Irreversible and Reversible Reactions

Irreversible and Reversible Reactions

Irreversible Reactions
Chemical reactions that take place in one direction only It goes on until at least one of the reactants is used up  complete reaction

Irreversible Reactions
Examples : - 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) HCl(aq) + H2O(l)  H3O+(aq) + Cl(aq) 2Mg(s) + O2(g)  2MgO(s) Cl2(g) + 2OH(aq)  ClO(aq) + Cl(aq) + H2O(l)

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
2Mg(s) + O2(g) MgO(s) Conditions for reversible reactions : - Closed reaction vessels to prevent escape of gases High temperature to favour the reversed processes

HCl(aq) + H2O(l) H3O+(aq) + Cl(aq)
Conditions for reversible reactions : - Concentration of HCl(aq) > 6 M

Cl2(g) + 2OH(aq) ClO(aq) + Cl(aq) + H2O(l)
Conditions for reversible reactions : - Closed reaction vessels to prevent escape of Cl2 Dilute OH(aq) at T  20C to prevent side reaction 3Cl2(g) + 6OH(aq)  ClO3(aq) + 5Cl(aq) + 3H2O(l) Hot and concentrated

Changes of physical phases are reversible
Reversible Processes Solid Liquid Vapour Changes of physical phases are reversible

Reversible Reactions Chemical reactions that can go in two opposite directions Incomplete reactions

Q.2 CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+(aq)
NH3(aq) + H2O(l) NH4+(aq) + OH(aq) Cl2(aq) + H2O(l) HCl(aq) + HOCl(aq) CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) 3H2(g) + N2(g) NH3(g)

Examples of reversible reactions
CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) yellow orange 1. When HCl(aq) is added to CrO42(aq) Observation : - The yellow solution turns orange.

CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) yellow orange 1. When HCl(aq) is added to CrO42(aq) Interpretation : - CrO42(aq) reacts with H+ to give Cr2O72(aq) There is no further colour change when rate of forward rx = rate of backward rx

CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) yellow orange 2. When NaOH(aq) is added to Cr2O72(aq) Observation : - The orange solution turns yellow.

CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) yellow orange 2. When NaOH(aq) is added to Cr2O72(aq) Interpretation : - H+ ions are being removed by NaOH rate of forward rx  rate of backward rx > rate of forward rx  a net change of Cr2O72(aq) to CrO42(aq)

CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) yellow orange 2. When NaOH(aq) is added to Cr2O72(aq) Interpretation : - There is no further colour change when rate of backward rx = rate of forward rx

BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq) colourless White ppt 1. When H2O(l) is added to BiCl3(aq) Observation : - The colourless solution turns milky.

BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq) colourless White ppt 1. When H2O(l) is added to BiCl3(aq) Interpretation : - BiCl3(aq) reacts with H2O(l) to give BiOCl(s) There is no further change when rate of forward rx = rate of backward rx

BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq) colourless White ppt 2. When HCl(aq) is added to BiOCl(s) Observation : - The milky solution becomes clear.

BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq) colourless White ppt 2. When HCl(aq) is added to BiOCl(s) Interpretation : - [HCl(aq)]   rate of backward rx > rate of forward rx  a net consumption of BiOCl(s)

BiCl3(aq) + H2O(l) BiOCl(s) + 2HCl(aq) colourless White ppt 2. When HCl(aq) is added to BiOCl(s) Interpretation : - There is no further change when rate of forward rx = rate of backward rx

Q.3 1. When NaOH(aq) is added to Br2(aq)
Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 1. When NaOH(aq) is added to Br2(aq) Prediction : - The red-orange solution turns colourless.

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 1. When NaOH(aq) is added to Br2(aq) Interpretation : - Before the addition, rate of forward rx = rate of backward rx

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 1. When NaOH(aq) is added to Br2(aq) Interpretation : - H+ ions are being removed by NaOH(aq)  rate of forward rx > rate of backward rx  a net consumption of Br2(aq)

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 1. When NaOH(aq) is added to Br2(aq) Interpretation : - There is no further change of colour when rate of forward rx = rate of backward rx

Q.3 Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 2. When HCl(aq) is added Prediction : - The colourless solution turns red-orange.

Q.3 Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 2. When HCl(aq) is added Interpretation : - Addition of HCl increases [H+(aq)]  rate of backward rx > rate of forward rx  a net production of Br2(aq)

Q.3 Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 2. When HCl(aq) is added Interpretation : - There is no further change of colour when rate of forward rx = rate of backward rx

Q.3 3. When AgNO3(aq) is added
Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 3. When AgNO3(aq) is added Prediction : - A pale yellow ppt is formed. The red-orange solution turns colourless.

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 3. When AgNO3(aq) is added Interpretation : - Ag+(aq) react with Br-(aq) to give pale yellow ppt of AgBr(s).  rate of backward rx < rate of forward rx  a net consumption of Br2(aq)

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq) Red-orange colourless 3. When AgNO3(aq) is added Interpretation : - There is no further change of colour when rate of forward rx = rate of backward rx

Phenolphthlein is a weak acid that ionizes slightly in water to give H3O+(aq)
Colourless Red

What is the colour of phenolphthalein when pH < 8.3 ?
+ H3O+(aq) Colourless Red

Colourless Red When pH < 8.3 (e.g. deionized water),
The colourless form predominates + H3O+(aq) Colourless Red

Colourless Red When NaOH(aq) is added, [H3O+(aq)] 
 rate of forward rx > rate of backward rx  a net production of the red form + H3O+(aq) Colourless Red

Colourless Red There is no further colour change when
rate of forward rx = rate of backward rx + H3O+(aq) Colourless Red

When pH > 10, The red form predominates + H3O+(aq) Colourless Red

Colourless Red When 8.3 < pH < 10,
Both forms have similar concentrations  pink + H3O+(aq) Colourless Red

Reversible reactions and dynamic equilibrium
For a reversible reaction, reactants products a state of dynamic equilibrium is said to be established when rate of forward rx = rate of backward rx Apparently, there is no change in the concentrations of reactants and products. Reactions continues at molecular level.

An example of dynamic equilibrium
No change in the position of the girl

Reversible reactions and chemical equilibrium
ALL chemical reactions are considered as reversible processes with different extents of completion.

At equil., k[reactant]eq = k’[product]eq  Ea’ > Ea k >> k’  [reactant]eq << [product]eq  Forward rx is more complete than backward rx Ea Ea’

 Ea > Ea’  Forward rx is less complete than backward rx Ea Ea’

Chemical Equilibrium vs Chemical Kinetics
Chemical equilibrium is about how far a reaction can proceed. Chemical kinetics is about how fast a reaction can proceed.

Chemical Equilibrium vs Chemical Kinetics
The rate of rx depends on Ea or Ea’ The extent of completion of rx depends on H Ea Ea’

Evidence for Dynamic Equilibrium
NaNO3(s) NaNO3(aq) saturated At fixed T, [NaNO3(aq)] is a constant Addition of 24NaNO3(s)  Detection of radioactivity in sat’d solution  Interchange of NaNO3 between the sat’d solution and the solid

Features of Chemical Equilibria
A system in chemical equilibrium consists of a forward reaction and a backward reaction both proceeding at the same rate. All macroscopic properties (such as temperature, pressure, concentration, density, colour, …etc.) of an equilibrium system remain unchanged.

Q.4 (i) H2(g) + I2(g) HI(g) Time taken to reach the equilibrium state

Q.4 (ii) H2(g) + I2(g) HI(g) Time taken to reach the equilibrium state The equilibrium concentrations need not be equal

Q.5 Constant flame colour and temperature

Q.5 Steady state CO2 and H2O Open system  Not at equilibrium state
Air Fuel

3. Equilibria can only be achieved in closed
3. Equilibria can only be achieved in closed systems with no exchange of matter with their surroundings.

The brown vapour escapes until all brown liquid disappears
Q.6 A Observation : - The brown vapour escapes until all brown liquid disappears Br2(g) Br2(l) Interpretation : - Br2 escapes from the system. Thus, the rate of condensation is always less than the rate of evaporation.

both forward and backward reactions stop due to absence of reactants.
Q.6 B Observation : - No observable change Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g) 500C Interpretation : - H2(g) and H2O(g) escape from the system leaving only Fe3O4(s) and Fe(s). Thus, both forward and backward reactions stop due to absence of reactants.

The amount of solid KCl 
Q.6 C Observation : - The amount of solid KCl  Interpretation : - Water escapes by evaporation. [KCl(aq)] , making the rate of precipitation greater than the rate of dissolution. KCl(s) KCl(aq)

A pleasant smell is detected. The volume of the mixture 
Q.6 D Observation : - A pleasant smell is detected. The volume of the mixture  Interpretation : - The more volatile ester escapes, causing a drop in volume of both reactants. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

4. The state of equilibrium can be attained from either the forward or the backward direction.
5. The equilibrium composition under a given set of conditions is independent of the direction from which the equilibrium is approached. In other words, the same set of equilibrium concentrations of reactants and products can be obtained from either side of the reversible reaction under the same set of conditions (See Q.7).

nH = nI = 1.0  nHI = 1.0 Q.7 H2(g) I2(g) HI(g) ninitial nequil n’initial /2 = 0.11 /2 = 0.11 0.78 ( )/2 = 0.11 ( )/2 = 0.11 0.78

Equilibrium position and equilibrium composition
For a system with a more complete forward reaction, A + B C the equilibrium position is said to lie more to the right hand side. The equilibrium composition is richer in C, i.e. [C]equil is much higher than [A]equil and [B]equil.

Equilibrium position and equilibrium composition
For a system with a less complete forward reaction, A + B C the equilibrium position is said to lie more to the left hand side. The equilibrium composition is richer in A and B, i.e. [A]equil and [B]equil are much higher than [C]equil.

Equilibrium Law

Equilibrium Law For any chemical system in dynamic equilibrium, the concentrations or partial pressures of all the substances present are related to one another by a mathematical expression which is always a constant at fixed temperature.

