Chapter 9: Chemical Equilibrium - R. M. FABICON's BLOG · 2010. 9. 14. · Consider the simple reaction A B » Examples d-alanine to l- alanine » Suppose an infintesimal amount of

LECTURE 9: Chemical Equilibrium

REACTION GIBBS ENERGY

Consider the simple reactionA B

» Examples d-alanine to l- alanine» Suppose an infintesimal amount of nA converts to nB ,

then dnA =-dnB

» If we define

(ksi) as the extent of reaction thend

= -dnA =dnB

Dimensions of

are molesFor finite reaction the change is

and nA goes to nA -

and nB goes to nB +

REACTIONS GIBBS ENERGY

Reaction Gibbs Energy is the slope of the Gibbs energy vs.

or r G = (G/ ) p,T

» Remember, dG = µA dnA + µB dnB so dG = -µA d

+ µB d

or r G = (dG/d ) p,T = µB -µA

» Reaction Gibbs energy is the difference of the product chemical potential and the reactant chemical potential

GIBBS ENERGY AND EQUILIBRIUM

Initially µB <µA so r G <0 and the reaction is spontaneous» Called an exergonic

reaction (work producing)

Gibbs Energy, G

0 Extent of Reaction,

G > 0r

G < 0r

G = 0r

If µB >µA so r G >0 and the reverse reaction is spontaneous» Called an endergonic

reaction (work consuming)

When µB =µA so r G = 0 or system is at equilibrium» Thus a minimum in the reaction Gibbs energy plot

represents equilibrium» At a minimum, the slope = 0

GIBBS ENERGY AND EQUILIBRIUM

IDEAL GAS EQUILIBRIA

Consider homogenous gas phase reactions» Reactants and products are gases» Gases to 1st approximation behave as ideal gases

Since r G = (dG/d ) p,T = µB -µA and µ = µ° + RT ln(p),r G = µB ° + RT ln(p B ) - (µA °+ RT ln(p A )) = µB °- µA ° + RT

ln(pB /pA )rG = r G ° + RT ln(Q) where

rG ° = µB ° - µA °, the standard reaction Gibbs energyLike std reaction enthalpy it is the difference in

standard free energies of formation of the products - that of reactants

» Q = pB /pA , the reaction quotient which ranges from 0 (pure A) to infinity (pure B)

At equilibrium, r G = 0, so r G ° = - RT ln(K) or K = exp(- rG °/RT)» K is the equilibrium value of Q or K = (pB /pA )equilibrium



The minimum in Gibbs energy arises from the mixing of reactant and products» If there were no mixing,

G would change linearly in proportion to the amount of B formed Slope of G vs. would

be rG °

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

xB


» In a previous chapter, we learned mixG ° = nRT(x A ln(x A ) + x B ln(x B ) which is a U shaped function (minimum at 50% B)

If r G ° <0 then ln(K) >0 so product is favored If r G ° >0 then ln(K) <0 so reactant is favored

IDEAL GAS EQUILIBRIAConsider now a complicated case,

baoa

A

bB

aoA

boB

p

p

po

)P(PP

)/P(P)/P(PK

by given constant, mequilibriu theis K whereKlnRTG

bB(g) aA(g)

r

IDEAL GAS EQUILIBRIA: SAMPLE PROBLEM

From the given thermodynamic data below, calculate the equilibrium constant for the reaction at 298 K:

N2 (g) + 3H2 (g) 2NH3 (g)

f Go (NH3 ) = -16.6 kJ/mol


At 298 K, the partial pressures of gases in the reaction mixture described in the previous example are PN2 = 190 torr. PH2 = 418 torr, and PNH3 = 722 torr. Calculate the value of r G for the reaction.


If you have a number of products and reactants (all ideal gases), Dalton’s Law tells you Pi V = vi RT where vi is the number of moles of the ith component (stoichiometric factor in the equation)

For each component, the Gibbs energy difference Gi -Gi ° = RT ln(Pi )

At constant temperature, dG =VdP =nRTd lnP so» R G -R G° = RT

vi ln(Pi ) and @ equilibrium R G =0

Remember R Gi ° = R Gi ° (products ) - R Gi ° (reactants) = RT(

vi ln(Pi (prod) -

vi ln(Pi (reactants) )» Thus -R G° = RT

vi ln(Pi )equilibrium or -R G° /RT =

vi ln(Pi ) equilibrium

R G° is a constant at t=constant so exp(-R G° /RT ) = K

Further,

vi ln(Pi ) = ln(

Pi vi )equilibrium

So K =

Pi vi

– vi is positive for products and negative for reactants

In terms of activities (aA = pA /p*A ), K =(

ai vi )equilibrium

IDEAL GAS EQUILIBRIA: NOTES

For pure solids and liquids a = 1 so they don’t contribute to the reaction quotient (or K)

