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Lecture 409.317 2013.F N07(Before Class Version)
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Transcript of Lecture 409.317 2013.F N07(Before Class Version)
Physical Chemistry
for Energy Engineering
(7th: 2013/10/8)
Takuji Oda
Assistant Professor, Department of Nuclear Engineering
Seoul National University
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
We consider a system consisting of two phases of a pure substance
in equilibrium each other. (liquid and gas, for example here)
The Gibbs energy of this system is given by
๐บ = ๐บ๐ + ๐บ๐
where ๐บ๐ and ๐บ๐ are the Gibbs energies of the liquid and the gas phase.
Now, suppose ๐๐ mole are transferred from the liquid to the solid phase,
where T and P are kept constant. The infinitesimal change in Gibbs energy
for this process is:
๐๐บ =๐๐บ๐
๐๐๐ ๐,๐๐๐๐ +
๐๐บ๐
๐๐๐ ๐,๐๐๐๐
As ๐๐๐ = โ๐๐๐, then
๐๐บ =๐๐บ๐
๐๐๐ ๐,๐โ
๐๐บ๐
๐๐๐ ๐,๐๐๐๐
Here, we define chemical potentials, ๐๐ =๐๐บ๐
๐๐๐ ๐,๐and ๐๐ =
๐๐บ๐
๐๐๐ ๐,๐, then
๐๐บ = ๐๐ โ ๐๐ ๐๐๐ (constant T and P)
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
We consider a system consisting of two phases of a pure substance
in equilibrium each other. (liquid and gas, for example here)
๐๐บ = ๐๐ โ ๐๐ ๐๐๐ (constant T and P)
where chemical potentials are ๐๐ =๐๐บ๐
๐๐๐ ๐,๐and ๐๐ =
๐๐บ๐
๐๐๐ ๐,๐.
If the two phases are in equilibrium with each other, then ๐๐บ = 0.
And because we suppose some amount of the liquid phase is
transferred to the gas phase (๐๐๐) here, to make ๐๐บ = 0 in this
condition, ๐๐ = ๐๐ is needed.
If ๐๐ < ๐๐, [๐๐ โ ๐๐] < 0. Thus, the process of ๐๐๐ > 0 (transfer
from the liquid phase to the gas phase) spontaneously takes place
as it holds ๐๐บ < 0.
Likewise, if ๐๐ > ๐๐, ๐๐๐ < 0.
Hence, in general, the transfer occurs from the phase with higher
chemical potential to the phase with lower chemical potential.
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
In general, the chemical potential is defined as:
ฮผ =๐๐บ
๐๐ ๐,๐
Because ๐บ is proportional to the size/amount of a system ๐บ โ ๐,
namely an extensive thermodynamic function, we can express it as:
๐บ = ๐๐ ๐, ๐ Here, this equation is consistent with the definition of chemical
potential:
ฮผ =๐๐บ
๐๐ ๐,๐=
๐ ๐๐ ๐,๐
๐๐ ๐,๐= ๐ ๐, ๐
which means ๐ ๐, ๐ , the chemical potential, is the same quantity as
the molar Gibbs energy and it is an intensive quantity.
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
As they are in equilibrium,
ฮผ๐ผ(๐, ๐) = ฮผ๐ฝ(๐, ๐)Now take the total derivatives of both sides
๐ฮผ๐ผ
๐๐ ๐๐๐ +
๐ฮผ๐ผ
๐๐ ๐๐๐ =
๐ฮผ๐ฝ
๐๐ ๐๐๐ +
๐ฮผ๐ฝ
๐๐ ๐๐๐
Since ๐ is simply the molar Gibbs energy for a single substance, utilizing ๐๐บ
๐๐ ๐= ๐ and
๐๐บ
๐๐ ๐= โ๐ (*these were previously derived along Maxwell
relations)๐ฮผ
๐๐ ๐=
๐ ๐บ
๐๐ ๐= ๐ and
๐ฮผ
๐๐ ๐=
๐ ๐บ
๐๐ ๐= โ ๐
where ๐ and ๐ are the molar volume and the molar entropy. Then, ๐๐ผ๐๐ โ ๐๐ผ๐๐ = ๐๐ฝ๐๐ โ ๐๐ฝ๐๐
Since we consider the two phases are in equilibrium each other๐๐
๐๐=
๐๐ฝโ ๐๐ผ
๐๐ฝโ ๐๐ผ=โ๐ก๐๐ ๐
โ๐ก๐๐ ๐=
โ๐ก๐๐ ๐ป ๐
โ๐ก๐๐ ๐=
โ๐ก๐๐ ๐ป
๐โ๐ก๐๐ ๐
We consider two phases (ฮฑ and ฮฒ) are in equilibrium each other.
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
In the previous slide, we obtain for two phases in equilibrium each other:
๐๐
๐๐=โ๐ก๐๐ ๐ป
๐โ๐ก๐๐ ๐
This equation is called the Clapeyron equation, and relates โthe slope
of the two-phase boundary line in a phase diagram with the values of
โ๐ก๐๐ ๐ป and โ๐ก๐๐ ๐ for a transition between these two phasesโ.
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
Exampe-1) Solid-liquid coexistence curve at around 1 atm for Benzene
โ๐๐ข๐ ๐ป = 9.95 kJ mol-1 and โ๐๐ข๐ ๐ = 10.3 cm3 mol-1 at the normal
melting point (278.7 K).
Thus, ๐๐ ๐๐ at the normal melting point of benzene is:๐๐
๐๐=
โ๐ก๐๐ ๐ป
๐โ๐ก๐๐ ๐=
9.95 ๐๐ฝ ๐๐๐โ1
278.7 ๐พ (10.3 ๐๐3๐๐๐โ1)= 34.2 ๐๐ก๐ ๐พโ1
Here, by taking the reciprocal of this result:๐๐
๐๐= 0.0292 ๐พ ๐๐ก๐โ1
Using the above result (0.0292 K atm-1) and
assuming โ๐๐ข๐ ๐ป and โ๐๐ข๐ ๐ are independent
of pressure, we predict the melting point as:
308 K at 1000 atom (experimental value
is 306 K)
570 K at 10000 atom (exp. value ~ 460 K)
*as clearly in the left figure, โ๐๐ข๐ ๐ป and
โ๐๐ข๐ ๐ are not independent of pressure at
high-pressure region
II-2. Phase equilibria- $23-3: The chemical potentials of a pure substance in two phases in equilibrium
are equal -
Exampe-2) Solid-liquid coexistence curve at around 1 atm for water
โ๐๐ข๐ ๐ป = 6.01 kJ mol-1 and โ๐๐ข๐ ๐ = -1.63 cm3 mol-1 at the normal
melting point (273.15 K).
