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19/01/2011 1 TOPIC 4: CHEMICAL EQUILIBRIUM Lecturer and contacts Mr. Vincent Madadi Department of Chemistry, University of Nairobi P. O. Box 30197-00100, Nairobi, Kenya Chemistry Dept. Rm 114 Tel: 4446138 ext 2185 Email: [email protected], [email protected] Website: http://www.uonbi.ac.ke/staff/vmadadi 19/01/2011 1 mov Introduction Chemical equilibrium is based on the fact that reversible reactions do not go to completion Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. A chemical equilibrium is a reversible reaction where the forward reaction rate is equal to the reverse reaction rate. Thus, Molecules are continually reacting, even though the overall composition of the reaction mixture does not change. 19/01/2011 2 mov

Chemical equilibrium

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Page 1: Chemical equilibrium

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TOPIC 4: CHEMICAL EQUILIBRIUM

Lecturer and contacts

Mr. Vincent Madadi

Department of Chemistry, University of Nairobi

P. O. Box 30197-00100,

Nairobi, Kenya

Chemistry Dept. Rm 114

Tel: 4446138 ext 2185

Email: [email protected], [email protected]

Website: http://www.uonbi.ac.ke/staff/vmadadi

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Introduction

• Chemical equilibrium is based on the fact that reversible

reactions do not go to completion

• Chemical equilibrium exists when two opposing reactions occur

simultaneously at the same rate.

• A chemical equilibrium is a reversible reaction where the

forward reaction rate is equal to the reverse reaction rate.

• Thus, Molecules are continually reacting, even though the

overall composition of the reaction mixture does not change.

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Introduction cont.

• The RATE of the forward reaction equals the RATE

of the reverse reaction

• Symbolically, this is represented as:

• aA(g) + bB (g) ⇌ cC (g) + dD (g)

• The equilibrium constant can be expressed as:

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Introduction cont.

• Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction.

• aA(g) + bB (g) ⇌ cC (g) + dD (g)

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Law of mass action

• At constant temperature, the rate of chemical reaction is

proportional to active mass of the reacting substances

• Active mass = is a thermodynamic quantity defined as a =

fC,

• Where a = active mass; f = activity coefficient, C = molar

concentration

• For gaseous systems at low pressure or dilute solutions,

f = 1, thus a = C

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Equilibrium law

• For the reaction , aA(g) + bB (g) → cC (g) + dD (g)

• Rate of forward reaction R1 = k1[A]a[B]b

• Rate of backward reaction R2 = k2[C]c[D]d

• At equilibrium,

k1[A]a[B]b = k2[C]c[D]d

Thus,

K1/k2 = [C]c[D]d/[A]a[B]b = Kc

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Introduction cont.

• Therefore , for the general reaction:

a A(g) + bB (g) ⇌ c C (g) + d D (g)

• Eqiliburium law can define a constant

Kc is defined for a reversible reaction at a given temperature as the

product of the equilibrium concentrations (in M) of the products,

each raised to a power equal to its stoichiometric coefficient in the

balanced equation, divided by the product of the equilibrium

concentrations (in M) of the reactants, each raised to a power

equal to its stoichiometric coefficient in the balanced equation.19/01/2011 7mov

Introduction cont.

• For gaseous reactions, concentration can be expressed in terms of partial pressure

• Where, aA + Bb ⇌ lL + mM,

• If PA, PB, PL and PM are the partial pressures of the gaseous species, then at equilibrium

• Kp = (PL)i(PM)m/(PA)a(PB)b

• If concentration is in mole fraction (X), the equilibrium constant can be expressed in form of Kx

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Introduction cont.

• Where

Kx = (XL)I(XM)m/(XA)a(XB)b

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Relationship between Kc, Kp and Kx

• For ideal gas, partial pressure of the component in the mixture is given by,

• Pi = (m/v)RT

• Where m = number of moles of the gas occupying volume V,

• Thus m/v = Ci = Molar concentration

• Therefore, Pi = CiRT,

• Substituting the value of Pi into the expression for Kp

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Relationship between Kc, Kp and Kx ...

• Kp = (CLRT)l(CMRT)m/(CART)a(CBRT)b

• Kp = (CL)l(CM)m (RT)l(RT)m

(CA)a(CB)b(RT)a(RT)b

• Kp = (CL)l(CM)m (RT)l+m

(CA)a(CB)b(RT)a+b

• But

Thus Kp = Kc (RT)Δn

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Relationship between Kc, Kp and Kx ...

