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

𝐺 = 𝐺𝑙 + 𝐺𝑔

where 𝐺𝑙 and 𝐺𝑔 are the Gibbs energies of the liquid and the gas phase.

Now, suppose 𝑑𝑛 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:

𝑑𝐺 =𝜕𝐺𝑔

𝜕𝑛𝑔 𝑃,𝑇𝑑𝑛𝑔 +

𝜕𝐺𝑙

𝜕𝑛𝑙 𝑃,𝑇𝑑𝑛𝑙

As 𝑑𝑛𝑙 = −𝑑𝑛𝑔, then

𝑑𝐺 =𝜕𝐺𝑔

𝜕𝑛𝑔 𝑃,𝑇−

𝜕𝐺𝑙

𝜕𝑛𝑙 𝑃,𝑇𝑑𝑛𝑔

Here, we define chemical potentials, 𝜇𝑔 =𝜕𝐺𝑔

𝜕𝑛𝑔 𝑃,𝑇and 𝜇𝑙 =

𝜕𝐺𝑙

𝜕𝑛𝑙 𝑃,𝑇, then

𝑑𝐺 = 𝜇𝑔 − 𝜇𝑙 𝑑𝑛𝑔 (constant T and P)


are equal -

We consider a system consisting of two phases of a pure substance

in equilibrium each other. (liquid and gas, for example here)

𝑑𝐺 = 𝜇𝑔 − 𝜇𝑙 𝑑𝑛𝑔 (constant T and P)

where chemical potentials are 𝜇𝑔 =𝜕𝐺𝑔

𝜕𝑛𝑔 𝑃,𝑇and 𝜇𝑙 =

𝜕𝐺𝑙

𝜕𝑛𝑙 𝑃,𝑇.

If the two phases are in equilibrium with each other, then 𝑑𝐺 = 0.

And because we suppose some amount of the liquid phase is

transferred to the gas phase (𝑑𝑛𝑔) here, to make 𝑑𝐺 = 0 in this

condition, 𝜇𝑔 = 𝜇𝑙 is needed.

If 𝜇𝑔 < 𝜇𝑙, [𝜇𝑔 − 𝜇𝑙] < 0. Thus, the process of 𝑑𝑛𝑔 > 0 (transfer

from the liquid phase to the gas phase) spontaneously takes place

as it holds 𝑑𝐺 < 0.

Likewise, if 𝜇𝑔 > 𝜇𝑙, 𝑑𝑛𝑔 < 0.

Hence, in general, the transfer occurs from the phase with higher

chemical potential to the phase with lower chemical potential.


are equal -

In general, the chemical potential is defined as:

μ =𝜕𝐺

𝜕𝑛 𝑃,𝑇

Because 𝐺 is proportional to the size/amount of a system 𝐺 ∝ 𝑛,

namely an extensive thermodynamic function, we can express it as:

𝐺 = 𝑛𝜇 𝑇, 𝑃 Here, this equation is consistent with the definition of chemical

potential:

μ =𝜕𝐺

𝜕𝑛 𝑃,𝑇=

𝜕 𝑛𝜇 𝑇,𝑃

𝜕𝑛 𝑃,𝑇= 𝜇 𝑇, 𝑃

which means 𝜇 𝑇, 𝑃 , the chemical potential, is the same quantity as

the molar Gibbs energy and it is an intensive quantity.


are equal -

As they are in equilibrium,

μ𝛼(𝑇, 𝑃) = μ𝛽(𝑇, 𝑃)Now take the total derivatives of both sides

𝜕μ𝛼

𝜕𝑃 𝑇𝑑𝑃 +

𝜕μ𝛼

𝜕𝑇 𝑃𝑑𝑇 =

𝜕μ𝛽

𝜕𝑃 𝑇𝑑𝑃 +

𝜕μ𝛽

𝜕𝑇 𝑃𝑑𝑇

Since 𝜇 is simply the molar Gibbs energy for a single substance, utilizing 𝜕𝐺

𝜕𝑃 𝑇= 𝑉 and

𝜕𝐺

𝜕𝑇 𝑃= −𝑆 (*these were previously derived along Maxwell

relations)𝜕μ

𝜕𝑃 𝑇=

𝜕 𝐺

𝜕𝑃 𝑇= 𝑉 and

𝜕μ

𝜕𝑇 𝑃=

𝜕 𝐺

𝜕𝑇 𝑃= − 𝑆

where 𝑉 and 𝑆 are the molar volume and the molar entropy. Then, 𝑉𝛼𝑑𝑃 − 𝑆𝛼𝑑𝑇 = 𝑉𝛽𝑑𝑃 − 𝑆𝛽𝑑𝑇

Since we consider the two phases are in equilibrium each other𝑑𝑃

𝑑𝑇=

𝑆𝛽− 𝑆𝛼

𝑉𝛽− 𝑉𝛼=∆𝑡𝑟𝑠 𝑆

∆𝑡𝑟𝑠 𝑉=

∆𝑡𝑟𝑠 𝐻 𝑇

∆𝑡𝑟𝑠 𝑉=

∆𝑡𝑟𝑠 𝐻

𝑇∆𝑡𝑟𝑠 𝑉

We consider two phases (α and β) are in equilibrium each other.


are equal -

In the previous slide, we obtain for two phases in equilibrium each other:

𝑑𝑃

𝑑𝑇=∆𝑡𝑟𝑠 𝐻

𝑇∆𝑡𝑟𝑠 𝑉

This equation is called the Clapeyron equation, and relates “the slope

of the two-phase boundary line in a phase diagram with the values of

∆𝑡𝑟𝑠 𝐻 and ∆𝑡𝑟𝑠 𝑉 for a transition between these two phases”.


are equal -

Exampe-1) Solid-liquid coexistence curve at around 1 atm for Benzene

∆𝑓𝑢𝑠 𝐻 = 9.95 kJ mol-1 and ∆𝑓𝑢𝑠 𝑉 = 10.3 cm3 mol-1 at the normal

melting point (278.7 K).