Thus, ๐๐ ๐๐ at the normal melting point of water is:๐๐
๐๐=๐โ๐ก๐๐ ๐
โ๐ก๐๐ ๐ป=
273.15 ๐พ (โ1.63 ๐๐3๐๐๐โ1)
6.01 ๐๐ฝ ๐๐๐โ1= โ0.00751 ๐พ ๐๐ก๐โ1
As already mentioned, the melting point of
ice decreases with increasing pressure.
Hence, the solid-liquid coexistence curve in
P-T phase diagram has a negative slope.
This clearly comes from the fact that
โ๐๐ข๐ ๐ < 0 (the molar volume is larger in
solid than that in liquid at around melting
point, 1atm), because โ๐๐ข๐ ๐ป = ๐โ๐๐ข๐ ๐ must
be larger than 0 as โ๐๐ข๐ ๐ must be positive
(liquid is more disordered than solid)
II-2. Phase equilibria- $23-4: The Clausius-Clapeyron equation gives the vapor pressure of a substance
as a function of temperature -
The assumption we made for benzene two slides ago that โโ๐๐ข๐ ๐ป and
โ๐๐ข๐ ๐ are independent of pressureโ is indeed satisfactory for solid-liquid
and solid-solid transitions over a small โ ๐.
However, it is not satisfactory for liquid-gas and solid-gas transitions
because the molar volume of a gas varies strongly with pressure.
But, if the temperature is not too near the critical point, ๐๐
๐๐=
โ๐ก๐๐ ๐ป
๐โ๐ก๐๐ ๐can
be cast into a useful form for condensed phase-gas phase transitions.
II-2. Phase equilibria- $23-4: The Clausius-Clapeyron equation gives the vapor pressure of a substance
as a function of temperature -
We consider the Clapeyron equation in liquid-gas equilibrium.
First, we rewrite the equation:๐๐
๐๐=
โ๐ก๐๐ ๐ป
๐(โ๐ก๐๐ ๐)=
โ๐ฃ๐๐ ๐ป
๐( ๐๐โ ๐๐)
Here, if (1) the system is not too near the critical point (thus, ๐๐ โซ ๐๐ and ๐๐
in the denominator is negligible) and (2) the pressure is not too high (thus, we
can assume the vapor is an ideal gas and ๐๐ = ๐ ๐ ๐) ๐๐
๐๐๐=๐ ln ๐
๐๐=
โ๐ฃ๐๐ ๐ป
๐๐( ๐๐โ ๐๐)~โ๐ฃ๐๐ ๐ป
๐๐( ๐๐)~
โ๐ฃ๐๐ ๐ป
๐๐( ๐ ๐ ๐)=โ๐ฃ๐๐ ๐ป
๐ ๐2
which is know as Clausius-Clapeyron equation (derived by Clausius in 1850).
If โ๐ฃ๐๐ ๐ป does not vary with temperature over integration limits of T, then
ln๐2
๐1= โ
โ๐ฃ๐๐ ๐ป
๐
1
๐2โ1
๐1=โ๐ฃ๐๐ ๐ป
๐
๐2โ๐1
๐1๐2
which can be used to calculate the vapor pressure at some temperature
given the molar enthalpy of vaporization and the vapor pressure at some
other temperature.
II-2. Phase equilibria- $23-4: The Clausius-Clapeyron equation gives the vapor pressure of a substance
as a function of temperature -
Example-1) Vapor pressure of benzene at 373.2 K
We obtain an integral form of Clausius-Clapeyron equation as
ln๐2
๐1=โ๐ฃ๐๐ ๐ป
๐
๐2โ๐1
๐1๐2
Note that we assume โ๐ฃ๐๐ ๐ป does not (largely) vary with temperature.
โ๐ฃ๐๐ ๐ป = 30.8 kJ mol-1 for benzene at the normal boiling point (353.2 K),
and assume it is not so dependent on temperature.
The vapor pressure of benzene is 760 torr = 1 atm at 353.2 K (as the
definition of boiling point).
Then, using the above equation:
ln๐
760=30.8ร 103
8.31
373.2โ353.2
373.2ร353.2
This gives us P=1330 torr, close to the experimental value,1360 torr.
II-2. Phase equilibria- $23-4: The Clausius-Clapeyron equation gives the vapor pressure of a substance
as a function of temperature -
If we take an integral of Clausius-Clapeyron equation indefinitely rather
than between definite limits, we obtain (still assuming โ๐ฃ๐๐ ๐ป is constant) ๐๐
๐๐๐=๐ ln ๐
๐๐=โ๐ฃ๐๐ ๐ป
๐ ๐2โ ln ๐ = โ
โ๐ฃ๐๐ ๐ป
๐ ๐+ constant
This means that a plot of the logarithm of the vapor pressure against the
reciprocal of temperature is a straight line with a slope of โโ๐ฃ๐๐ ๐ป ๐ .
The slope of line for
benzene (left figure) over
313-353 K gives us
โ๐ฃ๐๐ ๐ป = 32.3 ๐๐ฝ ๐๐๐โ1,
close to 30.8 kJ mol-1
which is the experimental
value at the normal boiling
point (353 K).
<Benzene case>
II-2. Phase equilibria- $23-4: The Clausius-Clapeyron equation gives the vapor pressure of a substance
as a function of temperature -
If we consider the temperature dependence of โ๐ฃ๐๐ ๐ป, for example, as
โ๐ฃ๐๐ ๐ป = ๐ด + ๐ต๐ + ๐ถ๐2 +โฏ
where A, B, C are constants.
Then, Clausius-Clapeyron equation, ๐๐
๐๐๐=๐ ln ๐
๐๐=โ๐ฃ๐๐ ๐ป
๐ ๐2, gives us
ln ๐ = โ๐ด
๐ ๐+๐ต
๐ ln ๐ +
๐ถ
๐ ๐ + ๐ + ๐ ๐2
This equation works a wider range accurately, more than the equation that
we obtain assuming โ๐ฃ๐๐ ๐ป is constant.
For example, for the vapor pressure of solid ammonia over 146-195 K is:
ln ๐ ๐ก๐๐๐ = โ4124.4 ๐พ
๐+ 1.8163 ln
๐
๐พ+ 34.4834
II-2. Phase equilibria- $23-4: The Clausius-Clapeyron equation gives the vapor pressure of a substance
as a function of temperature -
The Clausius-Clapeyron equation also proves that the slope of the solid-gas
coexistence curve must be greater than the slope of the liquid-gas coexistence
curve near the triple pint (where these curves meet), as follows.