For the chemical system in equilibrium,
aA + bB cC + dD Equilibrium constant expressed in concentration Kc depends on temperature and the nature of reaction

Equilibrium constant and reaction quotient
aA + bB cC + dD Equilibrium constant Reaction quotient

 the system is at equilibrium
aA + bB cC + dD Equilibrium constant Reaction quotient Qc = Kc  the system is at equilibrium

 the system is NOT at equilibrium
aA + bB cC + dD Equilibrium constant Reaction quotient Qc > Kc  the system is NOT at equilibrium The reaction proceeds from right to left until Qc = Kc.

 the system is NOT at equilibrium
aA + bB cC + dD Equilibrium constant Reaction quotient Qc < Kc  the system is NOT at equilibrium The reaction proceeds from left to right until Qc = Kc.

Equilibrium constant Reaction quotient aA + bB cC + dD Large Kc
The forward reaction is more complete The equilibrium position lies to the right. The equilibrium mixture is richer in the substances on the R.H.S. of the equation.

Equilibrium constant Reaction quotient aA + bB cC + dD Small Kc
The forward reaction is less complete The equilibrium position lies to the left. The equilibrium mixture is richer in the substances on the L.H.S. of the equation.

Kc gives no indication about the rate of reaction
Q.8 The rate of reaction depends on Ea

units mol1 dm3 A + B C C A + B mol dm3
Relationship of Kc to the Stoichiometry of Equations units A + B C mol1 dm3 C A + B mol dm3

units mol1 dm3 A + B C xA + xB xC molx dm3x
Relationship of Kc to the Stoichiometry of Equations units A + B C mol1 dm3 xA + xB xC molx dm3x

Relationship of Kc to the Stoichiometry of Equations
units A + B C mol1 dm3 A B C

Q.9 (1) A B (2) B C (3) C D A C (4) = K1K2 (4) = (1) + (2)  K4 = K1  K2

Q.9 (1) A B (2) B C (3) C D A D (5) = K1K2K3 (5) = (1) + (2) + (3)  K5 = K1  K2  K3

(4) = [(3) – (2) – (1)]  ½ = 5.11015 mol2 dm6 Q.10
(1) H2(g) + Cl2(g) HCl(g) (2) N2(g) + 3H2(g) NH3(g) (3) N2(g) + 4H2(g) + Cl2(g) NH4Cl(s) (4) NH3(g) + HCl(g) NH4Cl(s) (4) = [(3) – (2) – (1)]  ½ = 5.11015 mol2 dm6

Determination of Equilibrium Constants

Equilibrium System: (TAS Expt. 11)
CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)
Equilibrium System: CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) Reactant/ Product Experiment 1 Experiment 2 Amount at beginning (mol) Amount at equilibrium (mol) Amount at beginning (mol) Amount at equilibrium (mol) CH3COOH(aq) 0.250 0.083 0.198 0.000 0.020 0.098 0.296 CH3CH2OH(aq) 0.250 0.083 H2SO4(l) 0.020 0.020 CH3COOCH2 CH3(aq) 0.000 0.250 – 0.083 = 0.167 H2O(l) 0.000 0.167

Equilibrium System: CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) For experiment 1:

Equilibrium System: CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) For experiment 2:

Equilibrium System: CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) = (no unit)

Equilibrium System: CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) Conc. H2SO4 acts as a positive catalyst It can shorten the time taken to reach the state of equilibrium but has no effect on the extent of completion of the reaction.

With catalyst Without catalyst Same extent of completion
Same equilibrium composition

Equilibrium Constant in Terms of Partial Pressures

For gaseous systems in dynamic equilibria,
it is more convenient to express the equilibrium constants in terms of partial pressures. aA(g) + bB(g) cC(g) + dD(g)

Relationship between Kc and Kp
PV = nRT = [Gas]RT At fixed T, P  [Gas]

Consider the equilibrium system :
aA(g) + bB(g) cC(g) + dD(g) If a + b = c + d, Kp = Kc

Simple Calculations Involving Kc and Kp
0.50 mole of CO2(g) and 0.50 mole of H2(g) are mixed in a 5.0 dm3 flask at 690 K and are allowed to establish the following equilibrium. CO2(g) + H2(g) CO(g) + H2O(g) Kp = 0.10 at 690 K R = atm dm3 K1 mol1 Find partial pressures of all gaseous components

x = 0.12 CO2(g) + H2(g) CO(g) + H2O(g) Initial no. of moles 0.50
No. of moles at equil. x x x = 0.12

CO2(g) + H2(g) CO(g) + H2O(g)
No. of moles at equil. x x 0.38 0.12 nT = = 1.00 = atm

CO2(g) + H2(g) CO(g) + H2O(g)
No. of moles at equil. x x 0.38 0.12 nT = = 1.00 = 4.30 atm = 4.30 atm = 1.36 atm

= 3 Q.11 At fixed V & T, P  n 2SO2(g) + O2(g) 2SO3(g) 3x x 1.5x
Initial partial pressure 3x x Partial pressure at equilibrium 1.5x x – 1.5x/2 1.5x = 0.25x

Q.11 2SO2(g) + O2(g) 2SO3(g) 1.5x 0.25x 1.5x PT = 373 kPa
Partial pressure at equilibrium 1.5x 0.25x 1.5x PT = 373 kPa = 1.5x x + 1.5x x = 115 kPa = kPa1

Q.12 At fixed V & T, P  n PCl5(g) PCl3(g) + Cl2(g) x 0.14x 0.86x
Initial partial pressure x Partial pressure at equilibrium 0.14x 0.86x 0.86x PT = 101 kPa = 0.14x x x x = 54.3 kPa

Q.12 PCl5(g) PCl3(g) + Cl2(g) 0.14x 0.86x 0.86x x = 54.3 kPa = 287 kPa
Partial pressure at equilibrium 0.14x 0.86x 0.86x x = 54.3 kPa = 287 kPa

Q.13 N2(g) + O2(g) 2NO(g) 2.0 1.0 2.0 - x 1.0 – x 2x
Initial no. of moles 2.0 1.0 No. of moles at equilibrium 2.0 - x 1.0 – x 2x Concentration at equilibrium

Q.13 N2(g) + O2(g) 2NO(g) = Kp = 1.2  102 x = 0.073
Concentration at equilibrium = Kp = 1.2  102 x = 0.073 [N2] = (2.0 – 0.073)/2.0 = 0.96 mol dm3

Q.13 N2(g) + O2(g) 2NO(g) = Kp = 1.2  102
Concentration at equilibrium = Kp = 1.2  102 ∵ x is small, – x  2.0 and 1.0 – x  1.0

Q.13 N2(g) + O2(g) 2NO(g) = Kp = 1.2  102 x = 0.077
Concentration at equilibrium = Kp = 1.2  102 x = 0.077 [N2] = (2.0 – 0.077)/2.0 = 0.96 mol dm3

Homogeneous Equilibrium
Equilibrium system involving ONE phase only N2(g) + 3H2(g) NH3(g) Cl2(aq) + 2Br(aq) Cl(aq) + Br2(aq) CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

Dissolve both products
Homogeneous Equilibrium Glacial ethanoic acid Absolute alcohol Dissolve both products CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(l) + H2O(l) Immiscible  Two phases

Q.14 Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq)

Q.14 N2(g) + 3H2(g) NH3(g)

Q.14 CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

At pH 7, density of water  1000 g dm3
Q.14 H+(aq) + OH(aq) H2O(l) At pH 7, density of water  1000 g dm3 = 55.5 mol dm3

[H2O(l)]  55.5 M  In large excess   a constant Q.14
H+(aq) + OH(aq) H2O(l) [H2O(l)]  55.5 M  In large excess   a constant

Q.15(a) Calculate the molarity of water in M hydrochloric acid. Given: Density of M hydrochloric acid is 1.19 g cm3 at 298 K Mass of 1 dm3 of M HCl(aq) = 1.19 g cm3  1000 cm3 = 1190 g Mass of HCl present = mol  ( ) g mol1 = g

Q.15(a) Calculate the molarity of water in M hydrochloric acid. Given: Density of M hydrochloric acid is 1.19 g cm3 at 298 K Mass of water present = (1190 – 452.2) g = g = 41.0 M

Q.15(b) = 41.0 M < 55.5 M At very high acid concentrations, H2O is NOT in large excess. It is NOT justified to consider [H2O]equil as a constant in ALL aqueous solutions.

Q.15(b) HCl(aq) H+(aq) + Cl(aq) 12.38 M H+(aq) + OH(aq) H2O(l) < M 41.0 M

Heterogeneous Equilibrium
Equilibrium systems involving two or more phases H2O(l) H2O(g) ∵ Kp depends on temperature only ∴ at fixed T, vapour pressure of water (Kp) is a constant, irrespective of the amount of water present.

∴ [H2O(l)]equil   = constant (at fixed T)

= Kp (at fixed T)

In a solution, and in a gas, the concentration changes as the particles (molecules, atoms or ions) become closer together or further apart. In a solid or a liquid, the particles are at fixed distance from one another; this means that the ‘concentration’ is also fixed. In effect, the concentration of a solid or a liquid is equivalent to its density (also known as the effective reacting concentration).

In heterogeneous equilibria, the effective reacting concentration of a pure liquid or a pure solid is a constant and is independent of the amount of liquid or solid present Since collisions of reacting particles occur at the boundary of phases. A change in the surface area of a solid or a liquid (by changing the amount) affects the rates of forward and backward reactions to the same extent.

Changing the amount of a pure solid or a pure liquid in a heterogeneous equilibrium mixture does NOT disturb the equilibrium. Conclusion : - [X(s)] and [X(l)] do NOT appear in the equilibrium constant expressions of heterogeneous equilibria.