• From Kp = Kc (RT)Δn

• When Δn = +ve, the number of moles of reactants are

less than the number of moles of products , thus Kp >Kc

• When Δn = -ve, the number of moles of reactants are

more than the number of moles of products , thus Kp <Kc

• When Δn = 0, the number of moles of reactants equals

the number of moles of products , thus Kp =Kc

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Relationship between Kc, Kp and Kx ...

• For Kp and Kx,

• Note that the mole fraction of a component of a given

gas mixture is given by Xi = Pi/P

• Or Pi = XiP

• Where P = total pressure of the gaseous mixture

• Substituting the value of partial pressure into the

expression for Kp

• Then,

Kp = (PL)l(PM)m/(PA)a(PB)b

• Or Kp = (XLP)l(XMP)m/(XAP)a(XBP)b

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Relationship between Kc, Kp and Kx ...

• This can be expressed as,

Kp = (XL)l (XM)m(P)l (P)m

(XA)a (XB)b(P)a (P)b

• Or Kp = (XL)l (XM)m(P)(l+m)

(XA)a (XB)b(P)(a+b)

This can be reduced to,

• Or Kp = Kx (P)(l+m) = Kx PΔn

(P)(a+b)19/01/2011 14mov

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Relationship between Kc, Kp and Kx ...

• Therefore,

Kp = Kx PΔn = Kc (RT)Δn

• Hence, Kc = Kx (P/RT) Δn

• Equivalent to,

Kc = Kx (n/v) Δn

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Factors That Affect Chemical Equilibrium

1) concentration

2) pressure

3) volume

4) temperature

5) Catalysts- have no effect on position of

equilibrium

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Le Chatelier’s Principle

• If an external stress is applied to a system at

equilibrium, the system adjusts itself in such a way

that the stress is partially offset.

1) Changes in Concentrations

• Increase the yield of product by

– increasing concentration of reactant

– removing the product from the equilibrium

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Le Chatelier’s Principle cont.

• Eaxample

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

0.683 M 8.80 M 1.05 M

• Increase the concentration of NH3

to 3.65 M, the position of equilibrium shifts to the left

Q = [NH3]2 = ( 3.65 )2 = 0.0286

[N2] [H

2]3 (.683) (8.80)3

But Kc = [NH3]2 = ( 1.05 )2 =

[N2] [H

2]3 (.683) (8.80)3

• Q >Kc19/01/2011 18mov

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Le Chatelier’s Principle cont.

BaSO4

(s) Ba2+(aq) + SO42- (aq)

Kc = [Ba2+] [SO42-]

• By adding Ba2+(aq), [SO4 2-] decreases, but BaSO4 (s)

increases

• By add SO42-

(aq), [Ba2+] decreases but BaSO4 (s)

increases

• Add BaSO4 (s) no change19/01/2011 19mov

Changes in Volume and Pressure

• Little effect on reactions in solution

• Effect can be large on reactions in the gas phase

• Increase in pressure shifts the equilibrium to the side with the fewer moles of gas

• Example:

• How does the position of equilibrium change as the pressure is increased?

five moles of gaseous reactant two moles of gaseous product

• Increase in pressure causes an increase in products at the expense of reactants

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Changes in Temperature

• Changes the value of the equilibrium constant

– A temperature increase favours an endothermic reaction,

and a temperature decrease favours an exothermic reaction

• Example:

1) Endothermic reactions

• Shifts in the equilibrium position for the reaction:

58 kJ + N2O4 (g) 2NO2

(g)

• Increase temperature: Favours forward reaction

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The Effect of a Catalyst• Increases rate, but has no effect on position of

equilibrium

� effect on forward and reverse processes is the same

� a catalyst increases both the rate of both the forward and reverse reactions

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The Haber Process

• Maximize the yield of ammonia by: carrying out the

reaction at high pressure

ΔHo = -92.6 kJ/mol

1.

2.

• Maximize the yield of ammonia by: carrying out the

reaction at low temperatures

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The Haber Process

• In practice the reaction is carried out at 500 ºC

because the rate is too slow at lower temperatures

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Disruption and restoration of equilibrium

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At the left, the concentrations of the three components do not change with time because the system is at equilibrium.

By adding more hydrogen to the system, disrupts the equilibrium. A net reaction then ensues that moves the system to a new equilibrium state (right) in which the quantity of hydrogen iodide has increased; in the process, some ofthe I2 and H2 are consumed.

The new equilibrium state contains more hydrogen than did the initial state, but not as much as was added; as the LeChâtelier principle predicts, the change wemade (addition of H2) has been partially counteracted by the "shift to the right".

Equilibrium constant calculations

• One litre of equilibrium mixture from the following

system at a high temperature was found to contain 0.172

mole of phosphorus trichloride, 0.086 mole of chlorine,

and 0.028 mole of phosphorus pentachloride. Calculate

Kc for the reaction.

