Transcript Energy and Chemical Reactions

15. Energetics
1
Calorimetry
Calorimetry involves the measurement of
heat changes that occur in chemical
processes or reactions.
The heat change that occurs when a
substance absorbs or releases energy is
really a function of three quantities:
– The mass
– The temperature change
– The heat capacity of the material
2
Heat Capacity and Specific Heat
The ability of a substance to absorb or retain heat
varies widely.
The heat capacity depends on the nature of the
material.
The specific heat of a material is the amount of heat
required to raise the temperature of 1 gram of a
substance 1 oC (or Kelvin)
3
Specific Heat values for
Some Common Substances
Substance
CJ g-1 K-1
C J mol-1K-1
Water (liquid)
4.184
75.327
Water (steam)
2.080
37.47
Water (ice)
2.050
38.09
Copper
0.385
24.47
Aluminum
0.897
24.2
Ethanol
2.44
112
Lead
0.127
26.4
4
Heat Exchange
When two systems
are put in contact
with each other,
there will be a net
exchange of energy
between them
unless they are at
thermal equilibrium,
i.e. at the same
temperature.
Heat will flow from the substance at the higher temperature
to that at a lower temperature
5
Heat Changes
The heat equation may be stated as
DQ = m C DT
where:
DQ
m
C
DT
= Change in heat
= mass in grams
= specific heat in J g-1 oC-1
= Temperature change
6
Temperature Changes
Measuring the temperature
change in a calorimetry
experiment can be difficult
since the system is losing
heat to the surroundings
even as it is generating
heat.
By plotting a graph of time
v temperature it is possible
to extrapolate back to what
the maximum temperature
would have been had the
system not been losing
heat to the surroundings.
A time v temperature graph
7
Heat Transfer Problem 1
Calculate the heat that would be required an aluminum
cooking pan whose mass is 400 grams, from 20oC to
200oC. The specific heat of aluminum is 0.902 J g-1 oC-1.
Solution
DQ = mCDT
= (400 g) (0.902 J g-1 oC-1)(200oC – 20oC)
= 64,944 J
8
Heat Transfer Problem 2
What is the final temperature when 50 grams of water at
20oC is added to 80 grams water at 60oC? Assume
that the loss of heat to the surroundings is negligible.
The specific heat of water is 4.184 J g-1 oC-1
Solution: DQ (Cold) = DQ (hot)
Let T = final temperature
mCDT= mCDT
(50 g) (4.184 J g-1 oC-1)(T- 20oC)
= (80 g) (4.184 J g-1 oC-1)(60oC- T)
(50 g)(T- 20oC) = (80 g)(60oC- T)
50T -1000 = 4800 – 80T
130T =5800
T = 44.6 oC
9
Phase Changes & Heat
Energy is required to change the phase of a substance
The amount of heat necessary to melt a substance is
called the Heat of fusion (DHfus).The heat of fusion is
expressed in terms of 1 mole or 1 gram
It takes 6.00 kJ of energy to melt 1 mole (18 grams)
of ice into liquid water. This is equivalent to about
335 J per gram
The amount of heat necessary to boil a substance is
called the Heat of vaporization (DHvap)
It may be expressed in terms of 1 mole or 1 gram
It takes 40.6 kJ of energy to boil away 1 mole (18
grams) of water. This is equivalent to about 2240
J per gram.
10
Molar Heat Data for Some
Common Substances
Substance
DQfus
DQvap
Mercury, Hg
2.29kJ/mol
59.1kJ/mol
Ethanol, C2H5OH
5.02kJ/mol
38.6kJ/mol
Water, H2O
6.00kJ/mol
40.6kJ/mol
Ammonia, NH3
5.65kJ/mol
23.4kJ/mol
Helium, He
0.02kJ/mol
0.08kJ/mol
Acetone
5.72kJ/mol
29.1kJ/mol
Methanol, CH3OH
3.16kJ/mol
35.3kJ/mol
11
Heat Transfer Problem 3
How much energy must be lost for 50.0 g of liquid wax at
85.0˚C to cool to room temperature at 25.0˚C?
