Equilibrium

Brief OutlineWhat is reversible reaction?Examples of reversible reactionDynamic EquilibriumLe Chatelier’s PrincipleThe Haber Process

What is reversible reaction? A reversible reaction is a chemical reaction in

which the products can be converted back to reactants under suitable conditions.

A reversible reaction is shown by the sign ( ) a half-arrow to the right (forward reaction) and a half-arrow to the left (reverse reaction).

HgO HgO

HgOHgO

HgO

HgO

HgOHg

O2

Hg

Hg

HgO

Hg

HgO

O2

HgO

2 HgO (s) 2 Hg (l) + O2 (g) 2 Hg (l) + O2 (g) 2HgO (g)

Reversibe Reactions

Mercury and oxygen combine to form mercury oxide just as fast as mercury oxide decomposes into mercury and oxygen

2 HgO (s) 2 Hg (l) + O2 (g)

Both reactions continue to occur, but there is

no net change in the composition of the system.

The amounts of mercury(II)oxide, mercury(Hg),

and oxygen (O2) remain constant. There is a

state of equilibrium between two chemical reactions.

6

Dynamic Equilibrium

Evaporation (No Equilibrium)

Evaporation

Liquid Gas(No Equilibrium) (No Equilibrium)

Liquid Gas Liquid Gas(Equilibrium)

Closed System

Open System

ExampleFor the decomposition,

CaCO3 CaO + CO2

Would the reaction be the same when temperature is kept constantly high in an open and closed system?

Dynamic Equilibrium

Ag + + Cl -

AgCl (s) ChemicalEquilibrium

Ag + Cl -

Cl - Ag +

AgCl (s)

Rate of Precipitation = Rate of Dissolving

HC2H3O2 (aq) H + + C2H3O2 -

Rate of Dissociation (ionization) = Rate of Association

HC2H3O2

H + C2H3O2 -

H +

HC2H3O2C2H3O2 -

Dynamic equilibrium is the state in which the forward and reverse reactions are occurring at the same rate. The forward and reverse reactions both continue. The concentration of all reactants and products remains constant)

Dynamic equilibrium vs Static Equilibrium

Dynamic Equilibrium Static Equilibrium

Characteristics of Dynamic Equilibrium

The concentrations of the reactants and products (macroscopic properties) remain the same but the reactions don't stop!

The reactants are still reacting to form the product and the product is still being converted back to the reactants (microscopic processes).

These two sets represent the same chemical reaction system, but with the reactions occurring in opposite directions. Horizontal part : the concentrations of reactants and products remain constant when the system reaches equilibrium.

2 HI H2 + I2 Dissociation of Hydrogen Iodide

H2 + I2 2 HI Synthesis of Hydrogen Iodide

The equilibrium state is independent of the direction from which it is approached. Whether we start with an equimolar mixture of H2 and I2 (left) or a pure sample of hydrogen iodide (shown on the right, using twice the initial concentration of HI to keep the number of atoms the same), the composition after equilibrium is attained (shaded regions on the right) will be the same.

H2 + I2 2HI

For more information, Click Here

The adjacent graph shows the changes in the reaction rates of the forward and backward reactions:A + B C +

DInitially (t = 0), [A] and [B] were maximum, while [C] and [D] were zero. The rate of the forward reaction decreases as A and B are used up.The rate of the reverse reaction increases as C and D are formed. Equilibrium is attained when the two rates become equal

[A], [B], [C], and [D] remain constant at equilibrium.

Chemical Equilibrium

If NO2 is reddish brown and N2O4 is colorless:What is happening here?

What properties are changing?What is happening over time? After a long time?

Consider this reaction:

The Equilibrium Constant (homogeneous reactions)

The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

For a reaction of the type

the following is a CONSTANT (at a given T)

• Kc provides information about how far a reaction proceeds at a particular temperature. (can predict the conc. of reactants & products)• Only write the Kc for homogeneous reversible reactions (all in the same physical state)

Kc : equilibrium constant[C] and [D] : concentration of products[A] and [B] : concentration of reactants

Your Turn

.

Equilibrium mixWhat is the equilibrium

expression for the reaction?

UA GenChem

Kc ~ 92/(1 x 1) = 81

Your Turn

Your Turn

. .

