An1 Derivat.ro Chimie Curs Chimie Cap2

8/12/2019 An1 Derivat.ro Chimie Curs Chimie Cap2

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Thermodynamics of chemical process 39

2. THERMODYNAMICS OF

CHEMICAL PROCESSES

Thermodynamics (Greek: therm = heat; dynamys = power) studies the

effects of changes in temperature, pressure and volume on physical systems at the

macroscopic scale by analysing the collective motion of their particles by using

statistical methods.

Chemical thermodynamics has as main objectives:

- the thermal effects of chemical transformation;- the quantity of work, accompanying thermodynamic process which can

include a chemical reaction;- the laws describing the processes in which the energy pass from one

state to another;

- the possibility, the direction and the limits of natural and industrialprocess;

- the organization or disorganization state of thermodynamic systems;- the chemical system and thermal equilibrium after physical chemical

transformations;

- establish the determining conditions for equilibrium.


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40 GENERAL CHEMISTRY

2.1. Thermodynamic systems

Thermodynamic system: a part of the universe separated by a real or

imaginary boundary from the rest of the universe. The surroundingscomprises the

region outside the system, where we make our measurements.

Figure 13. Schematic representation of a thermodynamic system

Phase part of a system having the same properties in all its points and

separated from the rest of the system by a separation surface.

Thermodynamic systems are classified as follows:

- fromphysical point of view homogenous formed from only one phase. It has the

same properties in all its points (gas, solution, mixture of

gases)

heterogenous formed from 2 phases or many (ex.Physical equilibrium H2OL H2OV the properties

vary in at least 2 points)

- from point of view of interaction with the environment isolatedsystem, there is noenergy or mass exchange with

the environment;

System

Surrounding

Boundary


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no isolatedor open system it changes energy and masswith the environment;

closed system it does not exchange mass but it canexchange energy (when the system allows energy to

escape as heat system is calleddiathermic);

adiabatic system there is no change of heat with theenvironment.

The properties of a system could be:

- intensive they do not depend on quantity of substance: pressure,temperature;

- extensive they depend on quantity of substance: internal energy (U),enthalpy (H), Gibbs free energy (G), Helmholtz free energy (A or F).

In order to describe a thermodynamic process state equationsare used:

f(P,V,T) = 0 for a monocomponent system and f(P,V,T,xi) = 0 for multicomponent system.where P is the pressure, T is the temperature, V is the volume and xiis the

molar ratio of the ith

component -n

nx ii= ; ni is the number of moles of i

th

component and nis the total number of moles.

Thermodynamic process the energetic evolution of a thermodynamic

system proceeding from an initial state to a final state. This evolution is determined

by the variation of one or more state parameters.

The thermodynamic processes are classified as follows:

- reversible the process passes spontaneously from the initialstate to the final state and from the final state to the initial state

through the same sequence of intermediate states;

- irreversible the process in which the system can not comeback from the final state to the initial state through the same

sequence of intermediate states.


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2.2. The Thermodynamics laws

2.2.1 Zeroth law of thermodynamics

If two systems are in thermal equilibrium with a third system, then they

must be in thermal equilibrium with each other.This principle introduces the temperature parameter, T. When two

different systems are in thermal contact, the big body called also thermostat

preserves unchanged its properties while the small body, called also thermometer

varies its properties.

Figure 14. Schematic representation of zeroth law of thermodynamic

For the temperature there are several notations and units:

- t C (Celsius degrees)- T K (Kelvin degrees) (T = t + 273,16)- F = 9/5 C + 32 (Fahrenheit degrees) (anglosaxon system)

2.2.2. First law of thermodynamics

Energy is neither created nor destroyed; it changes from one form to

another.

Thermostatthermometer


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Consequences

- It is impossible to create a perpetual motion machine of I type (machinewhich produces energy from nothing) Helmholtz 1847;

- If a quantity of energy disappears from a system, a new form of energy,in an equivalent quantity appears in its place;

- The internal energy of an isolated system is constant EquivalenceLaw of Joule

calJconstQ

W

Q

W/18,4

2

2

1

1 === [1]

where:

W The Work is done when an object is moved against an opposing force;

Q The Heat: the system gives or receives heat.