Thus, 𝑑𝑃 𝑑𝑇 at the normal melting point of benzene is:𝑑𝑃

𝑑𝑇=


𝑇∆𝑡𝑟𝑠 𝑉=

9.95 𝑘𝐽 𝑚𝑜𝑙−1

278.7 𝐾 (10.3 𝑐𝑚3𝑚𝑜𝑙−1)= 34.2 𝑎𝑡𝑚 𝐾−1

Here, by taking the reciprocal of this result:𝑑𝑇

𝑑𝑃= 0.0292 𝐾 𝑎𝑡𝑚−1

Using the above result (0.0292 K atm-1) and

assuming ∆𝑓𝑢𝑠 𝐻 and ∆𝑓𝑢𝑠 𝑉 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, ∆𝑓𝑢𝑠 𝐻 and

∆𝑓𝑢𝑠 𝑉 are not independent of pressure at

high-pressure region


are equal -

Exampe-2) Solid-liquid coexistence curve at around 1 atm for water

∆𝑓𝑢𝑠 𝐻 = 6.01 kJ mol-1 and ∆𝑓𝑢𝑠 𝑉 = -1.63 cm3 mol-1 at the normal

melting point (273.15 K).

Thus, 𝑑𝑃 𝑑𝑇 at the normal melting point of water is:𝑑𝑇

𝑑𝑃=𝑇∆𝑡𝑟𝑠 𝑉

∆𝑡𝑟𝑠 𝐻=

273.15 𝐾 (−1.63 𝑐𝑚3𝑚𝑜𝑙−1)

6.01 𝑘𝐽 𝑚𝑜𝑙−1= −0.00751 𝐾 𝑎𝑡𝑚−1

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

∆𝑓𝑢𝑠 𝑉 < 0 (the molar volume is larger in

solid than that in liquid at around melting

point, 1atm), because ∆𝑓𝑢𝑠 𝐻 = 𝑇∆𝑓𝑢𝑠 𝑆 must

be larger than 0 as ∆𝑓𝑢𝑠 𝑆 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 “∆𝑓𝑢𝑠 𝐻 and

∆𝑓𝑢𝑠 𝑉 are independent of pressure” is indeed satisfactory for solid-liquid

and solid-solid transitions over a small ∆ 𝑇.

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, 𝑑𝑃

𝑑𝑇=


𝑇∆𝑡𝑟𝑠 𝑉can

be cast into a useful form for condensed phase-gas phase transitions.



We consider the Clapeyron equation in liquid-gas equilibrium.

First, we rewrite the equation:𝑑𝑃

𝑑𝑇=


𝑇(∆𝑡𝑟𝑠 𝑉)=

∆𝑣𝑎𝑝 𝐻

𝑇( 𝑉𝑔− 𝑉𝑙)

Here, if (1) the system is not too near the critical point (thus, 𝑉𝑔 ≫ 𝑉𝑙 and 𝑉𝑙

in the denominator is negligible) and (2) the pressure is not too high (thus, we

can assume the vapor is an ideal gas and 𝑉𝑔 = 𝑅𝑇 𝑃) 𝑑𝑃

𝑃𝑑𝑇=𝑑 ln 𝑃

𝑑𝑇=


𝑃𝑇( 𝑉𝑔− 𝑉𝑙)~∆𝑣𝑎𝑝 𝐻

𝑃𝑇( 𝑉𝑔)~


𝑃𝑇( 𝑅𝑇 𝑃)=∆𝑣𝑎𝑝 𝐻

𝑅𝑇2

which is know as Clausius-Clapeyron equation (derived by Clausius in 1850).

If ∆𝑣𝑎𝑝 𝐻 does not vary with temperature over integration limits of T, then

ln𝑃2

𝑃1= −


𝑅

1

𝑇2−1

𝑇1=∆𝑣𝑎𝑝 𝐻

𝑅

𝑇2−𝑇1

𝑇1𝑇2

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.



Example-1) Vapor pressure of benzene at 373.2 K

We obtain an integral form of Clausius-Clapeyron equation as

ln𝑃2

𝑃1=∆𝑣𝑎𝑝 𝐻

𝑅

𝑇2−𝑇1

𝑇1𝑇2

Note that we assume ∆𝑣𝑎𝑝 𝐻 does not (largely) vary with temperature.

∆𝑣𝑎𝑝 𝐻 = 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:

ln𝑃

760=30.8× 103

8.31

373.2−353.2

373.2×353.2

This gives us P=1330 torr, close to the experimental value,1360 torr.



If we take an integral of Clausius-Clapeyron equation indefinitely rather

than between definite limits, we obtain (still assuming ∆𝑣𝑎𝑝 𝐻 is constant) 𝑑𝑃


𝑑𝑇=∆𝑣𝑎𝑝 𝐻

𝑅𝑇2→ ln 𝑃 = −


𝑅𝑇+ 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 −∆𝑣𝑎𝑝 𝐻 𝑅.

The slope of line for

benzene (left figure) over

313-353 K gives us

∆𝑣𝑎𝑝 𝐻 = 32.3 𝑘𝐽 𝑚𝑜𝑙−1,

close to 30.8 kJ mol-1

which is the experimental

value at the normal boiling

point (353 K).

<Benzene case>



If we consider the temperature dependence of ∆𝑣𝑎𝑝 𝐻, for example, as

∆𝑣𝑎𝑝 𝐻 = 𝐴 + 𝐵𝑇 + 𝐶𝑇2 +⋯

where A, B, C are constants.

Then, Clausius-Clapeyron equation, 𝑑𝑃


𝑑𝑇=∆𝑣𝑎𝑝 𝐻

𝑅𝑇2, gives us

ln 𝑃 = −𝐴

𝑅𝑇+𝐵

𝑅ln 𝑇 +

𝐶

𝑅𝑇 + 𝑘 + 𝑂 𝑇2

This equation works a wider range accurately, more than the equation that

we obtain assuming ∆𝑣𝑎𝑝 𝐻 is constant.