Equilibrium mix

What is the equilibrium expression for the

reaction?

Your Turn

.

Equilibrium mix

Kc ~ 22/(6 x 4) = 0.17

What Does the Value of K Mean?

If K >> 1, the reaction proceeds almost totally towards the products.

If K << 1, the reaction hardly proceeds at all towards the products.

What Does the Value of K Mean?

If K >> 1, the conc of the product is greater than the conc of the reactants at equilibrium.

• If K << 1, the conc of the product is less than the conc of the reactants at equilibrium.

If K = 1, the conc of the product is = the conc of the reactants at equilibrium.

The reaction goes about half way to completion.

Reactants

Products

The equilibrium constantcan also be expressed as partial pressure of

gases. (Keq)when aqueous solutions are used,

concentrations are used (Kc)

The Equilibrium Constant (heterogeneous reactions)The concentrations of solids and liquids are constant. They do not appear in the equilibrium expression.

Keq = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

H2O

Water is not included• Liquid water is in lower entropy phase than ions (ions are

more spread out)• Concentration of water is a constant – factored into the Keq

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

Keq = [CO2] CO2 (g) is in the highest entropy

NH3(g) + HCl(g) NH4Cl(s) 1 [NH3] [HCl]Keq =

As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

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

What Are the Equilibrium Expressions for these equilibria?(a) 2SO2(g) + O2(g) 2SO3(g)

(b) Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq)(c) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Page 283 Qn 4

Give the expression of the equilibrium constant

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

Given [N2]=0.1M,[H2]=0.125M, [NH3]=0.11M, find the equilibrium constant for the reaction.

Example

Find the relationship between the equilibrium constants of the forward and backward reaction for:

H2 (g) + I2 (g) 2HI (g)

Example

The following reaction is an esterification reaction producing ethyl ethanoate.CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l)

The value of Kc at 250C is 4.0. This equilibrium can be approached experimentally from the opposite direction. What is the value for Kc for the reaction, the hydrolysis of ethyl ethanoate.

CH3CO2C2H5(l) + H2O(l) CH3CO2H(l) + C2H5OH(l)

Example

How far will a reaction go?Kc tells us how far a reaction has gone towards

completion under given conditions.E.g. H2 (g) + I2 (g) 2HI (g) Kc= 2 at 2770C

H2 (g) + Cl2 (g) 2HCl (g) Kc= 1018 at 2770C

The large difference in magnitude for Kc values illustrates how stable HCl is at 2770C compared to HI.The bondingin HI is weaker and therefore the reverse reaction is more evident with an equilibrium being established.

The following reaction takes place at 460ºC, where the equilibrium constant K has a value of 85. SO2(g) + NO2(g) NO (g) + SO3 (g) At a certain moment, the concentrations of the reactant and products were measured to be: [SO2] = 0.04, [NO2] = 0.5M, [NO] = 0.3M, [SO3] = 0.02M(a) Is this system at equilibrium?(b) If not, in which direction must the reaction go to reach equilibrium?Note: Use reaction quotient, Qc since we are unsure if it is in equilibrium.

Example

Since Qc < Kc , the reaction will move to the right to generate more product to reach equilibrium.

Relationship between Qc & Kc

If Qc < Kc , the reaction will move to the right to generate more product to reach equilibrium.

If Qc = Kc , the reaction is at equilibrium.

If Qc > Kc , the reaction will move to the left to generate more reactants to reach equilibrium.

The value of Qc in relation to Kc indicates the direction in which the net reaction must proceed as the system moves towards its equilibrium.

Le Chatelier’s Principle if a system at equiibrium is subjected to some changes,

the position of equilibrium will shift to minimise the change.

The possible changes in conditions includeChanges in concentration o f either the reactants or

products.Changes in pressure for gas phase reactionsChanges in temperaturePresence of a catayst

Has industrial significance as it allows chemists to alter the reaction conditions to produce an increased amount of product, hence increase the profitability of the chemical process.

ConcentrationPCl5 (g) PCl3(g) + Cl 2(g)

Kc = Qc

If we double the conc. of PCl5 , what

happens?

[PCl3][Cl2] [PCl5]

Kc =

Concentration If one of the products is removed from the mixture, the

equilibrium will shift to the right to replace the substance removed.