The first law of thermodynamic postulate that U the change in internal

energy of the system is given by:

WQU += [2]

Or in other words:

The change in internal energy of a closed system is equal to the energy that

passes through its boundary as heat or/and work.

All these dimensions are expressed in J (Joule) or cal (calorie):

1J = 4,186 cal

The internal energy is a state function. It is depending on the initial and

final state and not on the way, while the heat and the work are functions which

depend on the followed way. So we can write for the change in internal energy wecan write:

12 UUU = [3]

12

2

1

UUdUU == [4]

= 0dU for a reversible cycle [5]


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Sign convention:

(+) any quantity which increases the energy of the system (the heat

absorbed by the system, the work done on the system);

(-) any energy lost by the system.

Examples:

Heat:

(+) received by the system (evaporation, endothermic reaction)

(-) lost by the system (solidification, exothermic reaction)

Work:

(+) performed by the surrounding on the system (compression)

(-) performed by the system (expansion)

In these conditions the internal energy is given by:

WQU += - for a finite process

And:

WQdU += - for an elemental process [6]

Q , W are infinite small quantities.

Exemplification for expansion work:

pdVSdlS

FFdlW ===

pdVW = [7]

That means that the system receives work when its volume is reduced.

In these equations Fis the force, lis the distance, Sand Vare the surface and the

volume respectively.

Examples Reversible processes(perfect gas)

a) isotherm process(constant temperature)12

UUU = [8]


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==

2

1

)(

T

T

vdTCUTfU [9]

(where Cvis the molar heat capacity at constant volum)

0,12 == UUU [10]

WQ = [11]

1

2

2

1

2

1

lnV

VnRTdV

V

nRTpdVW === [12]

nRTpV= [13]

V

nRTp= [14]

constant temperature

2211 VpVp = [15]

2

1

1

2

p

p

V

V= [16]

2

1lnp

pnRTW = [17]

b) isochoric process (constant volum)00 == WdV [18]

QU= [19]

c) adiabatic process 0=Q WU=

)( 12 TTCW V =

=

2

1

T

T

VdTCU [20]

This means that the temperature will increase when the system is

compressed and the temperature will decrease when the system will undergoes an

expansion.


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d) isobar process(constant pressure)A new state function is introduced in order to characterize the isobar

process:

H Enthalpy, calorific function of Gibbs

PVUH + [21]

HHHdH ==2

1

12 [22]

0= dH for a reversible cycle [23]

Because the enthalpy is a state function its elemental change is an exact

differential.

VdPPdVdUdH ++= [23]

If the system is in mechanical equilibrium with its surroundings at a given

pressure Pand undergoes only expansion work, we can write: PdVdW =

VdPPdVPdVQdH ++= [25]

VdPQdH += ,0=dP (constant pressure)

( )p

QH = [26]

Qp= H [27]

The thermal effects of all process which take place at constant pressure

(chemical reactions, phase transitions, solubilization) represent the change in

enthalpy for that system.

2.2.2.1. Applications of the first law to the material constants

Heat capacity represents the amount of heat necessary to increase the

temperature of the system by 1C.

Specific heat capacity(specific heat) of a substance represents the heat

capacity divided by the mass, usually in grams or, in other words, it represents the


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calculations of quantities such as the heat capacity, heat of absorption, heat of

formation, etc.

Considering the following chemical equation:

HAAAAAAA r

iiii ++++++++ .............'''

2

'

2

'

1

'

1332211

where Hr is (+) for endothermic reactions and () for exothermic

reactions

At a given temperature and pressure:

=react

ii

prod

iiPT

rHHH ''

, [31]

where vi is the number of moles for ith

reactant; Hi is the molar enthalpy of ith

reactant; viis the number of moles for ith

product;Hiis the molar enthalpy of ith

product.