Q.16 Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)

Q.16 CaCO3(s) CaO(s) + CO2(g)

Q.16 Ag+(aq) + Cl(aq) AgCl(s)

Q.16 Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g)

Q.16 Br2(l) Br2(g)

Partition Equilibrium of a Solute Between Two Immiscible Solvents

Partition (Distribution) Equilibrium
The equilibrium established when a non-volatile solute distributes itself between two immiscible liquids A(solvent 2) A(solvent 1)

Water and hexane are immiscible with each other. Hexane Water

I2 dissolves in both solvents to different extent. Hexane Water

When dynamic equilibrium is established, rate of  movement = rate of  movement Hexane Water

Suppose the equilibrium concentrations of iodine in H2O and hexane are x and y respectively,
Water

KD : partition coefficient or distribution coefficient
Suppose the equilibrium concentrations of iodine in H2O and hexane are x and y respectively, When dynamic equilibrium is established the ratio of concentrations of iodine in water and in hexane is always a constant. KD : partition coefficient or distribution coefficient

Changing the concentrations by the same extent does not affect the quotient.
∵ the rates of the two opposite processes are affected to the same extent. However, changing the concentrations by the same extent changes the colour intensity of the solutions.

Partition Coefficient
The partition law can be represented by the following equation: (no unit) Units of concentration : - mol dm-3, mol cm-3, g dm-3, g cm-3

The partition coefficient of a solute between solvent 2 and solvent 1 is given by The partition coefficient of a solute between solvent 1 and solvent 2 is given by

Depends on temperature ONLY. Not affected by the amount of solute added and the volumes of solvents used. TAS Experiment No. 12

TAS Expt 12 = KD [CH3COOH]water [CH3COOH]2-methylpropan-1-ol

Partition law holds true
at fixed temperature for dilute solutions ONLY For concentrated solutions, interactions between solvent and solute have to be considered and the concentration terms should be expressed by ‘activity’ (effective concentration)

Partition law holds true
3. when the solute exists in the same form in both solvents. i.e. no association or dissociation of solute C6H5COOH(benzene) C6H5COOH(aq) C2 C1 C1 and C2 are determined by titrating the acid in each solvent with standard sodium hydroxide solution.

Not a constant [C6H5COOH]water(C1) / mol dm-3
[C6H5COOH]benzene(C2) / mol dm-3 C1/C2 0.06 0.483 0.124 0.12 1.92 0.063 0.14 2.63 0.053 0.20 5.29 0.038

Interpretation : -  Violation of Partition law
Benzoic acid tends to dimerize (associate) in non-polar solvent to give (C6H5COOH)2 Benzoic acid dimer The solute does not have the same molecular form in both solvents  Violation of Partition law

 = degree of association of benzoic acid
Interpretation : - 2C6H5COOH(benzene) (C6H5COOH)2(benzene)  = degree of association of benzoic acid [C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated C2 C2(1-) C2 Determined by titration with NaOH

Q.17(a) The interaction between benzoic acid and benzene molecules are weaker than the hydrogen bonds formed between benzoic acid molecules. Thus benzoic acids tend to form dimers when dissolved in benzene. In aqueous solution, benzoic acid molecules form strong H-bond with H2O molecules rather than forming dimer.

Q.17(b) In aqueous solution, there is no association as explained in (a). Also, dissociation of acid can be ignored since benzoic acid is a weak acid (Ka = 6.3  10-5 mol dm-3).

Q.17(c) C1 2C6H5COOH(benzene) (C6H5COOH)2(benzene)
C6H5COOH(benzene) C6H5COOH(aq) C1

 is a constant at fixed T

~constant [C6H5COOH]water(C1) / mol dm-3
[C6H5COOH]benzene(C2) / mol dm-3 0.06 0.483 0.087 0.086 0.12 1.92 0.14 2.63 0.20 5.29

Applications of partition law
Solvent extraction Chromatography Two classes of separation techniques based on partition law.

Solvent extraction I2 in KI(aq) I2 in hexane I2 in KI(aq) I2 in hexane
Colourless Hexane I2 in KI(aq) I2 in hexane I2 in KI(aq) I2 in hexane + hexane I2 in KI(aq) It dissolves I2 but not KI.  Organic solvents are preferred. It is immiscible with water.  Organic solvents are preferred. To remove I2 from an aqueous solution of KI, a suitable solvent is added. It can be recycled easily (e.g. by distillation)  Organic (volatile) solvents are preferred. What feature should the solvent have? At equilibrium, rate of  movement of I2 = rate of  movement of I2 By partition law,

Solvent Extraction Hexane layer Aqueous layer Before shaking
After shaking Hexane layer Aqueous layer Iodine can be extracted from water by adding hexane, shaking and separating the two layers in a separating funnel

Determination of I2 left in both layer
Titrated with standard sodium thiosulphate solution I S2O3  I + S4O62

Determination of I2 left in the KI solution
For the aqueous layer, starch is used as the indicator. For the hexane layer, starch is not needed because the colour of I2 in hexane is intense enough to give a sharp end point.

In solvent extraction, it is more efficient (but more time-consuming) to use the solvent in portions for repeated extractions than to use it all in one extraction. Worked example

Worked example : - 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution 50g X in 40 cm3 ether solution By partition law, (a) Calculate the partition coefficient of X between ether and water at 298 K. M is the molecular mass of X

Or simply, Worked example : - 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution 50g X in 40 cm3 ether solution Or simply,

(b)(i) xg of X in 30 cm3 ether solution
5g of X in 30 cm3 aqueous solution 30 cm3 ether (5-x)g of X in 30 cm3 aqueous solution Determine the mass of X that could be extracted by shaking a 30 cm3 aqueous solution containing 5 g of X with a single 30 cm3 portion of ether at 298 K

3.79 g of X could be extracted.
(b)(i) xg of X in 30 cm3 ether solution 5g of X in 30 cm3 aqueous solution 30 cm3 ether (5-x)g of X in 30 cm3 aqueous solution 3.79 g of X could be extracted.

(b)(ii) First extraction
x1g of X in 15 cm3 ether solution 5g of X in 30 cm3 aqueous solution 15 cm3 ether (5-x1)g of X in 30 cm3 aqueous solution

(b)(ii) Second extraction
x2g of X in 15 cm3 ether solution (5-x1)g of X in 30 cm3 aqueous solution 15 cm3 ether (5-x1-x2)g of X in 30 cm3 aqueous solution

total mass of X extracted
= ( ) g = 4.24 g > 3.79 g. Repeated extractions using smaller portions of solvent are more efficient than a single extraction using larger portion of solvent. However, the former is more time-consuming

Important extraction processes : -
1. Products from organic synthesis, if contaminated with water, can be purified by shaking with a suitable organic solvent. 2. Caffeine in coffee beans can be extracted by Supercritical carbon dioxide fluid (decaffeinated coffee) 2. Impurities such as sodium chloride and sodium chlorate present in sodium hydroxide solution can be removed by extracting the solution with liquid ammonia. Purified sodium hydroxide is the raw material for making soap, artificial fibre, etc.

Q.18(a) Alcohol layer 200 cm3 alcohol Aqueous layer
100 cm3 of M ethanoic acid Calculate the % of ethanoic acid extracted at 298 K by shaking 100 cm3 of a M aqueous solution of ethanoic acid with 200 cm3 of 2-methylpropan-1-ol;

Q.18(a) Let x be the fraction of ethanoic acid extracted to the alcohol layer No. of moles of acid in the original solution =  = mol x = = 39.6%

Q.18(b) 1st extraction 2nd extraction
Let x1, x2 be the fractions of ethanoic acid extracted to the alcohol layer in the 1st and 2nd extractions respectively. 1st extraction 2nd extraction % of acid extracted= =0.433=43.3%

x = 7.5 Q.19 ∴ 7.5 cm3 of solvent X is required
Let x cm3 be the volume of solvent X required to extract 90% of iodine from the aqueous solution and y be the no. of moles of iodine in the original aqueous solution. x = 7.5 ∴ 7.5 cm3 of solvent X is required

Chromatography A family of analytical techniques for separating the components of a mixture. Derived from the Greek root chroma, meaning “colour”, because the original chromatographic separations involved coloured substances.

Chromatography In chromatography, repeated extractions are carried out successively in one operation (compared with fractional distillation in which repeated distillations are performed) which results, (as shown in the worked example and Q.18), in an effective separation of components.

All chromatographic separations are based upon differences in partition coefficients of the components between a stationary phase and a mobile phase.

The stationary phase is a solvent (often H2O) adsorbed (bonded to the surface) on a solid.
The solid may be paper or a solid such as alumina or silica gel, which has been packed into a column or spread on a glass plate. The mobile phase is a second solvent which seeps through the stationary phase.

Three main types of chromatography
1. Column chromatography 2. Paper chromatography 3. Thin layer chromatography

Column chromatography
Stationary phase : - Water adsorbed on the adsorbent (alumina or silica gel) Mobile phase : - A suitable solvent (eluant) that seeps through the column

Eluant Column chromatography Sample Partition of components takes place repeatedly between the two phases as the components are carried down the column by the eluant. The components are separated into different bands according to their partition coefficients.

The component with the highest coefficient between mobile phase and stationary phase is carried down the column by the mobile phase most quickly and comes out first.

Suitable for large scale treatment of sample For treatment of small quantities of samples, paper or thin layer chromatography is preferred.

Paper chromatography Water adsorbed on paper. Mobile phase : -
Stationary phase : - Water adsorbed on paper. Mobile phase : - A suitable solvent The best solvent for a particular separation should be worked out by trials-and-errors X(adsorbed water) X(solvent) stationary phase mobile phase

Paper chromatography The solvent moves up the filter paper by capillary action Components are carried upward by the mobile solvent  Ascending chromatography

Different dyes have different KD between the mobile and stationary phases
They will move upwards to different extent

Paper chromatography The components separated can be identified by their specific retardation factors, Rf, which are calculated by

Using chromatography to separate the colours in a sweet.
filter paper spot of coloured dye solvent separated colours Using chromatography to separate the colours in a sweet.

