Chemical thermodynamics

Chapter 01

Thermodynamics: crude concept of 1st and 2nd law and their significance

CxHy (fuel)

O2

CxHy + O2 CO2 + H2O (combustion)

CO2 (exhaust)

Car moves fwd.

Combustion of fuel converts chemical energy into mechanical energy to run a car

Introduction:

+ H2O

In this example:

“Energy is not created, neither destroyed, total energy of the universe is constant.”

Above statement is the simplest outcome or the major significance of 1st lawof thermodynamics.

Conservation of energy is not enough to understand the laws of the universe

CxHy + O2CO2 + H2O

O2

CO2 pushing car backwd.+ H2O

?

? (never)

By feeding the car CO2 and H2O through the exhaust and pushing the car backwardscan we generate fuel in the car and produce O2 in the air?

Note: the absurd situation above is not violating the 1st law of thermodynamics!

Therefore in thermodynamics, 1st law is not good enough to understand the directionof any natural process of universe.

We require 2nd law of thermodynamics to understand the direction of any natural process

Significance of thermodynamics in Chemistry:

A + B C + D

•To understand whether a reaction will be exothermic or endothermic

•To understand whether a reaction or a physical change will attain an equilibrium

•To understand the direction of a spontaneous chemical change

•To understand the potential (voltage) of a chemical cell or a battery

•To understand the mechanism of chemical reactions in certain cases

•To understand how a chemical or physical change respond with change of T and P

•To understand the extent of a chemical change

First Law of Thermodynamics and Chemistry

First law of thermodynamics: (internal energy, heat, work, enthalpy)

We know that the total energy of the universe is constant.

Universe = system + surroundings

A system is defined as the particular region of space that we are making measurements on. Example of a system :

test tube

In chemistry, first law of thermodynamics is all about gaining and losing of energy by a chemical system. In the context of thermodynamics we refer this energy as ‘internal energy’ (U) and its unit is the joule (J).

The absolute value of U of a system cannot be measured but the difference (ΔU) can be.

ΔU = U final state – U initial state = Uf - Ui

Internal energy (ΔU) is a state function. It does not depend on the process of change, butonly depends on the initial and final states.

First law of thermodynamics:

Ui

A system withsome initial U

Uf

work done by the system in the surroundings

Q

(supplied heat)

the system withfinal U

If a system having initial internal energy (Ui) is supplied with heat Q, then the absorbedheat may be used to increase the internal energy of the system or perform work or both.

If the final internal energy after absorbing heat is (Uf) then change in internal energy= ΔU = Uf - Ui

According to the first law of thermodynamics, energy lost = energy gained

Q = ΔU + work done by the system

or, Q = ΔU + (-W) ……………………eq. (1)

Please note: according to IUPAC convention, work done by the system is expansion work and taken as (-W)

+

First law of thermodynamics: From equation (1) we get,

Q = ΔU + (– W)

or, ΔU = Q + W ……………….eq. (2)

or, dU = đQ + đW (in the differential form) ……………….eq. (3)

dU ≡ exact differential (we can apply an exact differential to any state function, which is path independent)

đQ, đW ≡ inexact differential (we can apply an inexact differential to any path function, contrary to state function, path function is path dependent)

Interpretation of work (W): work done by the system can be expansion and non-expansion.In the context of first law, we will ignore any non-expansion work.

Expansion work: In thermodynamics it is classically idescribed as the movement of a pistonby expansion or compression of a gas.

So, W = Wexp + Wnon-exp

Here, W = Wexp by setting Wnon-exp = 0

Pin

Poutcylinder

piston

gasB

gasA

Expansion work of a gas:If Pin > Pout then gasA expands by performing workagainst constant pressure Pout to obtain volume V2

from initial volume V1

So work done by the system = (-W) = Pout (V2-V1) = Pout ΔV …….. eq. (4)

In differential form, we can write eq. (4) as

(-đW) = Pout dV = Pdv …….. eq. (5)assuming Pout = P (any arbitrary opposing pressure)

For a finite change where the gas expands from volume V1 to V2, the total work done will be:

-∫đW = ∫Pdv (if P is variable) …….. eq. (6) V1

V2

V1

V2

-∫đW = P∫dv (if P is constant) …….. eq. (7) V1

V2

V1

V2

For ideal gas, P = nRT/V so, we can replace the value of P in eq. (6) and integrate to find W.

The first law of thermodynamicsIn differential form:

dU = đQ - PdVexpansion work done

by the system

Irreversible and reversible expansion work:

Pi TVi P TVf

Single step expansion against a constant pressure Pout

Stage A Stage B

Pout

Pout

Irreversible transformation!

-∫đW = P∫dv ( Pout = P = constant) = P(Vf – Vi)Vi

Vf

Vi

Vf

irreversible work done =

Pi TVi

Stage A

Pout

P TVf

Stage B

Pout

P2 TV2

Pout

P3 TV3

Pout

P4 TV4

Pout

infinite steps

…. ….

Pout = P

1 2 3

Pout = P2 Pout = P3 Pout = P4 Pout = P

multiple step expansion against a variable pressure

-∫đW = ∫Pdv = nRTln(Vf/Vi) = maximum work doneVi

Vf

Vi

Vf

reversible work done =

Reversible transformation!