For example, for the vapor pressure of solid ammonia over 146-195 K is:

ln 𝑃 𝑡𝑜𝑟𝑟 = −4124.4 𝐾

𝑇+ 1.8163 ln

𝑇

𝐾+ 34.4834



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 as𝑑𝑃𝑠

𝑑𝑇= 𝑃𝑠

∆𝑠𝑢𝑏 𝐻

𝑅𝑇2and

𝑑𝑃𝑙

𝑑𝑇= 𝑃𝑙


𝑅𝑇2

here 𝑃𝑠 and 𝑃𝑙 are the vapor pressure of the solid and the liquid, respectively.

Because 𝑃𝑠 = 𝑃𝑠 at the triple point, 𝑑𝑃𝑠 𝑑𝑇

𝑑𝑃𝑙 𝑑𝑇=∆𝑠𝑢𝑏 𝐻


Since enthalpy is a state function, the transition between “s→g” and ““s→l→g”

having the same initial and final states should hold the same ∆𝑡𝑟𝑎 𝐻. Thus,

∆𝑠𝑢𝑏 𝐻 = ∆𝑓𝑢𝑠 𝐻 + ∆𝑣𝑎𝑝 𝐻 𝑑𝑃𝑠 𝑑𝑇

𝑑𝑃𝑙 𝑑𝑇=∆𝑠𝑢𝑏 𝐻

∆𝑣𝑎𝑝 𝐻= 1 +

∆𝑓𝑢𝑠 𝐻


which indicates the slope of the solid-gas curve is greater than that of the

liquid-gas curve at the triple point.

*the subscription “sub”

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.

𝜈𝐴A(g) + 𝜈𝐵B(g) ⇌ 𝜈𝑌Y(g) + 𝜈𝑍Z(g)

The amount of species 𝑖 is 𝑛𝑖 [mol]. The Gibbs energy for this multi-component

system is a function of 𝑇, 𝑃, 𝑛𝐴, 𝑛𝐵 , 𝑛𝑌 𝑎𝑛𝑑 𝑛𝑍, then the total derivative is:

dG =𝜕𝐺

𝜕𝑇𝑃,𝑛𝐴,𝑛𝐵,𝑛𝑌,𝑛𝑍

𝑑𝑇 +𝜕𝐺

𝜕𝑃𝑇,𝑛𝐴,𝑛𝐵,𝑛𝑌,𝑛𝑍

𝑑𝑃 +𝜕𝐺

𝜕𝑛𝐴 𝑇,𝑃,𝑛𝐵,𝑛𝑌,𝑛𝑍

𝑑𝑛𝐴

+𝜕𝐺

𝜕𝑛𝐵 𝑇,𝑃,𝑛𝐴,𝑛𝑌,𝑛𝑍

𝑑𝑛𝐵 +𝜕𝐺

𝜕𝑛𝑌 𝑇,𝑃,𝑛𝐴,𝑛𝐵,𝑛𝑍

𝑑𝑛𝑌 +𝜕𝐺

𝜕𝑛𝑍 𝑇,𝑃,𝑛𝐴,𝑛𝐵,𝑛𝑌

𝑑𝑛𝑍

Using some equations derived around the Maxwell relation derivation (of 𝐺):

𝑑𝐺 = −𝑆𝑑𝑇 + 𝑉𝑑𝑃 + 𝜇𝐴𝑑𝑛𝐴 + 𝜇𝐵𝑑𝑛𝐵 + 𝜇𝑌𝑑𝑛𝑌 + 𝜇𝑍𝑑𝑛𝑍

𝜇𝐴 =𝜕𝐺

𝜕𝑛𝐴 𝑇,𝑃,𝑛𝐵,𝑛𝑌,𝑛𝑍

, 𝑒𝑡𝑐

If the reaction takes place in constant T and P,

dG = 𝜇𝐴𝑑𝑛𝐴 + 𝜇𝐵𝑑𝑛𝐵 + 𝜇𝑌𝑑𝑛𝑌 + 𝜇𝑍𝑑𝑛𝑍 (constant T and P)





We define a quantity 𝜉, called as the “extent of reaction”. Here 𝑛𝑖0 is the

initial number of moles for species 𝑖, then :

𝑛𝐴 = 𝑛𝐴0 − 𝜈𝐴𝜉 𝑛𝐵 = 𝑛𝐵0 − 𝜈𝐵𝜉

𝑛𝑌 = 𝑛𝑌0 − 𝜈𝑌𝜉 𝑛𝑍 = 𝑛𝑍0 − 𝜈𝑍𝜉

(reactants)

(products)

In this case, 𝜉 has units of moles. Then, the variations of 𝑛𝑖 is:

𝑑𝑛𝐴 = −𝜈𝐴𝑑𝜉 𝑑𝑛𝐵 = 𝜈𝐵𝑑𝜉

𝑑𝑛𝑌 = −𝜈𝑌𝑑𝜉 𝑑𝑛𝑍 = 𝜈𝑍𝑑𝜉

(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 = 𝜇𝐴𝑑𝑛𝐴 + 𝜇𝐵𝑑𝑛𝐵 + 𝜇𝑌𝑑𝑛𝑌 + 𝜇𝑍𝑑𝑛𝑍= −𝜈𝐴𝜇𝐴 − 𝜈𝐵𝜇𝐵 + 𝜈𝑌𝜇𝑌 + 𝜈𝑍𝜇𝑍 𝑑𝜉 (constant T and P)





dG = −𝜈𝐴𝜇𝐴 − 𝜈𝐵𝜇𝐵 + 𝜈𝑌𝜇𝑌 + 𝜈𝑍𝜇𝑍 𝑑𝜉 (constant T and P)

𝜕𝐺

𝜕𝜉𝑇,𝑃

= 𝜈𝑌𝜇𝑌 + 𝜈𝑍𝜇𝑍 − 𝜈𝐴𝜇𝐴 − 𝜈𝐵𝜇𝐵

Here, we define 𝜕𝐺

𝜕𝜉 𝑇,𝑃= ∆𝑟𝐺, which is the change in Gibbs energy

when the extent of reaction changes by one mole, and its unit is 𝐽 𝑚𝑜𝑙−1.