The Clausius-Clapeyron eq. gives the solid-gas and liquid-gas curve slopes as๐๐๐
๐๐= ๐๐
โ๐ ๐ข๐ ๐ป
๐ ๐2and
๐๐๐
๐๐= ๐๐
โ๐ฃ๐๐ ๐ป
๐ ๐2
here ๐๐ and ๐๐ are the vapor pressure of the solid and the liquid, respectively.
Because ๐๐ = ๐๐ at the triple point, ๐๐๐ ๐๐
๐๐๐ ๐๐=โ๐ ๐ข๐ ๐ป
โ๐ฃ๐๐ ๐ป
Since enthalpy is a state function, the transition between โsโgโ and โโsโlโgโ
having the same initial and final states should hold the same โ๐ก๐๐ ๐ป. Thus,
โ๐ ๐ข๐ ๐ป = โ๐๐ข๐ ๐ป + โ๐ฃ๐๐ ๐ป ๐๐๐ ๐๐
๐๐๐ ๐๐=โ๐ ๐ข๐ ๐ป
โ๐ฃ๐๐ ๐ป= 1 +
โ๐๐ข๐ ๐ป
โ๐ฃ๐๐ ๐ป
which indicates the slope of the solid-gas curve is greater than that of the
liquid-gas curve at the triple point.
*the subscription โsubโ
stand for sublimation.
II-3. Chemical Equilibrium- $26: Introduction -
Thermodynamics enables us to predict the equilibrium pressures or
concentrations of reaction mixtures.
In this chapter, we will derive a relation between the standard Gibbs
energy change and the equilibrium constant for a chemical reaction.
We will also learn how to predict the direction in which a chemical reaction
will proceed if we start with arbitrary concentrations (thus, not equilibrium)
of reactants and products.
II-3. Chemical Equilibrium- $26-1: Chemical equilibrium results when the Gibbs energy is a minimum with
respect to the extent of reaction -
Consider a general gas phase reaction, described by a balanced equation.
๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
The amount of species ๐ is ๐๐ [mol]. The Gibbs energy for this multi-component
system is a function of ๐, ๐, ๐๐ด, ๐๐ต , ๐๐ ๐๐๐ ๐๐, then the total derivative is:
dG =๐๐บ
๐๐๐,๐๐ด,๐๐ต,๐๐,๐๐
๐๐ +๐๐บ
๐๐๐,๐๐ด,๐๐ต,๐๐,๐๐
๐๐ +๐๐บ
๐๐๐ด ๐,๐,๐๐ต,๐๐,๐๐
๐๐๐ด
+๐๐บ
๐๐๐ต ๐,๐,๐๐ด,๐๐,๐๐
๐๐๐ต +๐๐บ
๐๐๐ ๐,๐,๐๐ด,๐๐ต,๐๐
๐๐๐ +๐๐บ
๐๐๐ ๐,๐,๐๐ด,๐๐ต,๐๐
๐๐๐
Using some equations derived around the Maxwell relation derivation (of ๐บ):
๐๐บ = โ๐๐๐ + ๐๐๐ + ๐๐ด๐๐๐ด + ๐๐ต๐๐๐ต + ๐๐๐๐๐ + ๐๐๐๐๐
๐๐ด =๐๐บ
๐๐๐ด ๐,๐,๐๐ต,๐๐,๐๐
, ๐๐ก๐
If the reaction takes place in constant T and P,
dG = ๐๐ด๐๐๐ด + ๐๐ต๐๐๐ต + ๐๐๐๐๐ + ๐๐๐๐๐ (constant T and P)
II-3. Chemical Equilibrium- $26-1: Chemical equilibrium results when the Gibbs energy is a minimum with
respect to the extent of reaction -
Consider a general gas phase reaction, described by a balanced equation.
๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
We define a quantity ๐, called as the โextent of reactionโ. Here ๐๐0 is the
initial number of moles for species ๐, then :
๐๐ด = ๐๐ด0 โ ๐๐ด๐ ๐๐ต = ๐๐ต0 โ ๐๐ต๐
๐๐ = ๐๐0 โ ๐๐๐ ๐๐ = ๐๐0 โ ๐๐๐
(reactants)
(products)
In this case, ๐ has units of moles. Then, the variations of ๐๐ is:
๐๐๐ด = โ๐๐ด๐๐ ๐๐๐ต = ๐๐ต๐๐
๐๐๐ = โ๐๐๐๐ ๐๐๐ = ๐๐๐๐
(reactants)
(products)
which means that as the reaction (left to right) proceeds, the reactants
decrease and the products increase according to the stoichiometry.
Using these equations:
dG = ๐๐ด๐๐๐ด + ๐๐ต๐๐๐ต + ๐๐๐๐๐ + ๐๐๐๐๐= โ๐๐ด๐๐ด โ ๐๐ต๐๐ต + ๐๐๐๐ + ๐๐๐๐ ๐๐ (constant T and P)
II-3. Chemical Equilibrium- $26-1: Chemical equilibrium results when the Gibbs energy is a minimum with
respect to the extent of reaction -
Consider a general gas phase reaction, described by a balanced equation.
๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
dG = โ๐๐ด๐๐ด โ ๐๐ต๐๐ต + ๐๐๐๐ + ๐๐๐๐ ๐๐ (constant T and P)
๐๐บ
๐๐๐,๐
= ๐๐๐๐ + ๐๐๐๐ โ ๐๐ด๐๐ด โ ๐๐ต๐๐ต
Here, we define ๐๐บ
๐๐ ๐,๐= โ๐๐บ, which is the change in Gibbs energy
when the extent of reaction changes by one mole, and its unit is ๐ฝ ๐๐๐โ1.
Assuming each species behaves as ideal gas, as the pressure dependence
of chemical potential is written as ๐๐ ๐, ๐ = ๐ยฐ๐ ๐ + ๐ ๐ ln ๐๐ ๐ยฐ , then:
โ๐๐บ = โ๐๐บยฐ + ๐ ๐ ln๐
โ๐๐บยฐ = ๐๐๐ยฐ๐(๐) + ๐๐๐ยฐ๐(๐) โ ๐๐ด๐ยฐ๐ด(๐) โ ๐๐ต๐ยฐ๐ต(๐)
๐ = ๐๐ ๐ยฐ
๐๐ ๐๐ ๐ยฐ๐๐
๐๐ด ๐ยฐ๐๐ด ๐๐ต ๐ยฐ
๐๐ต
๐ยฐ is the pressure of standard
state (1 bar) and ๐๐ด is the
partial pressure of species A.
II-3. Chemical Equilibrium- $26-1: Chemical equilibrium results when the Gibbs energy is a minimum with
respect to the extent of reaction -
Consider a general gas phase reaction, described by a balanced equation.
๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
(ideal gas, constant T and P)โ๐๐บ = โ๐๐บยฐ + ๐ ๐ ln๐
โ๐๐บยฐ = ๐๐๐ยฐ๐ ๐ + ๐๐๐ยฐ๐ ๐โ๐๐ด๐ยฐ๐ด(๐) โ ๐๐ต๐ยฐ๐ต(๐)
๐ = ๐๐ ๐ยฐ
๐๐ ๐๐ ๐ยฐ๐๐
๐๐ด ๐ยฐ๐๐ด ๐๐ต ๐ยฐ
๐๐ต
Here, the quantity โ๐๐บยฐ is the change in standard Gibbs energy for the
reaction between unmixed reactants to form unmixed products. All
species in their standard states at ๐ and ๐ยฐ. Note that ๐ยฐ = 1 bar.
When the reaction system is equilibrium, the Gibbs energy must
minimum with respect to any change from the equilibrium state, thus ๐๐บ
๐๐ ๐,๐= โ๐๐บ = โ๐๐บยฐ + ๐ ๐ ln๐๐๐ = 0 at an equilibrium state. Thus:
โ๐๐บยฐ = โ๐ ๐ ln ๐๐๐๐๐๐
๐๐
๐๐ด๐๐ด๐๐ต
๐๐ต๐๐
= โ๐ ๐ ln๐พ๐(๐)
II-3. Chemical Equilibrium- $26-1: Chemical equilibrium results when the Gibbs energy is a minimum with
respect to the extent of reaction -
Consider a general gas phase reaction, described by a balanced equation.
๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
โ๐๐บยฐ = โ๐ ๐ ln๐พ๐(๐)
๐พ๐ ๐ = ๐๐๐ = ๐๐ ๐ยฐ
๐๐ ๐๐ ๐ยฐ๐๐
๐๐ด ๐ยฐ๐๐ด ๐๐ต ๐ยฐ
๐๐ต๐๐
= ๐๐๐๐๐๐
๐๐
๐๐ด๐๐ด๐๐ต
๐๐ต๐๐
*the subscript eq emphasizes that the
partial pressures are in an equilibrium.
๐พ๐ ๐ is called as equilibrium constant. Be sure that ๐พ๐ ๐ has no unit.
As seen in the definition, this constant is defined after the target equation
is given. For example, if the ๐๐ด in the equation is changed (even keeping
the same meaning of reaction, like 2๐๐ดA(g) + 2๐๐ตB(g) โ 2๐๐Y(g) +
2๐๐Z(g) ), ๐พ๐ ๐ value is changed.
๐ยฐ = 1 ๐๐๐
II-3. Chemical Equilibrium- $26-1: Chemical equilibrium results when the Gibbs energy is a minimum with
respect to the extent of reaction -
(Example-1a) For reaction โ3 H2(g) + N2(g) โ 2 NH3(g)โ, the
equilibrium pressures are given as ๐๐ป2, ๐๐2, and ๐๐๐ป3 .
๐พ๐, ๐ = ๐๐๐ป32
๐๐ป23 ๐๐2 ๐๐
*Be sure that these pressures are
pressures at equilibrium, as in the
definition of equilibrium constant.
(Example-1b) For reaction โ3/2 H2(g) + ยฝ N2(g) โ NH3(g)โ, the
equilibrium pressures are given as ๐๐ป2, ๐๐2, and ๐๐๐ป3 .
๐พ๐ ๐ = ๐๐๐ป3
๐๐ป23/2๐๐21/2
๐๐
โ ๐พ๐ ๐ ๐๐ ๐๐ฅ๐๐๐๐๐ โ 1๐
Although the reactions themselves are identical, the equilibrium
constants are not the same, because the equilibrium constant
depends on the expression of chemical reaction equation.
II-3. Chemical Equilibrium- $26-2: An equilibrium constant is a function of temperature only -
Consider a general gas phase reaction, described by a balanced equation.
PCl5(g) โ PCl3(g) + Cl2(g)
The equilibrium-constant expression for this reaction is:
๐พ๐, ๐ = ๐๐๐ถ๐3๐๐ถ๐2
๐๐๐ถ๐5 ๐๐
Suppose we have 1 mol of PCl5 (g) and no PCl3 or Cl2 at the beginning.
When the reaction occurs to an extent ๐, PCl5: 1 mol โ (1- ๐) mol
PCl3: 0 mol โ ๐ mol, Cl2: 0 mol โ ๐ mol
Total: 1 mol โ (1+ ๐) mol
If ๐๐๐ is the extent of reaction at equilibrium, then the partial pressures are:
๐๐๐ถ๐3 = ๐๐ถ๐2 = ๐๐๐๐ 1 + ๐๐๐ , ๐๐๐ถ๐5 = 1 โ ๐๐๐ ๐ 1 + ๐๐๐
where ๐ is the total pressure. Then, the equilibrium constant is:
๐พ๐, ๐ = ๐๐๐
2
1 โ ๐๐๐2 ๐
II-3. Chemical Equilibrium- $26-2: An equilibrium constant is a function of temperature only -
Consider a general gas phase reaction, described by a balanced equation.
PCl5(g) โ PCl3(g) + Cl2(g)
๐พ๐ ๐ = ๐๐๐
2
1 โ ๐๐๐2 ๐
PCl5: 1 mol โ (1- ๐) mol
PCl3 , Cl2: 0 mol โ ๐ mol,
Total: 1 mol โ (1+ ๐) mol
๐พ๐ ๐ only depends on ๐, but
not ๐. So, if ๐ (total pressure)
is changed, ๐๐๐ must be
changed so that ๐พ๐ ๐ is kept
constant. For example, ๐พ๐ ๐of this reaction is 5.4 (no unit)
at 200ยบC.
II-3. Chemical Equilibrium- $26-2: An equilibrium constant is a function of temperature only -
Consider a general gas phase reaction, described by a balanced equation.
๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
So far, we express the equilibrium constant regarding pressure. We can
also express the equilibrium constant in terms of concentrations, etc, by
using the ideal-gas relation โ๐ = ๐๐ ๐โ, where ๐ = ๐ ๐ is the concentration:
โ๐๐บยฐ = โ๐ ๐ ln๐พ๐(๐)
๐พ๐ ๐ = ๐๐๐ = ๐๐ ๐ยฐ
๐๐ ๐๐ ๐ยฐ๐๐
๐๐ด ๐ยฐ๐๐ด ๐๐ต ๐ยฐ
๐๐ต๐๐
=๐ถ๐๐๐๐ถ๐
๐๐
๐ถ๐ด๐๐ด๐ถ๐ต
๐๐ต๐๐
๐ ๐
๐ยฐ
๐๐+๐๐โ๐๐ดโ๐๐ต
*As the same with ๐พ๐,
๐พ๐ has also no unit.