(P is the variable pressure at any step)

First law of thermodynamics:

đQ + đW = (∂U/∂T)V dT + (∂U/∂V)T dV [from eq. (3)]

Properties of internal energy (U) in mathematical form:

Internal energy of a system having fixed mass is a function of temperature and volume

U = f(T, V) this relationship can be expressed in the form of a differential equation:

dU = (∂U/∂T)V dT + (∂U/∂V)T dV ……………..eq. (8)

partial derivative partial derivative

Please note: any state function (exact differential) can be expressed in the form of eq. (8)

or, đQ - PdV = (∂U/∂T)V dT + (∂U/∂V)T dV [from eq. (5)] ……………..eq. (9)

At constant volume (isochoric) process, we know that dV = 0, so

đQV = (∂U/∂T)V dT = CVdT = dU ……………..eq. (10)

We define (∂U/∂T)V as Cv (heat capacity at constant volume)

So, under constant volume process,

dU = CvdT [where Cv = đQV/dT = (∂U/∂T)V ] Or, ΔU = CvΔT (for a finite change)

First law of thermodynamics:

Therefore, we can write the following equations

dU = CV dT + (∂U/∂V)T dV ……………..eq. (11)

đQ - PdV = CV dT + (∂U/∂V)T dV ……………..eq. (12)

For ideal gas, the derivative (∂U/∂V)T = 0

So for ideal gas, following relationships are true

dU = CV dT (under any process, U is a function of T only)

or, đQ = CV dT + PdV

First law of thermodynamics: For any isothermal process:\]=6543 `Temperature remains the same, that is dT = 0

So in general, the first law of thermodynamics becomes

đQ = (∂U/∂V)T dV + PdV ……………..eq. (13)

and for ideal gas,

đQ = PdV ……………..eq. (14)

For any adiabatic process:When no heat is exchanged during a process, we call the process as adiabatic.

It means đQ = 0So the general statement of first law becomes:

dU = đW = -PdV ……………..eq. (15) CV dT + (∂U/∂V)T dV = -PdV ……………..eq. (16)

First law of thermodynamics:

In chemistry, generally reactions take place at constant pressure than at constant volume!

But we have seen so far that heat absorbed at constant volume is change in internal energy!

So, what happens when heat is absorbed at constant pressure (this happens in chemistry)?

We define a state function (H) that gets changed when heat is absorbed at constant pressure.

Just like U, we can write H = f(T, P)

dH = (∂H/∂T)V dT + (∂H/∂P)T dP ……………..eq. (17)

At constant pressure process, we know that dP = 0, so

đQP = (∂H/∂T)P dT = CPdT = dH ……………..eq. (18)

We define (∂H/∂T)V as CP (heat capacity at constant pressure)

We call this state function (H) as enthalpy

For ideal gas, (∂H/∂P)T = 0

So heat of reaction or heat exchanged during a chemical reaction is actually enthalpy of the reaction since the chemical reaction takes place under constant pressure.

First law of thermodynamics: Enthalpy:

Relationship between enthalpy (H) and internal energy (U):

H = U + PV ……………..eq. (19) (P & V stand for pressure and volume of the system, respectively)

dH = dU + PdV + VdP

dH = dU + PdV (under constant pressure) ……………..eq. (20)

For 1 mole of ideal gas, H = U + RT

In differential form,

The derivation of the above relationship can be done elsewhere.

In words, change of enthalpy at constant pressure is change of internal energy plus productof pressure and change of volume of the system.

First law of thermodynamics: Relationship between QP and Qv

[from eq. (20)]

Under constant pressure,

dH = dU + PdV

For a finite change under constant pressure,ΔH = ΔU + PΔV = Qp [from eq. (18)] and Qv = ΔU [from eq. (10)]

So, we can write, Qp = Qv + PΔV or, Qp - Qv = PΔV ……………..eq. (21)

Relationship between CP and Cv for ideal gas

đQ = CV dT + (∂U/∂V)T dV + PdV …………….. from eq. (12)

đQP = CV dT + (∂U/∂V)T dV + PdV …………….. (under constant pressure)

đQP /dT = CV + (∂U/∂V)T (dV/dT) + P(dV/dT) ….(under constant pressure) or,

CP = CV + (∂U/∂V)T (∂V/∂T)P + P(∂V/∂T)P ….(under constant pressure) or,CP = CV + P(∂V/∂T)P ………………………………….[for real gas, the term (∂U/∂V)T = 0] or,CP = CV + nR ………………………………….[using V= nRT/P at constant P] or,

or, CP /n - CV /n = R ……………eq. (22)

First law of thermodynamics: CP /n - CV /n = R ……………eq. (22) [for ideal gas]

or, CP - CV = R …………eq. (23) [where CP & CV are respective molar heat capacities]

Adiabatic changes of state for ideal gas

CV dT = -PdV …………..eq. (24) [from eq. (15) & (16)]

We have seen earlier that for an ideal gas, when đQ = 0

The negative sign denotes that if the gas expands adiabatically, it’s temperature woulddecrease.

Irreversible adiabatic change for an ideal gas:

For a finite change in a single step, if the gas expands from V1 to V2, against a constant opposingpressure P, eq. (24) can be written as:

CV ΔT = -PΔV

CV (T2-T1) = -P(V2-V1)

Here T2 < T1 and V2 > V1

First law of thermodynamics: Reversible adiabatic change for an ideal gas:

Since, CV dT = -PdV

the temperature drop becomes maximum if the gas expands in infinite number of stepsagainst a variable pressure P at every step, i. e. gas expands in a reversible adiabatic manner.