Assuming each species behaves as ideal gas, as the pressure dependence

of chemical potential is written as 𝜇𝑗 𝑇, 𝑃 = 𝜇°𝑗 𝑇 + 𝑅𝑇 ln 𝑃𝑗 𝑃° , then:

∆𝑟𝐺 = ∆𝑟𝐺° + 𝑅𝑇 ln𝑄

∆𝑟𝐺° = 𝜈𝑌𝜇°𝑌(𝑇) + 𝜈𝑍𝜇°𝑍(𝑇) − 𝜈𝐴𝜇°𝐴(𝑇) − 𝜈𝐵𝜇°𝐵(𝑇)

𝑄 = 𝑃𝑌 𝑃°

𝜈𝑌 𝑃𝑍 𝑃°𝜈𝑍

𝑃𝐴 𝑃°𝜈𝐴 𝑃𝐵 𝑃°

𝜈𝐵

𝑃° is the pressure of standard

state (1 bar) and 𝑃𝐴 is the

partial pressure of species A.





(ideal gas, constant T and P)∆𝑟𝐺 = ∆𝑟𝐺° + 𝑅𝑇 ln𝑄

∆𝑟𝐺° = 𝜈𝑌𝜇°𝑌 𝑇 + 𝜈𝑍𝜇°𝑍 𝑇−𝜈𝐴𝜇°𝐴(𝑇) − 𝜈𝐵𝜇°𝐵(𝑇)

𝑄 = 𝑃𝑌 𝑃°



𝜈𝐵

Here, the quantity ∆𝑟𝐺° is the change in standard Gibbs energy for the

reaction between unmixed reactants to form unmixed products. All

species in their standard states at 𝑇 and 𝑃°. Note that 𝑃° = 1 bar.

When the reaction system is equilibrium, the Gibbs energy must

minimum with respect to any change from the equilibrium state, thus 𝜕𝐺

𝜕𝜉 𝑇,𝑃= ∆𝑟𝐺 = ∆𝑟𝐺° + 𝑅𝑇 ln𝑄𝑒𝑞 = 0 at an equilibrium state. Thus:

∆𝑟𝐺° = −𝑅𝑇 ln 𝑃𝑌𝜈𝑌𝑃𝑍

𝜈𝑍

𝑃𝐴𝜈𝐴𝑃𝐵

𝜈𝐵𝑒𝑞

= −𝑅𝑇 ln𝐾𝑃(𝑇)





∆𝑟𝐺° = −𝑅𝑇 ln𝐾𝑃(𝑇)

𝐾𝑃 𝑇 = 𝑄𝑒𝑞 = 𝑃𝑌 𝑃°



𝜈𝐵𝑒𝑞

= 𝑃𝑌𝜈𝑌𝑃𝑍

𝜈𝑍


𝜈𝐵𝑒𝑞

*the subscript eq emphasizes that the

partial pressures are in an equilibrium.

𝐾𝑃 𝑇 is called as equilibrium constant. Be sure that 𝐾𝑃 𝑇 has no unit.

As seen in the definition, this constant is defined after the target equation

is given. For example, if the 𝜈𝐴 in the equation is changed (even keeping

the same meaning of reaction, like 2𝜈𝐴A(g) + 2𝜈𝐵B(g) ⇌ 2𝜈𝑌Y(g) +

2𝜈𝑍Z(g) ), 𝐾𝑃 𝑇 value is changed.

𝑃° = 1 𝑏𝑎𝑟



(Example-1a) For reaction “3 H2(g) + N2(g) ⇌ 2 NH3(g)”, the

equilibrium pressures are given as 𝑃𝐻2, 𝑃𝑁2, and 𝑃𝑁𝐻3 .

𝐾𝑃, 𝑇 = 𝑃𝑁𝐻32

𝑃𝐻23 𝑃𝑁2 𝑒𝑞

*Be sure that these pressures are

pressures at equilibrium, as in the

definition of equilibrium constant.

(Example-1b) For reaction “3/2 H2(g) + ½ N2(g) ⇌ NH3(g)”, the

equilibrium pressures are given as 𝑃𝐻2, 𝑃𝑁2, and 𝑃𝑁𝐻3 .

𝐾𝑃 𝑇 = 𝑃𝑁𝐻3

𝑃𝐻23/2𝑃𝑁21/2

𝑒𝑞

≠ 𝐾𝑃 𝑇 𝑜𝑓 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 − 1𝑎

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 -


PCl5(g) ⇌ PCl3(g) + Cl2(g)

The equilibrium-constant expression for this reaction is:

𝐾𝑃, 𝑇 = 𝑃𝑃𝐶𝑙3𝑃𝐶𝑙2

𝑃𝑃𝐶𝑙5 𝑒𝑞

Suppose we have 1 mol of PCl5 (g) and no PCl3 or Cl2 at the beginning.

When the reaction occurs to an extent 𝜉, PCl5: 1 mol → (1- 𝜉) mol

PCl3: 0 mol → 𝜉 mol, Cl2: 0 mol → 𝜉 mol

Total: 1 mol → (1+ 𝜉) mol

If 𝜉𝑒𝑞 is the extent of reaction at equilibrium, then the partial pressures are:

𝑃𝑃𝐶𝑙3 = 𝑃𝐶𝑙2 = 𝜉𝑒𝑞𝑃 1 + 𝜉𝑒𝑞 , 𝑃𝑃𝐶𝑙5 = 1 − 𝜉𝑒𝑞 𝑃 1 + 𝜉𝑒𝑞

where 𝑃 is the total pressure. Then, the equilibrium constant is:

𝐾𝑃, 𝑇 = 𝜉𝑒𝑞

2

1 − 𝜉𝑒𝑞2 𝑃



PCl5(g) ⇌ PCl3(g) + Cl2(g)

𝐾𝑃 𝑇 = 𝜉𝑒𝑞

2

1 − 𝜉𝑒𝑞2 𝑃

PCl5: 1 mol → (1- 𝜉) mol

PCl3 , Cl2: 0 mol → 𝜉 mol,

Total: 1 mol → (1+ 𝜉) mol

𝐾𝑃 𝑇 only depends on 𝑇, but

not 𝑃. So, if 𝑃 (total pressure)

is changed, 𝜉𝑒𝑞 must be

changed so that 𝐾𝑃 𝑇 is kept

constant. For example, 𝐾𝑃 𝑇of this reaction is 5.4 (no unit)

at 200ºC.