Solvent front a d c b

a chromatogram separated colours

Paper Chromatography The Rf value of any particular substance is about the same when using a particular solvent at a given temperature The Rf value of a substance differs in different solvents and at different temperatures

Mixture of phenol and ammonia Mixture of butanol and ethanoic acid
Paper Chromatography Amino acid Solvent Mixture of phenol and ammonia Mixture of butanol and ethanoic acid Cystine 0.14 0.05 Glycine 0.42 0.18 Leucine 0.87 0.62 Rf values of some amino acids in two different solvents at a given temperature

Two-dimensional paper chromatography

Two-dimensional paper chromatography
All spots (except proline) appears visible (purple) when sprayed with ninhydrin (a developing agent)

Thin layer chromatography
Stationary phase : - Water adsorbed on a thin layer of solid adsorbent (silica gel or alumina). Mobile phase : - A suitable solvent X(adsorbed water) X(solvent) stationary phase mobile phase

Q.20 Suggest any advantage of thin layer chromatography over paper chromatography. A variety of different adsorbents can be used. The thin layer is more compact than paper, more equilibrations can be achieved in a few centimetres (no. of extraction ).  A microscope slide is long enough to provide effective separation

Factors Affecting Equilibrium

THREE factors affecting chemical equilibria.
1. Changes in pressure 2. Changes in concentration No effect on Kc or Kp Alter the equilibrium position by changing the equilibrium composition 3. Changes in temperature Alter the equilibrium position by changing the equilibrium constant

THREE ways of interpretation
1. Kc or Kp approach 2. Kinetic approach 3. Le Chatelier’s Principle

Effects of changes in pressure
increase in pressure A(g) B(g) C(g) decrease in pressure P  by reducing V Equilibrium position shifts to the right P  by increasing V Equilibrium position shifts to the left

Interpretation : Kp approach
A(g) B(g) C(g) Pequil 1 PA PB PC 2PA 2PB 2PC

A(g) B(g) C(g) Pequil 1 PA PB PC Equilibrium position shifts to the right until Qp = Kp 2PA 2PB 2PC

A(g) B(g) C(g) Pequil 1 PA PB PC Equilibrium position shifts to the right until Qp = Kp 2PA - x 2PB – x 2PC + x

Interpretation : kinetic approach
A(g) B(g) C(g) Both the rates of forward and backward reactions are increased by doubling the partial pressures of all gaseous components of the system. However, the rate of forward reaction is increased more. There is a net forward reaction Equilibrium position shifts to the right

A(g) + B(g) C(g) R1 = k1[A][B] R-1 = k-1[C] V½
Q.21 k1 A(g) B(g) C(g) k-1 R1 = k1[A][B] R-1 = k-1[C] V½ More affected R1’ = k1(2[A])(2[B]) = 4R1 R-1’ = k-1(2[C]) = 2R-1

A(g) + B(g) C(g) R1 = k1[A][B] R-1 = k-1[C] V2
Q.21 k1 A(g) B(g) C(g) k-1 R1 = k1[A][B] R-1 = k-1[C] V2 More affected R1’ = k1(½[A])(½[B]) = ¼R1 R-1’ = k-1(½[C]) = ½R-1

Le Chatelier’s Principle
If a system at equilibrium is forced to change, the equilibrium position of the system will shift in a way to reduce (or oppose) the effect of the change.

Q.22(a) A(g) B(g) C(g) Change : PT  Response : PT 

Q.22(b)/(c) A(g) B(g) C(g) Two moles One mole

A(g) + B(g) C(g) Q.22(d) Two moles One mole
One mole of gas exert less pressure.

A(g) + B(g) C(g) Q.22(e) Two moles One mole
One mole of gas exert less pressure. The forward reaction lowers the pressure of the system.

A(g) + B(g) C(g) Q.22(f) Two moles One mole
One mole of gas exert less pressure. The forward reaction lowers the pressure of the system. Equilibrium position shifts to the right.

Q.22(g) A(g) B(g) C(g) Change : PT  Response : PT 

Immediately after pushing in the plunger
N2O4(g) NO2(g) pale yellow brown Sealed nozzle N2O4(g), NO2(g) Syringe Immediately after pushing in the plunger The gas mixture turns darker brown due to a sudden increase in concentrations of both gases

The gas mixture turns paler
N2O4(g) NO2(g) pale yellow brown Sealed nozzle N2O4(g), NO2(g) Syringe After a few seconds The gas mixture turns paler because the system reduces the pressure by shifting the equilibrium position to the left (the side with less gas molecules).

Immediately after pulling out the plunger
N2O4(g) NO2(g) pale yellow brown Sealed nozzle N2O4(g), NO2(g) Syringe Immediately after pulling out the plunger The gas mixture turns paler due to a sudden decrease in concentrations of both gases

The gas mixture turns darker brown because
N2O4(g) NO2(g) pale yellow brown Sealed nozzle N2O4(g), NO2(g) Syringe After a few seconds The gas mixture turns darker brown because the system  the pressure by shifting the equilibrium position to the right (the side with more gas molecules).

Q.23(a)/(b) H2(g) CO2(g) H2O(g) CO(g) Cause :  in PT by reducing VT Effect : No effect on the equilibrium position Cause :  in PT by increasing VT Effect : No effect on the equilibrium position

Q.23(c)/(d) H2(g) CO2(g) H2O(g) CO(g) Cause :  in PT by increasing Effect : Equilibrium position shifts to the right Cause :  in PT by increasing PCO Effect : Equilibrium position shifts to the left

Q.23(e) H2(g) CO2(g) H2O(g) CO(g) Cause :  in PT by introducing He(g) Effect : No effect on equilibrium position Reason : The partial pressures of reactants and products remain unchanged.

Q.23(a)/(b) PCl5(g) PCl3(g) Cl2(g) Cause :  in PT by reducing VT Effect : Equilibrium position shifts to the left Cause :  in PT by increasing VT Effect : Equilibrium position shifts to the right

Q.23(c)/(d) PCl5(g) PCl3(g) Cl2(g) Cause :  in PT by increasing Effect : Equilibrium position shifts to the right Cause :  in PT by increasing Effect : Equilibrium position shifts to the left

Q.23(e) PCl5(g) PCl3(g) Cl2(g) Cause :  in PT by introducing He(g) Effect : No effect on equilibrium position

Q.23(a)/(b) Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) Cause :  in PT by reducing VT Effect : No effect on equilibrium position Cause :  in PT by increasing VT Effect : No effect on equilibrium position

Q.23(c)/(d) Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) Cause :  in PT by increasing Effect : Equilibrium position shifts to the right Cause :  in PT by increasing Effect : Equilibrium position shifts to the left

Q.23(e) Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) Cause :  in PT by introducing He(g) Effect : No effect on equilibrium position

For the systems H2(g) CO2(g) H2O(g) CO(g) Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) Changing PT by altering VT has no effect on the equilibrium position Interpretation : Kp approach

For the systems H2(g) CO2(g) H2O(g) CO(g) Pequil 1 = Kp

For the systems Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) Pequil 1 = Kp

For the systems H2(g) CO2(g) H2O(g) CO(g) Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) Kinetic approach The rates of forward and backward reactions are affected to the same extent.

For the systems H2(g) CO2(g) H2O(g) CO(g) Fe3O4(s) + 4H2(g) Fe(s) + 4H2O(g) By Le Chatelier’s principle Since the system has the same no. of gas molecules on either side, No adjustment made by the system can reduce the change.  No shifting of equil. position

For the systems H2(g) CO2(g) H2O(g) CO(g) PCl5(g) PCl3(g) Cl2(g)  in PT by changing VT has no effect on the equilibrium position However, the partial pressures and thus the equilibrium composition change by altering VT

Q.24 CO2(aq) CO2(g) Once the bottle is opened, CO2 escapes from the system and its partial pressure drops. The system responds by releasing CO2 from the aqueous solution.

Effects of pressure changes on equilibrium systems involving ONLY solids and/or liquids are negligible since solids and liquids are incompressible (with fixed density at fixed T) H2O(s) H2O(l)

Q.25 Extremely high P H2O(s) H2O(l) More open More closely packed The great increase in pressure causes the more open structure of ice to collapse to give the more closely packed structure of liquid water.

The Effect of Changes in Concentration on Equilibrium
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Test 1 : Add HCl Result : The white ppt disappears The equil. position shifts to the left

Interpretation : Kc approach
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Interpretation : Kc approach Since H2O is in large excess

Both [H+(aq)] and [Cl(aq)]  to the same extent
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Addition of HCl(aq) Both [H+(aq)] and [Cl(aq)]  to the same extent The equilibrium position shifts to the left to restore the Kc

 A net backward reaction is observed
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Kinetic approach : - Both the rates of forward and backward reactions increase but the backward reaction increases more.  A net backward reaction is observed  The equilibrium position shifts to the left

The system responds in such a way as to reduce the amount of HCl added
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) Bi3+(aq) + 3Cl(aq) + H2O(l) BiOCl(s) + 2HCl(aq) Addition of HCl(aq) The system responds in such a way as to reduce the amount of HCl added  the equilibrium position shifts to the left

The Effect of Changes in Concentration on Equilibrium
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Test 2 : Add large excess of H2O Result : The white ppt reappears The equil. position shifts to the right

Interpretation : Kc approach
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Interpretation : Kc approach Addition of large excess of H2O Equilibrium position shifts to the right such that *Qc = *Kc

Interpretation : Kinetic approach
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt Interpretation : Kinetic approach [H2O(l)]   rate of forward rx > rate of backward rx  equilibrium position shifts to the right

By Le Chatelier’s principle : -
Bi3+(aq) + Cl(aq) + H2O(l) BiOCl(s) + 2H+(aq) colourless white ppt By Le Chatelier’s principle : - [H2O(l)]  The system shifts to the right to reduce the water added.

a new equilibrium position with a new equilibrium constant.
The Effect of Changes in Temperature on Equilibrium A change in temperature of an equilibrium system results in an adjustment of the equilibrium system to a new equilibrium position with a new equilibrium constant.