• Ans: Reaction PCl5 PCl3 + Cl2

• Expression for Kc

• Substitute the values

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Equilibrium constant calculations cont.

• The decomposition of PCl5 was studied at another temperature.

One mole of PCl5 was introduced into an evacuated 1.00 litre

container. The system was allowed to reach equilibrium at the

new temperature. At equilibrium 0.60 mole of PCl3 was present in

the container. Calculate the equilibrium constant at this

temperature.

Ans: Reaction PCl5 ⇌ PCl3 + Cl2

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Equilibrium constant calculations cont.

• Kc = [PCl3][Cl2]

[PCl5],

• Kc = (0.60)(0.60) = 0.90

• (0.40)

Significance of Kc

• Large equilibrium constants indicate that most of the reactants are converted to products.

• Small equilibrium constants indicate that only small amounts of products are formed.Kc > 1 mostly productsKc < 1 mostly reactantsKc ~ 1 equal amounts of products and reactants19/01/2011 28mov

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Equilibrium constant calculations cont.

• The mass action expression or reaction quotient has the

symbol Q.

• Q has the same form as Kc, but the difference between

them is that the concentrations used in Q are not

necessarily equilibrium values.

• The Reaction Quotient

aA + bB ⇌ cC + dD

• Q will help us predict how the equilibrium will respond

to an applied stress.

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Equilibrium constant calculations cont.

• To make this prediction we compare Q with Kc.

Q = [C]c[D]d

[A]a[B]b

• Q < K Products are formed

• Q = K System is at equilibrium

• Q > K Reactants are formed

• Example

• The equilibrium constant for the following reaction is

49 at 450°C. If 0.22 mole of I2, 0.22 mole of H

2, and

0.66 mole of HI were put into an evacuated 1.00-liter

container, would the system be at equilibrium?19/01/2011 30mov

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Equilibrium constant calculations cont.

• Ans.

H2(g) + I2(g) → 2 HI(g)

0.22 M 0.22 M 0.66 M

• The Reaction Quotient

• Q= [HI]2 = (0.66)2 = 9.0

H2 [ ] [I2] (0.22)(0.22)

• Q = 9.0 but Kc = 49

• Thus Q < Kc

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Equilibrium constant calculations cont.

• The equilibrium constant, Kc, is 3.00 for the following

reaction at a given temperature. If 1.00 mole of SO2

and 1.00 mole of NO2 are put into an evacuated 2.00 L

container and allowed to reach equilibrium, what will

be the concentration of each compound at

equilibrium?

SO2(g) + NO2 (g) ⇌ SO3(g) + NO (g)

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Equilibrium constant calculations cont.

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Equilibrium constant calculations cont.

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Equilibrium 0.0800+ x 0.0600+ x 0.790 -2x

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Equilibrium constant calculations cont.

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Types of chemical equlibria

• There are two types of chemical equilibria:

1) Homogeneous equilibrium

• All the reactants and products are in the same phase

2) Heterogeneous equilibrium

• The reactants and products are in two or more

different phases

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Types of chemical equiliburia cont.

• Heterogeneous equilibria have more than one phase

present.

• For example, a gas and a solid or a liquid and a gas.

CaCO3(s) ⇌ CaO(s) + CO2(g)

• How does the equilibrium constant differ for

heterogeneous equilibria?

• Pure solids and liquids have activities of unity.

• Solvents in very dilute solutions have activities that are

essentially unity.

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Heterogeneous equilibrium

• The thermodynamic equilibrium, Ka is given by:

Ka = (aCaO )(aCO2 ) Eq.1

(aCaCO3)

• Where a = activity of various species

• For pure solids and liquids, activity are taken as unity at

all temperatures

• Thus, = aCaCO3 = 1 = aCaO

• The equation reduces to Ka = aCO2 Eq. 219/01/2011 38mov

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Heterogeneous equilibrium cont.

• If the gas behaves ideally, activity reduces to pressure. Thus the

equation 2 becomes:

Kp = P CO2

• The equilibrium constant for heterogeneous reactions is

independent of the amount of a pure solid (liquid) phase

• Thus equilibrium constant involves gaseous constituents but does

not include any term for concentration of either pure solid or

liquid

• Thus Kp of heterogeneous reaction is generally called condensed

equilibrium

39mov

Example- calculations involving Heterogeneous

equilibrium

• At a certain temperature, Kp for the dissociation of

the solid calcium carbonate is 4 x 10-2 atm and for the

reaction Cs + CO2 ⇌ 2CO is 2 atm respectively.