(Csolid wax= 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, Cliquid wax=2.31
J/g˚C; MM = 352.7 g/mol, DHfusion=70,500 J/mol)
DQ =
DQtota=
mCliquid waxDT
(50g)(2.31J g-1˚C-1)(62˚C-85˚C)
+ (50g/352.7gmol-1)(-70,500J mol-1)
+ (50g)(2.18J g-1˚C-1)(25˚C-62˚C)
+ n(DQfusion)
+ mCsolid waxDT
DQtotal= (-2656.5 J) + (-9994.3 J)+ (-4033 J)
DQtotal= -16,683.8 J
DQtotal
=
DQliquid wax + DQsolidification + DQsolid wax
DQtotal= = mCliquid waxDT +n(DQfusion) +
mCsolid waxDT
12
Heat Transfer Problem 4
Steam at 175°C that occupies a volume of 32.75 dm3 and a
pressure of 2.60 atm. How much energy would it need to lose
to end as liquid water at 20 oC?
Solution:
= (2.60 atm)(32.75 dm3)
(0.0821 dm3 atm mol-1 K-1)(448 K-1)
= 2.315 mol
DQ = (2.315 mol) (37.47 J mol-1K-1)(175oC-100oC)
+(2.315 mol)(40600 J mol-1)
+(2.315 mol)(75.327 J mol-1K-1)(100oC-20oC)
n = PV/RT
DQ = 6505.7J + 93989 J + 13950.6 J
= 114.445 kJ
= 114445.3 J
Chemical Reactions
In a chemical reaction
Chemical bonds are broken
Atoms are rearranged
New chemical bonds are formed
These processes always involve
energy changes
14
Energy Changes
Breaking chemical bonds requires energy
Forming new chemical bonds releases
energy
15
Exothermic and Endothermic
Processes
Exothermic processes release energy
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4H2O (g)
+ 2043 kJ
Endothermic processes absorb energy
C(s) + H2O (g) +113 kJ  CO(g) + H2 (g)
16
Energy Changes in endothermic
and exothermic processes
In an
endothermic
reaction there is
more energy
required to
break bonds
than is released
when bonds are
formed.
The opposite is
true in an
exothermic
reaction.
17
Enthalpy Calculations
18
Enthalpy
Enthalpy is the heat absorbed or released
during a chemical reaction where the only
work done is the expansion of a gas at
constant pressure
19
Enthalpy
Not all energy changes that occur as a result
of chemical reactions are expressed as heat
Energy = Heat + Work
Work is a force applied over a distance.
Most energy changes resulting from
chemical reactions are expressed in a
special term known as enthalpy
20
Enthalpy
It is nearly impossible to set up a
chemical reaction where there is no
work performed.
The conditions for a chemical reaction
are often set up so that work in
minimized.
Enthalpy and heat are nearly equal
under these conditions.
21
Enthalpy Changes
The change in enthalpy is designated by the
symbol DH.
– If DH < 0 the process is exothermic.
– If DH > 0 the process is endothermic.
– Sometimes the symbol for enthalpy (DH) is used
for heat (DQ)
– In many cases where work is minimal heat is a
close approximation for enthalpy.
– One must always remember that while they are
closely related, heat and enthalpy are NOT
identical
22
Energy and Enthalpy Changes
It is impractical to measure absolute
amounts of energy or enthalpy.
Hence we measure changes in enthalpy
rather than total enthalpy
Enthalpy is always measured relative to
previous conditions.
Enthalpy is measured relative to the system.
23
Measuring Enthalpy
The amount of heat absorbed or released
during a chemical reaction depends on the
conditions under which the reaction is
carried out including:
– the temperature
– the pressure
– the physical state of the reactants and
products
24
Standard Conditions
For most thermodynamic measurements
standard conditions are established as
– 25 oC or 298 K
– 1.0 atmosphere of pressure
– Note this is a change from the gas laws
where the standard temperature was 0oC
25
Standard State
The pure form of a substance at
standard conditions (25oC and 1
atmosphere) is said to be in the
standard state.