Therefore, for a reversible adiabatic expansion where gas expands from V1 to V2 , we can write:

CV dT = -nRTdV/V

T1

V2

V1

T2

CV dT/T = -nRdV/V or, CV dlnT = -nRdlnV or,

CV ∫dlnT = -nR∫dlnV or, CV ln(T2/T1) = -R ln(V2/V1) or, ln(T2/T1) =-(Cp – CV) ln(V2/V1)

CV

or, ln(T2/T1) = -(γ – 1) ln(V2/V1) or, ln(T2/T1) = -(γ – 1) ln(V2/V1)

or, (T2/T1) = (V1/V2)(γ-1)

[CP/CV = γ, a constant]

…………..eq. (25)

Using ideal gas law, the above equation can be transformed into:

P1V1γ = P2V2

γ or, PVγ = constant …………..eq. (25)

First law of thermodynamics: Graphical representation of adiabatic and isothermal change of an ideal gas

PV = constant

PVγ = constant

First law of thermodynamics: application in chemical reaction

Heat of reaction (ΔH)The heat of reaction or enthalpy of reaction is governed by two fundamental laws:

1) Lavoisier and Laplace’s law (1782)

2) Hess’ Law (1840)

The Heat (ΔH) exchanged in a transformation is equal and opposite in the reverse direction.

A BΔH1

ΔH2According to this law: ΔH1 = - ΔH2

The Heat (ΔH) exchanged in a transformation is same whether the process occurs in one or several steps.

A DΔHtotal

A BΔH1

CΔH2

DΔH3

single step transformation mutiple step transformation

According to this law: ΔHtotal = ΔH1 + ΔH2 + ΔH3(This is an obvious outcome since ΔH is a state function)

Application of Hess’s law

First law of thermodynamics: application in chemical reaction

(1) C (s) + O2 (g) → CO2 (g) ΔH1 = -394 kJmol-1

(2) H2 (g) + ½ O2 (g) → H2O (l) ΔH2 = -286 kJmol-1

(3) C2H6 (g) + 7/2 O2 (g) → 2CO2 + 3H2O (l) ΔH3 = -1542 kJmol-1

Calculate the enthalpy of formation of ethane in the given reaction from the following:

2C (s) + 3H2 (g) → C2H6 (g) ΔHf = ?

The enthalpy of formation of ethane 2C (s) + 3H2 (g) → C2H6 (g) may be calculated by:

2C (s) + 2O2 (g) → 2CO2 (g)

3H2 (g) + (3/2)O2 (g) → 3H2O (l)

2CO2 (g) + 3H2O (l) → C2H6 (g) + (7/2)O2 (g)

2C (s) + 3H2 (g) → C2H6 (g)

2 x (1)

3 x (2)

Reverse (3)

Add:

ΔH = 2(-394) kJmol-1

ΔH = 3(-286) kJmol-1

ΔH = +1542 kJmol-1

ΔH = -104 kJmol-1

Therefore, ΔHf = -104 kJmol-1

Second Law of Thermodynamics

Second law of thermodynamics:

Ocean’s hot waterQH

Can the ship move fwd??

First law:QH = (-W) if ΔU = 0

work doneby ship’s engine

(-W)

According to the first law of thermodynamics, a ship’s engine can extract heat from ambientOcean water and transform the heat into work under isothermal condition.

Do we see such situation in real life?

The hypothetical situation shown above is in perfect agreement with first law, but the situation is impossible in reality!

(hot reservoir)

Second law of thermodynamics: The impossible situation described in previous slide is actually a statement of second of thermodynamics in a particular form.

It is impossible for a system operating in a cycle and connected to a single heat reservoir to produce work in the surroundings.

Kelvin statement of second law (1850):

IMPOSSBLE POSSIBLE

Hot reservoir

Cold reservoir

QC

TC

TH

QH

(-W) = QH - QC

-

So, second law states that under isothermal condition no engine can convert heat into work!

engine

Efficiency (η) of the real engine = work done/ heat supplied= (QH - QC)/QH

= 1 – QC/QH ………….. eq. (26)

Second law of thermodynamics:

Hot reservoir

Cold reservoir

QC

TC

TH

QH

W

Kelvin statement of second law is also true in reverse way: known as Clausius statementof second law.

IMPOSSIBLE

Hot reservoir

Cold reservoir

QC

TC

TH

QH

POSSIBLE

A refrigerator cannot work itself by removing heat from a colder compartment and dumping it to a hotter environment. We need to spend energy (work) to achieve refrigeration.

Engine/compressor

Earlier, at the introduction we appreciated that second law determines the direction of a natural process. This statement can be connected with Kelvin’s statement by a historictheorem known as “Carnot cycle”.

Second law of thermodynamics: The Carnot cycle From the work of French Engineer Sadi Carnot (1824) what extent of work can be extracted from heat by an engine. Carnot is often referred as father of thermodynamics.

Carnot’s theorem was later developed by Clausius and Kelvin to frame second law.

In his theorem, 1 mole of ideal gas contained in a cylinder, fitted with a piston was used as an engine.

This engine was subjected to cyclic transformation between two reservoirs: hot and cold, kept at constant temperature.

In every step, the engine was subjected to reversible transformation to obtain maximum work out of it.

P1 THV1P2 THV2

P3 TCV3 P4 TCV4 P1 THV1

isothermal

exp.

adiabatic

exp.

isothermal

compr.

adiabatic

compr.

Step 1 Step 2 Step 3 Step 4

Second law of thermodynamics: the Carnot cycle

W

work done = RTH ln(V2 / V1)

work done = RTC ln(V4 / V3)

P1 THV1

P2 THV2

P3 TCV3

P4 TCV4

work done = CV (TC - TH)

work done =CV (TH - TC)

Total work done (-W) in Carnot cycle:

Second law of thermodynamics:

= RTH ln(V2 / V1) + CV (TC - TH) + RTC ln(V4 / V3) + CV (TH - TC)

= RTH ln(V2 / V1) - RTC ln(V3 / V4)

= RTH ln(V2 / V1) - RTC ln(V2 / V1) [from the graph as well as mathematically V2/V1 = V3/V4]

= R(TH – TC) ln(V2/V1)

Efficiency of Carnot engine (η) = Total work done (-W) by engine

Heat withdrawn by engine (QH)=

R(TH – TC) ln(V2/V1)

RTH ln(V2 / V1)

η = 1 – TC/TH As TH > TC, efficiency of Carnot engine is always less than unity.