Here, we consider some standard concentration ๐ยฐ (like ๐ยฐ ), often taken to
be โ1 mol L-1โ. Then:
๐พ๐ ๐ = ๐พ๐ถ ๐๐ยฐ๐ ๐
๐ยฐ
๐๐+๐๐โ๐๐ดโ๐๐ต
๐พ๐ ๐ = ๐๐ ๐ยฐ
๐๐ ๐๐ ๐ยฐ๐๐
๐๐ด ๐ยฐ๐๐ด ๐๐ต ๐ยฐ
๐๐ต๐๐
II-3. Chemical Equilibrium- $26-2: An equilibrium constant is a function of temperature only -
(Example-2) For reaction โNH3(g) โ 3/2 H2(g) +1/2 N2(g)โ, ๐พ๐ ๐ =1.36 ร 10โ3 at 298.15 K. Determine the corresponding ๐พ๐ถ ๐ .
๐พ๐ ๐ = ๐พ๐ถ ๐๐ยฐ๐ ๐
๐ยฐ
3/2+1/2โ1
= ๐พ๐ถ ๐๐ยฐ๐ ๐
๐ยฐ
1
๐พ๐ถ ๐ = ๐พ๐ ๐๐ยฐ๐ ๐
๐ยฐ
โ1
= 1.36 ร 10โ3 ร1 ๐๐๐ ๐ฟโ1 ร 0.0831 ๐ฟ ๐๐๐ ๐๐๐โ1๐พโ1 ร 298.15 ๐พ
1 ๐๐๐
โ1
= 5.49ร 10โ5
II-3. Chemical Equilibrium- $26-3: Standard Gibbs energies of formation can be used to calculate
equilibrium constants -
โ๐๐บยฐ = ๐๐๐ยฐ๐ ๐ + ๐๐๐ยฐ๐ ๐ โ ๐๐ด๐ยฐ๐ด(๐) โ ๐๐ต๐ยฐ๐ต(๐)
โ๐๐บยฐ = โ๐ ๐ ln๐พ๐(๐)
As already derived, ๐พ๐ is related to the difference between the standard
chemical potentials of the products and the reactants.
Because a chemical potential is an energy (it is the molar Gibbs energy of
a pure substance), we need to define a โzeroโ value.
As in the same manner with โstandard molar enthalpy of formationโ, we
can define โthe standard molar Gibbs energy of formationโ according to
โ๐๐บยฐ = โ๐๐ปยฐ โ ๐โ๐๐ยฐreferring to standard molar entropies.
So, for โ๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)โ, for example, we have
โ๐๐บยฐ = ๐๐โ๐๐บยฐ ๐ + ๐๐โ๐๐บยฐ ๐ โ ๐๐ดโ๐๐บยฐ ๐ด โ ๐๐ตโ๐๐บยฐ[๐ต]
where โ๐๐บยฐ ๐ is the standard molar Gibbs energy of formation for
substance ๐.
II-3. Chemical Equilibrium- $26-3: Standard Gibbs energies of formation can be used to calculate
equilibrium constants -
โ๐๐บยฐ = ๐๐๐ยฐ๐ ๐ + ๐๐๐ยฐ๐ ๐ โ ๐๐ด๐ยฐ๐ด(๐) โ ๐๐ต๐ยฐ๐ต(๐)
โ๐๐บยฐ = โ๐ ๐ ln๐พ๐(๐)
As already derived, ๐พ๐ is related to the difference between the standard
chemical potentials of the products and the reactants.
Because a chemical potential is an energy (it is the molar Gibbs energy of
a pure substance), we need to define a โzeroโ value.
As in the same manner with โstandard molar enthalpy of formationโ, we
can define โthe standard molar Gibbs energy of formationโ according to
โ๐๐บยฐ = โ๐๐ปยฐ โ ๐โ๐๐ยฐreferring to standard molar entropies.
So, for โ๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)โ, for example, we have
โ๐๐บยฐ = ๐๐โ๐๐บยฐ ๐ + ๐๐โ๐๐บยฐ ๐ โ ๐๐ดโ๐๐บยฐ ๐ด โ ๐๐ตโ๐๐บยฐ[๐ต]
where โ๐๐บยฐ ๐ is the standard molar Gibbs energy of formation for
substance ๐.
II-3. Chemical Equilibrium- Appendix) how to calculate โ๐๐ฎยฐ -
We can find thermodynamic database, usually the one at 298.15 K (but not
necessarily and not limited to):
โ๐๐ปยฐ: Standard molar enthalpy (heat) of formation (kJ mol-1)
โ๐๐บยฐ: Standard molar Gibbs energy of formation (kJ mol-1)
๐ยฐ: Standard molar entropy at 298.15 K in (J mol-1 K-1)
๐ถ๐: Molar heat capacity at constant pressure (J mol-1 K-1)
*As definition, the value is for 1 bar.
If the Gibbs energy is not available, you can calculate from enthalpy
and entropy. For example, for H2O(g) formation: "๐ป2 +1
2๐2 โ ๐ป2๐โ
โ๐๐บยฐ T K ๐ป2๐ g
= โ๐๐ปยฐ T K ๐ป2๐ g โ โ๐๐ปยฐ T K ๐ป2 g โ1
2โ๐๐ปยฐ T K ๐2 g
โ T ๐ยฐ T K ๐ป2๐ g โ ๐ยฐ T K ๐ป2 g โ1
2๐ยฐ T K ๐2 g
By this way, as the same with the standard molar enthalpy of formation,
pure elemental substances that appear as the equilibrium phase at the
temperature have โ๐๐บยฐ T K = 0, and the standard molar Gibbs energies
of other chemicals are aligned to them.
II-3. Chemical Equilibrium- Appendix) how to calculate โ๐๐ฎยฐ -
In addition, temperature dependence of โ๐๐บยฐ T K can be evaluated
using ๐ถ๐ (for constant pressure process) as:
๐ปยฐ ๐2๐พ = ๐ปยฐ ๐1๐พ + ๐1
๐2
๐ถ๐๐๐
๐ยฐ ๐2๐พ = ๐ยฐ ๐1๐พ + ๐1
๐2 ๐ถ๐๐๐๐
๐บยฐ ๐2๐พ = ๐ปยฐ ๐2๐พ โ ๐2๐ยฐ ๐2๐พ
where ๐ปยฐ, ๐ยฐ and ๐บยฐ are molar enthalpy, molar entropy, and molar Gibbs
energy. For evaluation of the standard Gibbs energy of formation, they
can be replaced with โ๐๐ปยฐ, ๐ยฐ and โ๐๐บยฐ.