The most stable form of an element at
standard conditions represents the
standard state for that element.
26
Bond Enthalpies
27
Bond Enthalpies
One approach to determining an enthalpy
change for a chemical reaction is to compute
the difference in bond enthalpies between
reactants and products
The energy to required to break a covalent
bond in the gaseous phase is called a bond
enthalpy.
Bond enthalpy tables give the average
energy to break a chemical bond. Actually
there are slight variations depending on the
environment in which the chemical bond is
located
28
Bond Enthalpy Table
The average bond enthalpies for several types of
chemical bonds are shown in the table below:
29
Bond Enthalpies
Bond enthalpies can be used to calculate the
enthalpy change for a chemical reaction.
Energy is required to break chemical bonds.
Therefore when a chemical bond is broken
its enthalpy change carries a positive sign.
Energy is released when chemical bonds
form. When a chemical bond is formed its
enthalpy change is expressed as a negative
value
By combining the enthalpy required and the
enthalpy released for the breaking and
forming chemical bonds, one can calculate
the enthalpy change for a chemical reaction
30
Bond Enthalpy Calculations
Example 1: Calculate the enthalpy change for the
reaction N2 + 3 H2  2 NH3
Bonds broken
1 N=N:
3 H-H:
= 945
3(435) = 1305
Total = 2250 kJ
Bonds formed
2x3 = 6 N-H:
6 (390) = - 2340 kJ
Net enthalpy change
= + 2250 - 2340 = - 90 kJ
31
Hess Law and Enthalpy
Calculations
32
Standard Enthalpy of Formation
Why is the formation of a gas less
exothermic than formation of a liquid?
Enthalpies of Formation
Some enthalpies of formation for common compounds
BaCO3
-1219
H2O (g) -242
HCl (g)
-93
Ba(OH)2
-998
H2O (l)
-286
HCl (aq) -167
BaO
-554
H2O2
-188
NH3 (g)
-46
CaCO3
-1207
C3H8
-104
NO
+90
CaO
-636
C4H10
-126
NO2
+33.8
Ca(OH)2
-987
CO
-110
SO2
-297
CaCl2
-796
CO2
-394
Al2O3(s)
-1670
See your text: Brown, LeMay and Bursten, Chemistry the Central Science,
7th edition pages 984-987 for addition values
39
Calculating Enthalpy from tables
Enthalpies of formation represent the
enthalpy changes when compound
forms from its elements
The enthalpy of formation for a
chemical reaction can be expressed as
the difference between the enthalpy
state of the products and that of the
reactants
DHreaction = S DHoproducts –SDHoreactants
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Sample Problem 1
Calcium carbonate reacts with hydrochloric acid
according to the following equation:
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
Calculate the enthalpy change for this reaction
DHoreaction = S DHoproducts –SDHoreactants
DHoCaCO3
-1207
DHo HCl (aq) -167
DHoCaCl2
2O (l)
-796
-286
DHo CO2 (g)
-394
DHo H
Solution
SDHoproducts =(-796)+(-286)+(-394)
= -1476 kJ
SDHoreactants =(-1207)+(2)(-167)
= -1541 kJ
DHoreaction = -1476-(-1541) = +75 kJ
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Sample Problem 2
Calculate the enthalpy change for the burning
of 11 grams of propane
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
DHoreaction = S DHoproducts –SDHoreactants
DHo
C 3H 8
DHo O2 (g)
DHo H2O
-104
0
-242
(g)
DHo CO2 (g)
-394
Solution
SDHoproducts =(3)(-394)+(4)(-242)
= -2150 kJ
SDHoreactants =(-104)+(5)(0)
= -104 kJ
DHoreaction = -2150-(-104) = -2046 kJmol-1
Now 11 grams = 0.25 mole of propane (11 g/44 g mol-1)
(0.25 mol )(-2046 kJ mol-1) = - 511.5 kJ
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Some things to Remember
The enthalpy of formation table is stated in
kJ mol-1.
