The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. C
The first law: transformation of energy into heat and work
Chemical reactions can be used to provide heat and for
doing work.
Compare fuel value of different compounds.
What drives these reactions to proceed in the direction they
do?
Why does combustion proceed spontaneously?
Why does NaCl dissolve spontaneously in water?
Spontaneous Processes
A spontaneous process is one that occurs by itself, given
enough time, without external intervention
Spontaneous for
T > 0oC
Spontaneous for
T < 0oC
Reversible vs Irreversible Processes
Reversible processes
Are at equilibrium
Driving force is only infinitesimally greater than the opposing
force
Process occurs in a series of infinitesimal steps, and at each
step the system in at equilibrium with the surroundings
Would take an infinite amount of time to carry out
Irreversible Process
Not at equilibrium; a spontaneous process
Reversal cannot be achieved by changing some variable by
an infinitesimal amount
Which direction will an irreversible process proceed to
establish equilibrium?
What thermodynamic properties determine the direction of
spontaneity?
The change in enthalpy during a reaction is an important
factor in determining whether a reaction is favored in the
forward or reverse direction.
Are exothermic reactions more likely to be spontaneous than
endothermic?
Not necessarily - for example, melting of ice is spontaneous
above 0oC and is endothermic.
Entropy
Consider the following expansion of a gas into a vacuum.
When the valve is open, there are four possible arrangements
or STATES; all states are equal in energy
Opening the valve allows this system of two particles, more
arrangements; higher degree of DISORDER.
As the number of particles increases in the system, the
number of possible arrangements that the system can be in
increases
Processes in which the disorder of the system increases
tend to occur spontaneously.
Ice melts spontaneously at T>0oC even though it is an
endothermic process.
The molecules of water that make up the ice crystal lattice are
held rigidly in place.
When the ice melts the water molecules are free to move
around, and hence more disordered than in the solid lattice.
Melting increases the disorder of the system.
A similar situation arises in the dissolution of solid NaCl in
water.
The entropy, S, of a system quantifies the degree of disorder
or randomness in the system; larger the number of
arrangements available to the system, larger is the entropy of
the system.
If the system is more disordered, entropy is larger
Like enthalpy, entropy is a state function; ∆
S depends on the
initial and final entropies of the system
∆
S = Sfinal
- Sinitial
If ∆
S > 0 => Sfinal
> Sinitial
If ∆
S < 0 => Sfinal
< Sinitial
So melting or vaporization increases entropy; freezing or
condensation decrease entropy.
Likewise, expansion of a gas increases entropy, compression
of a gas decreases entropy.
For a molecule: Sgas
> Sliquid
> Ssolid
Comparing molecules, larger molecules tend to have higher
entropy than smaller.
Example: entropy of H2
(g) < entropy of CCl4
(g)
Entropy and Temperature
Third law of thermodynamics :
The entropy of a perfect crystalline substance at equilibrium
approaches zero as the temperature approaches absolute
zero.
∆
S = qrev
T
For an isothermal process (constant temperature),
change in entropy is defined as:
The change in entropy, ∆
S, accompanying a phase transition,
at constant pressure, can be expressed as:
∆
S =
∆
H
T
For example, the ∆
S accompanying vaporization at the
normal boiling point of the liquid (Tb
):
∆
Svap
=
∆
Hvap
Tb
Entropies of Reactions
Consider the following gas phase reaction:
2NO(g) + O2
(g) --> 2NO2
(g)
Any reaction which results in a decrease in the number of
gas phase molecules is accompanied by a decrease in
entropy.
Reactions that increase the number of gas phase molecules
tend to be accompanied by positive changes in entropy.
Dissolution reactions are typically accompanied by an
increase in entropy since the solvated ions are more
disordered than the ions in the crystal lattice.
However, there are examples where the opposite is true:
MgCl2
(s) --> Mg2+(aq) + 2Cl-(aq)
Is accompanied by a negative change in entropy.
The standard molar entropy, So, is the absolute entropy of
one mole of substance at 298.15K.
Units of molar entropy - J K-1mol-1
For the general reaction
aA + bB --> cC + dD
the standard entropy change is
∆
So = cSo (C) + dSo (D) - aSo (A) - bSo (B)
The tabulated standard molar entropies of compounds can
be used to calculate the standard entropy change
accompanying a reaction.
Note: the standard molar entropies of the most stable forms
of elements are not zero at 298.15 K.
As dictated by the 3rd law, entropy of a compound is zero at
absolute zero (0K)
Calculate the ∆
So for the reaction:
N2
(g) + 3H2
(g) --> 2NH3
(g)
∆
So = 2So (NH3
(g)) - So (N2
(g)) - 3So (H2
(g))
Using the tabulated values
∆
So = -198.3 J/K
Reaction is accompanied by a decrease in entropy - why?