Example:322

2NH3HN +

22332

, NHNHPT

r HHHH =

a) Heat of formationThe enthalpy change for a reaction at a given temperature and pressure can

be derived from heat of formations of the reactants and products in the following

way:

=react

i

f

i

prod

i

f

iPT

rHHH ''

, [32]

The values for standard enthalpy0

298Hf (standard conditions, 1 atm and

298 K (reference state)) are collected in thermodynamic tables.

The standard enthalpy of formation0

Hf of a substance is the standard

reaction enthalpy for the formation of the compound from its elements in their

reference states.

Example:

)(2)()()(32323

gsgs COFeCOOFe ++


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molkcalHsOFe

f /2.1960)(,298 32

=

molkcalHgCO

f /4.260)(,298 =

molkcalHgCO

f/968.930 )(,298 2 =

00,298

= FefH (generally 0

298Hf is equal with zero for elements)

( ) ( ) ( ) kcalHr 504.64.2632.196968.9330298

==

b)Heat of combustion

The heat of combustion is the energy released as heat when one mol of a

compound undergoes complete combustion to upper stable oxides.

molkcalHc

/79.212O2HCO2OCH 0298(l)2(g)22(g)4(g)

=++

='

''

,

i

ic

i

i

i

c

iPT

r HHH [33]

Other heat type:

- heat of solubilization- heat of decomposition- heat of neutralization- heat from bond energy

2.2.2.3 The laws of thermochemistry

a) Hesss law (1840)The heat exchange accompanying a transformation is the same,

independently whether the process occurs in one or several steps.

The standard enthalpy of an overall reaction is the sum of the standard

enthalpies of the individual reactions into which a reaction may be divided.

The thermal effects of one reaction depend only on the nature and the state

of the reactants and products and not on the way in which takes place the reaction.


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Example:

calHCOOCO

calHCOOC

calHCOOC

ggg

ggs

ggs

676402/1

263902/1

94030

3)(2)(2)(

2)()(2)(

1)(2)(2)(

=+

=+

=+

321 HHH +=

b) Kirchhoff s lawThe variation with temperature of the thermal effects of a chemical

reaction is equal to the stoichiometric molar heat capacity of the system (constant

pressure).

+=2

1

12 ,,

T

T

P

r

PT

r

PT

r dTCHH [34]

Equation [34] represent the integral form of Kirchhoff equation

Usually:

...2

2

+++=

++=

cTbTaC

cTbTaC

rrr

P

r

P

Generally:

T1= 298K and P = 1 atm. (standard heat of formation)

( ) ++++=2

2

298

0

298

0 ...

T

rrrr

T

rdTcTbTaHH [35]

HCp > 0 increases when the temperature increases

HCp


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If phase transitions take place in the temperature range considered, the

enthalpies of transition should be taken into account.

++=2

2

1

112

00

T

T

P

rtr

T

T

P

r

T

r

T

r

tr

tr

dTCHdTCHH [36]

where represents the mole numbers of the substance undergoing the phase

transition. The transition enthalpy will be positive when a product is phase

transformed and will be negative when a reactant undergoes a phase transition.

1PC and

2PC are the change in heat capacity calculated with the values of heat

capacity function of the phase.

Example:

Calculate the thermal effect for the reaction:

Al2O3+ 3CO 2Al + 3CO2

at 1000 K and 1 atm.

It is known: calHr 1770802980 = , melting temperature of aluminium is

930 K, molcalHAl

melt /25000 = and the values of heat capacities at constant

pressure, given in the following table:

Substance Cp, cal/molK

Al2O3 22 + 810-3T

CO 6.6 + 1,210-3T

CO2 10.3 + 2.810-3T

Al(s) 4.8 + 3.2 10-3

Al(l) 7

Solution:

322)(2

332OAlCOCOsAl

PPPPP

r CCCCC +=

TTT

TTCPr

333

33

102.33.1)10822()102.16.6(3

)108.26.10(3)102.38.4(21

+=++

+++=


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322)(2

332OAlCOCOlAl

PPPPP

r CCCCC +=

T

TTTCpr

3

333

2

102.31.3

)10822()102.16.6(3)108.26.10(372

=

=++++=

+=+

++++=

)298930(3.1177080)102.31.3(

25002)102.33.1(177080

3

1000

930

3

930

298

0

1000

dTT

dTTHr

++

+

2

9301000102.3)9301000(1.325002

2

298930102.3

22

3

22

3

calHr

18250101000

=

c) Lavoisier and Laplaces law (1782)The heat exchange accompanying a transformation is equal in absolute

value and opposite in sign to the heat exchange accompanying the reverse

transformation.