Calculate the pressure of CO at this temperature

when solid carbon, CaO and CaCO3 are mixed and

allowed to attain equilibrium

• Ans. CaCO3(s) ⇌ CaO(s) + CO2 (g) eq. 1

• Cs + CO2(g) ⇌ 2CO(g) eq.2

• Kp for the reaction 1 is:

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Example- calculations involving Heterogeneous

equilibrium cont.

• Kp = aCaO.aCO2/aCaCO3 = P CO2

• K’p for reaction 2

K’p = (aCO)2/(aC)(aCO2) = P2 CO/P CO2

K’pKp = P CO2. P2CO/P CO2 = P2 CO

Thus, P CO = (KpK’p)½ = ((4.0 x 10-2)(2.0))½

= 0.28 atm

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Example- calculations involving Heterogeneous

equilibrium cont.

• A solid NH4HS is placed in a flask containing 1.50 atmospheric

ammonia. What is the partial pressures of ammonia and H2S

when equilibrium is attained? (Kp = 0.11)

• Ans: Reaction,

NH4HS(S) ⇌ NH3 + H2S

P NH3 = 1.5 + P H2S

Kp = (P NH3) (P H2S)

Thus 0.11 = (1.50 + P H2S) (P H2S)

This implies,

P H2S = 0.06 atm or -1.56 atm

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Example- calculations involving Heterogeneous

equilibrium cont.

• But 0.06 atm is more correct,

• Hence P NH3 = 1.50+0.06 = 1.56 atm

19/01/2011 mov 43

Example involving Kc, Kp & Kx

• At 21.5 oC and total pressure of 0.787 atm, N2O4 is

48.3% converted into NO2. Calculate the value of Kp

and Kc for the reaction

• Ans:

• Reaction N2O4 (g) ⇌ 2NO2 (g)

• Initial: a 0

• At t= t, a-x 2x

• Let a= 100, then x = 48.319/01/2011 mov 44

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Example involving Kc, Kp & Kx

• Therefore,

(a-x) = 100 - 48.3 = 51.7

Thus 2x = 48.3 x 2 = 96.6

Hence (nT)aq. = 51.7 + 96.6 = 148.3

PNO2 = (96.6/148.3) X PT = 96.6X 0.0787/148.3

PN2O4 = 51.7x 0.0787/148.3

Kp = (PNO2)2 = 96.6x0.0787/148.3

PN2O4 51.7 x0.0787/148.3

19/01/2011 mov 45

Example involving Kc, Kp & Kx

• Kp = 9.58 x 10-3 atm

• From th first principles,

• Kp = Kc(RT)Δn but Δn = 2-1 = 1

• Thus Kc = Kp/RT (R = 8.31 J/mol K = 0.0821

Latmmol-1K-1)

= 9.58x10-3/0.082 x (273 +21.5)

= 0.00396 molL-1

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Shifts in chemical equilibrium

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Example: Le Chatelier’s principle in

biological systems• Carbon monoxide poisoning. Carbon monoxide, a product of

incomplete combustion that is present in automotive exhaust

and cigarette smoke, binds to hemoglobin 200 times more tightly

than does O2.

• This blocks the uptake and transport of oxygen by setting up a

competing equilibrium O2-hemoglobin hemoglobin CO-

hemoglobin

• Air that contains as little as 0.1 percent carbon monoxide can tie

up about half of the hemoglobin binding sites, reducing the

amount of O2 reaching the tissues to fatal levels.

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Le Chatelier’s principle in biological

systems cont.

• Carbon monoxide poisoning is treated by administration of

pure O2 which promotes the shift of the above equilibrium to

the left.

• This can be made even more effective by placing the victim in a

hyperbaric chamber in which the pressure of O2 can be made

greater than 1 atm.

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Examples of chemical equilibrium

calculations• The commercial production of hydrogen is carried out by treating

natural gas with steam at high temperatures and in the presence

of a catalyst (“steam reforming of methane”):

CH4 + H2O ⇌ CH3OH + H2

• Given the following boiling points: CH4 (methane) = –161°C, H2O

= 100°C, CH3OH = 65°, H2 = –253°C, predict the effects of an

increase in the total pressure on this equilibrium at 50°, 75° and

120°C.

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Ans:

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Examples of chemical equilibrium

calculations cont.

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Comment: This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding effect of acid rain on buildings and statues. The first reaction is “driven” by a second reaction having a large equilibrium constant.

From the standpoint of the LeChâtelier principle, the first reaction is “pulled to the right” by the removal of carbonate by the hydrogen ion. “Coupled” reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked in this way.

Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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Examples of chemical equilibrium

calculations cont.

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