Chapter 6 Reaction Kinetics

Question 1

The rate of reaction refers to the change in the amount or concentration of a reactant or product per unit time and can be found by:

Answer 1

• Measuring the decrease in the concentration of a reactant OR
• Measuring the increase in the concentration of a product over time
  
  • The units of rate of reaction are mol dm^-3 s^-1

Question 2

Rate equation

Answer 2

• The thermal decomposition of calcium carbonate (CaCO₃) will be used as an example to study the rate of reaction

CaCO₃ (s) → CaO (s) + CO₂ (g)

• The rate of reaction at different concentrations of CaCO₃ is measured and tabulated

Question 3

Rate of reactions table

Answer 3

Question 4

• Rate equations can only be determined experimentally and cannot be found from the stoichiometric equation

Answer 4

Rate of reaction = k [A]^m [B]ⁿ

[A] and [B] = concentrations of reactants

m and n = orders of the reaction

Question 5

Order of reaction

Answer 5

• The order of reaction shows how the concentration of a reactant affects the rate of reaction
  
  • It is the power to which the concentration of that reactant is raised in the rate equation
  • The order of reaction can be 0, 1,2 or 3
  • When the order of reaction of a reactant is 0, its concentration is ignored

Question 6

The overall order of reaction is

Answer 6

• the sum of the powers of the reactants in a rate equation
• For example, in the following rate equation, the reaction is:

Rate = k [NO₂]²[H₂]

• • Second-order with respect to NO
    
    • First-order with respect to H₂
    • Third-order overall (2 + 1)

Question 7

Half-life

Answer 7

• The half-life (t_1/2) is the time taken for the concentration of a limiting reactant to become half of its initial value

Question 8

The rate-determining step is the

Answer 8

• slowest step in a reaction
• If a reactant appears in the rate-determining step, then the concentration of that reactant will also appear in the rate equation

Question 9

a bimolecular reaction

Answer 9

• Bimolecular: two species involved in the rate-determining step

Question 10

Unimolecular:

Answer 10

one species involved in the rate-determining step

Question 11

The intermediate is derived from

Answer 11

substances that react together to form it in the rate-determining step

• For example, for the reaction above the intermediate would consist of CH₃Br and OH^-

Question 12

The order of reaction shows how the concentration of a

Answer 12

• reactant affects the rate of reaction

Rate = k [A]^m [B]ⁿ

• When m or n is zero = the concentration of the reactants does not affect the rate
• When the order of reaction (m or n) of a reactant is 0, its concentration is ignored
• The overall order of reaction is the sum of the powers of the reactants in a rate equation

Question 13

In a zero-order the concentration of the reactant is

Answer 13

inversely proportional to time

• This means that the concentration of the reactant decreases with increasing time
• The graph is a straight line going down

Question 14

In a first-order reaction the concentration of the reactant

Answer 14

• decreases with time
  
  • The graph is a curve going downwards and eventually plateaus

Question 15

In a second-order reaction the concentration of the reactant

Answer 15

decreases more steeply with time

• The concentration of reactant decreases more with increasing time compared to in a first-order reaction
• The graph is a steeper curve going downwards

Question 16

The progress of the reaction can be followed by measuring the

Answer 16

• initial rates of the reaction using various initial concentrations of each reactant
• These rates can then be plotted against time in a rate-time graph

Question 17

In a zero-order reaction the rate doesn’t depend on the

Answer 17

• concentration of the reactant
  
  • The rate of the reaction therefore remains constant throughout the reaction
  • The graph is a horizontal line
  • The rate equation is rate = k

Question 18

In a first-order reaction the rate is

Answer 18

directly proportional to the concentration of a reactant

• The rate of the reaction decreases as the concentration of the reactant decreases when it gets used up during the reaction
• The graph is a straight line
• The rate equation is rate = k [A]

