Thermodynamics - samal-lab.webnode.com · Laws of thermodynamics Introduction Zeroth law: Define temperature First law of thermodynamics: concept of conservation of energy Four laws

Thermodynamics

Reading

Physical ChemistryPeter Atkins

Oxford University Press, Oxford, 2007

Physical ChemistryIra N Levine

Tata McGraw Hill Education Pvt. Ltd., New York (2002).

Physical Chemistry a molecular approachDonald A. McQuarrie, John D. Simon

Viva student edition.

Thermodynamics deals with stability of systems. It tells us ‘what should happen?’. ‘Will it actually happen(?)’ is not the domain of thermodynamics and falls under the realm of kinetics.

Thermodynamics (TD): perhaps the most basic science

Classical thermodynamics considered as a ‘system level’ science- i.e. it deals with descriptions of the whole system and not with interactions (say) at individual particles level. I.e. it deals with quantities (like T,P) averaged over a large collection of entities (like molecules, atoms).

TD puts before us some fundamental laws which are universal in nature (and hence applicable to fields across disciplines).

entities (like molecules, atoms). This implies that questions like: “What is the temperature or entropy of an atom?”; do not make sense in the context of thermodynamics.

Concept of (classical) thermodynamics applied to macroscopic properties of matter. Statistical thermodynamics rationalize the microscopic properties of matter.

Thermodynamics is the science of equilibrium system.

Laws of thermodynamics

Introduction

Zeroth law: Define temperature

First law of thermodynamics: concept of conservation of energy

Four laws govern thermodynamics

First law of thermodynamics: concept of conservation of energy

Second law of thermodynamics: Concept of entropy, which way the reaction proceeds

Third law of thermodynamics: Numerical value of entropy

Significance of thermodynamics in Chemistry:

A + B C + D

To understand whether a reaction will be exothermic or endothermic

To understand the direction of a spontaneous chemical change

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

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 the extent of a chemical change

System is the part of the world we are investigating.

Surrounding is the rest of the universe.

Universe = System + Surrounding

The language of TD

To a thermodynamic system two ‘things’ may be added/removed: energy (heat, work)

matter.

Open, closed and isolated systems

matter.

An open system is one to which you can add/remove matter (e.g. a open beaker to which we can add water). When you add matter- you also end up adding heat (which is contained in that matter).

A system to which you cannot add matter is called closed. Though you cannot add/remove matter to a closed system, you can still add/remove heat (you can cool a closed water bottle in fridge).

A system to which neither matter nor heat can be added/removed is called isolated.A closed vacuum ‘thermos’ flask can be considered as isolated.

In TD the two modes of transfer of energy to the system considered are Heat and Work. Heat and work are modes of transfer of energy and not ‘energy’ itself. Once inside the system, the part which came via work and the part which came via heat, cannot be distinguished.

Heat and Work

Work is coordinated/organized flow of energy or matter.

Lowering of a weight can do work

Heat involves random motion of matter (or the constituent entities of matter).

Like gas molecules in a gas cylinder

Water molecules in a cup of water

Atoms vibrating in a block of Cu.

Lowering of a weight can do work

Motion of piston can do work

Flow of electrons in conductor can do work.

The transfer of energy as a result of existence of unbalanced forces between system

Definition and sign convention of Heat and Work

The transfer of energy as a result of a temperature difference between system and surroundings is called heat (Q).

Heat input to a system is considered as positive; heat evolved by a system is considered a negative quantity.

Heat

WorkThe transfer of energy as a result of existence of unbalanced forces between system

and its surrounding is called work.

Work (W) in mechanics is displacement (d) against a resisting force (F). W = F d

If the energy of the system is increased by the work, we say the work is done on the

system, and it is taken to be a positive quantity.

If the energy of the system is decreased by the work, we say the work is done by

the system, and it is taken to be negative quantity.

Work can be expansion work (PV), electrical work, magnetic work etc. (many sets

of stimuli and their responses).

Here is a brief listing of a few kinds of processes, which we will encounter in

TD:

Isothermal process → the process takes place at constant temperature

(e.g. freezing of water to ice)

Isobaric → constant pressure

(e.g. heating of water in open air→ under atmospheric pressure)

Isochoric → constant volume

Processes in TD

Isochoric → constant volume

(e.g. heating of gas in a sealed metal container)

Reversible process → the system is close to equilibrium at all times (and

infinitesimal alteration of the conditions can restore the universe (system +

surrounding) to the original state.