For ideal gases (aA = pA /p*A ), but for real gases (aA = fA /p*A ), where fA is the fugacity

K is a function of temperature and R G° is also a function of temperature

K is independent of total pressure and variation in partial pressures

K is called the thermodynamic equilibrium constant» Like activities it is dimensionless» You can approximate

K by replacing fugacities by partial pressures or molalities or molar concentrations

REAL GASESFugacity (f) – replaces partial pressure for real gases.

bar. 1 topressures low from ideally behavedgas the if bar 1 be would

fugacity the which at state :state standard fugacityPf :coeficient fugacity

fugacity fPfRTln o

o

REAL GASES:

For aA bB

paaA

bbB

f

aA

bB

f

KK)bar1/PA()bar1/PB(K

bar) 1 / (f)bar 1/f(K

REAL GASES: SAMPLE PROBLEM

What is the partial pressure of NO gas produced in the earth’s atmosphere from the reaction of atmospheric nitrogen and oxygen?

N2 (g) + O2 (g) 2NO(g)

Data: Gf °(N2 ) = 0; Gf °(O2 ) = 0; Gf °(NO ) = +86.6 kJ/molAt atmospheric pressure PN2 = 0.78 and PO2 = 0.21

SAMPLE PROBLEM:

For the equation, H2 O(g) H2 (g) + 1/2O2 (g), what is the mole fraction of oxygen resulting from passing steam through a tube at 2000 K if the Gibbs energy is 135.2 kJ/mol and P = 200kPa?

K = exp(-R G°/RT ) = exp ((-135.2 kJ/mol)/(8.3145*2000 J/mol)) =exp(-8.13)=2.945 x 10-4

If x moles of water dissociate then fraction left is 1-x, fraction of hydrogen = x and fraction of oxygen=x/2Total pressure = {(1-x) + x +x/2}p = {1 + 1/2x} [p=200 kPa]Partial pressures :water = (1-x)p/(1+1/2x); H2 = xp/(1+1/2x); O2 = 1/2xp/(1+1/2x)K = {(xp/(1+1/2x)( 1/2xp/(1+1/2x))0.5)/[ (1-x)p/(1+1/2x)]; if 1>>x this simplifies toK = (1/2) 0.5(xp) 1.5 /p = (p/2) 0.5(x) 1.5 = 2.945 x 10-4

Substituting for p (200 kPa/105 Pa)and solving for x , (x) 1.5 = 2.945 x 10-4 ;so x = 0.004426Mole fraction of oxygen = x/2= 0.00221

REACTIONS IN SOLUTION

aA(aq) bB(aq)

a

b

c

aoA

boB

m

mr

)M1/A()M1/B(K

Molarities inexpress we if or

molalities are m )m/m()m/m(K

KlnRTG o

HETEROGENOUS EQUILIBRIA

Solids and liquids, a = 1; do not include in equilibrium constant expressions.

Example:

CaCO3 (s) CaO(s) + CO2 (g)

SAMPLE PROBLEM:

Calculate the equilibrium constant for the following reaction at 298 K, using the following thermodynamic data:

2H2 (g) + O2 (g) 2H2 O(l)

Go (H2 O) = -237.2 kJ/mol

EFFECT OF TEMPERATURE

RS

RTHKln

RTGKln

T1

T1

RH

KKln

or

or

or

21

or

1

2



If reaction is exothermic H <0 slope of ln(K) vs. 1/T is positive (>0)» As T increases 1/T decreases so as T increases thus K gets smaller» Increase in temperature favors reactants.

If reaction is endothermic H >0 slope of ln(K) vs. 1/T is negative (>0)» As T increases 1/T decreases so as T increases thus K gets larger» Increase in temperature favors products.