๐ถ๐ is molar heat capacity at constant pressure. If ๐2 is enough close to
๐1, ๐ถ๐ can be regarded as a constant in most case.
II-3. Chemical Equilibrium- Appendix) how to calculate โ๐๐ฎยฐ -
y = -0.1323x + 10.192
(liquid)
y = -0.2483x + 49.356
(gas)
-50
-45
-40
-35
-30
-25
-20
300 320 340 360 380
Mo
lar
Gib
bs
ener
gy /
kJ
mo
l-1
Temperature / K
y = 0.2681x + 48.961
(liquid)
y = 0.149x + 193.83
(gas)
0
50
100
150
200
250
300
300 320 340 360 380
Mo
lar
entr
op
y /
J m
ol-1
Temperature / K
y = 0.0851x - 16.241
(liquid)
y = 0.0544x + 29.463
(gas)
0
10
20
30
40
50
60
300 320 340 360 380Mo
lar
enth
alp
y /
kJ
mo
l-1
Temperature / K
Boiling pointComparison among ๐ปยฐ (a molar
enthalpy of formation), ๐ยฐ (a molar
entropy) and ๐บยฐ (a molar Gibbs
energy of formation) for methanol
at 1 atm, around the boiling point.
Note that ๐บยฐ is a continuous
function, but ๐ปยฐ and ๐ยฐ are not.
๐ปยฐ๐ยฐ
๐บยฐ
II-3. Chemical Equilibrium- $26-3: Standard Gibbs energies of formation can be used to calculate
equilibrium constants -
โ๐๐บยฐ = 3 2 โ๐๐บยฐ ๐ป2(๐) + 1 2 โ๐๐บยฐ ๐2(๐) โ 1 โ๐๐บยฐ ๐๐ป3 ๐
= 3 2 0 + 1 2 0 โ 1 โ16.637 ๐๐ฝ ๐๐๐โ1 = 16.637 ๐๐ฝ ๐๐๐โ1
(Example-3) Using the standard molar Gibbs energies of formation, calculate
โ๐๐บยฐ and ๐พ๐ at 298.15 K for
NH3(g) โ 3/2 H2(g) + 1/2 N2(g)
โ๐๐บยฐ ๐๐ป3(๐) = โ16.637 ๐๐ฝ ๐๐๐โ1, โ๐๐บยฐ ๐ป2(๐) = โ๐๐บยฐ ๐2(๐) = 0 ๐๐ฝ ๐๐๐
โ1
ln๐พ๐(๐) = โโ๐๐บยฐ
๐ ๐= โ
16.637 ร 103 ๐ฝ ๐๐๐โ1
8.31 ๐ฝ ๐พโ1 ๐๐๐โ1 298.15 ๐พ= โ6.60
Hence, ๐พ๐ ๐ = 1.36 ร 10โ3 at 298.15 K.
II-3. Chemical Equilibrium- $26-4: A plot of the Gibbs energy of a reaction mixture against the extent of
reaction is a minimum at equilibrium -
Consider the thermal decomposition of N2O4 (g) at 298.15 K as an example
to treat the Gibbs energy of a reaction mixture:
N2O4 (g) โ 2 NO2 (g)
๐บ ๐ = 1 โ ๐ ๐บ๐2๐4 + 2๐ ๐บ๐๐2
= 1 โ ๐ ๐บยฐ๐2๐4 + ๐ ๐ ln๐๐2๐4 + 2๐ ๐บยฐ๐๐2 + ๐ ๐ ln๐๐๐2
Suppose the initial state is โN2O4 (g) = 1 mol and NO2 (g) = 0 molโ and the
reaction brings the system to โN2O4 (g) = 1- ๐ mol and NO2 (g) = 2 ๐ molโ.
Here we assume the reaction occurs at a constant pressure (๐๐ก๐๐ก๐๐ = 1 ๐๐๐).
Then, as the molar fraction is ๐ฅ๐2๐4 =1โ๐
1+๐for N2O4 and ๐ฅ๐๐2 =
2๐
1+๐for NO2:
๐๐2๐4 = ๐ฅ๐2๐4๐๐ก๐๐ก๐๐ = ๐ฅ๐2๐4, ๐๐๐2 = ๐ฅ๐๐2๐๐ก๐๐ก๐๐ = ๐ฅ๐๐2
๐บ ๐ = 1 โ ๐ ๐บยฐ๐2๐4 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ ๐บยฐ๐๐2 + ๐ ๐ ln
2๐
1 + ๐
= 1 โ ๐ โ๐๐บยฐ๐2๐4 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ โ๐๐บยฐ๐๐2 + ๐ ๐ ln
2๐
1 + ๐
II-3. Chemical Equilibrium- $26-4: A plot of the Gibbs energy of a reaction mixture against the extent of
reaction is a minimum at equilibrium -
Consider the thermal decomposition of N2O4 (g) at 298.15 K as an example
to treat the Gibbs energy of a reaction mixture:
N2O4 (g) โ 2 NO2 (g)
๐บ ๐ = 1 โ ๐ โ๐๐บยฐ๐2๐4 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ โ๐๐บยฐ๐๐2 + ๐ ๐ ln
2๐
1 + ๐
This equation gives the Gibbs energy of the reaction mixture, ๐บ, as a
function of the extent of the reaction, ๐. Substituting the values for โ๐๐บยฐ:
๐บ ๐ = 1 โ ๐ 97.8 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ 51.3 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln
2๐
1 + ๐
II-3. Chemical Equilibrium- $26-4: A plot of the Gibbs energy of a reaction mixture against the extent of
reaction is a minimum at equilibrium -
Consider the thermal decomposition of N2O4 (g) at 298.15 K as an example
to treat the Gibbs energy of a reaction mixture:
N2O4 (g) โ 2 NO2 (g)
๐บ ๐ = 1 โ ๐ 97.8 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ 51.3 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln
2๐
1 + ๐
II-3. Chemical Equilibrium- $26-4: A plot of the Gibbs energy of a reaction mixture against the extent of
reaction is a minimum at equilibrium -
Consider the thermal decomposition of N2O4 (g) at 298.15 K as an example
to treat the Gibbs energy of a reaction mixture:
N2O4 (g) โ 2 NO2 (g)
๐บ ๐ = 1 โ ๐ 97.8 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ 51.3 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln
2๐
1 + ๐
๐บ ๐ is minimized at ๐ = 0.1892 ๐๐๐, corresponding to the equilibrium state,
thus ๐ = ๐๐๐ = 0.1892 ๐๐๐.