Cyclic process → the final and initial state are the same. However, q and w need

not be zero.

Adiabatic process → dq is zero during the process (no heat is added/removed

to/from the system)

A combination of the above are also possible: e.g. ‘reversible adiabatic process’.

Properties (macroscopic) of matter can be defined by following variablesp, T, V, ni, m

Properties of matter are of two types:Extensive properties

These scale with size of the system or depends on the amount of substance in the sample. Example: m, V,..

Intensive properties

Macroscopic properties

Intensive propertiesThese do not scale with size of the system or independent of the amount of substance in the sample. Example: Temperature (T), pressure (P), etc

A property which depends only on the current state of the system (as defined by T, P, V

etc.) is called a state function. This does not depend on the path used to reach a particular

state.

Example: Internal energy, U or E.

Work and heat are not state functions.

State functions in TD

First Law of Thermodynamics

The energy can be neither created nor destroyed. The change in internal energy of a system is the amount of energy added to the system as heat energy minus the amount of the work done by the system as heat energy minus the amount of the work done by the system.

ΔU = Q − (−W) = Q + W

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

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 (by classical thermodynamics) 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, but only depends on the initial and final states.

First law of thermodynamics

Ui

A system withsome initial U

Uf

work done by the systemin the surroundings

Q

(supplied heat)

the system withfinal U

If the final internal energy after absorbing heat is (U ) then change in internal energy

+

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)

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,


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)

Expansion work: In thermodynamics it is classically described as the movement of a piston by expansion or compression of a gas.

Properties of internal energy (U) in mathematical form:

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

This relationship can be expressed in the form of a differential equation:

dU = (∂U/∂T)V dT + (∂U/∂V)T dV ……………..eq. (4)

partial derivative partial derivative


U = f(T, V)

đQ + đW = (∂U/∂T)V dT + (∂U/∂V)T dV [from eq. (3)] ………..eq. (5)

partial derivative partial derivative

or, đQ - PdV = (∂U/∂T)V dT + (∂U/∂V)T dV ………eq. (6) (since the

work done by the system = – đW = pdV

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

dU = đQ - PdV = (∂U/∂T)V dT ……………..eq. (7)

Or, ΔU = CvΔT (for a finite change)

(I) At constant volume (isochoric) process, dV = 0, so

đQV = (∂U/∂T)V dT ……………..eq. (8)

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


đQV = dU = CVdT ……………..eq. (9)

Or, ΔU = CvΔT (for a finite change)

(II) For any isothermal process: Temperature remains the same, dT = 0

By putting dT = 0,đQ = (∂U/∂V)T dV + PdV ……………..eq. (10)

and for ideal gas,

đQ = PdV ……………..eq. (11)

The general equation of first law of thermodynamics:

đQ − PdV = (∂U/∂T)V dT + (∂U/∂V)T dV


For any adiabatic process:

When no heat is exchanged during a process. đQ = 0

So the general statement of first law becomes:

The general equation of first law of thermodynamics:

đQ − PdV = (∂U/∂T)V dT + (∂U/∂V)T dV

So the general statement of first law becomes:

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

− PdV = (∂U/∂T)V dT + (∂U/∂V)T dV

In general, reactions take place at constant pressure (open to the atmosphere) than at constant volume!

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

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

dH = (∂H/∂T)p dT + (∂H/∂P) dP ……………..eq. (13)

For any isobaric process:


dH = (∂H/∂T)p dT + (∂H/∂P)T dP ……………..eq. (13)

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

đQP = dH = (∂H/∂T)P dT ……………..eq. (14)

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

Enthalpy is a practical and important thermodynamic function

đQP = dH = CPdT ……………..eq. (15)

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

Enthalpy

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

Q = ΔU + (-W) ………………From eq. (1)

đQ = dU + PdV

Δ H = Δ U + P Δ V ……………..eq. (16) Or , H = U + PV ……………..eq. (17) (P & V stand for pressure and volume of the system, respectively)

Qp = ΔU + P ΔV

The heat transfer at constant pressure is the enthalpy change of the system

dH = dU + PdV + VdP

dH = dU + PdV (under constant pressure) ……………..eq. (18)

In differential form,

Relationship between QP and Qv

[from eq. (18)]