Now, we have seen earlier that η = 1 – QC/QH for any real engine [from eq. (26)].

….. eq. (27)

We can write comparing eq. (26) & (27), TC/TH = QC/QH ….. eq. (28)

Therefore the conclusion of Carnot cycle is: no real heat engine is 100% efficient

Second law of thermodynamics: concept of entropy

From eq. (28), we get: QH/QC = TH/TC

or, QH/TH = QC/TC

or, đQH/TH = đQC/TC

or, đQH/TH - đQC/TC = 0

or, đQH/TH + (-đQC)/TC= 0

From Carnot cycle if we try to assess the value of đQ/T then combining all the stepsin the cyclic process, we can write:

∑ đQ/T = ∫ đQ/T = đQH/TH + 0 + (-đQC)/TC + 0 = 0 (Heat rejected in the cold reservoir is associated with negative sign)

In Carnot cycle each & every step is reversible, so heat exchanged in each step is also reversible

∫ đQ/T = ∫ đQrev/T = 0 …….. eq. (29)

∫ ≡ cyclic integral,

If cyclic integral of any function is zero,then the function is a state functionSo Qrev/T is a state function.

Second law of thermodynamics: concept of entropy The state function Qrev/T is called entropy and denoted by S

∫ dS = 0 …….. eq. (30) where dS = đQrev/T

So from eq. (29), we get

For any reversible cyclic process, a system’s entropy change is zero.

Also, it can be proved that for any irreversible cyclic process

∫ đQirrev/T < 0 …….. eq. (31)

Combined form of first and second law of thermodynamics:

đQrev = dU + (– đW)……….. from first law, eq. (3) for a reversible process

or, TdS = dU + (– đW)……….. Combined form of first and second law, eq. (32)

or, TdS = dU + PdV……….. eq. (32)

Second law of thermodynamics: properties of entropy Concept of systems: definition and analogy

Open system Mass as well as energy

can pass through

Closed systemOnly energy

can pass through but no mass

Isolated systemIdeally neither mass

nor energy canPass through

An open containerof boiling water

A closed containerof boiling water

A thermo flask

v v v

đQ = 0đQ ≠ 0 đQ ≠ 0

Second law of thermodynamics: properties of entropy The Clausius inequality

A system is transformed irreversibly from state 1 to state 2

Then it was restored reversibly from state 2 to state 1

∫ đQ/T = ∫ đQirrev/T + ∫ đQrev/T 1

12

2

So we can writeSince this is an irreversible cycle, this equation is less than zero [from eq. (31)]

or, ∫ đQirrev/T - ∫ dS < 0 1 1

2 2

or, ∫ dS > ∫ đQirrev/T 1 1

2 2

for any infinitesimal change, dS > đQirrev/T Clausius inequality ……………eq. (33)

This is the fundamental requirement for any real transformation.

Second law of thermodynamics: properties of entropy

System + Surroundings = Universe

If earth is a system and space is its surroundings then the entire universe can bethought of an isolated system. Because neither mass nor energy can be transported in or out of the universe.

For any change of state in an isolated system, đQirrev = 0 So the Clausius inequality becomes dS > 0 for an isolated system

Therefore, the requirement for a real transformation in an isolated system is that The entropy change must be positive or entropy must increase. This is true for any natural change.

The universe is an isolated system, so dS for universe > 0

Second law of thermodynamics: entropy and natural process

The entropy change of an isolated system attains the maximum value at en equilibrium situation.

In reality, whether the entropy change of the universe will at all reach the maximum value or the equilibrium, and if so, when is not known.

But entropy of the universe has been trying to attain a maximum value from its birth.

All the essence of first and second law of thermodynamics can be summed up by a Statement:

“The energy of the universe is constant; the entropy strives to reach a maximum”

- The aphorism of Clausius

Interpretation of Entropy and Chemistry

Entropy and natural processesFrom second law of thermodynamics (Clausius statement):

ΔS > 0 for any process that takes place by itself, naturally. In other words,natural processes proceed in a direction where entropy change is positive.

O2

CO2 pushing car backwd.+ H2O

?

Now, we can understand why fuel is not produced by performing the above operation,which is an unnatural process.

Natural processes are irreversible like the example above.

A (SA) B (SB)

In terms of entropy, if SB > SA, then formation of B or the fwd. direction is preferred

In terms of entropy, if SB < SA, then formation of A or the backwd. direction is preferredIf, if SB = SA, then neither direction is preferred. We call this situation an equilibriumsituation in terms of entropy (ΔS = 0).

Entropy and randomness:

Entropy is also interpreted as a measure of randomness or disorderdness or chaos.

less entropymore entropy

Entropy and probability:Imagine a person randomly tossing off bricks on the ground. Which situation would youexpect to see?

most improbable situation most probable situationordered array disordered pile

less entropy more entropy

A more probable event or arrangement is associated with higher entropy

ΔS increasing

Solid

most ordered stateLiquid

moderately ordered state

Gas

Least ordered state

Entropy and state of a substance:

For any substance, the entropy increases for its change of state from solid to gas

Entropy of a binary mixture

0 1x (mole fraction)

ΔS

In a binary mixture, the entropy change is maximum when the two components are presentin equal amount.