Conditions for spontaneous processes
The change in enthalpy does not necessarily determine
whether a process will be spontaneous or not.
How about change in entropy?
While many spontaneous processes occur with an increase
in entropy, there are examples of spontaneous processes
that occur with the system’s entropy decreasing.
For example, below 0oC, water spontaneously freezes even
though the process is accompanied by a decrease in entropy.
What happens to the entropy of the surroundings when the
entropy of the system changes?
For example, when water freezes, the heat liberated is taken
up by the surroundings, whose entropy increases.
Define the change in entropy of the universe, ∆
Suniverse
:
∆
Suniverse
= ∆
Ssystem
+ ∆
Ssurrounding
The second law of thermodynamics states:
For a process to be spontaneous, the entropy of the universe
must increase.
∆
Suniverse
= ∆
Ssystem
+ ∆
ssurrounding
> 0 for a spontaneous process
If ∆
Suniverse
< 0 => non-spontaneous process
If ∆
Suniverse
> 0 => spontaneous process
If ∆
Suniverse
= 0 => the process is at equilibrium
Need to know ∆
Suniverse
to determine if the process will
proceed spontaneously.
The Gibbs Free Energy Function
The Gibbs free energy function, G, allows us to focus on the
the thermodynamic properties of the system.
At constant pressure and temperature:
G = H -T S
Units of G - J
G is a state function, like H and S
The change in the free energy function of the system
accompanying a process at constant P and T is
∆
Gsyst
= ∆
Hsyst
- T∆
Ssyst
What is the sign of ∆
G for a spontaneous reaction?
At constant T
∆
Ssurr
=qsurr
T
The change in heat of the system during a process, is
equal in magnitude but opposite in sign to the heat change
of the surroundings.
qsurr
= -qsyst
Hence,
∆
Ssurr
=
qsyst
T
qsurr
T
= -
At constant pressure:
∆
Ssurr
= -
∆
Hsyst
T
qsyst
= qp
= ∆
H
∆
Suniverse
= ∆
Ssyst
+ ∆
Ssurrounding
> 0 for a spontaneous process
-
∆
Hsyst
T
∆
Suniverse
= ∆
Ssyst
> 0
Mutliplying by T
T ∆
Ssyst
- ∆
Hsyst
> 0 for a spontaneous reaction
=> ∆
Hsyst
- T ∆
Ssyst
< 0 for a spontaneous reaction
=> ∆
Gsyst
< 0 for a spontaneous reaction
-
∆
Hsyst
T
∆
Suniverse
= ∆
Ssyst
> 0
For a process at constant P and T
∆
Gsyst
< 0 => spontaneous
∆
Gsyst
> 0 => non spontaneous
∆
Gsyst
= 0 => equilibrium
The sign of ∆
G is determined by the relative magnitudes of
∆
H and T∆
S accompanying the process
∆
Gsyst
= ∆
Hsyst
- T∆
Ssyst
Or simply: ∆
G = ∆
H -T∆
S
H2
O(l) --> H2
O(s)
Normal freezing point of H2
O = 273.15K
The measured change in enthalpy is the enthalpy of fusion -
enthalpy change associated when one mole of liquid water
freezes at 1atm and 273.15K
∆
H = -6007J
Entropy change at 273.15K
∆
S=
∆
H
T
=
-6007 J
273.15K
= -21.99 J/K
∆
G = ∆
H -T∆
S = -6007J - (273.15K)(-21.99J/K) = 0J
Water and ice are at equilibrium at 273.15 K and 1atm
The Gibbs Free Energy Function and Phase Transitions
As water is cooled by 10K below 273.15K to 263.15K
In calculating ∆
G at 263.15K assume that ∆
H and ∆
S do not
change
∆
G = ∆
H -T∆
S = -6007J - (263.15K)(-21.99J/K) = -220J
Since ∆
G < 0 water spontaneously freezes
At 10K above the freezing point, 283.15K
∆
G = ∆
H -T∆
S = -6007J - (283.15K)(-21.99J/K) = 219J
Since ∆
G > 0 water does not spontaneously freeze at 283.15K,
10K above the normal freezing point of water
Above 273.15K, the reverse process is spontaneous - ice
melts to liquid water.
For phase transitions, at the transition temperature, the
system is at equilibrium between the two phases.
Above or below the transition temperature, the phase is
determined by thermodynamics.
∆
G is the driving force for a phase transition.
Plot of ∆
H and T∆
S vs T for the freezing of water. At 273.15 K
the system is at equilibrium, ∆
G = 0
freezing
spontaneous
Freezing not
spontaneous;
melting
spontaneous
Standard Free-Energy Changes
The standard molar free energy function of formation, ∆
Gf
o, is
the change in the free energy for the reaction in which one
mole of pure substance, in its standard state is formed from
the most stable elements of its constituent elements, also in
their standard state.