Under constant pressure,

dH = dU + PdV

For a finite change under constant pressure,

ΔH = ΔU + PΔV = Qp

Since dU = Qv, we can write, Qp = Qv + PΔV or, Qp − Qv = PΔV …………..eq. (19)

Relationship between C and C for ideal gasRelationship between CP and Cv for ideal gas

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

đQP = CV dT + (∂U/∂V)T dV + PdV

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

CP = CV + (∂U/∂V)T (∂V/∂T)P + P(∂V/∂T)P ….(under constant pressure)

CP = CV + P(∂V/∂T)P ……………[for ideal gas, the term (∂U/∂V)T = 0]

CP = CV + nR …………………………….[using V= nRT/P at constant P]

CP /n - CV /n = R ……………eq. (22) [for ideal gas]

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

Relationship between CP and Cv for ideal gas

Ans.

= 6.01 kJ. mol−1

Q2. The value of ΔH at 298 K and one bar for the reaction described by 2H2 (g) + O2 (g) → 2H2O (l)

is -572 kJ. Calculate ΔU for this reaction.

R = 0.08314 L.bar.K-1.mol-1; 1kJ = 10 bar.L

Ans. ΔU = -565 kJ

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)

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

A BΔH1

ΔH2

According to this law: ΔH1 = −ΔH2

2) Hess’ Law (1840)

ΔH2

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

A D

ΔHtotal

A B

ΔH1

C

ΔH2

D

ΔH3

single step transformation multiple step transformation

According to this law: ΔHtotal = ΔH1 + ΔH2 + ΔH3

(This is an obvious outcome since H is a state function)

Application of Hess’s law

(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

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

2C (s) + 3H2 (g) → C2H6 (g) ΔHf = ?

Enthalpy changes for chemical equations are additive.

Given that

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

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

Concept of Entropy and Second Law of Thermodynamics

Second law of thermodynamics tells us the spontaneous direction of change of a system.

Spontaneous process does not require any work to be done to bring it about.

Thermodynamics is silent on the rate at which a spontaneous change occurs.

The direction in which isolated system evolve to the equilibrium state is governed by entropy of the system. For an isolated system, entropy always increases as a result of any spontaneous process.

Second law of thermodynamics

No process is possible which involves only absorption of heat from a reservoir and its complete conversion into work.

Kelvin statement of second law (1850):

Hot reservoirTH

QH

IMPOSSBLE

-

POSSIBLE

Cold reservoir

QC

TC

QH

(-W) = QH - QCengine

It is impossible to construct an engine that draws heat from a reservoir and completely convert it to work.

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

Concept of entropy (randomness)

less entropymore entropy

Entropy is a state function.

Concept of entropy

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

The systems evolve spontaneously not only in a direction that lowers their energy but that they also seek to increase their disorder.

There is a competition between the tendency to minimize the energy and to

For any spontaneous (irreversible) processes, the change in entropy is positive.

There is a competition between the tendency to minimize the energy and to maximize disorder.

If disorder is not a factor, then the direction of any spontaneous process is that which minimizes energy. (Ex: mechanical systems)

If energy is not a factor, then the direction of any spontaneous process is that which increases disorder. (Ex: mixing of two gases)

Second law of thermodynamics: concept of entropy

Entropy is the measure of disorderness of the system and the change in entropy during a process is defined as:

For a measurable change between two states i and f is

Or, dqrev = TdS

The second law of thermodynamics can be expressed in terms of entropy:

The entropy of an isolated system increases in the course of spontaneous change: ΔStot > 0

Stot is the total entropy of the system and its surroundings.

Unlike energy, entropy is not necessarily conserved; it increases in a spontaneous process.

Consider a two compartment isolated system, separated by a rigid heat conducting wall.

The system A and B are not at equilibrium with each other.

TA and TB are the temperature of system A and B. Consider TA > TB


Entropy of an isolated system increases as a result of a spontaneous process.

A B

TA VA TB VB

For an isolated system, Internal energy remain constant.

UA + UB = constant

S = SA + SB

VA and VB are fixed. So, for each separate system,

dUA = dqrev + dwrev = TAdSA

dUB = dqrev + dwrev = TBdSB

TA and TB are the temperature of system A and B. Consider TA > TB

The entropy change of the two compartment system is given by

dS = dSA + dSB

= dUA/TA + dUB/TB

Since the two compartment system is isolated, so dUA = −dUB


Since TA > TB,

dS > 0

We may generalize this result as the spontaneous flow of energy as heat from a body at higher temperature to a body at a lower temperature is governed by the condition dS > 0.

