chapter 4 chemical reaction equilibria

5/26/2018 Chapter 4 Chemical Reaction Equilibria

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CHEMIC L RE CTIONEQUILIBRI


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PURPOSE

OF THIS CHAPTER

To determine the effect of of temperature, pressure and initial composition on the

equilibrium conversions of chemical reaction.

Example

The rate of oxidation of sulfur dioxide to sulfur trioxide catalyzed by vanadium pentoxide

increases at higher temperature.

However, the equilibrium conversion falls as temperature rises, decreasing from 90% at

520oC to 50% at 680oC

These values represent maximum possible conversion regardless of catalyst or reaction

rate.

CONCLUSION

Both the equilibrium and rate must be considered in the exploitation of chemical

reactions for commercial purposes.


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Reaction is from where Chemical Engineers getDerived ChemicalsfromFeed Stock (rawmaterial)

Examples of reactions:

Why is chemical reaction so important tochemical engineers ?

1. Production of Methanolfrom Carbon Monoxideand Hydrogen

CO (g) + 2H2 (g) = CH3OH (g)

2. Fischer Tropsch Synthesis

x CO (g) + y H2 (g) = Hydrocarbons


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Chemical Reaction

Kinetics - reaction rate Equilibria - extent of reaction

Composition

Catalyst

Temperature Pressure

we will focus here !

parameters

The difference between chemical reaction kinetics and the

chemical reaction equilibrium

To sum it all .


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Reaction Equilibrium

A measure to indicate the extent of reaction

especially for reversible reaction.

Combining both concept (reaction kinetic and equlibrium).

Indicate the speed of a reaction system achieving

its equilibriumwhich will be the end state.

It is worth looking at the thermodynamics of chemical equilibrium before

moving over to look at its kinetics.


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Forward Reaction

Reverse Reaction

CO (g) + H2O (g) CO2(g) + H2(g)

On a macroscopic view, the reaction will be seemed as if it has

achieved a final static condition.

In actual fact, at microscopic level, the reaction still proceeds but theforward and the reverse reaction have balanced each other in term of

the speed and therefore no net change could be viewed.


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The reaction coordinate

Consider the following reaction equation .

............ nNmMbBaA

The equation can be written in general form as

uaA + ub B + .. umM + unN +

where is known as stoichiometric coefficient. It has a positive value for

product but negative value for reactant.


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Iu1I A1 + Iu2I A2 + Iu3I A3 + ..... -----> IunI An + Iun+1I A n+1+......

Reactant Product

Applied to a differential amount of reaction, the equation below is derived:

dni = ui de

ui- stoichiometry numbers

An- chemical species

+ve for products

-ve for reactants

dn1 = dn2 = dn3 = ......... = de

Reaction coordinate-

Characterizes extent or degree

of reaction

as a general form (the differential change dni with de;

The general form of reaction structure could be written in the following equation;

u1 u2 u3

(i = 1,2,..N)


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by integrating the equation,

dn = ui de

ni = nio + ui e (i = 1,2,3 ...., n)

Summation over all species ;

n = Sni = S nio + e Sui

note : here e is expressed in mol

ni

nio

e

0

i i i

The mole fraction for each component at any time can be calculated using

e

e

0

0

n

n

n

nxory iiiii where

i

i

No. of moles initially for

component i

FOR SINGLE REACTION


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SEE EXAMPLE 13.1


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Consider a fuel cell based on the direct conversion of methanol to form hydrogen:

H2O(g) + CH3OH(g)CO2(g) + 3 H2(g)

The reaction is carried out at 60 C and low pressure, with a feed of twice as

much water as methanol. The equilibrium extent of reaction (e) is 0.87. How

many moles of H2can be produced per mole of CH3OH in the feed. What is themole fraction of H2.