Exothermic reaction: N2(g) + 3H2(g) 2NH3(g) Examples : -
Temperature (K) 500 600 700 800 Kp (atm-2) 90.0 3.0 0.3 0.04 Kp decreases as T increases

2C(graphite) + O2(g) 2CO(g)
Examples : - Exothermic reaction: 2C(graphite) + O2(g) CO(g) Temperature (K) 298 500 700 900 1100 Kp (atm) 1.5  1048 3.1  1032 1.2  1026 3.1  1022 1.5  1020 Kp decreases as T increases

Endothermic reaction:
Examples : - Endothermic reaction: N2O4(g) NO2(g) Temperature (K) 200 300 400 500 Kp (atm) 1.8  10-6 0.174 51 1510 Kp increases as T increases

Endothermic reaction:
Examples : - Endothermic reaction: N2(g) + O2(g) NO(g) Temperature (K) 700 1100 1500 Kp (no unit) 5  10-13 4  10-8 1  10-5 Kp increases as T increases

Van’t Hoff Equation C and C’ are constants related to

An increase in T shifts the equilibrium position to the left
If the forward process is exothermic, T   K  An increase in T shifts the equilibrium position to the left (in the endothermic direction)

An decrease in T shifts the equilibrium position to the right
If the forward process is exothermic, T   K  An decrease in T shifts the equilibrium position to the right (in the exothermic direction)

An increase in T shifts the equilibrium position to the right
If the forward process is endothermic, T   K  An increase in T shifts the equilibrium position to the right (in the endothermic direction)

An decrease in T shifts the equilibrium position to the left
If the forward process is endothermic, T   K  An decrease in T shifts the equilibrium position to the left (in the exothermic direction)

Conclusion : 1. An increase in temperature shifts the equilibrium position in the endothermic direction. 2. A decrease in temperature shifts the equilibrium position in the exothermic direction. Consistent with Le Chatelier’s principle

Q.26(a) Forward reaction is exothermic lnK (If C > 0)

Q.26(a) Forward reaction is exothermic lnK (If C < 0)

Q.26(a) Forward reaction is endothermic lnK (If C > 0)

Q.26(a) Forward reaction is endothermic (If C < 0) lnK

N2O4(g)(pale yellow) 2NO2(g)(brown)
50C Q.27 Increase in T Decrease in T

N2O4(g)(pale yellow) 2NO2(g)(brown)
Q.27(a) Increase in T N2O4(g)(pale yellow) NO2(g)(brown) Decrease in T  in T shift the equilibrium position to the right. Thus, the forward reaction is endothermic By Le Chatelier’s principle, the system tends to decrease the T by shifting in the endothermic direction.

Q.27(b)(i) Assume no change in equilibrium position  n is fixed PT inside the syringe = atomospheric pressure  PT is fixed

N2O4(g) 2NO2(g) Q.27(b)(ii) ∵ equilibrium position shifts to the right
 total no. of moles of gas molecules   total volume of the system  further

N2O4(g) 2NO2(g) Increase in T Q.27 V(dm3) Actual  in V
373 1.37 Ideal gas expansion 273 1.00

Interpretation of the Effects of Temperature Changes on Equilibrium in Terms of Chemical Kinetics
A–B + X A + B–X Ea Ea’ AB+X A+BX Potential energy Reaction co-ordinate

> 1 ∵ Ea > 0 & T2 – T1 > 0 Rate  as T 
For the forward reaction (exothermic) > 1 Ea Ea’ AB+X A+BX Potential energy Reaction co-ordinate ∵ Ea > 0 & T2 – T1 > 0 Rate  as T 

∵ Ea’ > Ea & T2 – T1 > 0 For the backward reaction (endothermic)
AB+X A+BX Potential energy Reaction co-ordinate ∵ Ea’ > Ea & T2 – T1 > 0

Conclusion : An  in temperature  the rates of endothermic and exothermic reactions to different extents. The rate of endothermic reaction is affected more by temperature changes.

X(l) X(g) Q.28 Prediction : -
An  in T shifts the equilibrium position to the right (in endothermic direction) Interpretation : - An  in T increases Kp Thus, more X(l) evaporate until Qp = Kp

X(s) X(g) Q.28 Prediction : -
An  in T shifts the equilibrium position to the right (in endothermic direction) Interpretation : - An  in T increases Kp Thus, more X(s) sublime until Qp = Kp

and the Extent of Completion of Reaction
if < 0 (forward reaction is exothermic) and C’ is negligibly small log10K > 0 K > 1 (the exothermic reaction is more complete

and the Extent of Completion of Reaction
if > 0 (forward reaction is endothermic) and C’ is negligibly small log10K < 0 K < 1 (the endothermic reaction is less complete)

Example : - Estimate the values of K at 298 K for the equilibrium systems in which the H of the forward reactions are (i) –100 kJ mol1 and (ii) 100 kJ mol1 respectively. (Given : R = J K1 mol1) (i) K  3  1017 (Units not known)

Example : - Estimate the values of K at 298 K for the equilibrium systems in which the H of the forward reactions are (i) –100 kJ mol1 and (ii) 100 kJ mol1 respectively. (Given : R = J K1 mol1) (ii) K  3  1018 (Units not known)

Conclusion : - Exothermic processes are Far More Complete than endothermic processes.

Since all gases arises from NH4HS(s)
Q.29 The total pressures of the following equilibrium system are 2.333104 Nm2 and 6.679104 Nm2 at K and K respectively. NH4HS(s) NH3(g) + H2S(g) Since all gases arises from NH4HS(s)

NH4HS(s) NH3(g) + H2S(g) (1) (2) At 282.5 K At 298.1 K (2) – (1)
= kJ mol1

Effects of catalysts on Equilibrium
It can be shown that catalysts have no effect on the equilibrium position since they affect the rates of both forward and backward reactions to the same extent. (Refer to Notes on Chemical Kinetics, p.37 Q.29) A catalyst has no effect on the equilibrium position but can change the time taken to attain the equilibrium state.

A B Q.30 Time taken to attain equilibrium
Less time to attain equilibrium Concentration t2 t1 Time

VT  T  (adiabatic compression)
Q.31 A(g) + B(g) C(g) H > 0 Time Rate of reaction Forward reaction Backward reaction  in T  in PT by reducing VT VT  T  (adiabatic compression) t1 e.g. expanding universe

Q.31 A(g) + B(g) C(g) Adding a +ve catalyst H > 0 Rate of reaction
Time Rate of reaction Forward reaction Backward reaction t2 Adding a +ve catalyst

Q.32 H2(g) + I2(g) 2HI(g) H < 0 t1 : 1. adding a catalyst
2.  in PT by adding an an inert gas at fixed VT t t t t4 Concentration [HI(g)] [H2(g)] [I2(g)] Time

Q.32 H2(g) + I2(g) 2HI(g) H < 0
 in PT by reducing VT has no effect on the equilibrium position but changes the equilibrium composition Concentration [HI(g)] [H2(g)] [I2(g)] Time t t t t4

Q.32 H2(g) + I2(g) 2HI(g) H < 0 t4 :  in PT by reducing VT
Concentration [HI(g)] [H2(g)] [I2(g)] Time t t t t4

Q.32 H2(g) + I2(g) 2HI(g) H < 0 t2 :  in T at fixed VT
Concentration [HI(g)] [H2(g)] [I2(g)] Time t2 :  in T at fixed VT t t t t4

Q.32 H2(g) + I2(g) 2HI(g) H < 0 t3 : Input of H2(g) at fixed VT
Concentration [HI(g)] [H2(g)] [I2(g)] Time t3 : Input of H2(g) at fixed VT t t t t4

Effect on equilibrium position
CO(g) + 2H2(g) CH3OH(g) H < 0 Changes Effect on equilibrium position Effect on Kp  in PT by reducing VT Shifts to the right No effect  in T Shifts to the left Kp 

CO(g) + 2H2(g) CH3OH(g) H < 0 Changes Effect on equilibrium position Effect on Kp Doubling PCO and No effect No effect Doubling and Shifts to the right No effect

Q.33 Soluble in water CO(g) + 2H2(g) CH3OH(g) H < 0 Changes
Effect on equilibrium position Effect on Kp A positive catalyst is added No effect No effect A little H2O(l) is added Shifts to the right No effect

A(g) + B(g) C(g) H = 0 Changes Effect on equilibrium position Effect on Kp  in T at fixed PT No effect No effect  in T at fixed PT No effect No effect

A(g) + B(g) C(g) H = 0 Changes Effect on equilibrium position Effect on Kp  in T at fixed VT Shifts to the right No effect  in T at fixed VT Shifts to the left No effect

Summary of the Effects of Changes of Various Factors on Equilibrium
aA(g) + bB(g) cC(g) + dD(g) Factor Equilibrium position Equilibrium constant Increase in concentration of reactants A or B Shifts to right No change Increase in concentration of products C or D Shifts to left

aA(g) + bB(g) cC(g) + dD(g) Factor Equilibrium position Equilibrium constant Increase in pressure by reducing the volume of the container Shifts to right if (c + d) < (a + b) Shifts to left to (a + b) < (c + d) No change if a + b = c + d No change Isothermal compression

aA(g) + bB(g) cC(g) + dD(g) Factor Equilibrium position Equilibrium constant Increase in temperature Shifts to right if the forward reaction is endothermic Shifts to left if the forward reaction is exothermic Kp  if the forward reaction is endothermic Kp  if the forward reaction is exothermic

aA(g) + bB(g) cC(g) + dD(g)
Summary of the Effects of Changes of Various Factors on Equilibrium aA(g) + bB(g) cC(g) + dD(g) Factor Equilibrium position Equilibrium constant Addition of a catalyst No change

Check Point 16-1 Answer Back
16.1 Irreversible and Reversible Reactions (SB p.89) Check Point 16-1 Back In the following reversible reaction: A B (a) Give the letter that represents the reactant of the forward reaction. (b) Give the letter that represents the reactant of the backward reaction. (c) Which is the forward reaction, A  B or B  A ? Answer A B A  B

List some characteristics of chemical equilibrium.
16.2 Dynamic Nature of Chemical Equilibrium (SB p.91) Back Let's Think 1 List some characteristics of chemical equilibrium. Answer Some characteristics of chemical equilibrium include: • It can only be achieved in a closed system. • It can be achieved from either forward or backward reactions. • It is dynamic in nature. • The concentrations of all chemical species present in a system at equilibrium state remain constant as long as the reaction conditions are unchanged.