LE CHATELIER’S PRINCIPLE

Henry Le Chatelier (1888) and F. Braun (1887)

“If a system in equilibrium is perturbed by subjecting it to a small variation in one of the variables that define the equilibrium, it will tend to return to an equilibrium state, which is usually somewhat different than the initial state”

EFFECT OF PRESSURE

4PKK

2B(g) A(g) reaction, the for

p

p

Response of Equilibria to Pressure

» Effect of additional inert gas - none because partial pressures unchanged (assumes ideality

» Compression (changing volume):Le Chatelier’s Principle says reaction will adjust to minimize

pressure increase

RESPONSE OF EQUILIBRIA TO PRESSURE

Example: A

2B; K = pB2/pA p°

On compression [A] will increase to compensate for pressure change» If extent of reaction is

then we start with n moles the A = (1- )n and b=2

n pb

Mole fraction of A, xA = (1-

)n/((1- )n + 2

n) = (1-

)/(1+ )

Mole fraction of B, xA = 2

/(1+ )

K = [(2

/(1+ ))p]2/(1-

)/(1+ )p = p[(4

2/(1+ )(1-

)] orK = 4

2 p/(1- 2 ) or 4

2 p + K 2 - K = (4p + K) 2 - K = 0– Note : p = p/p°

Solving for 2 = K/ (4p + K) = 1/(1 + 4p/K) or =1/(1 + 4p/K)1/2

This says as p increases then the extent of reaction decreases

Example: 3H2 + N2

2NH2

K = pNH32/p2

H2 pN2

K = xNH32p2/x3

H2 p3xN2 p where pA =xA pK = [xNH3

2 /x3H2 xN2 ] [p2/p 4 ] = [xNH3

2 /x3H2 xN2 ] [p2/p 4 ]

K = KX 1/ p2 where, KX = [xNH32 /x3

H2 xN2 ] or KX

p2

» Since K is constant, if p increases must increase as the square» Self test 9.3: If there is a ten-fold increase in pressure, what will be the effect on the

constituents

KX

p2 so a ten fold increase in pressure will increase KX by 100

This means that the products must increase or the reactants must decrease– Le Chatleier’s principle tells you that by inspection since reactants have the greatest contribution to

the total pressure

EFFECT OF CATALYST

SAMPLE PROBLEMConsider the following reaction:

4NH3 (g) + 3O2 (g) 2N2 (g) + 6H2 O(g) H = +42.6 kJ/mol

How does the equilibrium position shift when:a) 0.20 atm of Argon is added?b) The gaseous water product is liquified as soon as it is formed?c) 0.2 moles of oxygen is added?d) The temperature of the system is increased from 278K to 700K?e) The total system pressure is increased by 100 atm?f) If gaseous HCl is added? Note that NH3 reacts with HCl to form

NH4 Cl(s)

EQUILIBRIUM APPLICATIONS: LIME PRODUCTION FROM LIMESTONE

CaCO3 (s) CaO(s) + CO2 (g) H = +178 kJ/mole

Reaction is endothermic; so formation of lime is favored at high temperature.

One mole gas is formed; no gas in the reactant so decreasing the pressure will result in the formation of more CaO.

In normal practice the reaction is carried out at 900oC to 1000oC at atmospheric pressure in lime kilns that are well ventilated.

limestone lime

CO2p p K

EQUILIBRIUM APPLICATIONS:

LIME PRODUCTION FROM LIMESTONE


METHANOL PRODUCTIONMethanol is an important alcohol

used in fuel mixtures, making methyl esters and oxidation to methanol (formaldehyde) to make urea- formaldehyde resin glues.

It is manufactured directly from synthesis gas

CO(g) + 2H2

(g) CH3

OH(g) (∆H = -90

kJ mol-1)

Kp

= pCH3OH ---------------pCO

pH22


METHANOL PRODUCTION

The reaction is carried out at 250oC, over a Cu-ZnO-Al2

O3

(alumina) catalyst

at a pressure of 5-10 x 106 Pa

(5-10MPa, 50-100 atm).

Theoretically the reaction is favored by high pressure (3 gas mol ==> 1 gas mol) and low temperature.

In practice a high pressure is used to give an acceptable yield in accordance with Le Chatelier's Principle, but a moderately high temperature plus a catalyst, are employed to get an economic production rate.


SYNTHESIS OF AMMONIA -

THE HABER PROCESS

The three raw principal materials are (i) air, (ii) coal-coke/methane or higher hydrocarbons and (iii) water.

The nitrogen is obtained from air (80% N2) by two principal methods:

Air is cooled and compressed under high pressure to form liquid air (liquefaction). The liquid air is fractionally distilled at low temperature to separate oxygen (used in welding, hospitals etc.), nitrogen (for making ammonia), Noble Gases e.g. argon for light bulbs, helium for balloons).