Taking a basis of 1 mole CH3OH, the initial composition can be written as:

n 0CH3OH = 1 and n0H2O = 2 with the number of moles of product = 0

Using the equation: ni = nio + ui en (CH3OH) = 1 - e

n (H2O) = 2 - e

n (CO2) = e

n (H2) = 3e

n (total) = 3 + 2e

Example


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For a given e= 0.87,

n (H2) could be calculated i.e.,

n(H2) = 0.87 X 3 = 2.61 moles

The others can be calculated as well,

n(CH3OH) = 1 - 0.87 = 0.13 moles

n(H2O) = 2 - 0.87 = 1.13 molesn(CO2) = 0.87 X 1 = 0.87 moles

The mole fraction of H2is calculated by ;

y (H2) = 2.61 / (3 + 2(0.87))

= 0.55


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FOR MULTIPLE REACTION

See Example 13.3


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dG = 0T,P

Similar to physical equilibrium, the criteria for chemical reaction equilibrium from

thermodynamics point of view;

G

eee

dG = 0T,P

The total Gibbs Free Energy is at minimum

possible value at equilibrium condition

d(nG) = nV dP - nS dT + S m i dnii

Now, for physical equilibrium we can write ;

d(nG) = nV dP - nS dT + S vi mi dei

[ dG T,P / de]S v m = 0iii

= 0 at const. T,P

Equilibrium Criteria for Chemical Reaction

i

Similarly, for chemical reaction, the equation can be written as;

At equilibrium


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i

ii

PT

G0

,

me

At equilibrium, we can write

Substituting for chemical potential,

0

0ln

i

iii

f

fRTmm

i i

iii

PT f

fRT

G0)ln(

0

0

,

me

We know that:


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Simplify;

i

i

vi

i

iiif

fRTG 0ln 00

RT

Gv

f

fo

ii

i

vi

i

i

0

ln

Where signifies the product over all species i.


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Go= standard Gibbs-energy change of reaction

- fixed for a given reaction once the temperature is established and is

independent of the equilibrium pressure and composition.

K= equilibrium constant


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EFFECT OF TEMPER TURE ON THEEQUILIBRIUM CONST NT


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Relation of K and enthalpy of reaction (DhR)

From the following equation,

Equilibrium Constant

-RT ln K(T) = Svi mio= DgR

o

We can derive,

dT

RTgd

dT

Kd R )/(ln 0D

2

0001)/(

RT

g

dT

gd

RTdT

RTgd RRR DD

D

Since,

2

00

1lnRT

gdT

gdRTdT

Kd RR D

D

Therefore,

From earlier developed relation,

0

0

RP

R sT

gD

D

Recall,

dG = (dG/dT)PdT + (dG/dP)TdP

= - s dT + v dP


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From the definition;000

RRR sThg DDD

2

001ln

RT

g

dT

gd

RTdT

Kd RR DD

00

r

P

r sT

gD

D

We will obtain;

2

0ln

RT

h

dT

KdR

D

And the enthalpy of reaction can be calculated in similar manner to the Gibbs Free

energy for reaction but enthalpy of formation for the individual component is used.

Svi hio= Svi (Dhf

o)i= DhRo


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Effect of Temperature on Equilibrium Constant

From the following equation,

we can develop the relation that give the temperature dependence

of the equilibrium constant by integrating the above relation.


2

0ln

RT

h

dT

Kd RD

=

If Ho is independent of T

(constant)

If Ho is varies with T(dependent on T)


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See example 13.4

Problem 13.3 and 13.11


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Consider the following reaction equation,

Find the equilibrium constant at room temperature.