Then, the equilibrium constant is:
๐พ๐ =๐๐๐2
2
๐๐2๐4=
2๐๐๐ 1 + ๐๐๐2
1 โ ๐๐๐ 1 + ๐๐๐= 0.148
We can also calculate it from โ๐๐บยฐ:
ln๐พ๐ = โโ๐๐บยฐ
๐ ๐= โ
2 โ๐๐บยฐ ๐๐2 ๐ โ 1 โ๐๐บยฐ ๐2๐4 ๐
8.31 ๐ฝ ๐พโ1 ๐๐๐โ1 298.15 ๐พ= โ1.908
Hence, ๐พ๐ = 0.148
II-3. Chemical Equilibrium- $26-4: A plot of the Gibbs energy of a reaction mixture against the extent of
reaction is a minimum at equilibrium -
Consider the thermal decomposition of N2O4 (g) at 298.15 K as an example
to treat the Gibbs energy of a reaction mixture:
N2O4 (g) โ 2 NO2 (g)
๐บ ๐ = 1 โ ๐ 97.8 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln1 โ ๐
1 + ๐+ 2๐ 51.3 ๐๐ฝ ๐๐๐โ1 + ๐ ๐ ln
2๐
1 + ๐
We may also differentiate this equation with respect to ๐, then finally obtain:
๐๐บ
๐๐๐,๐
= โ๐๐บยฐ + ๐ ๐ ln๐๐๐2
2
๐๐2๐4
Since ๐๐บ
๐๐ ๐,๐= 0 at equilibrium,
โ๐๐บยฐ = โ๐ ๐ ln๐๐๐2
2
๐๐2๐4 ๐๐
= โ๐ ๐ ln๐พ๐
which is the same equation that we have derived several slides ago.
In addition, solving ๐๐บ
๐๐ ๐,๐= 0 explicitly, ๐๐๐ = 0.1892 ๐๐๐ is obtained as well.
II-3. Chemical Equilibrium- $26-5: The ratio of the reaction quotient to the equilibrium constant determines
the direction in which a reaction will proceed -
Consider a general reaction: ๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
โ๐๐บ ๐ = โ๐๐บยฐ ๐ + ๐ ๐ ln๐๐๐๐๐๐
๐๐
๐๐ด๐๐ด๐๐ต
๐๐ต
Realize that this equation is a general equation, so pressures are not
necessarily the equilibrium pressures, but are arbitrary.
Generally, this equation gives the value of โ๐๐บ when
โA(g) of [๐๐ด mol at ๐๐ด bar] react with B(g) of [๐๐ต mol at ๐๐ต bar] to produce
Y(g) of [๐๐ด mol at ๐๐ด bar] and Z(g) of [๐๐ mol at ๐๐ bar] โ.
1) If all the partial pressures are equl to 1 bar, โ๐๐บ ๐ = โ๐๐บยฐ ๐ . In
other words, the Gibbs energy change will be equal to the standard
Gibbs energy change.
2) If the pressures are the equilibrium pressures, โ๐๐บ ๐ = 0 and then
we obtain results that were derived in previous slides.
II-3. Chemical Equilibrium- $26-5: The ratio of the reaction quotient to the equilibrium constant determines
the direction in which a reaction will proceed -
Consider a general reaction: ๐๐ดA(g) + ๐๐ตB(g) โ ๐๐Y(g) + ๐๐Z(g)
โ๐๐บ ๐ = โ๐๐บยฐ ๐ + ๐ ๐ ln๐๐๐๐๐๐
๐๐
๐๐ด๐๐ด๐๐ต
๐๐ต
By introducing a quantity called the โreaction quotientโ ๐๐ =๐๐๐๐๐๐
๐๐
๐๐ด๐๐ด๐๐ต
๐๐ต, and
using an equation to correlate โ๐๐บยฐ with ๐พ๐, โ๐๐บยฐ = โ๐ ๐ ln๐พ๐, then
โ๐๐บ ๐ = โ๐ ๐ ln๐พ๐ + ๐ ๐ ln๐๐ = ๐ ๐ ln ๐๐ ๐พ๐
Note that this is not for the equilibrium; โ๐๐บ ๐ = 0 at equilibrium.
With this equation,
At equilibrium, โ๐๐บ ๐ = 0 and thus ๐๐ = ๐พ๐.
If ๐๐ < ๐พ๐, then ๐๐ must increase (because ๐พ๐ is a constant) as the
system proceeds toward equilibrium.
This is achieved by increasing the partial pressures of the
products and decreasing those of the reactants. Hence, the
reaction spontaneously proceeds from left to right.
If ๐๐ > ๐พ๐, ๐๐ decrease as the system proceeds toward equilibrium.
In the same though, the reaction is spontaneous from right to left.
II-3. Chemical Equilibrium- $26-5: The ratio of the reaction quotient to the equilibrium constant determines
the direction in which a reaction will proceed -
(Example-4: #26-5) Consider โ 2SO2(g) + O2(g) โ 2SO3(g) โ. The
equilibrium constant ๐พ๐ = 10 at 960 K. Calculate โ๐๐บ and check in which
direction the reaction will proceed spontaneously for
2 SO2 (1.0 ร 10โ3 ๐๐๐) + O2 (0.20 ๐๐๐) โ 2 SO3(1.0 ร 10
โ4 ๐๐๐)
๐๐ =๐๐๐32
๐๐๐22 ๐๐2
=1.0 ร 10โ4 2
1.0 ร 10โ3 2 0.20= 5.0 ร 10โ2
Note that the value for ๐๐ is unitless because the definition of ๐ is, for
example in a general form, ๐ = ๐๐ ๐ยฐ
๐๐ ๐๐ ๐ยฐ๐๐
๐๐ด ๐ยฐ ๐๐ด ๐๐ต ๐ยฐ ๐๐ตwhere ๐ยฐ = 1 ๐๐๐.
โ๐๐บ ๐ = ๐ ๐ ln ๐๐ ๐พ๐ = ๐ ๐ ln 5.0 ร 10โ2 10 < 0
Hence, due to โ๐๐บ ๐ < 0 (or due to ๐๐ < ๐พ๐), the reaction proceed
from left to right.
II-3. Chemical Equilibrium- $26-6: The sign of โ๐๐บ and not that of โ๐๐บยฐ ๐ determines the direction of reaction
spontaneity -
It should be clearly understood that โ๐๐บยฐ is the value of โ๐๐บ when all the
reactants and products are unmixed at partial pressures equal to 1 bar.
Namely, โ๐๐บยฐ is the standard Gibbs energy changes. (recall that โstandardโ
assumes 1 bar) Hence,
If โ๐๐บยฐ < 0, then ๐พ๐ > 1, meaning that the reaction will proceed from
reactants to products if all species are mixed at 1 bar (for each partial
pressure; not total pressure).