Example:

1(l)2(g)2(g)2(g)4OHCO2OCH Hr++

2(g)2(g)4(l)2(g)22OCHO2HCO Hr++

molkcalH

molkcalH

r

r

/7.212

/7.212

2

1

=

=

Consequence

When one mole of a chemical compound is decomposed into chemical

elements quantity of heat is absorbed or evolved a quantity of heat equal to the heat

evolved or absorbed when one mole is formed from the constituent elements.

HH df = [37]

where Hf represents heat of formation and Hd heat of decomposition.


15/32


2.2.3 The second law of thermodynamics

Similarly to the Ist law, the II

nd law of thermodynamics is the result of

generalization of the huge human experience. It cannot be demonstrated. At the

earth scale it was not infirmed by any particular case.

Clausius formulation (1850)

Heat cannot spontaneously flow from a colder to a hotter material.

Kelvin formulation

No process is possible in which the sole result is the absorption of heat

from a reservoir and its complete conversion into work. Such a kind of machine

will be a perpetual Motion Machine of IInd

type.

A heat engineis a mechanical device that provides useful work based on

the difference in temperature of two bodies. In other words, a heat engine is a

mechanical device that continuously transforms heat into an useful work. Therefore

in order to function such an engine necessarily has to be endowed with two heat

reservoirs: a hot source (T1) and a cold source (sink) (T2) (figure 15)

Figure 15. Schematic representation of the functioning of heat engine

During the functioning of this engine some heat Q1 is absorbed from the

hot source, a part is transformed into work and the rest Q2is transferred to the sink.

Sign convention

(+) for heat received by the system

(-) for heat lost by the system

2121)( QQQQW +== [38]

Hot body

T1

EngineQ1

W

Q2 Cold body

T2


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The efficiency is given by:

1

21

1 Q

QQ

Q

L +== [39]

Sadi Carnot (1824) established the ideal efficiency of heat engines for an

engine which work through cycles formed from two isotherm transformations

(constant temperature) and two adiabatic transformations (Q = ct). Such a cycle is

called Carnot cycle (figure 16).

1

21

T

TT= [40]

Figure 16. Carnot Cycle

Homework: demonstrate the eq. [40] knowing the heat for the isotherm

transformations:1

2

11ln

V

VRTQ = and

1

2

22ln

V

VRTQ =

Consequence

The maximum efficiency of a heat engine does not depend on the nature of

substance which follows the transformations. It is determined only by the

temperature of the two sources.

For an irreversible cycle the efficiency will be lower than the ideal

efficiency.

=

+processreversiblefor

processleirreversibfor

1

21

1

21

T

TT

Q

QQ [41]

V

AB

CD

T1

T2

Q1

Q2

P


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2.2.3.1. State functions introduced by the IInd

Law of Thermodynamics

a) The Entropy

1

21

1

21

T

TT

Q

QQ

+ [42]

1

2

1

2 11T

T

Q

Q+ [43]

1

1

2

2

TQ

TQ [44]

02

2

1

1 +T

Q

T

Q [45]

The ratioT

Q between the heat exchanged in an isotherm and reversible

process and the temperature corresponding to this exchange is called reduced heat.

For any elemental cycle it can be written in the following way:

02

2

1

1 +TQ

TQ [46]

By summing all the reduced heats one obtaines:

0= TQrev 0


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For anirreversible process

dST

Qirev VUdS exists. So the IInd

law of

thermodynamics can be formulate in terms of entropy as follows:

In one isolated system the entropy of the system can not decrease, it is

constant ( )0, =VUdS if in the system only reversible processes take place, and it

increases (dS>0) in any irreversible process. This is why the IInd

law of

thermodynamics is called also the entropy law.