In the reaction below, what happens when

(i) water is removed / added ?

(ii) ethanol , C2H5OH is added / removed ?

CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l)

ConcentrationConsider the reaction below:2[CrO4]2-(aq) + 2H+(aq) [Cr2O7]2-(aq) +

H2O(l)chromium(VI)ion from acid dichromate(VI) ion

When acid (H+ ) is added, acording to Le Chetelier’s principle, equilibrium shifts to the right to use up the acid. The solution changes to orange.

When alkali (OH- ) is added, the equilibrium shifts to the left because the OH- reacts with H+ to form water. This reduces the concentration of H+ ions. The equilibrium shifts to the left to reduce the effect by replacing the H+ ions.The solution changes to yellow.

PressureExample An increase in pressure will shift the position of the

equilibrium towards the side that involves fewer molecules.

In the reaction below, what happens when pressure is

(i) increased ? (ii) decreased ?

N2O4(g) 2NO2(g)

colourless brown

As the pressure is increased, the colour is initially darker as the same number of molecules are squeezed into a smaller space – the conc is increased . The mixture then becomes more colourless . According to Le Chatelier’s principle, the equilibrium shifts to the left , the side with fewer molecules. The colour lightens to almost colourless.

Temperature An increase in temperature will always favour the

endothermic process and will shift the equilibrium towards that direction. Lowering the temperature will favour the exothermic process.

A change in temperature will alter the Kc.

Example

In the reaction below, what happens when the temperature is

(i) increased ? (ii) decreased ?

N2O4(g) 2NO2(g) ΔHƟ = +57kJmol-

1

colourless brownIncrease in temp (putting in a hot water bath) cause the colour to be darker as the equilibrium shifts to the right that is endothermic in order to reduce the effect of the higher temperature.Decrease in temp (putting in ice bath) will cause the colour to be lighter.

The final state depends on:

1)The chemical nature of the reactants and products

2)The conditions of the system (temperature, pressure, volume).

Get time progressionCheck silberberg

N2O4 2NO2

Low T High T

Effect of pressure on KcAt 400K, N2O4 2NO2

2.00M 1.71M

At equilibrium , Kc = 1.46 moldm-3 at 400K. With pressure doubled,conc is doubled, [N2O4]= 4.00M , [NO2 ]=3.42M

The system is not in equilibrium.

NO2 must react to produce more N2O4 until the Kc becomes equal.

At higher pressure, the percentage of N2O4 is greater => the position of the equilibrium will shift to the left ( the side with fewer molecules of gas)

K is not affected by a change in pressure.

Kc = [1.71]2

[2.00]Kc = =1.46 moldm-

3

Kc = [3.42]2

[4.00]Kc = =2.92 moldm-

3

Effect of concentration on Kc

CH3COOH + C2H5OH CH3COOC2H5 + H2O

At equilibrium, 2.88M 2.88M 5.76M 5.76M

With water added, conc. of water increases but conc. of other species will drop as the total volume increased.

2.61M 2.61M 5.22M 10.4M

The system is not in equilibrium.

Ethyl ethanoate must react with water to use up the excess water.

At higher concentration, the position of the equilibrium will shift to the left in order to maintain the same Kc

K is not affected by a change in concentration.

Kc = [5.76][5.76][2.88][2.88]Kc = =

4

Kc = [5.22][10.4][2.61][2.61]Kc = =

8

The equilibrium will shift to the side so as to restore the value of K.

Effect of temperature on KcConsider the reaction

CO(g) + 2H2(g) CH3OH(g) ΔHƟ = -90kJmol-1

In an exothermic reaction, according to Le Chatelier’s principle, the equilibirum will shift to the side that is endothermic, that is to the left.

the concentration of the top of Kc will derease ad the concentration of the bottom will increase.

Kc decreases.

Kc = [CH3OH(g) ][CO(g)][H2(g)]2

CO(g) + 2H2(g) CH3OH(g) ΔHƟ = -90kJmol-1

Temperature /K

Kc / mol-2dm6

298 1.7 x 1017

500 1.1 x 1011

1000 2.1 x 106

Kc = [CH3OH(g) ][CO(g)][H2(g)]2

For exothermic reaction, the increase in temp decreases the Kc value.