H2O(g) + CO(g) H2(g) + CO2(g)
16.3 Examples of Chemical Equilibrium (SB p.92) Check Point 16-3 Back A trace amount of carbon monoxide labelled with radioactive carbon-14 is added to the following equilibrium system: H2O(g) + CO(g) H2(g) + CO2(g) Explain why radioactive carbon dioxide molecules are formed. Answer Chemical equilibrium is dynamic in nature. When a trace amount of carbon monoxide labelled with radioactive carbon-14 is added to the equilibrium system, the equilibrium position shifts to the right. Therefore, radioactive carbon dioxide molecules are formed.

16.4 Equilibrium Law (SB p.94) Back Let's Think 2 What is a closed system? Why can chemical equilibrium only be established in a closed system? Answer A closed system means that there is no transfer of matter between the system and the surroundings. If the system is open, some of the reactants or products can enter or leave the system. As a result, the equilibrium state can never be reached.

16.4 Equilibrium Law (SB p.94) Check Point 16-4 Write the equilibrium expression (for Kc) and unit of equilibrium constants for the following equilibrium system. (a) 2O3(g) O2(g) Answer (a) Unit of Kc: mol dm-3

16.4 Equilibrium Law (SB p.94) Check Point 16-4 Write the equilibrium expression (for Kc) and unit of equilibrium constants for the following equilibrium system. (b) N2(g) + 3H2(g) NH3(g) Answer (b) Unit of Kc: mol-2 dm6

Check Point 16-4 Answer Back
16.4 Equilibrium Law (SB p.94) Back Check Point 16-4 Write the equilibrium expression (for Kc) and unit of equilibrium constants for the following equilibrium system. (c) C(graphite) + H2O(g) CO(g) + H2(g) Answer (c) Unit of Kc: mol dm-3

Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag( s)
16.5 Determination of Equilibrium Constants (SB p.98) Check Point 16-5A In the determination of the equilibrium constant (Kc) of: Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag( s) 100 cm3 of M AgNO3(aq) and 100 cm3 of M FeSO4(aq) are mixed in a dry conical flask. The mixture is then allowed to stand overnight and filtered. The concentration of Ag+(aq) is found by titration cm3 of the filtrate is titrated with M KCNS(aq) and 6.10 cm3 of the KCNS( aq) is required for complete reaction.

16.5 Determination of Equilibrium Constants (SB p.98)
Check Point 16-5A (a) Calculate the equilibrium concentrations of Ag+(aq), Fe2+(aq) and Fe3+(aq). Answer (a) Ag+(aq) + CNS–(aq)  AgCNS(aq) Number of moles of KCNS(aq) = = 3.05  10-4 mol  Number of moles of Ag+(aq) in 25 cm3 of the filtrate at equilibrium = 3.05  10–4 mol [Ag+(aq)]eqm = = mol dm-3

Check Point 16-5A (a)  Fe2+(aq) and Ag+(aq) are consumed at the same rate.  [Fe2+(aq)]eqm = [Ag+(aq)]eqm = mol dm–3  [Fe2+(aq)]initial = = 0.05 mol dm-3 [Fe3+(aq)]eqm = [Fe2+(aq)]initial – [Fe2+(aq)]eqm = (0.05 – ) mol dm-3 = mol dm-3

Check Point 16-5A Answer (b) Calculate the equilibrium constant (Kc).
16.5 Determination of Equilibrium Constants (SB p.98) Check Point 16-5A (b) Calculate the equilibrium constant (Kc). Answer (b) = = mol-1 dm3

Check Point 16-5A Answer Back (c) What is the significance of
16.5 Determination of Equilibrium Constants (SB p.98) Check Point 16-5A Back (c) What is the significance of (i) using a dry conical flask? (ii) allowing the mixture to stand overnight? Answer (c) (i) The significance of using a dry conical flask is to make sure the reaction mixture in the conical flask is not diluted by the presence of water. (ii) The reaction mixture is allowed to stand overnight in order to give sufficient time for the reaction mixture to reach the equilibrium state.

Example 16-5A For the reversible reaction of hydrogen and iodine at equilibrium: H2(g) + I2(g) HI(g) If the initial amount of H2(g) is a mol, I2(g) is b mol and the amount of H2(g) or I2(g) reacted is x mol, express the equilibrium constant (Kc) in terms of a, b and x. Answer

Back Example 16-5A Let the volume of the reaction mixture be V dm3. Reactant / Product Initial number of moles (mol) Change in number of moles (mol) Number of moles at equilibrium (mol) H2(g) a -x a – x I2(g) b b – x HI(g) 2x

Example 16-5B Answer For the Haber process, N2(g) + 3H2(g) 2NH3(g)
16.5 Determination of Equilibrium Constants (SB p.99) Example 16-5B For the Haber process, N2(g) + 3H2(g) NH3(g) If the initial amount of N2(g) is a mol, H2(g) is b mol and the amount of N2(g) reacted is x mol, express the equilibrium constant (Kc) in terms of a, b and x. Answer

Back Example 16-5B Let the volume of the reaction mixture be V dm3. Reactant / Product Initial number of moles (mol) Change in number of moles (mol) Number of moles at equilibrium (mol) N2(g) a -x a – x H2(g) b -3x b – 3x NH3(g) 2x

Fe3+(aq) + NCS–(aq) [Fe(NCS)]2+(aq)
16.5 Determination of Equilibrium Constants (SB p.99) Example 16-5C A student mixed 10 cm3 of 2.0 × 10–3 M Fe(NO3)3(aq) with 10 cm3 of 2.0 × 10–3 M KNCS(aq). Fe3+(aq) + NCS–(aq) [Fe(NCS)]2+(aq) When the system reaches the equilibrium, the concentration of [Fe(NCS)]2+(aq) is 1.4 × 10–4 M. Determine the equilibrium constant (Kc) of the reaction. Answer

Example 16-5C 16.5 Determination of Equilibrium Constants (SB p.100)
Initial concentration of Fe3+(aq) = = 1.0  10-3 mol dm-3 Initial concentration of NCS-(aq) = Fe3+(aq) NCS-(aq) [Fe(NCS)]2+(aq) At start:  10-3 M  10-3 M M Amount changed: x M –x M x M At equilibrium: (1.0  10-3 – x) M (1.0  10-3 – x) M x M

Back Example 16-5C From the given data, the equilibrium concentration of [Fe(NCS)]2+(aq) is 1.4 × 10–4 M, thus x = 1.4 × 10–4 M. ∴ [Fe3+(aq)]eqm = (1.0 × 10–3 – 1.4 × 10–4) mol dm–3 = 0.86 × 10–3 mol dm–3 [NCS-(aq)]eqm = (1.0 × 10–3 – 1.4 × 10–4) mol dm–3 Kc = = dm3 mol-1

Back Let's Think 3 What is the implication for an equilibrium reaction having an equilibrium constant much smaller than 1.0? Answer The equilibrium constant of a reaction is related to the ratio of the concentration of products to the concentration of reactants at equilibrium. When the equilibrium constant of a reaction is much greater than 1, the reaction goes nearly to completion. Conversely, when the equilibrium constant of a reaction is much smaller than 1, the reaction hardly goes to completion.

Check Point 16-5B At 400 K, mole of PCl3(g) and mole of PCl5(g) were mixed in a 1 dm3 flask. After the system was left overnight, an equilibrium was established and mole of chlorine gas was found in the flask. Determine the equilibrium constant (Kc) of the reaction: PCl5(g) PCl3(g) + Cl2(g) Answer

Back Check Point 16-5B  At equilibrium, mole of Cl2(g) was found in the flask. X = mol Kc = = = mol dm-3 Reactant / Product Initial no. of moles (mol) Change in no. of moles (mol) No. of moles at equilibrium (mol) PCl5(g) 0.009 -x 0.009 – x PCl3(g) 0.250 +x x Cl2(g) x

Example 16-6A Answer The following equilibrium reaction
16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.101) Example 16-6A The following equilibrium reaction 2NOBr(g) NO(g) + Br2(g) is studied at 298 K. The partial pressures of NOBr(g), NO(g) and Br2(g) at equilibrium were found to be: PNOBr = 246 Nm–2 PNO = 450 Nm–2 PBr2 = 300 Nm–2 Calculate the value of Kp for the reaction at 298 K. Answer

16.6 Equilibrium Constant in Terms of Partial Pressures (SB p.101)
Example 16-6A 2NOBr(g) NO(g) + Br2(g) The expression of Kp is: Substituting the partial pressures into the expression, we have: = Nm–2 Back

Example 16-6B The decomposition of dinitrogen tetroxide to form nitrogen dioxide is a reversible reaction. N2O4(g) NO2(g) When the reaction reaches an equilibrium state, the partial pressure of N2O4(g) was found to be 2.71 atm. Calculate the partial pressure of NO2(g) at equilibrium given that the value of Kp is atm. Answer

Example 16-6B N2O4(g) NO2(g) The expression of Kp is: Substituting the value of Kp and the partial pressure of N2O4(g) into the expression, = Kp × = × 2.71 = atm2 ∴ = atm Back

Check Point 16-6 Equal amounts of hydrogen and iodine are allowed to reach an equilibrium at 298 K: H2(g) + I2(g) HI(g) If 80% of the hydrogen is converted to hydrogen iodide at the equilibrium, what is the value of Kp at this temperature? Answer

Check Point 16-6 Assume that the initial number of moles of H2(g) is 1 mol. H2(g) I2(g) HI(g) At start: mol mol mol At equilibrium: (1 – 0.8) mol (1 – 0.8) mol (0.8  2) mol = 0.2 mol = 0.2 mol = 1.6 mol Mole fraction of H2(g) = = 0.1 Mole fraction of I2(g) =

Check Point 16-6 Mole fraction of HI(g) = = 0.8 Let P be the total pressure of the system. = 64 Back

Equilibrium concentration (mol dm-3)
16.7 Equilibrium Position (SB p.103) Example 16-7 Determine the equilibrium constant (Kc) from the following data on the equilibrium system 2SO2(g) + O2(g) SO3(g) at 873 K. Experiment Equilibrium concentration (mol dm-3) [SO2(g)] [O2(g)] [SO3(g)] 1 1.60 1.30 3.62 2 0.71 0.50 1.00 Answer

Example 16-7 Back 16.7 Equilibrium Position (SB p.103)
The expression of Kc is: From experiment 1: = 3.94 dm3 mol-1 From experiment 2: = 3.97 dm3 mol-1 Since Kc is a constant at a specific temperature, the values of Kc from experiments 1 and 2 are very close, and the average value of Kc at 873 K is dm3 mol–1.