CH4O (g) H2CO (g)+ H2(g)

DgRo= S viDgf

o298= - (-162.0 kJ/mol) + (-110 kJ/mol) + (0 kJ/mol)

mio

= 52 kJ/molRefer to appendix data

K298 = exp ( - Dgfo

298/ RT ) = exp ( - 52,000 / 8.314 . 298.15)

K298

= 7.64 X 10-10


formaldehyde


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To demonstrate the effect of temperature on K, consider the reaction system ;

H2O(g)+ CH3OH(g)CO2(g)+ 3 H2(g) Given K298(25 C)= 4.69

Similarly to Gibbs Free Energy, the enthalpy of reaction could be determined from

the enthalpy of formation,

Dh0rxn= S vi(Dh0f)i = 1 (Dh0f)CO2+ 3 (Dh0f)H2- 1 (Dh0f)H2O - 1 (Dh0f)CH3OH

Dh0rxn= (-393.51) + 3 (0)(-241.82)(- 200.66) = 48.97 (kJ/mol)

ln (K2/K1) = - ( DhR/ R ) [ 1/T2- 1/T1]From equation :

ln (K333/K298) = - ( DhR/ R ) [ 1/333 - 1/298 ]

ln (K333/4.69) = - ( DhR/ 8.314 ) [ 1/333 - 1/298 ]

K333= 37.44

Find the K value at 60 oC



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and Cpi= [ai+ biT + ciT2 + diT

-2]R

Therefore, enthalpy of reaction at temp T is ;

integrate the above relation and substitute in the equation below,

In actual fact, the heat capacity is a function of temperature.

yielding the eqn. in the next slide


iP

P

i CdT

dh

DD

T

T

PiT

T dTCvhh

i0

0

00

D

D T

T

T

T

dT

T

Th

RRT

gd

00

2

00)(1


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the constants ai, biand ciare determined from Appendix 2,

Text Chem. Eng. Thermo. by J. Winnick.


2

0

2

2

0

2

0

00

0

11

262

ln11

TT

dvTT

cvTT

bv

T

Tav

TT

F iiiiii

ii

D ii

iiii

iiT dv

T

cvT

bvTavT

R

hF

0

3

0

2

000

1

32

0and

This will give )lnln( 00

00

TTTT KK

RT

g

RT

g

D

D


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Example : We wish to produce formaldehyde, CH2O by the gas phase pyrolysis of

methanol, CH3OH according to;

CH3OH(g)

CH2O(g)+ H2(g)

a. What is the equilibrium constant at room temperature? Would you expect

an appreciable amount of product?

b. Consider the reaction at 600 C and 1 bar. What is the equilibrium

constant?

i. Assuming enthalpy of reaction is constant

ii. If enthalpy of reaction varies with temperature

Data : CH3OH CH2O H2

Gibbs Free Energy at 298 -162 -110 0

(Formation)

Enthalpy of Formation at 298 -200.7 -116 0

Vi -1 1 1

ai 2.211 2.264 3.249

bi 1.222X10-2 7.022X10-3 0.422X10-3

ci -3.450X10-6 -1.877X10-6 -di - - 0.083X10

5


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a. The equilibrium constant can be calculated from the Gibbs Free Energy of Reaction

DgRo= S viDgf

o298 = 1 ( -110) + 1 (0)1 (-162.0) = 52 kJ/mol

-RT ln K(T) = Svi mio= DgR

o

ln K = - 52,000 / RT = -52,000 / (8.314 X 298.15) = 7.64 X 10-10

too small !

b. (i) To calculate K at 600 C, we need the enthalpy of reaction.

DhRo= S viDhf

o298 = 1 ( -116) + 1 (0)1 (-200.7) = 84.7 kJ/mol

D

12

0

1

2 11lnTTR

h

K

K R

K at 873 K = 4.63 produce noticeable amount of product

K = 2.07 x 10 -7


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b. (ii) To calculate K at 600 C while accounting for changes in enthalpy of reaction.

2

0

2

2

0

2

0

00

0

11

262ln

11

TT

dvTT

cvTT

bv

T

Tav

TTF

iiiiii

ii

D ii

iiii

iiT dv

T

cvT

bvTavT

R

hF

0

3

0

2

000

1

32

0

and

Substitute all the value for the constants and taking T0= 298 and T = 873, the valueof - ln K at 873 could be obtained

)lnln( 00

00

TTTT KK

RT

g

RT

g

D

D

K (873 K) = 8.67

The assumption of constant enthalpy has caused significant error !