If TA = TB, then the two compartment system is in equilibrium and dS = 0

S

t

Smax

Spontaneous process

Equillibrium


t

dS > 0, for spontaneous process in an isolated system.

dS = 0, for an reversible process in an isolated system.

In an isolated system, the entropy will continue to increase until no more spontaneous processes occur, in which case the system will be in equilibrium.

Entropy change and some chemical reactions:

850obutane (C4H10) methane (CH4) + propene (C3H6)

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 1 molecule of cyclohexane (C6H6)

ΔS = negative

The Clausius Inequality

We know more energy flows when work is done under reversible process than under irreversible process.

− dwrev ≥ − dw; or dw − dwrev ≥ 0

Since internal energy is a state function, does not depend on path (same for both reversible and irreversible paths between two states)

dU = dq + dw = dqrev + dwrev

This expression is the Clausius inequality

Third law of thermodynamics : Quantitative measurement of entropy

From First law of thermodynamics

dU = dQrev + dWrev

dQ = TdS and dW = −PdV

dU = TdS − PdV

Third law of thermodynamics: Entropy

We know that (dU/dT)V = CV

At Constant pressure

dH = d(U + PV) = dU +PdV + VdP

dH = TdS + VdP (since dU = TdS – PdV)

Most of the processes occur at constant pressure.


If we know CP as a function of T, then we can calculate entropy.

If T1 = 0 K, then the above equation becomes

One can calculate entropy of a substance, if one knows S at 0 K and CP from T = 0 K to any temperature.

Third law of thermodynamics says that the entropy of a perfect crystal (substance) is zero at 0 K.

S0 = 0

From statistical thermodynamicsS = kB ln W

W is the number of ways the total energy of the system may be distributed.Since at 0 K, the system is at lowest energy state, so W = 1


Since at 0 K, the system is at lowest energy state, so W = 1And hence S = 0

Now

Change of entropy in change of states

Solid Liquid

Isothermal condition at melting point (Tm)

ΔS = ΔQrev/Tm = ΔHfusion/Tm

(under constant pressure)

Melting of ice at Tm

ΔHfusion is the heat of fusion or melting

Liquid Vapor

Isothermal condition at boiling point (Tb)

ΔS = ΔQrev/Tb = ΔHvap/Tb

(under constant pressure)

Boiling of water at Tb

ΔHvap is the heat of vaporization

Absolute value of entropy

Tfus = melting point

Csp = heat capacity of the solid phase

Tvap = boiling point

Clp = heat capacity of liquid phase

Cgp = heat capacity of the gaseous phase

ΔHfus and ΔHvap are the enthalpies of fusion and vaporization

Substance So/R (at 298 K)

C (diamond) 0.286

C (graphite) 0.690

Standard entropies of few substances at 298 K:

C (graphite) 0.690

Hg 9.129

H2O 8.41

H2 15.7

CO2 25.699

Free energy and Spontaneity

The criterion that dS > 0 for a spontaneous process applied only to an isolated system.

In most of the chemical reactions, the system is not isolated (many a times temperature is applied to the system).(many a times temperature is applied to the system).

What is the criteria for a spontaneous process in a non isolated systems??

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!

at 298 K

Reversibility and transformations

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

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

dS = 0

spontaneous (irreversible) processes, dS > 0

Condition for spontaneous transformation in terms of entropy is dS > 0 or TdS > đQ irreversible

Therefore, combining the two conditions in previous slide, we get:

TdS ≥ đQ (for any transformation reversible or irreversible)

Let’s consider a (not isolated) system at constant temperature and pressure. Constant temperature: a system must be in thermal contact with a thermal reservoir.

dU = đQ + đW

We know that TdS ≥ đQ and đW = − PdV

Substituting these values in above equation, we get

dU ≤ TdS − PdV

Or, dU − TdS + PdV ≤ 0

Gibb’s energy

Or, dU − TdS + PdV ≤ 0

Since both T and P are constant, we can write this expression as

d(U − TS + PV) ≤ 0

A new thermodynamic state function is defined by G = U − TS + PV

dG ≤ 0G is called as Gibbs energy

In a system at constant P and T, the Gibbs energy will decrease as the result of any spontaneous process until the system reaches equilibrium, where dG = 0.