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 “𝑃 = 𝑐𝑅𝑇”, where 𝑐 = 𝑛 𝑉 is the concentration:


𝐾𝑃 𝑇 = 𝑄𝑒𝑞 = 𝑃𝑌 𝑃°



𝜈𝐵𝑒𝑞

=𝐶𝑌𝜈𝑌𝐶𝑍

𝜈𝑍

𝐶𝐴𝜈𝐴𝐶𝐵

𝜈𝐵𝑒𝑞

𝑅𝑇

𝑃°

𝜈𝑌+𝜈𝑍−𝜈𝐴−𝜈𝐵

*As the same with 𝐾𝑃,

𝐾𝑐 has also no unit.

Here, we consider some standard concentration 𝑐° (like 𝑃° ), often taken to

be “1 mol L-1”. Then:

𝐾𝑃 𝑇 = 𝐾𝐶 𝑇𝑐°𝑅𝑇

𝑃°

𝜈𝑌+𝜈𝑍−𝜈𝐴−𝜈𝐵

𝐾𝑐 𝑇 = 𝑐𝑌 𝑐°

𝜈𝑌 𝑐𝑍 𝑐°𝜈𝑍

𝑐𝐴 𝑐°𝜈𝐴 𝑐𝐵 𝑐°

𝜈𝐵𝑒𝑞


(Example-2) For reaction “NH3(g) ⇌ 3/2 H2(g) +1/2 N2(g)”, 𝐾𝑃 𝑇 =1.36 × 10−3 at 298.15 K. Determine the corresponding 𝐾𝐶 𝑇 .

𝐾𝑃 𝑇 = 𝐾𝐶 𝑇𝑐°𝑅𝑇

𝑃°

3/2+1/2−1

= 𝐾𝐶 𝑇𝑐°𝑅𝑇

𝑃°

1

𝐾𝐶 𝑇 = 𝐾𝑃 𝑇𝑐°𝑅𝑇

𝑃°

−1

= 1.36 × 10−3 ×1 𝑚𝑜𝑙 𝐿−1 × 0.0831 𝐿 𝑏𝑎𝑟 𝑚𝑜𝑙−1𝐾−1 × 298.15 𝐾

1 𝑏𝑎𝑟

−1

= 5.49× 10−5

II-3. Chemical Equilibrium- $26-3: Standard Gibbs energies of formation can be used to calculate

equilibrium constants -

∆𝑟𝐺° = 𝜈𝑌𝜇°𝑌 𝑇 + 𝜈𝑍𝜇°𝑍 𝑇 − 𝜈𝐴𝜇°𝐴(𝑇) − 𝜈𝐵𝜇°𝐵(𝑇)


As already derived, 𝐾𝑃 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 “zero” value.

As in the same manner with “standard molar enthalpy of formation”, we

can define “the standard molar Gibbs energy of formation” according to

∆𝑟𝐺° = ∆𝑟𝐻° − 𝑇∆𝑟𝑆°referring to standard molar entropies.

So, for “𝜈𝐴A(g) + 𝜈𝐵B(g) → 𝜈𝑌Y(g) + 𝜈𝑍Z(g)”, for example, we have

∆𝑟𝐺° = 𝜈𝑌∆𝑓𝐺° 𝑌 + 𝜈𝑍∆𝑓𝐺° 𝑍 − 𝜈𝐴∆𝑓𝐺° 𝐴 − 𝜈𝐵∆𝑓𝐺°[𝐵]

where ∆𝑓𝐺° 𝑌 is the standard molar Gibbs energy of formation for

substance 𝑌.



∆𝑟𝐺° = 𝜈𝑌𝜇°𝑌 𝑇 + 𝜈𝑍𝜇°𝑍 𝑇 − 𝜈𝐴𝜇°𝐴(𝑇) − 𝜈𝐵𝜇°𝐵(𝑇)


As already derived, 𝐾𝑃 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 “zero” value.

As in the same manner with “standard molar enthalpy of formation”, we

can define “the standard molar Gibbs energy of formation” according to

∆𝑟𝐺° = ∆𝑟𝐻° − 𝑇∆𝑟𝑆°referring to standard molar entropies.

So, for “𝜈𝐴A(g) + 𝜈𝐵B(g) → 𝜈𝑌Y(g) + 𝜈𝑍Z(g)”, for example, we have

∆𝑟𝐺° = 𝜈𝑌∆𝑓𝐺° 𝑌 + 𝜈𝑍∆𝑓𝐺° 𝑍 − 𝜈𝐴∆𝑓𝐺° 𝐴 − 𝜈𝐵∆𝑓𝐺°[𝐵]

where ∆𝑓𝐺° 𝑌 is the standard molar Gibbs energy of formation for

substance 𝑌.

II-3. Chemical Equilibrium- Appendix) how to calculate ∆𝒇𝑮° -

We can find thermodynamic database, usually the one at 298.15 K (but not

necessarily and not limited to):

∆𝑓𝐻°: Standard molar enthalpy (heat) of formation (kJ mol-1)

∆𝑓𝐺°: Standard molar Gibbs energy of formation (kJ mol-1)

𝑆°: Standard molar entropy at 298.15 K in (J mol-1 K-1)

𝐶𝑃: 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: "𝐻2 +1

2𝑂2 → 𝐻2𝑂“

∆𝑓𝐺° T K 𝐻2𝑂 g

= ∆𝑓𝐻° T K 𝐻2𝑂 g − ∆𝑓𝐻° T K 𝐻2 g −1

2∆𝑓𝐻° T K 𝑂2 g

− T 𝑆° T K 𝐻2𝑂 g − 𝑆° T K 𝐻2 g −1

2𝑆° T K 𝑂2 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 ∆𝑓𝐺° T K = 0, and the standard molar Gibbs energies

of other chemicals are aligned to them.