To find the sum of enthalpies of formation
for reactants or products, multiply the
number of moles of each substance by the
enthalpy of formation for that substance.
Then find the difference: ProductsReactants
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Standard Enthalpy Change of Combustion
Hess’ Law – Indirect Enthalpy
Calculations by Rearranging
Reactions
Hess’ Law provides a way to calculate
enthalpy changes even when the
reaction cannot be performed directly.
If a series of reactions are added
together, the enthalpy change for the
net reaction will be the sum of the
enthalpy change for the individual
steps
46
Techniques
Equations may be multiplied, divided, or
reversed and then added together to form
a new equation
If an equation is multiplied or divided the
enthalpy of the reaction is multiplied or
divided by the same factor
If the direction an equation is reversed the
sign of the enthalpy is the opposite as well
When adding equations together the
enthalpies are added together as well
47
Hess’ Law: Example 1
N2 (g) + O2 (g)  2 NO (g)
DH1 = +181 kJ
2 NO (g) + O2 (g)  2 NO2 (g) DH2 = -113 kJ
Find the enthalpy change for
N2 (g) + 2 O2 (g)  2 NO2 (g)
48
Hess’ Law: Example 1
The required equation is really the sum of the two
given equations
Solution:
N2 (g) + O2 (g)  2 NO (g)
DH1 = +181 kJ
2 NO (g) + O2 (g)  2 NO2 (g) DH2 = -113 kJ
------------------------------------------------------------N2 (g) +2O2 (g)+ 2 NO (g)  2 NO (g) + 2 NO2 (g)
N2 (g) +2O2 (g)  + 2 NO2 (g)
DH = DH1 + DH2 = +181 kJ +(-113) = + 68 kJ
49
Hess Law: Example 2
From the following reactions and enthalpy changes:
2 SO2 (g) + O2 (g)  2 SO3 (g)
DH = -196 kJ
2 S (s) +3 O2 (g)  2 SO3 (g)
DH = -790 kJ
Find the enthalpy change for the following reaction:
S (s) + O2 (g)  SO2 (g)
Solution:
2 SO3 (g)  2 SO2 (g) + O2 (g)
2 S (s) +3 O2 (g)  2 SO3 (g)
DH = +196 kJ
DH = -790 kJ
--------------------------------------------------------------------------------------------------------------
Reversing the order of the first equation reverses the sign of
DH
50
Hess Law Example 2
From the following reactions and enthalpy changes:
2 SO2 (g) + O2 (g)  2 SO3 (g)
DH = -196 kJ
2 S (s) +3 O2 (g)  2 SO3 (g)
DH = -790 kJ
Find the enthalpy change for the following reaction:
S (s) + O2 (g)  SO2 (g)
2 SO3 (g)  2 SO2 (g) + O2 (g)
2 S (s) +3 O2 (g)  2 SO3 (g)
DH = +196 kJ
DH = -790 kJ
--------------------------------------------------------------------------------------------------------------
2 SO3(g) +2 S(s) + 2 3 O2 (g)  2 SO3 (g)+2 SO2 (g) + O2 (g)
DH = -594 kJ
2 S(s) + 2 O2 (g)  2 SO2 (g)
DH = -594 kJ
51
Hess Law: Example 2
From the following reactions and enthalpy changes:
2 SO2 (g) + O2 (g)  2 SO3 (g)
DH = -196 kJ
2 S (s) +3 O2 (g)  2 SO3 (g)
DH = -790 kJ
Find the enthalpy change for the following reaction:
S (s) + O2 (g)  SO2 (g)
2 SO3 (g)  2 SO2 (g) + O2 (g)
2 S (s) +3 O2 (g)  2 SO3 (g)
DH = +196 kJ
DH = -790 kJ
--------------------------------------------------------------------------------------------------------------
2 SO3(g) +2 S(s) + 2 3 O2 (g)  2 SO3 (g)+2 SO2 (g) + O2 (g)
DH = -594 kJ
2 S(s) + 2 O2 (g)  2 SO2 (g)
S(s) +
O2 (g) 
SO2 (g)
DH = -594 kJ
DH = -297 kJ
52
Born Haber Cycle
53
Born-Haber Cycle
When an ionic compound is formed from
its elements, the overall reaction can be
broken down into separate steps.