Entropy for a chemical reaction

For a general chemical reaction like:

.......................... '''2

'

2

'

1

'

12211 +++++++ iiii AAAAAA

the variation of the entropy is:

=react

ii

prod

ii

r SSS '' [55]


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where vi is the number of moles for ith reactant; Si is the molar entropy of i

th

reactant; vi is the number of moles for ithproduct; Si is the molar entropy of i

th

product.

When some phase transitions take place:

T

QdS

= ; at constant pressure QP =H so for elemental processes

dHQ= .

T

dHdS= [56]

For a finite process:

tr

tr

tr

T

HSSS

==

12 [57]

Ex.v

vap

vap

T

HS

=

For a chemical reaction:

+

+=2

2

1

1

12

T

T

P

tr

tr

i

T

T

P

T

r

T

r

tr

tr

dTT

C

T

HdT

T

CSS [58]

Example:

Calculate the entropy for the reaction:

3CO2Al3COAl32

++

At 1000 K and 1 atm. It is known grdcalSr /88,1380

298 = , the melting

temperature for aluminium 930K, molcalHAl

melt /2500= and the heat capacities

listed in the table below:

Substance Cp, cal/mol K

Al(s) 4.8 + 3.2 10-2T

Al(l) 7

CO(g) 6.6 + 1,2 10-3T

Al2O3(s) 22 + 9 10-3T

C(s) 2.7 + 2.6 10-3T


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Solution:

+

++=1000

930

930

298

2981000 212

T

dTC

T

H

T

dTCSS

p

r

m

m

p

rrr

)3(32)((32)()(1

) sCsOAlgCOsAl ppppp

r CCCCC ++=

TCpr 3

1108.67.0 =

)3(32)()(32)()(2 sCsOAlgCOlAl

pppppr CCCCC ++=

TCp

r 3102.137.3

2

=

+

+

+=1000

930

3930

298

3

1000

102.137.3

930

25002108.67.088.138 dT

T

TdT

T

TSr

grdcalSr /51.138

1000=

b) Thermodynamic potentials Helmholtz Free Energy (A or F), GibbsFree Energy (G)

According to the correlation between Istand II

ndlaws:

LdUTdS [59]

LTdSdU [60]

( ) LT TS-Ud [61]

= reversible process


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( )VTAA ,= - state function

= 0dA [63]

==2

1

12 AAAdA [64]

( ) LdA T [65]

The Helmholtz free energy is a thermodynamic potential which measuresthe useful work obtainable from a closed thermodynamic system at a constant

temperature.

Similarly:

pdVdUTdS + [66]

PVUH += [67]

VdPPdVdUdH ++= [68]

VdPPdVdHdU = [69]

PdVVdPPdVdHTdS + [70]

VdPdHTdS [71]

VdPTdSdH [72]

( ) VdPTSHd T [73]

By definition:

TSHG [74]

G Gibbs free energy

( )PTGG ,= state function [75]

= 0dG [76]

GGGdG == 122

1

[77]

( ) VdPdG T [78]

if the pressure is constant then

( ) 0, PTdG [79]


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In a system, at constant temperature and pressure can take place

spontaneously only the processes in which the Gibbs free energy is

decreasing ( ) 0,


23/32


2.2.4. Third Law of thermodynamics

The Nernst heat Theorem

The entropy change accompanying any physical or chemical

transformation approaches zero as the temperature approaches zero: S0 as T0

provided all the substances involved are perfectly ordered.

Planck affirmation:

At 0 K not only S is zero but also the molar entropy (S) of any crystalline

pure element is zero.

00 =S [87]

S0= 0 [88]

Equation [88] represents the mathematical expression of the IIIrd

law of

thermodynamic.

Summarized Third Law of thermodynamics is: The entropy of all perfect

crystalline substances is zero at T = 0 K.

2.2.5. Chemical Potential

In open systems formed from many substances, (so the composition is

varying due to the changes with the surroundings), the expressions of the

thermodynamics functions established in the frame of Istand II

ndLaws should be

completed with composition variables:

.....),......,,,,(21 innnSVUU= [89]

.....),......,,,,(21 i

nnnSPHH= [90]

.....),......,,,,(21 i

nnnSVFF= [91]

.....),......,,,,(21 i

nnnPTGG= [92]

The elementary variation for the Gibbs free energy will be:


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...