D

DRT

g

RT

g TT000


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REL TION OF EQUILIBRIUM CONST NTS TO COMPOSITION

Gas-phase Reactions

The standard state for the gas is ideal gas state of the pure gas at the standard-state pressure

(Po) of 1 bar. Since

For ideal solution,

Each for a pure species can be evaluated from a generalized correlation once the equilibrium T and P

are specified .

For pressures sufficiently low or temperatures sufficiently high, the equilibrium mixture behaves essentially

as an ideal gas. In this case, = 1, so above equation becomes;


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EXAMPLES


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LIQUID-PHASE REACTIONS

Where, is the fugacity of pure liquid i at the temperature of system and at 1 bar.

The fugacity ratio can be expressed as:

By using equation 13.29 and 13.30, equation 13.10 can be written as;

At high pressure, the equation is reduced to;

If the equilibrium mixture is an ideal solution, = 1, therefore,

=

=

Equilibrium for Non Ideal Gases Reaction System


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Some reactions of industrial importance are carried out at high pressure therefore

the impact causing from the non ideality behaviour becomes more apparent.

P ( ai)vi = exp [ - S vi mi O / RT ] = K (T)

Recall the equation

The modification required here is to account for the fterm.

ai= yi fiP instead of = yiP as used previously

Therefore,

K = P yi fi Pvi vi v

In order to determine fi, use method as given in earlier section on real gas.

= (z - 1)/P dP

P

ln fP*

For high density region, use generalised compressibility chart

For low density region, use virial equation of lowest order

Equilibrium for Non-Ideal Gases Reaction System

ii


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For our earlier example

in mol

fractionx/(2x+1) 2x/(2x+1) (1-x)/(2x+1)

CO(g) + 2H2(g) CH3OH(g)

Equilibrium composition x 2x 1-x

will now become

Pyy

yK

HCO

OHCH

2

2

3

2

2

22

33 Pyy

yK

HHCOCO

OHCHOHCH

ff

f


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Example : Consider reaction to produce ammonia as below;

N2 + 3 H2 2 NH3 High pressure reaction and exothermic

ii vii

v

i PyaK f

2

3

2

2222

33

Pyy

y

KHHNN

NHNH

ff

f

Using the free energy data, calculation on equilibrium constant give K = 8.93 X 10-6

Knowing K value then the composition could be calculated. However for non ideal

gas case the fugacity coefficient (f) has to be additionally determined.

One way of doing so is by calculating it based on pure component basis (Lewis

and Randall Rule)

RTBP

i ef

B = BRRTc/ Pc

BR= B0

+ wB1

ffor ammonia calculated is 0.975

ffor other components nearly 1


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To calculate the equilibrium concentrations,

N2 + 3 H2

2 NH3

At equilibrium x 3x 2(1-x) N total = 2 + 2x

2

3

2

2222

33

P

yy

yK

HHNN

NHNH

ff

fSubstitute in

2

23

2

2

23

2

1.31.

975.0).22(

.3.

).22(

22

3

P

nxx

xP

nxx

xK

TTHN

NH

ff

f

2

23

2

6

1.31.

975.0).22(1093.8

P

nxx

x

Ty (NH3)= 0.144

y(H2)

= 0.642

y (N2) = 0.214

Answer :


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EQUILIBRIUM CONVERSIONS FOR

SINGLE REACTIONS

SINGLE-PHASE REACTIONS

For homogeneous system, and K is known, the calculation for phase composition at equilibrium is straightforward if the gas ph ase is assumed an ideal gas and liquid mixture is ideal solution.

For non-ideal solution, activity coefficient must be determined first.

See example 13.5 to 13.9.

chapter 4 chemical reaction equilibria

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