In addition, temperature dependence of ∆𝑓𝐺° T K can be evaluated

using 𝐶𝑃 (for constant pressure process) as:

𝐻° 𝑇2𝐾 = 𝐻° 𝑇1𝐾 + 𝑇1

𝑇2

𝐶𝑃𝑑𝑇

𝑆° 𝑇2𝐾 = 𝑆° 𝑇1𝐾 + 𝑇1

𝑇2 𝐶𝑃𝑇𝑑𝑇

𝐺° 𝑇2𝐾 = 𝐻° 𝑇2𝐾 − 𝑇2𝑆° 𝑇2𝐾

where 𝐻°, 𝑆° and 𝐺° are molar enthalpy, molar entropy, and molar Gibbs

energy. For evaluation of the standard Gibbs energy of formation, they

can be replaced with ∆𝑓𝐻°, 𝑆° and ∆𝑓𝐺°.

𝐶𝑃 is molar heat capacity at constant pressure. If 𝑇2 is enough close to

𝑇1, 𝐶𝑃 can be regarded as a constant in most case.


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 𝐻° (a molar

enthalpy of formation), 𝑆° (a molar

entropy) and 𝐺° (a molar Gibbs

energy of formation) for methanol

at 1 atm, around the boiling point.

Note that 𝐺° is a continuous

function, but 𝐻° and 𝑆° are not.

𝐻°𝑆°

𝐺°



∆𝑟𝐺° = 3 2 ∆𝑓𝐺° 𝐻2(𝑔) + 1 2 ∆𝑓𝐺° 𝑁2(𝑔) − 1 ∆𝑓𝐺° 𝑁𝐻3 𝑔

= 3 2 0 + 1 2 0 − 1 −16.637 𝑘𝐽 𝑚𝑜𝑙−1 = 16.637 𝑘𝐽 𝑚𝑜𝑙−1

(Example-3) Using the standard molar Gibbs energies of formation, calculate

∆𝑟𝐺° and 𝐾𝑃 at 298.15 K for

NH3(g) ⇌ 3/2 H2(g) + 1/2 N2(g)

∆𝑓𝐺° 𝑁𝐻3(𝑔) = −16.637 𝑘𝐽 𝑚𝑜𝑙−1, ∆𝑓𝐺° 𝐻2(𝑔) = ∆𝑓𝐺° 𝑁2(𝑔) = 0 𝑘𝐽 𝑚𝑜𝑙

−1

ln𝐾𝑃(𝑇) = −∆𝑟𝐺°

𝑅𝑇= −

16.637 × 103 𝐽 𝑚𝑜𝑙−1

8.31 𝐽 𝐾−1 𝑚𝑜𝑙−1 298.15 𝐾= −6.60

Hence, 𝐾𝑃 𝑇 = 1.36 × 10−3 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) ⇌ 2 NO2 (g)

𝐺 𝜉 = 1 − 𝜉 𝐺𝑁2𝑂4 + 2𝜉 𝐺𝑁𝑂2

= 1 − 𝜉 𝐺°𝑁2𝑂4 + 𝑅𝑇 ln𝑃𝑁2𝑂4 + 2𝜉 𝐺°𝑁𝑂2 + 𝑅𝑇 ln𝑃𝑁𝑂2

Suppose the initial state is “N2O4 (g) = 1 mol and NO2 (g) = 0 mol” and the

reaction brings the system to “N2O4 (g) = 1- 𝜉 mol and NO2 (g) = 2 𝜉 mol”.

Here we assume the reaction occurs at a constant pressure (𝑃𝑡𝑜𝑡𝑎𝑙 = 1 𝑏𝑎𝑟).

Then, as the molar fraction is 𝑥𝑁2𝑂4 =1−𝜉

1+𝜉for N2O4 and 𝑥𝑁𝑂2 =

2𝜉

1+𝜉for NO2:

𝑃𝑁2𝑂4 = 𝑥𝑁2𝑂4𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑥𝑁2𝑂4, 𝑃𝑁𝑂2 = 𝑥𝑁𝑂2𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑥𝑁𝑂2

𝐺 𝜉 = 1 − 𝜉 𝐺°𝑁2𝑂4 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 𝐺°𝑁𝑂2 + 𝑅𝑇 ln

2𝜉

1 + 𝜉

= 1 − 𝜉 ∆𝑓𝐺°𝑁2𝑂4 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 ∆𝑓𝐺°𝑁𝑂2 + 𝑅𝑇 ln

2𝜉

1 + 𝜉





N2O4 (g) ⇌ 2 NO2 (g)

𝐺 𝜉 = 1 − 𝜉 ∆𝑓𝐺°𝑁2𝑂4 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 ∆𝑓𝐺°𝑁𝑂2 + 𝑅𝑇 ln

2𝜉

1 + 𝜉

This equation gives the Gibbs energy of the reaction mixture, 𝐺, as a

function of the extent of the reaction, 𝜉. Substituting the values for ∆𝑓𝐺°:

𝐺 𝜉 = 1 − 𝜉 97.8 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 51.3 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln

2𝜉

1 + 𝜉





N2O4 (g) ⇌ 2 NO2 (g)

𝐺 𝜉 = 1 − 𝜉 97.8 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 51.3 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln

2𝜉

1 + 𝜉





N2O4 (g) ⇌ 2 NO2 (g)

𝐺 𝜉 = 1 − 𝜉 97.8 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 51.3 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln

2𝜉

1 + 𝜉

𝐺 𝜉 is minimized at 𝜉 = 0.1892 𝑚𝑜𝑙, corresponding to the equilibrium state,

thus 𝜉 = 𝜉𝑒𝑞 = 0.1892 𝑚𝑜𝑙.