In a 50% – 50% mixture, the randomization or the disorderdness becomes maximum

Entropy change and some chemical reactions:butane (C4H10) methane (CH4) + propene (C3H6)850o Cv

one reactant two products (net increase = +1)

More disorderdness in product side!

ΔS > 0

due to formation of one extra gas molecule (CO2), the ΔS is a large positive value!

3 H2C=CH2

3 molecules of ethylene

850o C

1 molecule of cyclohexane (C6H6)

ΔS = negative

Properties of entropy: mathematical equations

Partial differential: dS = (∂S/∂T)V dT + (∂S/∂V)T dV ……………..eq. (34)

S = S (T, V)

Partial differential: dS = (∂S/∂T)P dT+ (∂S/∂P)T dP ……………..eq. (35)

S = S (T, P)

Like any other state function, S can be expressed as function of T, V or T, P

Also, from eq. (32) dS = dU/T + (P/T)dV ……………………….eq. (36)

For ideal gas, eq. (36) becomes dS = (Cv/T)dT + (nR/V)dV ………… eq. (37)

v

vv

dS = dH/T – (V/T)dP …………… eq. (38) since dU = dH – PdV – VdP v

Under constant pressure, for any system

dS = dH/T = (CP/T)dT…………………………….. Eq. (39) at constant pressurev

Properties of entropy: Third law of thermodynamics

dS = (CP/T)dT

Under constant pressure:

∫ dS = ∫ (CP/T)dT 0 0

T T

or, ΔS = ST – S0 =

or, ST = S0 + ∫ (CP/T)dT 0

T

ST = ∫ (CP/T)dT 0

T

S0 is the entropy at absolute zero, if the temperature is taken in Kelvin scale

S0 is zero at absolute zero. Because at absolute zero, any matter is supposed to be solid and hence perfectly crystalline or packed. There is absolutely no degrees of freedomin those molecules. This statement is known as the Third law of thermodynamics.

v ……………. eq. (40)

If the pressure P = 1 atm, then ST = STo called as the standard enthalpy of a substance

The statement of third law of thermodynamics is also known as Planck Statement (1913)

Change of entropy in change of states:

Liquid Vapor

Isothermal condition at boiling point (Tb)

ΔS = ΔQrev/Tb = ΔHvap/Tb

(under constant pressure)

Boiling of water at Tb

Solid Liquid

Isothermal condition at melting point (Tm)

ΔS = ΔQrev/Tm = ΔHfusion/Tm

(under constant pressure)

Melting of ice at Tm

ΔHvap is the heat of vaporization

ΔHfusion is the heat of fusion or melting

Properties of entropy: standard entropyThe standard entropy of a liquid at a temperature above its melting point:

SoT = ∫ (Cp

o (s)/T)dT 0

Tm

∫ (Cpo (l)/T)dT

Tm

T

+ ΔHOfusion /Tm +

The standard entropy of a gas at a temperature above its boiling point:

SoT = ∫ (Cp

o (s)/T)dT 0

Tm

∫ (Cpo (l)/T)dT

Tm

Tb

+ ΔHOfusion /Tm + + ΔHO

vap /Tb + ∫ (Cpo (g)/T)dT

Tb

Tm

………. eq. (41)

…. eq. (42)

Substance So/R (at 298 K)C (diamond) 0.286C (graphite) 0.690

Hg 9.129H2O 8.41

H2 15.7

CO2 25.699

Standard entropies of few substances at 298 K:

Free energy and Spontaneity

Reversibility and transformations:

From our knowledge of thermodynamics so far, we learnt:

For irreversible processes, the Clausius inequality tells usTdS > đQ irreversible

Natural processes take place by itself in a spontaneous manner, these processesare irreversible in nature and we have seen that for such processes, dS > 0

Reversible processes consists of infinitesimal reversible steps, where in each stepcondition of equilibrium is valid.

For such reversible processes, dS = đQrev/T = 0 since, Qrev = 0 for isolated system

So condition for equilibrium in terms of entropy isdS = 0 or TdS = đQ rev

1

So condition for spontaneous transformation in terms of entropy isdS > 0 or TdS > đQ irreversible

2

Condition for equilibrium and spontaneity:Therefore, combining the two conditions in previous slide, we get:

TdS ≥ đQ (for any transformation reversible or irreversible)

Where TdS = đQ (equilibrium) TdS > đQ (spontaneity) ………………… eq. (43)

For isolated system, we can take đQ = 0

However, the condition dS > 0 is not enough to predict the whether a chemical reaction would be spontaneous!

NH3 (g) + HCl (g) NH4Cl (s) ΔSo = - 284 JK-1mol-1

Although the change of entropy is a large negative value, the above reaction isspontaneous!

In the remaining portion of this section, we will try to learn what is the state functionother than the entropy that predicts the condition of spontaneity and equilibrium.

at 298 K

Gibb’s energy or free energy (ΔG): Let’s consider transformations at constant temperature and constant pressure:

-dU + TdS ≥ (-đW)-dU + TdS ≥ PdV + đWa (where Wa is the non-expansion work)

-dU - PdV + TdS ≥ đWa

-dU - dPV + dTS ≥ đWa (under constant temperature and pressure)

-d(U + PV - TS) ≥ đWa (under constant temperature and pressure)

-dG ≥ đWa (Where G ≡ U + PV –TS, a state function, known as Gibbs energy/Free energy )

-ΔG ≥ Wa ………………….. eq. (44)

-ΔG = Wa (reversible) (condition for equilibrium)-ΔG > Wa (condition for spontaneity)

For any transformation at constant temp. & pressure, the decrease of Gibbs energy is greater than equal to the maximum non-expansion work available over and above the expansion work in that transformation.