By burning natural gas in air

and separating the nitrogen from the water and carbon dioxide combustion products.



THE HABER PROCESSSynthesis gas is the primary source of hydrogen, though 'syn-gas' or 'syngas' is a mixture of hydrogen and carbon monoxide and is made by three main methods.

1. Historically synthesis gas was made by passing steam over white hot

coke at 800-1200oC. Coke is made by heating coal at high temperature and is mainly carbon. This process is still used in many parts of the world and if coal is available in a country without oil, then it will be cheaper than using hydrocarbons from oil.



THE HABER PROCESSC(s) + H2

O(g) CO(g) + H2

(g) (∆H = +131

kJ mol-1)

The very high temperature favors the endothermic production of hydrogen and of course considerably increases the rate of reaction. Water and carbon show little reaction at lower temperatures.

The reaction can be carried out at normal pressure since higher pressures favour the reverse reaction (1 gas mol <== 2 gas mol).



THE HABER PROCESS2.

Methane can be partially oxidized

by reaction with the oxygen in air.

CH4

(g) + 1/2O2

(g)

CO(g) + 2H2

(g)

(∆H = -36

kJ mol-1)

Low pressure (1.5 gas mol ==> 3.0 gas mol) and low temperatures (exothermic) favor this reaction.

This reaction can be carried out with nearly 100% conversion at atmospheric pressure (so low pressure) at 800-1000oC using cobalt/nickel based catalysts e.g. Ni-ThO2 or Ni-SiO2 . Very little water or carbon dioxide is formed if the reaction conditions are carefully controlled.



THE HABER PROCESSThese days hydrogen is primarily made by reacting methane

(natural gas) and water (steam), and the process is called steam- methane reforming.

CH4

(g) + H2

O(g) 3H2

(g) + CO(g) (∆H = +206

kJ mol-1)

Kp

= pH23 pCO

-----------------pCH4 pH2O

The reaction is carried out at 700-1000oC, over a nickel catalyst, at a pressure of 1-2 x 106

Pa

(1-2MPa, 10-20 atm).



THE HABER PROCESS

This reaction is favoured by high temperature (endothermic, heat absorbing) and low pressure (2 gas moles ==> 4 gas moles).

In practice a high temperature is indeed used, and combined with a catalyst, both factors will increase the rate of reaction to ensure the process is economic.

However, although a low pressure is advantageous, in practice a moderately high pressure is used, which effectively increases the concentration of the reactants and provides a greater bulk of material passing through the reactor in a given time.



THE HABER PROCESS

The yield of hydrogen can be further increased by using the 'shift' reaction.

CO(g) + H2

O(g) CO2

(g) + H2

(g)

(∆H = -41

kJ mol-1)

Since 2 gas mol ==> 2 gas mol, changing pressure offers no advantage. However, the lower the temperature, the greater the yield of this exothermic reaction.

But, too low a temperature leads to an uneconomic low rate of reaction, so research continues into low temperature catalysts

The carbon dioxide can be absorbed under pressure into water or potassium carbonate solution.

HABER PROCESS:



THE HABER PROCESSThe Bosch-Haber

synthesis equation for ammonia is

N2

(g) + 3H2

(g) 2NH3

(g)

(∆H = -92 kJ mol-1)

Kp

= pNH32---------------

pN2 pH23 Since a dynamic equilibrium will form, there is no chance of 100% yield even if you use, as you actually do, the theoretical stoichiometric reactant volume/mole ratio of N2 :H2 of 1 : 3 !

In forming ammonia, heat energy is given out to the surroundings

Four moles of 'reactant' gas form two moles of 'product' gas, so there is net decrease in gas molecules on forming ammonia.



THE HABER PROCESSSo applying the equilibrium rules, the formation of ammonia

is favored by:

(a)

Using high pressure

because you are going from 4 to 2 gas molecules (the high pressure also speeds up the reaction because it effectively increases the concentration of the gas molecules) and allows a greater bulk flow rate material through the reactor, but

very high pressure means more dangerous and more costly engineering to address the health and safety issues involved.

(b)

Carrying out the reaction at a low temperature, because it is an exothermic reaction favored by low temperature, but this may produce too slow a rate of reaction,



THE HABER PROCESS

So, the idea is to use a set of optimum conditions to get the most efficient yield of ammonia

and this may well, as in this case, be a low % yield (e.g. 8% conversion) but fast!!!