Then, the equilibrium constant is:

𝐾𝑃 =𝑃𝑁𝑂2

2

𝑃𝑁2𝑂4=

2𝜉𝑒𝑞 1 + 𝜉𝑒𝑞2

1 − 𝜉𝑒𝑞 1 + 𝜉𝑒𝑞= 0.148

We can also calculate it from ∆𝑟𝐺°:

ln𝐾𝑃 = −∆𝑟𝐺°

𝑅𝑇= −

2 ∆𝑓𝐺° 𝑁𝑂2 𝑔 − 1 ∆𝑓𝐺° 𝑁2𝑂4 𝑔

8.31 𝐽 𝐾−1 𝑚𝑜𝑙−1 298.15 𝐾= −1.908

Hence, 𝐾𝑃 = 0.148





N2O4 (g) ⇌ 2 NO2 (g)

𝐺 𝜉 = 1 − 𝜉 97.8 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln1 − 𝜉

1 + 𝜉+ 2𝜉 51.3 𝑘𝐽 𝑚𝑜𝑙−1 + 𝑅𝑇 ln

2𝜉

1 + 𝜉

We may also differentiate this equation with respect to 𝜉, then finally obtain:

𝜕𝐺

𝜕𝜉𝑇,𝑃

= ∆𝑟𝐺° + 𝑅𝑇 ln𝑃𝑁𝑂2

2

𝑃𝑁2𝑂4

Since 𝜕𝐺

𝜕𝜉 𝑇,𝑃= 0 at equilibrium,

∆𝑟𝐺° = −𝑅𝑇 ln𝑃𝑁𝑂2

2

𝑃𝑁2𝑂4 𝑒𝑞

= −𝑅𝑇 ln𝐾𝑃

which is the same equation that we have derived several slides ago.

In addition, solving 𝜕𝐺

𝜕𝜉 𝑇,𝑃= 0 explicitly, 𝜉𝑒𝑞 = 0.1892 𝑚𝑜𝑙 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: 𝜈𝐴A(g) + 𝜈𝐵B(g) ⇌ 𝜈𝑌Y(g) + 𝜈𝑍Z(g)

∆𝑟𝐺 𝑇 = ∆𝑟𝐺° 𝑇 + 𝑅𝑇 ln𝑃𝑌𝜈𝑌𝑃𝑍

𝜈𝑍


𝜈𝐵

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 ∆𝑟𝐺 when

“A(g) of [𝜈𝐴 mol at 𝑃𝐴 bar] react with B(g) of [𝜈𝐵 mol at 𝑃𝐵 bar] to produce

Y(g) of [𝜈𝐴 mol at 𝑃𝐴 bar] and Z(g) of [𝜈𝑍 mol at 𝑃𝑍 bar] ”.

1) If all the partial pressures are equl to 1 bar, ∆𝑟𝐺 𝑇 = ∆𝑟𝐺° 𝑇 . In

other words, the Gibbs energy change will be equal to the standard

Gibbs energy change.

2) If the pressures are the equilibrium pressures, ∆𝑟𝐺 𝑇 = 0 and then

we obtain results that were derived in previous slides.



Consider a general reaction: 𝜈𝐴A(g) + 𝜈𝐵B(g) ⇌ 𝜈𝑌Y(g) + 𝜈𝑍Z(g)

∆𝑟𝐺 𝑇 = ∆𝑟𝐺° 𝑇 + 𝑅𝑇 ln𝑃𝑌𝜈𝑌𝑃𝑍

𝜈𝑍


𝜈𝐵

By introducing a quantity called the “reaction quotient” 𝑄𝑃 =𝑃𝑌𝜈𝑌𝑃𝑍

𝜈𝑍


𝜈𝐵, and

using an equation to correlate ∆𝑟𝐺° with 𝐾𝑃, ∆𝑟𝐺° = −𝑅𝑇 ln𝐾𝑃, then

∆𝑟𝐺 𝑇 = −𝑅𝑇 ln𝐾𝑃 + 𝑅𝑇 ln𝑄𝑃 = 𝑅𝑇 ln 𝑄𝑃 𝐾𝑃

Note that this is not for the equilibrium; ∆𝑟𝐺 𝑇 = 0 at equilibrium.

With this equation,

At equilibrium, ∆𝑟𝐺 𝑇 = 0 and thus 𝑄𝑃 = 𝐾𝑃.

If 𝑄𝑃 < 𝐾𝑃, then 𝑄𝑃 must increase (because 𝐾𝑃 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 𝑄𝑃 > 𝐾𝑃, 𝑄𝑃 decrease as the system proceeds toward equilibrium.

In the same though, the reaction is spontaneous from right to left.



(Example-4: #26-5) Consider “ 2SO2(g) + O2(g) ⇌ 2SO3(g) ”. The

equilibrium constant 𝐾𝑃 = 10 at 960 K. Calculate ∆𝑟𝐺 and check in which

direction the reaction will proceed spontaneously for

2 SO2 (1.0 × 10−3 𝑏𝑎𝑟) + O2 (0.20 𝑏𝑎𝑟) ⇌ 2 SO3(1.0 × 10

−4 𝑏𝑎𝑟)

𝑄𝑃 =𝑃𝑆𝑂32

𝑃𝑆𝑂22 𝑃𝑂2

=1.0 × 10−4 2

1.0 × 10−3 2 0.20= 5.0 × 10−2

Note that the value for 𝑄𝑃 is unitless because the definition of 𝑄 is, for

example in a general form, 𝑄 = 𝑃𝑌 𝑃°


𝑃𝐴 𝑃° 𝜈𝐴 𝑃𝐵 𝑃° 𝜈𝐵where 𝑃° = 1 𝑏𝑎𝑟.

∆𝑟𝐺 𝑇 = 𝑅𝑇 ln 𝑄𝑃 𝐾𝑃 = 𝑅𝑇 ln 5.0 × 10−2 10 < 0

Hence, due to ∆𝑟𝐺 𝑇 < 0 (or due to 𝑄𝑃 < 𝐾𝑃), the reaction proceed

from left to right.

II-3. Chemical Equilibrium- $26-6: The sign of ∆𝑟𝐺 and not that of ∆𝑟𝐺° 𝑇 determines the direction of reaction

spontaneity -

It should be clearly understood that ∆𝑟𝐺° is the value of ∆𝑟𝐺 when all the

reactants and products are unmixed at partial pressures equal to 1 bar.

Namely, ∆𝑟𝐺° is the standard Gibbs energy changes. (recall that “standard”

assumes 1 bar) Hence,

If ∆𝑟𝐺° < 0, then 𝐾𝑃 > 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 ∆𝑟𝐺° > 0, then 𝐾𝑃 < 1, thus reaction proceeds from products to

reactants in the same condition.