Most of the steps are endothermic but the
overall is exothermic due to the high lattice
enthalpy.
Born Haber Cycle for formation
of NaCl
Some Definitions
The enthalpy of atomization is the enthalpy change
that occurs when one mole of gaseous atoms is
formed from the element in the standard state under
standard conditions
Example: ½ Cl2 (g)  Cl (g)
DHoat = 121 kJ mol-1
The electron affinity is the enthalpy change that
occurs when an electron is added to an isolated atom
in the gaseous state:
O (g) + e-  O- (g)
DHo = -142 kJ mol-1
O- (g) + e-  O2- (g)
DHo = +844 kJ mol-1
The lattice enthalpy is the enthalpy change that
occurs from the conversion of an ionic compound in
the gaseous state into its gaseous ions
LiCl (g)  Li+ (g) + Cl- (g)
DHo = +846 kJ mol-1
59
Magnesium chloride
http://chemistry.tutorvista.com/inorganicchemistry/reaction-pathways-born-habercycle.html
Lattice Energy
•
Ionic compounds are usually solids. The release of
energy on forming the solid, called the lattice energy is
the driving force for the formation of ionic compounds.
•
Because of high lattice energies, ionic solids tend to be
hard and have high melting points. Ionic compounds are
insulators in the solid state, because electrons are
localized on the ions, but conduct when molten or in
solution, due to flow of ions (not electrons).
•
Lattice energies can be calculated using Hess’s law, via a
Born-Haber Cycle.
•
•
Step 1: Convert elements to atoms in the gas state:atomization and bond
energy
e.g. for Li, Li (s)  Li (g)
DH1 = DHatomization
for F, 1/2 F2 (g)  F (g)
DH2 = 1/2 (Bond Energy)
Step 2: Electron transfer to form (isolated) ions: Ionization Energy and
Electron Affinity
Li (g)  Li+ (g) + e–
DH3 = IE1
F (g) + e–  F– (g)
•
DH4 = EA1
Step 3: Ions come together to form solid: lattice enthalpy
Li+ (g) + F– (g)  LiF (s)
DH5 = Lattice Energy
•
Overall: Li (s) + 1/2 F2 (g)  LiF (s)
DH = DHf = S(DH1–5)
•
Lattice Energy = DHf – (DH1 + DH2 + DH3 + DH4)
Born Haber Cycle Diagram: Finding the Enthalpy of Formation of NaCl
The stepwise energy changes for the formation of NaCl
63
Born Haber Cycle for NaCl
The formation of NaCl can be considered as a five
step process
Na (s) + 1/2 Cl2 (g) NaCl (s)
1. The vaporization of sodium metal to form the gaseous
2.
3.
4.
5.
element.
The dissociation of chlorine gas to gaseous chlorine
atoms is equal to one half of the bond energy for a Cl-Cl
covalent bond
The ionization of gaseous sodium atoms to Na(g) Na+
The ionization of chlorine atoms. (This quantity is the
negative electron affinity for the element chlorine.)
The lattice energy on the formation of sodium chloride
from the gaseous ions
64
Born-Haber Cycle for NaCl
The stepwise energy changes for the formation of NaCl:
The vaporization of sodium metal to form the gaseous
element.
Na (s)  Na (g)
∆H°sublimation = + 109 kJ mol-1
The dissociation of chlorine gas to gaseous chlorine
atoms is equal to one half of the bond energy for a ClCl covalent bond
1/2 Cl2 (g)
 Cl (g)
∆H°diss = + 122 kJ mol-1
The ionization of gaseous sodium atoms to:
Na (g) 
Na+ (g) + e-
∆H°ionization = + 496 kJ mol-1
The ionization of chlorine atoms. (This quantity is the
negative electron affinity for the element chlorine.)