Initial no. of moles (mol)
16.7 Equilibrium Position (SB p.104) Check Point 16-7 Answer For the following reversible reaction: CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) Calculate the equilibrium constant (Kc) using the following data. (Assume that the equilibrium is established in a container of 1 dm3.) Expt Initial no. of moles (mol) No. of moles at eqm (mol) CH3CHOOH(l) CH3CH2OH(l) CH3COOH(l) 1 1.00 0.33 2 4.00 0.07

Check Point 16-7 16.7 Equilibrium Position (SB p.104)
The equilibrium constant for the equilibrium is expressed as: For experiment 1: CH3COOH(l) + CH3CH2OH(l) At start: mol mol At eqm: mol mol CH3COOCH2CH3(l) + H2O(l) At start: mol mol At eqm: (1.00 – 0.33) mol (1.00 – 0.33) mol = 0.67 mol = 0.67 mol

Check Point 16-7 Back 16.7 Equilibrium Position (SB p.104)
For experiment 1: CH3COOH(l) + CH3CH2OH(l) At start: mol mol At eqm: mol (4.00 – 0.93) mol = 3.07 mol CH3COOCH2CH3(l) + H2O(l) At start: mol mol At eqm: (1.00 – 0.07) mol (1.00 – 0.07) mol = 0.93 mol = 0.93 mol Since Kc is a constant at a specific temperature, the average value of Kc from experiments 1 and 2 is 4.07.

16. 8 Partition Equilibrium of a Solute Between Two Immiscible
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104) Example 16-8A An organic compound X has a partition coefficient of 30 in ethoxyethane and water. There is 3.1 g of X in 50 cm3 of water. 50 cm3 of ethoxyethane is then added to extract X from water. How much X is extracted using ethoxyethane? Answer

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8A Let a g be the mass of X extracted using 50 cm3 of ethoxyethane, then the mass of X left in water is (3.1 – a) g. a = 3.0 Back

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8B At 298 K, 50 cm3 of an aqueous solution containing 6 g of solute Y is in equilibrium with 100 cm3 of an ether solution containing 108 g of Y. Calculate the mass of Y that could be extracted from 100 cm3 of an aqueous solution containing 10 g of Y by shaking it with (a) 100 cm3 of fresh ether at 298 K; (b) 50 cm3 of fresh ether twice at 298 K. Answer

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8B = 1.08 g cm-3 = 0.12 g cm-3 = 9

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8B Let m g be the mass of Y extracted using 100 cm3 of ether, then the mass of Y left in the aqueous layer is (10 – m) g. m = 9  9 g of Y can be extracted using 100 cm3 of fresh ether.

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8B (b) Let m1 g be the mass of Y extracted using the first 50 cm3 of ether, then the mass of Y left in the aqueous layer is (10 – m1) g. m1 = 8.182  Mass of Y extracted using the first 50 cm3 of ether = g Mass of Y left in the aqueous layer = (10 – 8.182) g = g

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8B Let m2 g be the mass of Y extracted using the second 50 cm3 of ether, then the mass of Y left in the aqueous layer is (1.818 – m2) g. m2 = 1.487  Mass of Y extracted using the second 50 cm3 of ether = g Mass of Y left in the aqueous layer = (1.818 – 1.487) g = g

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.106) Example 16-8B ∴ Total mass of Y extracted = m1 + m2 = ( ) g = g Back

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.108) Check Point 16-8A The partition coefficient (KD) of an unknown organic compound A between 1,1,1-trichloroethane and water is expressed as: Calculate the mass of A that can be extracted from 60 cm3 of an aqueous solution initially containing 6 g of A using 100 cm3 of fresh 1,1,1-trichloroethane. Answer

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.108) Check Point 16-8A Let m be the mass of A extracted using 100 cm3 of 1,1,1-trichloroethane, then the mass of A left in 60 cm3 of aqueous layer is (6 – m). m = 5.77 g  5.77 g of A is extracted using 100 cm3 of 1,1,1- trichloroethane. Back

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110) Check Point 16-8B A student wrote the following explanation for the different Rf values found in the separation of two amino acids, leucine (Rf value = 0.5) and glycine (Rf value = 0.3), by paper chromatography using a solvent containing 20% of water. “Leucine is a much lighter molecule than glycine.” Do you agree with this explanation? Explain your answer. Answer

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110) Check Point 16-8B (a) The difference in Rf value of leucine and glycine is due to the fact that they have different partition between the stationary phase and the mobile phase. Therefore, they move upwards to different extent. The Rf value is not related to the mass of the solute.

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110) Check Point 16-8B (b) Draw a diagram to show the expected chromatogram of a mixture of A, B, C and D using a solvent X, given that the Rf values of A, B, C and D are 0.15, 0.40, 0.70 and 0.75 respectively. Answer

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110) Check Point 16-8B Back (b)

16.9 Significances of Equilibrium Constants (SB p.111)
Check Point 16-9 The reaction N2(g) + 3H2(g) NH3(g) has an equilibrium constant of dm6 mol–2 at 500 oC. Predict the net chemical change, if there is any, for the following concentrations of reactant(s) and product(s). (a) [NH3(g)] = mol dm–3, [N2(g)] = mol dm–3 and [H2(g)] = mol dm–3 Answer

Check Point 16-9 (a) = mol-2 dm6 Since Qc > Kc, the reaction proceeds from the right (product side) to the left (reactant side) until the equilibrium is reached.

Check Point 16-9 The reaction N2(g) + 3H2(g) NH3(g) has an equilibrium constant of dm6 mol–2 at 500 oC. Predict the net chemical change, if there is any, for the following concentrations of reactant(s) and product(s). (b) [NH3(g)] = mol dm–3, [N2(g)] = 1 mol dm–3 and [H2(g)] = 0.08 mol dm–3 Answer

Check Point 16-9 Back (b) = mol-2 dm6 Since Qc < Kc, the reaction proceeds from the left (reactant side) to the right (product side) until the equilibrium is reached.

As4O6(s) + 6C(s) As4(g) + 6CO(g)
Factors Affecting Equilibrium (SB p.113) Example 16-10A Back For the equilibrium system: As4O6(s) + 6C(s) As4(g) + 6CO(g) predict how the equilibrium position will shift in response to the following changes: (a) removing CO(g) (b) adding more As4(g) Answer (a) According to Le Chatelier’s principle, the equilibrium position will shift to the right. (b) According to Le Chatelier’s principle, the equilibrium position will shift to the left.

16.10 Factors Affecting Equilibrium (SB p.114)
Example 16-10B (a) For the reaction H2(g) + I2(g) HI(g), the following data are determined at 490 oC, [H2(g)] = 0.22 mol dm–3; [I2(g)] = 0.22 mol dm–3 and [HI(g)] = 1.56 mol dm–3 Calculate the equilibrium constant (Kc) at 490 oC. Answer H2(g) + I2(g) HI(g)

Example 16-10B (b) If an additional mol dm–3 of H2(g) is added to the above equilibrium mixture while keeping volume and temperature constant, what will happen? Calculate the equilibrium concentrations of all species when equilibrium is reached. Answer

Example 16-10B 16.10 Factors Affecting Equilibrium (SB p.114)
(b) H2(g) I2(g) HI(g) At eqm: mol dm mol dm mol dm-3 Now: ( ) mol dm mol dm mol dm-3 Since the value of the reaction quotient (Qc) is less than that of the equilibrium constant (Kc), the system is not at equilibrium. In order to re-establish the equilibrium, the value of the reaction quotient should be increased until it equals Kc. It can be predicted that more H2(g) and I2(g) will react to form more HI(g).

Example 16-10B Back 16.10 Factors Affecting Equilibrium (SB p.114)
(b) H2(g) I2(g) HI(g) Now: ( ) mol dm mol dm mol dm-3 At eqm: ( x) mol dm-3 (0.22 – x) mol dm ( x) mol dm-3 Since Kc remains constant, we obtain: By solving the quadratic equation, x = 0.06 or 0.77. If x equals 0.77, the concentration of H2(g) and I2(g) at equilibrium will be negative. Therefore, the correct answer of x is 0.06. ∴ [H2(g)]eqm = 0.42 mol dm– mol dm–3 = 0.36 mol dm–3 [I2(g)]eqm = 0.22 mol dm– mol dm–3 = 0.16 mol dm–3 [HI(g)]eqm =  0.06 mol dm–3 = 1.68 mol dm–3

2CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l)
Factors Affecting Equilibrium (SB p.115) Check Point 16-10A Consider the following reaction at equilibrium: 2CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) Explain the changes of the graph at time t0, t1, t2 and t3 respectively. Answer

Check Point 16-10A Back 16.10 Factors Affecting Equilibrium (SB p.115)
When CrO42–(aq) and H+(aq) are mixed at t0, they react continuously to form Cr2O72–(aq) and H2O(l ). At t1, an equilibrium between them is established. At t2, when more H+(aq) is added to the system, the equilibrium can no longer be maintained. In order to attain the equilibrium again (i.e. at t3), the additional H+(aq) must be removed by shifting the equilibrium to the right to form more Cr2O72–(aq) and H2O(l). Back

Example 16-10C The diagram on the right shows the effect of increasing pressure on the equilibrium 2NO2(g) N2O4(g). The equilibrium constant Kp for the reaction is 0.92 atm–1 at a given temperature.