)(3132 ,,

2

,..,,.,2

1

,...,,,,1

+

+

+

++=

i

nPTinnnPTnnnPT

dnn

Gdn

n

Gdn

n

GVdPSdTdG

ijjii

[93]

The partial molar function:

( )

i

nPTi

i

ijj

n

GG =

=

,,

[94]

is called chemical potential (Gibbs)

++=i

iidnVdPSdTdG [95]

2.3. Chemical equilibrium

The experience showed that the chemical reactions occur in forward

direction as well as in opposite direction.

Example:

H2+ I2 2 HI

In a chemical process, chemical equilibrium is the state in which the

chemical activities or concentrations of the reactants and products have no net

charge over time. Usually, this state results when the forward chemical process

proceeds at the same rate as their reverse reaction. The reactions rates of the

forward and reverse reactions are generally not zero but, being equal, there are no

net charges in any of the reactant or product concentrations.

2.3.1. Thermodynamical condition of chemical equilibrium.

Reaction isotherm

Let take a chemical reaction in general form:

......2211 ++++ iiAvAvAv ......''

2'

2'

1'

1' ++++ iiAvAvAv


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The chemical potentials (free partial molar Gibbs energy) are:

,,21

,..., i for reactants ,''

2

'

1...,,,lyrespective

i for products.

The expression:

( ) 0'', == react

ii

prod

iiPT

rvvG [96]

is the thermodynamical condition of the chemical equilibrium.

For p = 1 atm

=react

ii

prod

iiTr

vvG ''0 [97]

representing the standard Gibbs energy of the reaction.

The expression:

aT

rKRTG ln0 = [98]

links thermodynamic data with chemical important equilibrium constant K.

Ka is the equilibrium constant, calculated with activities:

=react

v

i

prod

v

i

ai

i

a

a

K

'

'

[99]

where : ai is the activity of ith component at equilibrium. It is a kind of

effective mole fraction:

*

i

i

ip

pa = ; [100]

wherepiis the vapor pressure of ithcomponent (at equilibrium) when it is a

component of solution and*

ip vapor pressure of pure ith

component at

equilibrium;

- notation for product operator.

For gases instead of activities fugacities f are used which have pressure

dimensions: 1lim0

= p

f

p.

For an ideal gas mixture pf = :


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=

react

v

i

prod

v

i

pi

i

p

p

K

'

'

[101]

For very diluted solutions activities can be replaced by corresponding

concentrations:

xa KK or ma KK or ca KK

=react

v

i

prod

v

i

xi

i

x

x

K

'

'

=react

v

i

prod

v

i

mi

i

m

m

K

'

'

=react

v

i

prod

v

i

ai

i

c

c

K

'

'

[102]

where: xi molar ratio at equilibrium

t

i

in

nx = [103]

mi molal concentration (moles of substance in 1 kg of solvent) at

equilibrium

ci molar concentration (mols of substance in 1L of solution) atequilibrium

In this case the free Gibbs energies of reaction will be:

( ) ( ) ( )cT

r

mT

r

T

r GGG 000 [104]

wherexT

rKRTG ln0 =

mmT

r KRTG ln)( 0 =

ccT

r KRTG ln)( 0 =

2.3.2 Expression of equilibrium constants for reactions among

gases

i

i

v

i

vv

v

i

vv

p

ppp

pppK

...

...

21

''2

'1

21

''

2

'

1

= [105]


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a) ii Pxp = ,

where: ip is partial pressure of ithcomponent at equilibrium

Pis the total pressure

xi is molar ratio of ithcomponent at equilibrium

v

xv

v

v

i

vvv

v

i

vvv

p PK

P

P

xxxx

xxxxK

i

i

i

i

i

i

=

=

'

321

''3

'2

'1

......

......

321

''

3

'

2

'

1 [106]

= ii vvv' [107]

b) RTnVp ii = [108]

RTcRTV

np

i

i

i == [109]

( )

( )( )

=

= RTKRT

RT

cccc

ccccK

c

i

i

pi

i

i

i '

321

''3

'2

'1

......