Cl (g) + e-  Cl- (g)
∆H°elect.affinity = - 368 kJ mol-1
The lattice energy on the formation of sodium chloride
from the gaseous ions
Na+ (g) + Cl- (g)  NaCl (s)
∆H°lattice = - 770 kJ mol-1
65
The lattice enthalpy is a measure of the
strength of the attraction between
ions.Depends on:
The charge of the ions
The size of the ions
The lattice enthalpy of magnesium oxide is
much larger than sodium chloride because
the ion is smaller
Periodic Trends in Lattice Energy
Coulomb’s Law
electrostatic force a
charge A X charge B
distance2
http://hyperphysics.phyastr.gsu.edu/hbase/electric/elefor.html
(since energy = force X distance)
•
So, lattice energy increases, as ionic radius decreases
(distance between charges is smaller).
•
Lattice energy also increases as charge increases.
Figure 9.7
Entropy
Entropy
Entropy is defined
as a state of
disorder or
randomness.
In general the
universe tends to
move toward
release of energy
and greater entropy.
70
Entropy
The statistical
interpretation of
thermodynamics was
pioneered by James
Clerk Maxwell (1831–
1879) and brought to
fruition by the Austrian
physicist Ludwig
Boltzmann (1844–1906).
71
Entropy
Spontaneous chemical
processes often result in a
final state is more Disordered
or Random than the original.
The Spontaneity of a
chemical process is related to
a change in randomness.
Entropy is a thermodynamic
property related to the
degree of randomness or
disorder in a system.
Reaction of potassium
metal with water. The
products are more
randomly distributed
than the reactants
72
Entropy and
Thermodynamics
According to the second law or
thermodynamics the entropy of the
universe is always increasing.
This is true because there are many more
possibilities for disorder than for order.
Entropy is Disorder
Disorder in a system can take many forms.
Each of the following represent an increase in
disorder and therefore in entropy:
1. Mixing different types of particles. i.e.
dissolving salt in water.
2. A change is state where the distance between
particles increases. Evaporation of water.
3. Increased movement of particles. Increase in
temperature.
4. Increasing numbers of particles. Ex.
2 KClO3  2 KCl + 3O2
74
Entropy States
The greatest increase in entropy is usually
found when there is an increase of
particles in the gaseous state.
The symbol for the change in disorder or
entropy is given by the symbol, DS.
The more disordered a system becomes
the more positive the value for DS will be.
Systems that become more ordered have
negative DS values.
75
Entropy, S
The entropy of a substance depends
on its state:
S (gases) > S (liquids) > S (solids)
So (J/K-1mol-1)
H2O (liquid)
69.95
H2O (gas)
188.8
76
Entropy and States of Matter
S˚(Br2 liquid) < S˚(Br2 gas)
S˚(H2O solid) < S˚(H2O liquid)
77
Entropy, Phase & Temperature
S increases
slightly with T
S increases a
large amount
with phase
changes
78
Entropy and Temperature
The Entropy of a substance increases with
temperature.
Molecular motions
of heptane, C7H16
Molecular motions of
heptane at different temps.
79
Standard Entropy Values
The standard entropy, DSo, of a substance
is the entropy change per mole that occurs
when heating a substance from 0 K to the
standard temperature of 298 K.
Unlike enthalpy, absolute entropy changes
can be measured.
Like enthalpy, entropy is a state function.
The change in entropy is the difference
between the products and the reactants
DSo = S So (products) - S So (reactants)
80
Standard Entropy Values
Some standard enthalpy values
The amount of entropy in a
pure substance depends on
the temperature, pressure,
and the number of molecules
in the substance.
Values for the entropy of
many substances at have
been measured and tabulated.
The standard entropy is also
measured at 298 K.
81
Factors That Determine
Entropy States
1.