C(s) + O
2
(g) --> CO
2
(g)
Calculate ∆
Gf
o, for this reaction.
∆
Gf
o = ∆
Hf
o - T ∆
So
∆
Hf
o(CO
2
(g)) = -393.51 kJ
∆
So = So(CO
2
(g)) - So(C(s)) - So(O
2
(g)) = 2.86 J K-1
∆
Gf
o = -394.36 kJ at 298K
Standard Free Energies of Reactions
For a reaction: aA + bB --> cC + dD
∆
Go = c ∆
Gf
o (C) + d ∆
Gf
o(D) - a ∆
Gf
o(A) -b ∆
Gf
o(B)
∆
Go = Σ ∆
Gf
o (products) - Σ ∆
Gf
o (reactants)
The standard free energy of formation of elements in their
most stable form at 298.15K has been set to zero.
∆
Go = ∆
Ho - T ∆
So
∆
Ho = Σ ∆
Hf
o (products) - Σ ∆
Hf
o (reactants)
∆
So = Σ
So (products) - Σ
So (reactants)
Problem: Calculate the standard free-energy change for the
following reaction at 298K:
N2
(g) + 3H2
(g) --> 2NH3
(g)
Given that ∆
Gf
o[NH3
(g)] = -16.66 kJ/mol.
What is the ∆
Go for the reverse reaction?
Since the reactants N2
(g) and H2
(g) are in their standard state
at 298 K their standard free energies of formation are defined
to be zero.
∆
Go = 2 ∆
Gf
o[NH3
(g)] - ∆
Gf
o[N2
(g)] - 3 ∆
Gf
o[H2
(g)]
= 2 x -16.66 = -33.32 kJ/mol
For the reverse reaction: 2NH3
(g) --> N2
(g) + 3H2
(g)
∆
Go = +33.32 kJ/mol
Problem:
C3
H8
(g) + 5O2
(g) --> 3CO2
(g) + 4H2
O(l) ∆
Ho = - 2220 kJ
a) Without using information from tables of thermodynamic
quantities predict whether ∆
Go for this reaction will be less
negative or more negative than ∆
Ho.
b) Use thermodynamic data calculate ∆
Go for this reaction.
Whether ∆
Go is more negative or less negative than ∆
Ho
depends on the sign of ∆
So since ∆
Go = ∆
Ho - T ∆
So
The reaction involves 6 moles of gaseous reactants and
produces 3 moles of gaseous product
=> ∆
So < 0 (since fewer moles of gaseous products
than reactants)
Since ∆
So < 0 and ∆
Ho < 0 and ∆
Go = ∆
Ho - T ∆
So
=> ∆
Go is less negative than ∆
Ho
b)
∆
Go = 3∆
Gf
o(CO2
(g))+4∆
Gf
o(H2
O(l)-∆
Gf
o(C3
H8
(g))-5∆
Gf
o(O2
(g))
= 3(-394.4) + 4(-237.13) - (-23.47) - 5(0)
= -2108 kJ
Note: ∆
Go is less negative than ∆
Ho
Effect of temperature on ∆
Go
Values of ∆
Go calculated using the tabulated values of ∆
Gf
o
apply only at 298.15K.
For other temperatures the relation:
∆
G= ∆
Ho - T ∆
So
can be used, as long as ∆
Ho and ∆
So do not vary much with
temperature
∆∆∆∆
H
o
,
∆∆∆∆
G
o
∆∆∆∆
S
o
T (K)
∆
G= ∆
Ho - T ∆
So
If both ∆
Ho and ∆
So are > 0, then temperatures above Teq
the reaction is spontaneous, but below Teq
reaction is non-
spontaneous.
If both ∆
Ho and ∆
So are negative, reaction is spontaneous
below Teq
.
Teq
=
∆
Ho
∆
So
At equilibrium ∆
Go = 0
Problem: The normal boiling point is the temperature at
which a pure liquid is in equilibrium with its vapor at 1atm.
a) write the chemical equation that defines the normal
boiling point of liquid CCl4
b) what is the value of ∆
Go at equilibrium?
c) Use thermodynamic data to estimate the boiling point of
CCl4
.
a) CCl4
(l) CCl4
(g)
b) At equilibrium ∆
G = 0. In any equilibrium for a normal
boiling point, both the liquid and gas are in their standard
states
Hence, ∆
Go = 0.
c) ∆
Go = ∆
Ho - T ∆
So and since for the system is at
equilibrium, ∆
Go = 0
∆
Ho - T ∆
So = 0
T =
∆
Ho
∆
So
where for this system T is the boiling point
To determine the boiling point accurately we need the values
of ∆
Ho and ∆
So for the vaporization process at the boiling
point of CCl4
However, if we assume that these values do not change
significantly with temperature we can use the values of ∆
Ho
and ∆
So at 298K from thermodynamic tables.