......

321

''

3

'

2

'

1 [110]

R = 0.082 Latm / molK

c) = )(RTKPK cx [111]

=

RT

PKK

Xc [112]

If there is no change of mole numbers in a reaction, 0=v ,

cxp KKK == [113]

These equilibrium constants are dimensionless.

Example

For the reaction COCl2 CO + Cl2, at equilibrium, at 550C and 1 atm

77% of COCl2is dissociated. Calculate Kx, Kpand Kc.

Solution:

COCl2 CO + Cl2

Initial: 1 0 0 moles

Equilibrium: 1 moles Total 1 + + = 1 + moles


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1100

77= moles of COCl2dissociated.

+=

1COx

+=

12Clx

+

=

1

12COCl

x

( )( )

+=

+

+

+=

11

1

111

2

xK

( )( )( )

456.177.0177.01

77.02

=+

=xK

= PKK xp 1111 =+=

456.11456.1 ==pK

( ) = RTKK pc

0215.0823082.0

1456.1 =

=cK

2.3.3 Estimation of the equilibrium constant from

thermodynamic data

It is known that aTr KRTG ln0 =

But

000

T

r

T

r

T

r STHG = [114]

dTT

CSS

Tp

rr

T

r +=298

0

298

0 [115]

+=T

p

rr

T

r dTcHH298

0

298

0 [116]

SoRT

GK T

r

a

0

ln

= [117]


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orR

S

RT

HK T

r

T

r

a

00

ln

+

= [118]

or

+

=

Tp

rT

p

rT

rr

a dT

T

cdTc

TRR

S

RT

HK

298298

00

298 11ln [119]

The equation [119] solves the most important problem of chemical

equilibrium: the integral calculus of equilibrium constant from thermodynamical

data - reaction heat, molar heat capacities, and entropies.

2.3.4. The shift of chemical equilibrium; Le Chatelier Brawn

principle

If a dynamic equilibrium is disturbed by changing the conditions, the

position of equilibrium shifts in order to counteract the change.

A system at equilibrium, when subjected to a disturbance, responds in a

way that tends to minimize its the effect.

2.3.4.1 Temperature influence on the equilibrium constant. Equation

Vant Hoff

T

G

RK

r

a

01ln

= [120]

Differentiation of aKln with respect to temperature gives:

dTT

Gd

RdTKd

r

a

=

0

1ln [121]

From the Gibbs Helmholtz equation we have:

2

0

0

RT

H

dT

T

Gd

r

r

=

[122]

So:


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2

ln

RT

H

dT

Kd ra = [123]

The equation [123] is called vant Hoff equation

From the equation [123] one can see the following:

- for endothermic reactions ar KH > 00 increases with theincreasing temperature. So the increase the temperature will have as

effect the shift of the reaction to the right side;

- for exothermic reaction ar KH 0 decreases with the increases of pressureExample: CO2 CO +

2

1O2

2

1=v

xKv


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kcalGr

72.440

298 = . Calculate 0

400G

r and Kp at 400 K, supposing that

0= pC cal/Kfor the temperature range 298 - 400K, and an ideal gases behavior.

Solution:

2

0ln

RT

H

dT

Kd rp =

=2

1

2

1

2

0

ln

T

T

rK

K

pTdT

RHKd

P

P

=

21

0 11ln

1

2

TTR

H

K

K r

P

P

0

2981298 GKT

r=

0

4002400 GKT

r=

pT

r KRTG ln0 =

2

0400

1

0298

21lnln

RT

GK

RT

GK

r

P

r

P==

=

+

21

0

1

0

298

2

0

400 11

TTR

H

RT

G

RT

G rrr

=

211

0

298

2

0

400

11

TTH

T

GTG

r

r

+=

400

1

298

114.44

298

72.444000

400Gr

kcalGr 450

400 =

2

0

400lnRT

GK

r

p

=

51.56400987.1

45000ln

400 ==pK

24105.3 =pK .


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An1 Derivat.ro Chimie Curs Chimie Cap2

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