2.
3.
4.
The greater the disorder or randomness in a system
the larger the entropy.
Some generalizations
The entropy of a substance always increases as it
changes from solid to liquid to gas and vice versa.
When a pure solid or liquid dissolves in a solvent,
the entropy of the substance increases.
When gas molecules escape from a solvent, the
entropy increases.
Entropy generally decreases with increasing
molecular complexity
Gibbs Free Energy
83
Spontaneity
A chemical reaction is spontaneous if it
results in the system moving form a less
stable to a more stable state.
Decreases in enthalpy and increases in
entropy move a system to greater stability.
The combination of the enthalpy factor and
the entropy factor can be expressed as the
Gibbs Free Energy.
84
Gibbs Free Energy
The standard free energy change is defined
by this equation
DGo = DHo – T DSo
Where
DHo = the enthalpy change
DSo = the entropy change
T = Kelvin temperature
A chemical reaction is
spontaneous if it results
in a negative free energy change.
85
Gibbs Free Energy
Possible Combinations for free energy change:
DGo = DHo – T DSo
DG
DH
DS
DH-TDS
Always
Spontaneous
< 0 (-) < 0 (-) > 0 (+)
Always (-)
Never
Spontaneous
> 0 (+) > 0 (+) < 0 (-)
Always (+)
Spontaneous at
High Temperature
< 0 (-)
> 0 (+) > 0 (+) > 0 (+)
Spontaneous at
Low Temperature
> 0 (+)
< 0 (-) < 0 (-)
< 0 (-)
(-) if T large
(+) if T small
(+) if T large
(-) if T small
86
Free Energy Problem 1
A certain chemical reaction is exothermic with a
standard enthalpy of - 400 kJ mol-1.
The entropy change for this reaction is +44 J mol-1 K1. Calculate the free energy change for this
reaction at 25 oC.
Is the reaction spontaneous?
87
Free Energy Problem 1
A certain chemical reaction is exothermic with a
standard enthalpy of - 400 kJ mol-1. The entropy
change for this reaction is +44 J mol-1 K-1.
Calculate the free energy change for this reaction
at 25 oC. Is the reaction spontaneous?
Solution
Convert the entropy value to kJ. 44 J mol-1 K-1 = 0.044
kJ mol-1 K-1
DG = - 400 kJ mol-1 – (298 K)(0.044 kJ mol-1 K-1)
DG = - 400 kJ mol-1 – 13.1 kJ mol-1
DG = - 413.1 kJ mol-1 .
Since DG is negative the
reaction is spontaneous.
Note. Because DH <0 and DS >0, this reaction is spontaneous
at all temperatures.
88
Free Energy Problem 2
A certain chemical reaction is endothermic with a
standard enthalpy of +300 kJ mol-1. The entropy
change for this reaction is +25 J mol-1 K-1.
Calculate the free energy change for this reaction
at 25 oC. Is the reaction spontaneous?
89
Free Energy Problem 2
A certain chemical reaction is endothermic with a
standard enthalpy of +300 kJ mol-1. The entropy
change for this reaction is +25 J mol-1 K-1.
Calculate the free energy change for this reaction
at 25 oC. Is the reaction spontaneous?
Solution
Convert the entropy value to kJ. 25 J mol-1 K-1 = 0.025
kJ mol-1 K-1
DG = + 300 kJ mol-1 – (298 K)(0.025 kJ mol-1 K-1)
DG = + 300 kJ mol-1 – 7.45 kJ mol-1
DG = + 292.55 kJ mol-1 .
Since DG is positive the
reaction is non-spontaneous.
Note. Because DH >0 and DS >0, this reaction is nonspontaneous at low temperatures. It the temperature were
substantially increased it would become spontaneous.
90
http://ww2.chemistry.gatech.edu/~wilkinson/Class_notes/spring_2004_
1311_page/slides/Solubility%20of%20ionic%20compounds%20and%2
0intermolecular%20forces%202%20up.pdf
Why NaCl dissolves in water?