G = U − TS + PV

Gibb’s energy

G = H − TS where H = U + PV

ΔG = ΔH − TΔS ≤ 0

The equality holds for reversible (equilibrium) process, whereas the inequality holds for an irreversible (spontaneous) process.

Properties of ΔG and Gibbs fundamental equation

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)

G = H - TS

reversible process)

dG = – SdT + VdP ……………..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 on P and T:

dG = – SdT + VdP ……………..eq. (52)

(∂G/∂T)P = −S ……………..eq. (53)

(∂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. (54)

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

Gibb’s (Free) energy (ΔG) and spontaneity

All natural processes proceed in a direction in which a system’s free energy gets lowered.

For any spontaneous chemical reaction, ΔG = negative

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

ΔGo = −92 kJmol-1

298 K, 1 atm

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

ΔG = negative

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

ΔGo = + 65 kJmol-1

298 K, 1 atmpressure.

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

Under standard condition, we can write:

ΔGo = ΔHo – TΔSo

The temperature dependence of the Gibb’s energy

dG = −SdT + VdP

(∂G/∂T)P = −S

We know that

Also we know that G = H − TS; or −S = (G − H)/T

(∂G/∂T)P = (G − H)/T

The equilibrium constant of a reaction depends on G/T (we will see later)

Putting the value of (∂G/∂T)P from above, we get

or

Gasn moles,

P = pi

T = 298 K

Gasn moles,

P = pf atm, T = 298 K

Change of pressure

under constant temp.

Pressure dependence of Gibb’s energy

Let’s consider n moles of gas at constant temperature expand or contract from initial pressure pi to a final pressure pf

dG = −SdT + VdP

At constant temperature, dT = 0

dG = VdP

Integrating both sides,

For an ideal gas, V = nRT/p

Pressure dependence of Gibb’s energy

If the initial pressure is the standard pressure, 1 bar; pi = p° and pf = p

G° is the Gibb’s free energy at standard pressure and temperature.

G(p) = G° + nRT lnpOr, Since p° = 1 bar

Chemical potential andchemical equilibrium

Chemical Potential of a pure substance

The chemical potential (μ) of a pure substance is defined as

Chemical potential shows how the Gibb’s energy of a system changes as a substance is added to it.

Chemical potential is same as the molar Gibb’s energy.

G(p) = G° + nRT lnpWe know that

G(p) = G° + nRT lnp

μ = μ° + RT lnp

G(p)/n = G°/n + RT lnp

For one mole, the above equation becomes

μ° is the standard chemical potential of a pure gas.

Gibbs energy (Joules) is a extensive property, depends on number of moles

Chemical potential (J mol-1) is an intensive property independent of number of moles.

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 …,

At 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

= ∑i µidni

Chemical potential

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

µ = (G/n) = molar Gibbs energy

Overall for a varying composition, T and P

dG = −SdT + VdP + ∑i µi dni

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

µ = (G/n) = molar Gibbs energy

dG = ∑iμidni

G = nμ

Chemical equilibrium in a reaction mixture

Consider a gas phase reaction

After a certain time the number of moles of each species is given by:

reactants products

Differentiating above equations

reactants products

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

Gib

b’s

en

ergy


At equilibrium,

This partial derivative commonly called as ΔG reaction

Therefore, at equilibrium:

ΔG reaction = ∑i μi νi, eq = 0

Gib

b’s

en

ergy

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 gasand μo

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


ΔG = ∑i μi νi

We know that the ΔG for the reaction:

= νY μo

Y + νY RTlnpY + νZ μo

Z + νZ RTlnpZ − νA μo

A − νA RTlnpA − νB μo

B − νB RTlnpB

= (νY μo

Y + νZ μo

Z − νA μo

A − νB μo

B ) + RTln (pZνZ pY

νY / pAνA pB

νB) ΔG

ΔG = ΔGo + RTln (pZνZ pY

νY / pAνA pB

νB)

Chemical equilibrium in a reaction mixture of ideal gases

Here ΔGo denotes the standard Gibbs energy of the reaction

ΔG = ΔGo + RTlnQp Qp is known as quotient of pressure

ΔG = 0

ΔG = ΔGo + RTln (pZνZ pY

νY / pAνA pB

νB)

At equilibrium, we know: ΔG = 0

Kp ≡ quotient of equilibrium partial pressure≡ pressure equilibrium constant

0 = ΔGo + RTln [(pZνZ)eq (pY

νY)eq / (pAνA)eq (pB

νB)eq]