Described below are the conditions to give the most economic production of ammonia.

These arguments make the point that the yield* of an equilibrium reaction depends on the conditions used



THE HABER PROCESS



THE HABER PROCESS

The above chart completely illustrates and vindicates the predictions from Le Chatelier's

Principle.

In industry moderate-high pressures of 2.5-25 x 106

Pa

(2.5-25MPa, 25-250 atm) in line with the theory but to high to raise H&S engineering costs.

Theoretically a low temperature would give a high yield of ammonia BUT

... Nitrogen is very stable molecule

and not very reactive i.e. chemically inert

and the rate of reaction at high % yield low temperatures is much to slow to be economic.



THE HABER PROCESS

To speed up the reaction an iron oxide catalyst

(Fe3 O4 ) is used as well as a higher temperature (e.g. 400 -

450oC).

The higher temperature is an economic compromise, i.e. it is more economic to get a low yield very fast, than a high yield very slowly!

Note: a catalyst does NOT affect the yield of a reaction, i.e. the equilibrium position BUT you do get there faster!



THE HABER PROCESSAt the end of the process, ammonia is liquified

and is so can be removed and stored in cylinders.

Any unreacted

nitrogen or hydrogen (NOT liquified), is recycled back through the reactor chamber, nothing is wasted!

Nitrogen (bp: -196oC) and hydrogen (bp: -252oC) have much lower boiling points than ammonia (bp: -33oC).

To sum up: A low % yield of ammonia is produced quickly at moderate high temperatures and pressures, and is more economic than getting a higher % equilibrium yield of ammonia at a more costly high pressure or

slower lower temperature reaction.


MANUFACTURE OF SULFURIC ACID

The Contact Process of sulphur

trioxide production must be economically efficient for the manufacture of the important industrial chemical sulphuric

acid.

In the Contact Process reactor the sulphur dioxide is mixed with air (the required stoichiometric volume/mole SO2:O2 ratio is 2:1, in practice 1-2:1 is used) and the mixture passed over a catalyst of vanadium(V) oxide V2 05 at a relatively high temperature of about 450°C and at a pressure of between 1-2 atm.



(1) In the reactor the sulphur

dioxide is oxidised

in the reversible exothermic reaction.

2SO2

(g) + O2

(g) 2SO3

(g) (∆H = -196

kJ mol-1)

Kp

= pSO32 ----------------pSO22 pO2

The reaction forms sulphur trioxide and the equilibrium is very much to the right hand side because



Despite the reaction being exothermic a relatively high temperature is used which favours the reverse reaction R to L, from the energy change equilibrium rule, i.e. increasing temperature shifts the equilibrium in the endothermic direction. However the value of Kp is high enough to give a 99% yield. The reaction is favoured by high pressure (pressure equilibrium rule, 3 => 2 gas molecules), but only a small increase in pressure is used to give high yields of sulphur trioxide, because the right hand side is energetically very favourable (quite exothermic and high Kp)



The use of the V2O5 catalyst ensures a fast reaction

without having to use too a higher temperature which would begin to favor the left hand side too much (energy change equilibrium rule), but remember a catalyst does not affect the % yield or equilibrium concentration of SO3 , you just get there more economically faster. Multiple reactor beds are used to ensure the maximum % conversion and heat exchange systems are used to control the temperature, and pre-heat incoming reactant gases. Good anti-pollution measures

need to be in place since the sulphur oxides are harmful and would cause local acid rain! To help this situation AND help the economics of the process the residual SO2 is kept to the minimum by the reaction conditions describe above.



(2) The sulphur trioxide is dissolved in concentrated sulphuric acid to form fuming sulphuric acid (oleum).

SO3

(g) + H2

SO4

(l) ==> H2

S2

O7

(l)

(3) Water is then carefully added to the oleum to produce concentrated sulphuric acid (98% H2 SO4 ).

H2

S2

O7

(l) + H2

O(l) ==> 2H2

SO4

(l)

If the sulphur trioxide is added directly to water an acid mist forms which is difficult to contain because the reaction to form sulphuric acid solution is very exothermic with a big K value!



THE HABER PROCESS



THE HABER PROCESS



THE HABER PROCESS

Documents

Chapter 9: Chemical Equilibrium - R. M. FABICON's BLOG · 2010. 9. 14. · Consider the simple reaction A B » Examples d-alanine to l- alanine » Suppose an infintesimal amount of