∆
Ho = (1mol)(-106.7 kJ/mol) - (1mol)(-139.3kJ/mol) = 32.6 kJ
∆
So = (1mol)(309.4 J/mol-K) -(1mol)(214.4 J/mol-K) = 95.0 J/K
T
b
~
∆
Ho
∆
So
= 32.6 kJ
0.095 kJ/K
= 343 K
(normal boiling point of CCl
4
is 338 K)
The Gibbs Function and the Equilibrium Constant
For any chemical reaction, the free-energy change under
non standard conditions, ∆
G, is
∆
G = ∆
Go + R T ln Q
Q - reaction quotient
aA + bB -> cC + dD
Q =
p
C
c
p
D
d
p
A
A
p
B
b
Under standard conditions, reactants and products are in
their standard states => Q = 1; hence lnQ = 0 and ∆
G = ∆
Go
Q =
[C]
c
[D]
d
[A]
a
[B]
b
At equilibrium Q = K => ∆
G = 0
∆
Go = - R TlnK or K = e- ∆
G
o
/RT
∆
G = R Tln Q
K
If Q < K => ∆
G < 0
If Q > K => ∆
G > 0
If Q = K => ∆
G = 0
∆
G = ∆
Go + R T ln Q = - RT ln K + RT ln Q
∆
G can be considered to be the slope of the line at any
point along the curve. At equilibrium when Q=K, the slope
= 0 and hence ∆
G = 0
∆
Gforward
< 0
∆
Gforward
> 0
Temperature Dependence of K
ln K = - ∆
G
R T
= - ∆
Ho
∆
So
R T
R
+
If the equilibrium constant of a reaction is known at one
temperature and the value of ∆
Ho is known, then the
equilibrium constant at another temperature can be
calculated
ln K1
= - ∆
Ho
∆
So
R T1
R
+
ln K2
= - ∆
Ho
∆
So
R T2
R
+
ln K2
- ln K1
= ln
K1
= ∆
Ho
R
1
T2
1
T1
(
)
-
K2
For the equilibrium between a pure liquid and its vapor,
the equilibrium constant is the equilibrium vapor
pressure
Liquid
Gas K = Pgas
Hence
ln
p2
=
∆
Hvap
R
1
T2
1
T1
(
)
-
p1
Clausius-Clapeyron
equation
The Clausius-Clapeyron equation indicates the variation of
vapor pressure of a liquid with temperature
Driving Non-spontaneous Reactions
Because so many chemical and biological reactions are
carried out under conditions of constant pressure and
temperature, the magnitude of ∆
G is a useful tool for
evaluating reactions.
A reaction for which ∆
G is large and negative (like the
combustion of gasoline), is much more capable of doing
work on the surroundings than a reaction for which ∆
G is
small and negative (like the melting of ice).
The change in free energy for a process equals the
maximum amount of work that can be done by the system
on its surroundings in a spontaneous process at constant
pressure and temperature
wmax
= ∆
G
For non-spontaneous reactions (∆
G > 0), the magnitude of
∆
G is a measure of the minimum amount of work that must
be done on the system to cause the process to occur.
Many chemical reactions are non-spontaneous.
For example: Cu can be extracted from the mineral
chalcolite which contains Cu2
S.
The decomposition of Cu2
S is non-spontaneous
Cu2
S --> 2Cu(s) + S(s)
∆
Go = +86.2 kJ
Work needs to be done on this reaction, and this is done by
coupling this non-spontaneous reaction with a spontaneous
reaction so that the overall reaction is spontaneous.
Consider the following reaction:
S(s) + O2
(g) --> SO2
(g) ∆
Go = -300.4kJ
Coupling this reaction with the extraction of Cu from Cu2
S
Cu2
S --> 2Cu(s) + S(s)
∆
Go = +86.2 kJ
S(s) + O2
(g) --> SO2
(g)
∆
Go = -300.4kJ
Net: Cu2
S(s) + O2
(g) --> 2Cu(s) + SO2
(g) ∆
Go = -214.2kJ
Biological systems employ the same principle by using
spontaneous reactions to drive non-spontaneous reactions.
The metabolism of food is the usual source of free energy
needed to do the work to maintain biological systems.
C6
H12
O6
(s) + 6O2
(g) --> 6CO2
(g) + 6H2
O(l)
∆
Ho = -2803 kJ
The free energy released by the metabolism of glucose is
used to convert lower-energy ADP (adenine diphosphate) to
higher energy ATP (adenine triphosphate).
C6
H12
O6
6CO2
(g) + 6H2
O(l)
ADP
ATP
Free energy
released by
oxidation of glucose
converts ADP to
ATP
Free energy released
by ATP converts
simple molecules to
more complex
molecules