If โ๐๐บยฐ > 0, then ๐พ๐ < 1, thus reaction proceeds from products to
reactants in the same condition.
So, the sing of โ๐๐บยฐ just indicates the spontaneous reaction direction
for the condition of ๐๐๐๐๐ก๐๐๐ = 1 bar (for all species of both reactants
and products); not necessarily for all conditions.
II-3. Chemical Equilibrium- $26-6: The sign of โ๐๐บ and not that of โ๐๐บยฐ ๐ determines the direction of reaction
spontaneity -
(Example-5) Consider โ N2O4 (g) โ 2 NO2 (g) โ,
for which โ๐๐บยฐ = 4.729 kJ ๐๐๐โ1 and ๐พ๐ = 0.148 at 298.15 K.
First, make sure that this โ๐๐บยฐ > 0 does NOT mean โno N2O4 dissociate at
298.15 K when we place some of N2O4 in a reaction vesselโ.
To correctly consider, we have to calculate โ๐๐บ as:
โ๐๐บ = โ๐๐บยฐ + ๐ ๐ ln๐๐ = 4.729 ๐๐ฝ ๐๐๐โ1 + 2.479 ๐๐ฝ ๐๐๐โ1 ln
๐๐๐22
๐๐2๐4
If we just โplace some of N2O4 in a reaction vesselโ, ln ๐๐๐22 ๐๐2๐4 has a
large negative value, thus โ๐๐บ < 0. Accordingly, some N2O4 dissociate.
The equilibrium state is achieved by the condition โ๐๐บ = 0 (note โ๐๐บยฐ is a
constant while โ๐๐บ changes depending on ๐๐๐2 and ๐๐2๐4 ), at which point
๐๐ = ๐พ๐. Until this point is achieved, ๐๐๐2 increases and ๐๐2๐4 decreases.
II-3. Chemical Equilibrium- $26-6: The sign of โ๐๐บ and not that of โ๐๐บยฐ ๐ determines the direction of reaction
spontaneity -
(Example-6) Consider โ H2 (g) + ยฝ O2 (g) โ H2O (l) โ,
for which โ๐๐บยฐ = โ237 kJ ๐๐๐โ1 at 298.15 K.
In this case, โ๐๐บยฐ has a large negative value, thus basically H2O (l) is
much more stable than the reactants at 298.15 K. However, a mixture of
H2 (g) and O2 (g) remains unchanged.
If a spark or a catalyst is introduced, then the reaction occurs explosively.
The โnoโ of the thermodynamics is emphatic: If thermodynamics
insists that a certain process will not occur spontaneously, then it will
not occur.
On the other hand, the โyesโ is actually โmaybeโ. The fact that a
process will occur spontaneously does not imply that it will occur at a
detectable rate.
Diamond remains its form, although a graphite is more favorable
energetically, is another example.
The speed of reaction can be analyzed in the framework
of rate theory. (next topic)
II-3. Chemical Equilibrium- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
We utilize the Gibbs-Helmohltz equation:
Substitute โ๐๐บยฐ = โ๐ ๐ ln๐พ๐ (note this is a definition of โ๐๐บยฐ, not a condition
achieved at equilibrium state) for this equation:
๐โ๐บยฐ ๐
๐๐๐
=โ๐ปยฐ
๐2
๐ ln๐พ๐(๐)
๐๐๐
=๐ ln๐พ๐(๐)
๐๐=โ๐๐ปยฐ
๐ ๐2
This means that
If โ๐๐ปยฐ > 0 (endothermic reaction), ๐พ๐(๐) increases with temperature.
If โ๐๐ปยฐ < 0 (exothermic reaction), ๐พ๐(๐) decreases with temperature.
Integrate the equation:ln๐พ๐(๐2)
๐พ๐(๐1)=
๐1
๐2 โ๐๐ปยฐ(๐)
๐ ๐2๐๐
If the temperature range is small enough to consider โ๐๐ปยฐ constant:
ln๐พ๐(๐2)
๐พ๐(๐1)= โ
โ๐๐ปยฐ
๐
1
๐2โ1
๐1
II-3. Chemical Equilibrium- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
(Example-7) Consider โ PCl3 (g) + Cl2 (g) โ PCl2 (g) โ.
Given that โ๐๐ปยฐ has an average value of -69.8 kJ mol-1 over 500-700 K and
๐พ๐ is 0.0408 at 500K, evaluate ๐พ๐ at 700K.
As โ๐๐ปยฐ can be assumed as a constant over the concerned temperatures,
ln๐พ๐(๐2)
๐พ๐(๐1)= โ
โ๐๐ปยฐ
๐
1
๐2โ1
๐1
Substituting provided values gives:
ln๐พ๐(700 ๐พ)
๐พ๐(500 ๐พ)= ln
๐พ๐(700 ๐พ)
0.0408= โ
โ69.8 ร 103
๐
1
700โ1
500
๐พ๐ 700 ๐พ = 3.36 ร 10โ4
Note that since the reaction is exothermic, ๐พ๐ 700 ๐พ is less than ๐พ๐ 700 ๐พ ;
namely less product (PCl2) at higher temperatures as ๐พ๐ = ๐๐๐ถ๐2
๐๐๐ถ๐3๐๐ถ๐2.
II-3. Chemical Equilibrium- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
ln๐พ๐(๐2)
๐พ๐(๐1)=
๐1
๐2 โ๐๐ปยฐ(๐)
๐ ๐2๐๐
Finally, we consider the temperature dependence of โ๐๐ปยฐ constant in:
This equation can write down as:
ln๐พ๐(๐) = ln๐พ๐(๐1) + ๐1
๐ โ๐๐ปยฐ(๐โฒ)
๐ ๐โฒ2๐๐โฒ
For example, as we learned, the temperature dependence of โ๐๐ปยฐ may
be written as:
โ๐๐ปยฐ ๐2 = โ๐๐ปยฐ ๐1 + ๐1
๐2
โ๐ถยฐ๐ ๐ ๐๐
or expanded in respect to temperature as:
โ๐๐ปยฐ ๐ = ๐ผ + ๐ฝ๐ + ๐พ๐2 + ๐ฟ๐3 +โฏ
In this latter case, ln๐พ๐(๐) becomes:
ln๐พ๐(๐) = โ๐ผ
๐ ๐+๐ฝ
๐ ln ๐ +
๐พ
๐ ๐ +
๐ฟ
2๐ ๐2 + (๐๐๐ก๐๐๐๐๐ก๐๐๐_๐๐๐๐ ๐ก๐๐๐ก)โฆ