So, the sing of ∆𝑟𝐺° just indicates the spontaneous reaction direction

for the condition of 𝑃𝑝𝑎𝑟𝑡𝑖𝑎𝑙 = 1 bar (for all species of both reactants

and products); not necessarily for all conditions.


spontaneity -

(Example-5) Consider “ N2O4 (g) ⇌ 2 NO2 (g) ”,

for which ∆𝑟𝐺° = 4.729 kJ 𝑚𝑜𝑙−1 and 𝐾𝑃 = 0.148 at 298.15 K.

First, make sure that this ∆𝑟𝐺° > 0 does NOT mean “no N2O4 dissociate at

298.15 K when we place some of N2O4 in a reaction vessel”.

To correctly consider, we have to calculate ∆𝑟𝐺 as:

∆𝑟𝐺 = ∆𝑟𝐺° + 𝑅𝑇 ln𝑄𝑃 = 4.729 𝑘𝐽 𝑚𝑜𝑙−1 + 2.479 𝑘𝐽 𝑚𝑜𝑙−1 ln

𝑃𝑁𝑂22

𝑃𝑁2𝑂4

If we just ”place some of N2O4 in a reaction vessel”, ln 𝑃𝑁𝑂22 𝑃𝑁2𝑂4 has a

large negative value, thus ∆𝑟𝐺 < 0. Accordingly, some N2O4 dissociate.

The equilibrium state is achieved by the condition ∆𝑟𝐺 = 0 (note ∆𝑟𝐺° is a

constant while ∆𝑟𝐺 changes depending on 𝑃𝑁𝑂2 and 𝑃𝑁2𝑂4 ), at which point

𝑄𝑃 = 𝐾𝑃. Until this point is achieved, 𝑃𝑁𝑂2 increases and 𝑃𝑁2𝑂4 decreases.


spontaneity -

(Example-6) Consider “ H2 (g) + ½ O2 (g) ⇌ H2O (l) ”,

for which ∆𝑟𝐺° = −237 kJ 𝑚𝑜𝑙−1 at 298.15 K.

In this case, ∆𝑟𝐺° 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 “no” 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 “yes” is actually “maybe”. 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

Van’t Hoff Equation -

We utilize the Gibbs-Helmohltz equation:

Substitute ∆𝑟𝐺° = −𝑅𝑇 ln𝐾𝑃 (note this is a definition of ∆𝑟𝐺°, not a condition

achieved at equilibrium state) for this equation:

𝜕∆𝐺° 𝑇

𝜕𝑇𝑃

=∆𝐻°

𝑇2

𝜕 ln𝐾𝑃(𝑇)

𝜕𝑇𝑃

=𝑑 ln𝐾𝑃(𝑇)

𝑑𝑇=∆𝑟𝐻°

𝑅𝑇2

This means that

If ∆𝑟𝐻° > 0 (endothermic reaction), 𝐾𝑃(𝑇) increases with temperature.

If ∆𝑟𝐻° < 0 (exothermic reaction), 𝐾𝑃(𝑇) decreases with temperature.

Integrate the equation:ln𝐾𝑃(𝑇2)

𝐾𝑃(𝑇1)=

𝑇1

𝑇2 ∆𝑟𝐻°(𝑇)

𝑅𝑇2𝑑𝑇

If the temperature range is small enough to consider ∆𝑟𝐻° constant:

ln𝐾𝑃(𝑇2)

𝐾𝑃(𝑇1)= −

∆𝑟𝐻°

𝑅

1

𝑇2−1

𝑇1



(Example-7) Consider “ PCl3 (g) + Cl2 (g) ⇌ PCl2 (g) ”.

Given that ∆𝑟𝐻° has an average value of -69.8 kJ mol-1 over 500-700 K and

𝐾𝑃 is 0.0408 at 500K, evaluate 𝐾𝑃 at 700K.

As ∆𝑟𝐻° can be assumed as a constant over the concerned temperatures,

ln𝐾𝑃(𝑇2)

𝐾𝑃(𝑇1)= −

∆𝑟𝐻°

𝑅

1

𝑇2−1

𝑇1

Substituting provided values gives:

ln𝐾𝑃(700 𝐾)

𝐾𝑃(500 𝐾)= ln

𝐾𝑃(700 𝐾)

0.0408= −

−69.8 × 103

𝑅

1

700−1

500

𝐾𝑃 700 𝐾 = 3.36 × 10−4

Note that since the reaction is exothermic, 𝐾𝑃 700 𝐾 is less than 𝐾𝑃 700 𝐾 ;

namely less product (PCl2) at higher temperatures as 𝐾𝑃 = 𝑃𝑃𝐶𝑙2

𝑃𝑃𝐶𝑙3𝑃𝐶𝑙2.



ln𝐾𝑃(𝑇2)

𝐾𝑃(𝑇1)=

𝑇1

𝑇2 ∆𝑟𝐻°(𝑇)

𝑅𝑇2𝑑𝑇

Finally, we consider the temperature dependence of ∆𝑟𝐻° constant in:

This equation can write down as:

ln𝐾𝑃(𝑇) = ln𝐾𝑃(𝑇1) + 𝑇1

𝑇 ∆𝑟𝐻°(𝑇′)

𝑅𝑇′2𝑑𝑇′

For example, as we learned, the temperature dependence of ∆𝑟𝐻° may

be written as:

∆𝑟𝐻° 𝑇2 = ∆𝑟𝐻° 𝑇1 + 𝑇1

𝑇2

∆𝐶°𝑃 𝑇 𝑑𝑇

or expanded in respect to temperature as:

∆𝑟𝐻° 𝑇 = 𝛼 + 𝛽𝑇 + 𝛾𝑇2 + 𝛿𝑇3 +⋯

In this latter case, ln𝐾𝑃(𝑇) becomes:

ln𝐾𝑃(𝑇) = −𝛼

𝑅𝑇+𝛽

𝑅ln 𝑇 +

𝛾

𝑅𝑇 +

𝛿

2𝑅𝑇2 + (𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛_𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)…

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Lecture 409.317 2013.F N07(Before Class Version)