TdS ≥ đQ from eq. (43)

Gibb’s energy or free energy (ΔG):

-ΔG = 0 or ΔG = 0 (equilibrium) …………………………………………………………………………. eq. ( 45)-ΔG > 0 or ΔG < 0 (spontaneity, transformation is in natural direction) …………….. eq. (46)

A real process, where we encounter the non-expansion work (Wa):

Cu2+ (aq) + Zn (s) = Zn2+ (aq) + Cu (s) (electrochemical reaction in Daniell cell)

The electrochemical work done in a battery Ξ Wa

For any other process, where Wa = 0, the condition for equilibrium and spontaneity becomes:

and obviously,

ΔG > 0 (non-spontaneity, the transformation is not in natural direction) ………….. eq. (47)

Earlier, we have seen: G ≡ U + PV –TS = H - TS

Therefore, we can write: ΔG = ΔH – TΔS ………….. eq. (48)

dG = dH – TdS ………….. eq. (49)

Free energy and spontaneity (ΔG): All natural processes proceed in a direction in which a system’s free energy gets lowered.

For any spontaneous chemical reaction, ΔG = negative

ΔG = negative

NH3 (g) + HCl (g) NH4Cl (s)

ΔG = - ve

ΔG = + ve

ΔGo = - 92 kJmol-1

298 K, 1 atm

MgCO3 (s) MgO (s) + CO2 (g)

ΔGo = + 65 kJmol-1

298 K, 1 atm

ΔGo = standard free energy, or ΔG at 298 K and under 1 atm pressure.

A non-spontaneous reaction at 298 K, may become spontaneous at higher temperature!

Under standard condition, we can write:

ΔGo = ΔHo – TΔSo …….. eq. (50) bridging equation between first and second law of thermodynamics

C (diamond) C (graphite)

But we have never seen diamonds turning into graphite not even in thousand years!

ΔGo = negative does not guaranty that a reaction will take place spontaneously in standard condition!

Thermodynamics is unable to answer such anomaly.

That’s why we have to learn Chemical Kinetics later.

Free energy and spontaneity (ΔG):

ΔGo = - 2.9 kJmol-1

The free energy change of above transformation is negative under standard condition!

Properties of ΔG and Gibbs fundamental equation: Since G is a state function and G = G (T, P), we can write

dG = (∂G/∂T)P dT+ (∂G/∂P)T dP ……………..eq. (51)

dG = dH – TdS - SdT (T & P are variable here)

Also, by definition,

= dU + PdV + VdP – TdS – SdT (H = U + PV)

= TdS + VdP – TdS – SdT ( dU + PdV = đQ = TdS assuming reversible process) = – SdT + VdP

Above equation explains the dependence of G on T & P when both of these vary

G = H - TS

dG ……………..eq. (52)

This equation (52) is known as Gibbs fundamental equation as it relates the change of free energy with the two most natural variable on earth: T and P.

Dependence of ΔG and on P and T:

dG = (∂G/∂T)P dT+ (∂G/∂P)T dP ……………..eq. (51)

= – SdT + VdP dG ……………..eq. (52)

Comparing the above two equations, we get:

(∂G/∂T)P = - S

(∂G/∂P)T = V

Since entropy of a substance is positive, increase of T at constant P will decrease the free energy of a substance

……………..eq. (53)

……………..eq. (54)

Since volume of a substance is positive, increase of P at constant T will increase the free energy of a substance

S gas > S liquid > S solid

At constant pressure

V gas > V liquid > V solid

At constant temperature

Increasing T at constant P would lead to maximum decrease of free energy for gas!Increasing P at constant T would lead to maximum increase of free energy for gas!

Chemical potential and chemical equilibrium

Free energy change of a pure ideal gas:Gas

n moles, P = po = 1 atm pressure, temperature T = 298 K

Gas n moles,

P = p atm, temperature T = 298 K

Change of pressure

under constant temp.

= – SdT + VdP dG ……………..from eq. (52)

= VdP (under constant temperature)

Therefore, for the above change, we can write:

∫dG = ∫VdP = ∫nRTdP/P = nRT∫dlnp = nRTln(p/po)po

p

po

p p

popo

p

or G – Go = nRTlnp

or (G/n) – (Go/n) = RTlnp ……………..eq. (53)

It is customary to introduce a symbol μ, to replace free energy per mole (G/n)

Therefore, μ – μo = RTlnp (here μ is called molar Gibbs energy) ……..eq. (54)

Chemical change and free energy:

Chemical reactions generally occur under constant temperature and constant ambient pressure!

= – SdT + VdP = 0 for any chemical reaction at constant T & P? dG

Does that mean

Definitely this is not the case because, in any chemical reaction, the compositionof the substances (reactants, products, etc) vary!

Therefore G = f (T, P, composition of substances)Because, a chemical reaction is a mixture of varying composition

Varying composition means: mole numbers of the constituents (n1, n2, n3,………..) vary

So we can write that overall G is a function of T, P and mole numbers

Partial molar Gibbs energy (chemical potential):

For a pure substance or for a mixture of fixed composition, G = G (T, P)

If the mole numbers, n1, n2,…., of the substances in a composition varythen G = G (T, P, n1, n2…..)