Exothermic Reaction

Endothermic reaction N2 (g) + O2(g) 2NO(g) ΔHƟ = +181kJmol-1

Temperature /K

Kc / mol-2dm6

298 4.3 x 10-31

500 2.7 x 10-18

1000 7.5 x 10-9

2000 4.0 x 10-4

3000 0.015

Kc = [NO(g) ] 2

[N2(g)][O2(g)]

For endothermic reaction, the increase in temp increases the Kc value.

Catalyst A catalyst has no effect on the position of an equilibrium

at a particular temperature as it increases the rate of the forward and the reverse reaction equaly.

It reduces the time it takes the system to reach equilibrium.

The equilibrium constant and rateKc provides information about the extent of

a reactionIt does not provide information on the rate

of reaction.

Page 287 Qn 5

ExampleDraw a table showing how the position of equilibrium in the

reaction A,B and C would be affected by the followng changes:

(a) Increased temperature

(b) Increased pressure

Reaction A : interconversion of oxygen and ozone

3O2(g) 2O3(g) ΔHƟ = +284 kJmol-1

Reaction B : the reaction between sulfur dioxide and oxygen in the presence of a platinium/rhodium catalyst.

SO2(g) + 2O3(g) SO3(g) ΔHƟ = -197 kJmol-1

Reaction C : the reaction between hydrogen and carbon dioxide.

CO2(g) + 2H2(g) CO(g) + H2O(g) ΔHƟ = +41 kJmol-1

Conditions affecting the position of Equilibrium• Increase the

concentration of a species• The equilibrium

moves to the side without the species, reducing its concentration

• Kc : no change

• Increase pressure:• The equilibrium

moves to the side with fewer gaseous moles, reducing pressure

• Kc : no change

• Increase in temperature:• The equilibrium

moves in the endothermic direction, taking in the extra energy

• Kc changes

• A catalyst does not alter the position of equilibrium but speeds up the forward and reverse reactions equally

• Kc : no change

Concentration Pressure

TemperatureCatalyst

An equilibrium

system opposes changes

Haber Process

N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g)

from air from natural gas, CH4

ConditionsPressure: 20000kPa (200 atm)Temperature: about 500°CCatalyst: Iron

According to Le Chatelier’s principle, at higher pressure, there should be greater yield of ammonia, however, there will be an increased cost in building a plant that can operate at a much higher pressure, and other safety and maintenance issues.Hence, a pressure of 200 atm with about 55.8% yield of ammonia is ideal.

Haber Process

N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g) ΔH = -92 kJmol-1

With a higher temperature at constant pressure, there will bea drop in the yield of ammonia.

Ideally, the reaction should be carried outat a lower temperature. However, the rate ofreaction will be slower at a lower temperature.Hence, a compromise temperature of about 500°C is used.

Ammonia and the Haber Process

The Haber Process

• N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g)

Typical Conditions

• Pressure: 20000kPa (200 atm)• Temperature: 380-450°C• Catalyst: Iron

Equilibrium Theory Favours

• Low temperature: exothermic reaction – higher yield at low temperature• High pressure: decrease in number of gaseous molecules

Kinetic Theory Favours

• High temperature: greater average energy + more frequent collisions• High pressure: more frequent collisions for gaseous molecules• Catalyst: lower activation energy

Compromise Conditions

• The conditions used are a compromise with the catalyst enabling the rate to be kept up, even at a lower temperature

Example 1Given that the equilibrium constant for the

reaction

H2 (g) + I2 (g) 2HI (g) at 700K is 54 and the conc of H2 and I2 at equilibrium are 0.25 moldm-3 and 0.50 moldm-3 what is the equilibrium conc. of HI?

0.12 moldm-3

Example 25.00mol H2 and 3.00mol I2 are mixed

together in a vessel of volume 10.0dm and allowed to come to equilibrium at 1100K. At equilibrium there were 0.43 mol I2 present in the reaction mixture. Calculate the value of of the equilibrium constant.

25.3

Example 33.00mol NO2 and 1.00mol N2O4 are

mixed together in a vessel of volume 1.0dm and allowed to come to equilibrium at 398K. At equilibrium there were 1.74 mol N2O4 present in the reaction mixture. Calculate the value of of the equilibrium constant.

The reaction is 2 NO2(g) N2O4(g)

0.753 mol-1dm3