Example 16-10C (a) Calculate the partial pressures of NO2(g) and N2O4(g) at equilibrium if the total pressure is 1 atm. Answer (a) Let the partial pressure of NO2(g) at equilibrium be p atm, then the partial pressure of N2O4(g) at equilibrium is (1 – p) atm. p = or –1.719 (rejected)  PNO2 = atm PN2O4 = (1 – 0.632) atm = atm

Example 16-10C (b) Calculate the partial pressures of NO2(g) and N2O4(g) if the total pressure at equilibrium is 2 atm. Answer Using the same method as in (a), p = or –2.115 (rejected)  PNO2 = atm PN2O4 = (2 – 1.028) atm = atm

Example 16-10C (c) Compare the results of (a) and (b), and state the effect of an increase in pressure on the equilibrium. Answer (c) Comparing the results in (a) and (b), PN2O4 is more than doubled while PNO2 is less than doubled when the total pressure increases from 1 atm to 2 atm. Thus, the equilibrium position shifts to the side with a smaller number of molecules when the pressure increases.

Example 16-10C (d) Explain why the brown colour of the equilibrium mixture fades out when the pressure of the equilibrium system is increased. Assume there is no temperature change. (Hint: The colour of NO2(g) is dark brown and that of N2O4(g) is pale brown or colourless.) Answer (d) The impact of the increased pressure is reduced by shifting the equilibrium position to the right-hand side of 2NO2(g) N2O4(g). More NO2(g), which is brown in colour, is used up. More N2O4(g), which is colourless, is formed. A colour change from brown to pale brown (or colourless) can be observed.

Back Example 16-10C (e) Given that the enthalpy change for the reaction 2NO2(g) N2O4(g) is –58 kJ, predict the colour change when a glass syringe containing the equilibrium mixture is put into a beaker of hot water for about 30 seconds, and then a beaker of water with a large amount of crushed ice for another 30 seconds. Answer (e) When the equilibrium mixture is put into hot water (the temperature increases), the equilibrium will shift to the left and more NO2(g) will be formed. Thus, the colour of the mixture will change to a darker brown. When the equilibrium mixture is put into ice water (the temperature decreases), the equilibrium will shift to the right and more N2O4(g) will be formed. As a result, the colour of the mixture will change to pale brown (or colourless).

3NO2(g) + H2O(g) 2HNO3(g) + NO(g)
Factors Affecting Equilibrium (SB p.118) Check Point 16-10B 1. Consider the following reaction at equilibrium: 3NO2(g) + H2O(g) HNO3(g) + NO(g) (a) Referring to the chemical equation above, write a mathematical expression for the equilibrium constant, Kp. Answer (a)

3NO2(g) + H2O(g) 2HNO3(g) + NO(g)
Factors Affecting Equilibrium (SB p.118) Check Point 16-10B 1. Consider the following reaction at equilibrium: 3NO2(g) + H2O(g) HNO3(g) + NO(g) (b) If the partial pressure of H2O(g) is increased at constant temperature, what changes, if any, occur in the partial pressures of (i) NO2(g)? (ii) HNO3(g)? (iii) NO(g)? (i) Decrease (ii) Increase (iii) Increase Answer

3NO2(g) + H2O(g) 2HNO3(g) + NO(g)
Factors Affecting Equilibrium (SB p.118) Check Point 16-10B 1. Consider the following reaction at equilibrium: 3NO2(g) + H2O(g) HNO3(g) + NO(g) (c) If the partial pressure of H2O(g) is increased at constant temperature, will the value of Kp increase, decrease or remain the same? Answer (c) The value of Kp will remain the same.

Check Point 16-10B 2. The equilibrium partial pressures of N2O4(g) and NO2(g) were found to be atm and atm respectively for the following reversible reaction at 100 oC. 2NO2(g) N2O4(g) (a) Calculate the equilibrium constant, Kp, for the reaction. Answer

Check Point 16-10B 2. The equilibrium partial pressures of N2O4(g) and NO2(g) were found to be atm and atm respectively for the following reversible reaction at 100 oC. 2NO2(g) N2O4(g) (b) The vessel containing the equilibrium mixture is compressed to one-half original volume suddenly. Predict what would happen. Calculate the equilibrium partial pressures of N2O4(g) and NO2(g). Answer

Check Point 16-10B 16.10 Factors Affecting Equilibrium (SB p.118)
After compression to one-half the original volume, all the gas pressures will be doubled. Therefore, the partial pressures of N2O4(g) and NO2(g) will be atm and atm respectively. 2NO2(g) N2O4(g)

Check Point 16-10B 16.10 Factors Affecting Equilibrium (SB p.118)
Since the value of the reaction quotient is less than that of the equilibrium constant, the system is not at equilibrium. The reaction proceeds from the left to the right until the equilibrium is reached. As a result, more N2O4(g) will be formed. 2NO2(g) N2O4(g) At start: At eqm: – 2x x

Check Point 16-10B Back 16.10 Factors Affecting Equilibrium (SB p.118)
(b) By solving the quadratic equation, x = or  x =  PNO2 = – 2  = atm PN2O4 = = atm Back

Check Point 16-10C Predict how the equilibrium position is affected when the equilibrium system N2O4(g) NO2(g) ΔH = +58 kJ is subjected to the following changes: (a) addition of NO2(g) (b) removal of N2O4(g) (c) addition of He( g) (d) increase in volume of the container (e) decrease in temperature Answer

Check Point 16-10C Back 16.10 Factors Affecting Equilibrium (SB p.120)
(a) The equilibrium position shifts to the left. (b) The equilibrium position shifts to the left. (c) The equilibrium position remains unchanged. (d) The equilibrium position shifts to the right. (e) The equilibrium poistion shifts to the left. Back

Example 16-10D The equilibrium constant (Kp) of the following reaction is 1.6 × 10–4 atm–2 at 673 K and 1.4 × 10–5 atm–2 at 773 K. N2(g) + 3H2(g) NH3(g) Determine the mean enthalpy change of formation of 1 mole of ammonia from its elements in the temperature ranges from 673 K to 773 K. (Given: R = 8.31 J K–1 mol–1) Answer

Example 16-10D Back 16.10 Factors Affecting Equilibrium (SB p.121)
At 673 K, At 773 K, Combining (1) and (2), H = J mol-1 = kJ mol-1

Example 16-10E Determine graphically the enthalpy change of formation of NO2(g) from N2O4(g) using the following data: Temperature (K) Kp (atm) 298 0.115 350 3.89 400 47.9 500 1700 600 17 800 (Given: R = J K–1 mol–1) Answer

Example 16-10E 16.10 Factors Affecting Equilibrium (SB p.122) +9.79
1.67  10-3 +7.44 2.00  10-3 +3.87 2.50  10-3 +1.36 2.86  10-3 -2.16 3.36  10-3 ln Kp 1/T (K-1)

Example 16-10E 16.10 Factors Affecting Equilibrium (SB p.122)
A graph of ln Kp against produces a straight line with slope

Example 16-10E Back 16.10 Factors Affecting Equilibrium (SB p.122)
Slope = = H =  8.314 = J mol-1 = 59.2 kJ mol-1 Back

N2(g) + 3H2(g) 2NH3(g) H = –92.6 kJ
Factors Affecting Equilibrium (SB p.122) Example 16-10F Haber process is an important industrial process to manufacture ammonia with the use of nitrogen and hydrogen. Ammonia has numerous uses like making fertilizers and explosives. The reaction between nitrogen and hydrogen is a reversible reaction. It takes place with release of thermal energy. N2(g) + 3H2(g) NH3(g) H = –92.6 kJ (a) Based on your knowledge about “chemical equilibrium”, predict the necessary conditions to increase the yield of ammonia in the Haber process. Answer

Example 16-10F 16.10 Factors Affecting Equilibrium (SB p.122)
Since the reaction is exothermic, a lower temperature will shift the equilibrium to the right-hand side and hence increase the yield of ammonia. As shown in the chemical equation, there are totally four nitrogen and hydrogen molecules on the left-hand side of the equation and only two ammonia molecules on the right-hand side. A higher pressure will shift the equilibrium position to the right and more ammonia will be produced. Also, increasing the concentration of the reactants (i.e. nitrogen and hydrogen) or removing the product (i.e. ammonia) from the reaction mixture will shift the equilibrium position to the right and thus the yield of ammonia will be increased.

Example 16-10F (b) The actual operating conditions of the Haber process are a temperature of about 450 oC, a pressure of about 400 atm and the presence of a catalyst (e.g. iron). Justify the conditions used. Answer

Example 16-10F Back 16.10 Factors Affecting Equilibrium (SB p.122)
(b) The use of high pressure is as predicted in (a). This not only shifts the equilibrium position to the right but also increases the rate of the reaction. The use of catalysts shortens the time for the reaction to reach the equilibrium while it has no effect on the equilibrium constant. The use of a high temperature is contradictory to the prediction made in (a). It can be explained based on the rate of the reaction which in turn determines the rate of manufacture of ammonia. Although the equilibrium position shifts to the right at a lower temperature, the rate of the reaction is very low (i.e. a longer time is required to reach the equilibrium state). The use of a moderate temperature is a compromise between the rate and the yield of the reaction. At 450 °C, the reaction is reasonably fast and the yield of ammonia is optimum.

Irreversible and Reversible Reactions

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Irreversible and Reversible Reactions

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