So we can now write,

dG = (δG/δT)P,ni dT + (δG/δp)T,ni dp + (δG/δn1)T,P,nj dn1 + (δG/δn2)T,P,nj dn2 ……,

under constant temperature and pressure,

= (δG/δT)p,ni dT + (δG/δp)T,ni dp + ∑i(δG/δni)T,p,nj dni

dG = ∑i(δG/δni)T,p,nj dni = ∑i(µi)T,p,nj dni

µi = (δG/δni)T,p,nj

μi ≡ chemical potential of ith component, Joules per mole

(nj is the mole numberother than ni)

Ξ partial molar Gibbs energy

……..eq. (55)

……..eq. (56)

= ∑iµidni

Chemical potential is an intensive property:

For a mixture of substance of varying composition (at constant T & P):

µi = (δG/δni)T,p,nj = chemical potential/partial molar Gibbs energy

For a pure substance (at constant T & P):

µ = (G/n) = molar Gibbs energy

G = ∑iniμi ……..eq. (58)

G = nμ ……..eq. (57)

Overall for a varying composition, T and P

dG = -SdT + VdP + ∑i µi dni ……..eq. (59)

Gibbs energy (Joules) is a extensive property, depends on number of molesChemical potential (J mol-1) is an intensive property independent of number of moles.

Chemical equilibrium in a reaction mixture:Consider the following balanced reaction at constant T and P:

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

When ζ mole of reaction is completed: MnO2consumed = no - ζ HCl consumed: no - 4ζ MnCl2 produced = no + ζ H2O produced = no + 2ζ Cl2 produced = no + ζ

Let no be the initial number of moles of each reactant or product

ζ is called the advancement of the reaction.

Let’s imagine that in a reaction after a certain time if ni is the number of moleof any reactant or a product present at that time then we can write:

ni = noi + νiζ (νi is positive for products and negative for reactants)

At constant T and P, in a reaction mixture, for varying composition, we know:

dG = ∑i μi dni

dni = νi dζ (differentiating)

= ∑i (μi νi) dζ i (δG/δζ)T, P = ∑i μi νi ……..eq. (61) or,

Chemical equilibrium in a reaction mixture:The partial derivative: (δG/δζ)T, P

denotes rate of change of free energywith advancement of reaction.

(δG/δζ)T, P < 0 reaction moves fwdIf

(δG/δζ)T, P > 0 reaction moves bckwdIf

(δG/δζ)T, P = 0 equilibrium conditionIf

(δG/δζ)T, P, eq = ∑i μi νi, eq = 0

At equilibrium,

This partial derivative commonly calledas ΔG reaction

Therefore, at equilibrium:

ΔG reaction = ∑i μi νi, eq = 0 ……..eq. (62)

Gibbs energy of the reaction with extent of reaction:

(δG/δζ)T, P slope of G vs ζ plot at this point!

ζ

Chemical equilibrium in a reaction mixture of ideal gases:

μ = μo + RTlnp

For a pure ideal gas at 298 K:

For a mixture of ideal gases at 298 K, the chemical potential (μi) of the i th gas:

μi = μoi

+ RTlnpi

(where pi is the partial pressure of the ith gas and μo

i is the chemical potential of the ith gas in pure form)

……..from eq. (54)

Lets consider the following reaction mixture:

αA (g) + βB (g) → γC (g) + δD (g)

ΔG = ∑i μi νi

We know (from eq. 62) that the ΔG for the reaction:

= γμoC

+ γRTlnpC + δμoD

+ δRTlnpD - αμoA - αRTlnpA - βμo

B - βRTlnpB

= (γμoC

+ δμoD

- αμoA - βμo

B) + RTln (pcγ pD

δ/pAα pB

β) ΔG

= ΔGo + RTln (pcγ pD

δ/pAα pB

β) ΔG ……..eq. (63)

Chemical equilibrium in a reaction mixture of ideal gases:

αA (g) + βB (g) → γC (g) + δD (g)

= ΔGo + RTln (pcγ pD

δ/pAα pB

β) ΔG

Here ΔGo denotes the standard Gibbs energy of the reaction

= ΔGo + RTlnQpΔG Qp is known as proper quotient of pressure

At equilibrium, we know: ΔG = 0

= ΔGo + RTln0 (pc

γ)eq (pDδ)eq

(pAα)eq (pB

β)eq

So

ΔGo = - RTln(pc

γ)eq (pDδ)eq

(pAα)eq (pB

β)eq

Or,

Or,

ΔGo = - RTlnKp

……..eq. (64)

……..eq. (65) Kp ≡ quotient of equilibrium partial pressure ≡ pressure equilibrium constant

Expressions of equilibrium constants:

αA (g) + βB (g) → γC (g) + δD (g)

Ideal gas:

Kp = (pc

γ)eq (pDδ)eq

(pAα)eq (pB

β)eq

αA (aq) + βB (aq) → γC (aq) + δD (aq)

Ideal solution (infinitely dilute solution):

Kc = (ac

γ)eq (aDδ)eq

(aAα)eq (aB

β)eq

ai: thermodynamic activity of the reactants/productsKc ≡ concentration equilibrium constants

Non-ideal solution (example: ionic solutions):

Kc = (ac

γ)eq (aDδ)eq

(aAα)eq (aB

β)eq

= (cc

γ)eq (cDδ)eq

(cAα)eq (cB

β)eq

ai = ci. fi

x(fc

γ)eq (fDδ)eq

(fAα)eq (fB

β)eq

(c stands for concentration & f stands foractivity coefficient. For ideal solution, f = 1)

Here f ≠ 1

μi = μoi

+ RTlnpi

μi = μoi

+ RTlnai

μi = μoi

+ RTlncifi

……..eq. (66)

……..eq. (67)

Phase equilibrium and phase rule

= – SdT + VdP

dG

The fundamental equation of Gibbs energy states:

Dividing by the mole numbers (n) in both sides, we get:

for a pure substance (fixed composition)

= – SdT + VdP

dμ – – (since μ = G/n)

Under constant pressure: = – SdT dμ –

– (δμ/δT)P = - S ……..eq. (68)

The third law of thermodynamics tells us that a substance’s entropy will always be positiveIf the ambient temperature is greater than 0 K

So (δμ/δT)P = always negative

S gas >> S liquid > S solid

At constant pressure

At constant pressure

(δμ/δT)P of gas has large negative value

Chemical potential versus T at constant pressure:

Phase diagram in terms of μ:The partial derivative: (δμ/δT)P represents the slope of μ versus T plot

This is a qualitative phase diagram for anySubstance in its solid, liquid and gaseous form.

