Thermodynamics
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
Predicting spontaneity of a chemical
Reaction
• Second Law of Thermodynamics: Reactions
proceed in the direction that increases the entropy
of the system plus surroundings.
• A spontaneous process is one that proceeds on its
own without any continuous external influence.
• A nonspontaneous process takes place only in the
presence of a continuous external influence.
Introduction to Entropy 02
• The measure of molecular disorder in a system is
called the system’s entropy; this is denoted S.
• Entropy has units of J/K (Joules per Kelvin).
S = Sfinal – Sinitial
– Positive value of S indicates increased disorder.
– Negative value of S indicates decreased disorder.
Introduction to Entropy 03
Introduction to Entropy 05
• Predict whether S° is likely to be positive or
negative for each of the following reactions.
• a. 2 CO(g) + O2(g)  2 CO2(g)
b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
c. C2H4(g) + Br2(g)  CH2BrCH2Br(l)
d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
Introduction to Entropy 04
• To decide whether a process is spontaneous, both
enthalpy and entropy changes must be considered:
• Spontaneous process:
Decrease in enthalpy (–H), Increase in entropy (+S).
• Nonspontaneous process:
Increase in enthalpy(+H),Decrease in entropy (–S).
Introduction to Free Energy 01
• Gibbs Free Energy Change (G): Weighs the
relative contributions of enthalpy and entropy to
the overall spontaneity of a process.
G = H – TS
G < 0
Process is spontaneous
G = 0
Process is at equilibrium
G > 0
Process is nonspontaneous
Which of the following reactions are spontaneous under
standard conditions at 25°C?
•AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
G° = –55.7 kJ
2 C(s) + 2 H2(g)  C2H4(g)
G° = 68.1 kJ
N2(g) + 3 H2(g)  2 NH3(g)
H° =-92.2 kJ
S° = -199 J/K
ΔGo = ΔHo -TΔSo = (-92.2 kJ) - (298 K)(-0.199 kJ/K) = -32.9 kJ
Because ΔGo is negative, the reaction is spontaneous.
Introduction to Free Energy 04
• Equilibrium (G° = 0): Estimate the temperature
at which the following reaction will be at
equilibrium. Is the reaction spontaneous at room
temperature?
–
N2(g) + 3 H2(g)  2 NH3(g)
H° = –92.0 kJ
S° = –199 J/K
– Equilibrium is the point where G° = H° – TS° = 0
– T = 462 K