ΔGo = − RTln [(pZνZ)eq (pY

νY)eq / (pAνA)eq (pB

νB)eq]

ΔGo = − RTln Kp

Expressions of equilibrium constants

Ideal gas:

Kp = (pc

γ)eq (pDδ)eq

(pAα)eq (pB

β)eq

Ideal solution (infinitely dilute solution):

K = (ac

γ)eq (aDδ)eq Kc ≡ concentration equilibrium constants

μi = μoi + RTlnai

For any reaction of the following type

μi = μoi + RTlnpi

Kc = (aA

α)eq (aBβ)eq

ai is the activity of the reactants/products

Kc ≡ 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 + RTlncifi

Variation of equilibrium constant with temperature

We know that

By substituting ΔGo = − RTln Kp in above equation,

Van’t Hoff Equation

Van’t Hoff Equation

Phase equilibrium and phase rule

Phase diagram summarize the solid-liquid-gas behavior of a substance, which indicates under what conditions of pressure and temperature the various states of matter of a substance exist in equilibrium.

1 atm

dG = – SdT + VdP

The fundamental equation of Gibbs energy states:

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

for a pure substance (fixed composition)

(since μ = G/n)

Under constant pressure:

Chemical potential versus T at constant pressure

The third law of thermodynamics tells us that a substance’s entropy will always be positive.

So (δμ/δT)P = always negative

S gas >> S liquid > S solid

At constant pressure

Change of Chemical Potential (molar Gibb’s energy) μ with T

At melting point (Tm):

μ solid = μ liquid

Two phases are at equilibrium

At boiling point (T ):

Solid

At boiling point (Tb): μ liquid = μ gas

Two phases are at equilibrium

At any T, the phase having lowest μis the most stable phase!

Liquid

Gas

T→

μ

Tm TbT1T2

Under constant temperature:

V gas >> V liquid > V solidAt constant T

If pressure is decreased, μ is decreased at constant T. The decrease is greatest for gas and least for solid.

Change of Chemical Potential (molar Gibb’s energy) μ with P

μ versus T curves at lower pressure (dashed line) and higher pressure (solid line)

Pressure dependence of μ versus T curves:

At sublimation point (TS):

μ solid = μ gas

μ

S

L

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

μ solid = μ liquid = μ gas

Ts stands for sublimation temperature

T

L

G

This temperature is known astriple point

Tt

Generalized phase diagram: Clapeyron equation

It 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

μα (T, P) = μβ (T, P)

(at temperature ,T and at pressure, P)

μα (T, P) = μβ (T, P)

Assuming the equilibrium condition is still fulfilled if the temperature and pressure changesinfinitesimally 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 = dμβor,

Generalized phase diagram: Clapeyron equation

Rearranging, we have

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 Clapeyron equation

The equation denotes dependence of equilibrium pressure on Temperature

Conveniently, we will be able to draw phase diagrams by plotting equilibrium pressure vs. temperature.

P vs T phase diagram: solid-liquid equilibrium

Let’s consider the following transformation

(at equilibrium)

So, dP/dT (= ΔS/ΔV) (for solid → liquid) is a large constant value either +ve or -ve

T

Psolid

liquid

for ice → waterfor any s → l

T

P

solidliquid

locus of all points at which solid & liquid

coexist at eqm.

So, there is large change in pressure by a small change in temperature.

P vs T phase diagram: liquid-gas equilibrium

Lets consider the following transformation

(at equillibrium)

dP/dT (= ΔS/ΔV) is a small +ve value but

the resulting plot is not a straight line, it is a

curve!solid

liquid

gas

solid, liquid andgas coexists atequilibrium

P vs T phase diagram: solid-gas equilibrium

Lets consider the following transformation

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

(at equilibrium)

since ∆Vvap ≈ ΔVsub and ΔHsub > ΔHvap

dP/dT (= ΔS/ΔV) is a small +ve value, resulting plot is a curve but bit steeper than liquid-gas 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 one component 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) (three phases coexist)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

where P stands for number of phases

Number of phases present 1 2 3

Degrees of freedom 2 1 0

Thermodynamics - samal-lab.webnode.com · Laws of thermodynamics Introduction Zeroth law: Define temperature First law of thermodynamics: concept of conservation of energy Four laws

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