At melting point (Tm):

μ solid = μ liquid

Two phases are at equilibrium

At boiling point (Tb):

μ liquid = μ gas

Two phases are at equilibrium

T1 T2

At any T, the phase having lowest μ is the most stable phase!

Solid at T1 and at gas at T2 are the most stablephases at a constant pressure, respectively.

Pressure dependence of μ versus T curves:Under constant temperature: = VdP dμ –

– (δμ/δP)T = V ……..eq. (69) V gas >> V liquid > V solid

At constant T

If pressure is decreased, μ is decreased at constant T. The decrease is greatest for gas and least for solid. Below: μ versus T curves at lower pressure (dashed line) and higherpressure (solid line)

At a considerable lower pressure, the dashed Line of gas may crossthe melting point by shifting left!

That is the situationfor sublimation!

Pressure dependence of μ versus T curves:

Ts stands for sublimation temperature

At sublimation point (TS):

μ solid = μ gas

μ

T

S

L

G

At a certain T and Pall three lines may intersectat a common point.

μ solid = μ liquid = μ gas

This temperature is known astriple point

Tt

Generalized phase diagram: Clapeyron equationIt is bit difficult to ascertain μ of any substance to draw the μ vs T phase diagrams.

It is rather easy to draw P vs T phase diagram of any substance.

Clapeyron equation helps us to do that.

Derivation: Lets assume that α and β phases are in equilibrium with each other

α β

μα (T, P) = μβ (T, P)

(at temperature =T and at pressure = P)

Assuming the equilibrium condition is still fulfilled if the temperature and pressure changes infinitesimally to T + dT and P + dP, respectively. In that condition, the chemical potentialis assumed to change by an amount dμ.

Therefore,μα (T, P) + dμα = μβ (T, P) + dμβ

(at temperature =T + dT and at pressure = P + dP)

dμα = dμβ

– SαdT + VαdP =

– – – SβdT + VβdP

– – or,

or,

Generalized phase diagram: Clapeyron equation

– SαdT + VαdP =

– – – SβdT + VβdP

– –

Rearranging, we have

(Sβ – Sα)dT = (Vβ – Vα)dP – – – –

If the transformation is taken as α → β, then ΔS = Sβ – Sα and ΔV = Vβ – Vα

Therefore the above equation can be written as:

dP/dT = ΔS/ΔV ……..eq. (70A) This is known as Clapeyron equation

The equation denotes dependence of equilibrium pressure on Temperature

dT/dP = ΔV/ΔS ……..eq. (70B) This is also Clapeyron equation

This equation denotes dependence of equilibrium temperature on pressure

Conveniently, using eq. (70A), we will be able to draw phase diagrams by plottingequilibrium pressure vs. temperature.

P vs T phase diagram: solid-liquid equilibrium

solid liquidLets consider the following transformation

ΔS = Ssolid – Sliquid = ΔSfus = ΔHfus/T = +ve (always) – –

ΔV = Vsolid – Vliquid = ΔVfus = small +ve (mostly) = small -ve (for water)

– –

(at equilibrium)

Please note that, dP/dT (for solid → liquid) is a large constant value either +ve or -ve

T

P

T

Psolidliquid solid

liquid

for any s → l for ice → water

locus of all points at which solid & liquid

coexist at eqm.

P vs T phase diagram: liquid-gas equilibrium

liquid gasLets consider the following transformation

ΔS = Sliquid – Sgas = ΔSvap = ΔHvap/T = +ve (always) – –

ΔV = Vliquid – Vgas = ΔVvap = big +ve (always)

– –

(at equilibrium)

dP/dT is a small +ve value but the resulting plot is not a straight line, it is a curve!

solid

liquid

gas

solid, liquid and gas coexists at equilibrium

P vs T phase diagram: solid-gas equilibrium

solid gasLets consider the following transformation

ΔS = ΔHsub/T = +ve (always) ΔV = ΔVsub = big +ve (always)

(at equilibrium)

since ∆Vvap ≈ ΔVsub and ΔHsub > ΔHvap

dP/dT is a small +ve value, resulting plot is a curve but bit steeper than s-l curve near the triple point!

General phase diagramof a pure substance

triple point

limit at critical pressureand temperature point

P = 1 atm

normal boiling point

Phase diagram of few common substances:

CO2 H2O

The phase rule:

μα (T, P) = μβ (T, P)

The coexistence of two phases requires the following

That means here T and P are not independent, but related to each other.If we know T, P is automatically fixed and vice versa.So we say that this system has only one degrees of freedom (F) either T or P

At triple point:μα (T, P) = μβ (T, P) = μγ (T, P)

This happens at a particular T and P, which are fixed and unique for a pure substance

So in this case, degrees of freedom (F) = 0

For any phase (not at equilibrium), F = 2 (T as well P)

There is a general rule by which we can predict the degrees of freedom

F = 3 - P ……..eq. (71)

where P stands for number of phases

Number of phases present 1 2 3

Degrees of freedom 2 1 0

(three phases coexist)