Question 19

In a second-order reaction, the rate

Answer 19

• is directly proportional to the square of concentration of a reactant
  
  • The rate of the reaction decreases more as the concentration of the reactant decreases when it gets used up during the reaction
  • The graph is a curved line
  • The rate equation is rate = k [A]²

Question 20

• The order of a reaction can also be deduced from its half-life (t_1/2 )
• For a zero-order reaction the successive half-lives decrease

Answer 20

with time

• This means that it would take less time for the concentration of reactant to halve as the reaction progresses

Question 21

The half-life of a first-order reaction remains constant throughout the reaction

Answer 21

• The amount of time required for the concentration of reactants to halve will be the same during the entire reaction

Question 22

For a second-order reaction, the half-life increases with time

Answer 22

• This means that as the reaction is taking place, it takes more time for the concentration of reactants to halve

Question 23

Half-lives of zero, first and second-order reactions

Answer 23

Question 24

The initial rate can be calculated by using the

Answer 24

initial concentrations of the reactants in the rate equation

Question 25

The half-life of a first-order reaction is independent of the

Answer 25

concentration of reactants

• This means that despite the concentrations of the reactants decreasing during the reaction
• The amount of time taken for the concentrations of the reactants to halve will remain the same throughout the reaction
• The graph is a straight line going downwards

Question 26

Experimental data of the changes in concentration over time suggests that the half-life is

Answer 26

• constant
  
  • Even if the half-lives are slightly different from each other, they can still be considered to remain constant
• This means that no matter what the original concentration of the CH₃CN is, the half-life will always be around 10.0 minutes

Question 27

In a first-order reaction, the time taken for the concentration to halve remains constant

Answer 27

Question 28

The rate constant (k) of a reaction can be calculated using:

Answer 28

• The initial rates and the rate equation
• The half-life

Question 29

Calculating the rate constant from the initial rate

Answer 29

• The reaction of calcium carbonate (CaCO₃) with chloride (Cl^-) ions to form calcium chloride (CaCl₂) will be used as an example to calculate the rate constant from the initial rate and initial concentrations
• The reaction and rate equation are as follows:

CaCO₃ (s) + 2Cl^- (aq) + 2H⁺ (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Rate = k [CaCO₃] [Cl^-]

• The progress of the reaction can be followed by measuring the initial rates of the reaction using various initial concentrations of each reactant

Question 30

Calculating the rate constant from the half-life

Answer 30

• The rate constant (k) can also be calculated from the half-life of a reaction
• You are only expected to deduce k from the half-life of a first-order reaction as the calculations for second and zero-order reactions are more complicated

Question 31

• For a first-order reaction, the half-life is related to the rate constant by the following expression:

Answer 31

• Rearranging the equation to find k gives:
• So, for a first-order reaction such as the methyl (CH₃) rearrangement in ethanenitrile (CH₃CN) with a half-life of 10.0 minutes the rate constant is:

= 1.16 x 10^-3 dm³ mol^-1 s^-1

Question 32

The reaction mechanism of a reaction describes how many steps are involved in the

Answer 32

• making and breaking of bonds during a chemical reaction

Question 33

It is the slowest step in a reaction and includes the reactants that have an impact on the reaction rate when their concentrations are changed

• Therefore, ll reactants that therefore appear in the

Answer 33

• rate equation will also appear in the rate-determining step
• This means that reactants that have a zero-order and intermediates will not be present in the rate-determining step

Question 34

The overall reaction equation and rate equation can be used to

Answer 34

predict a possible reaction mechanism of a reaction

Question 35

The order of a reactant and thus the rate equation can be deduced from a

Answer 35

reaction mechanism given that the rate-determining step is known

Question 36

Identifying the rate-determining step

Answer 36

• from a rate equation given that the reaction mechanism is known
• For example, propane (CH₃CH₂CH₃) undergoes bromination under alkaline solutions
• The overall reaction is:

CH₃CH₂CH₃ + Br₂ + OH^- → CH₃CH₂CH₂Br + H₂O + Br^-

• The reaction mechanism is:
• The rate equation is:

Rate = k [CH₃CH₂CH₃] [OH^-]

• From the rate equation, it can be deduced that only CH₃COCH₃ and OH^- are involved in the rate-determining step and not bromine (Br₂)
• Since only in step 1 of the reaction mechanism are CH₃COCH₃ and OH^- involved, the rate-determining step is step 1 is the case for step 1 of the reaction mechanism

Question 37

Identifying intermediates & catalyst

Answer 37

• When a rate equation includes a species that is not part of the chemical reaction equation then this species is a catalyst

Question 38

Identifying intermediates & catalyst

Answer 38

• For example, the halogenation of butanone under acidic conditions
• The reaction mechanism is:
• The rate equation is:

Rate = k [CH₃CH₂COCH₃] [H⁺]

• The H⁺ is not present in the chemical reaction equation but does appear in the rate equation
  
  • H⁺ must therefore be a catalyst
• Furthermore, the rate equation suggest that CH₃CH₂COCH₃ and H⁺ must be involved in the rate-determining (slowest) step
• The CH₃CH₂COCH₃ and H⁺ appear in the rate-determining step in the form of an intermediate (which is a combination of the two species)

Question 39

At higher temperatures, a greater proportion of molecules have energy greater than than the

Answer 39

• activation energy
• Since the rate constant and rate of reaction is directly proportional to the fraction of molecules with energy equal or greater than the activation energy, then at higher temperatures:
  
  • The rate constant increases
  • The rate of reaction increases

Question 40

• The relationship between the rate constant and the temperature is given by the following equation:

Answer 40

ln k = natural logarithm of the rate constant

A = constant related to the collision frequency and orientation of the molecules

E_a = activation energy (joules, J)

R = gas constant (8.31 J K^-1 mol^-1)

T = temperature (kelvin, K)

Question 41

The graph of ln k over 1/T is a straight line with gradient -E_a/R

Answer 41

• A varies only a little bit with temperature, it can be considered a constant
• E_a and R are also constants
• The equation shows that an increase in temperature (higher value of T) gives a greater value of ln k (and therefore a higher value of k)
• Since the rate of the reaction depends on the rate constant (k) an increase in k also means an increased rate of reaction

Question 42

Catalysts increase the rate of reaction by providing an alternative

Answer 42

• pathway which has a lower activation energy
• Catalysts can be either homogeneous or heterogeneous
• Homogeneous catalysts are those that are in the same phase as the reaction mixture
• For example, in the esterification of ethanoic acid (CH₃COOH) wit

Question 43

In heterogeneous catalysis, the molecules react at the

Answer 43

• surface of a solid catalyst
• The mode of action of a heterogeneous catalyst consists of the following steps:
• Adsorption (or chemisorption) of the reactants on the catalyst surface
  
  • The reactants diffuse to the surface of the catalyst
  • The reactant is physically adsorbed onto the surface by weak forces
  • The reactant is chemically adsorbed onto the surface by stronger bonds
  • Chemisorption causes bond weakening between the atoms of the reactants
• Desorption of the products
  
  • The bonds between the products and catalyst weaken so much that the products break away from the surface

Question 44

In the Haber process ammonia (NH₃) is produced from

Answer 44

nitrogen (N₂) and hydrogen (H₂)

Question 45

An iron catalyst is used which

Answer 45

• speeds up the reaction by bringing the reactants close together on the metal surface
• This increases their likelihood to react with each other

Question 46

The mode of action of the iron catalyst is as follows:

Answer 46

• Diffusion of the nitrogen and hydrogen gas to the iron surface
• Adsorption of the reactant molecules onto the iron surface by forming bonds between the iron and reactant atoms
  
  • These bonds are so strong that they weaken the covalent bonds between the nitrogen atoms in N₂ and hydrogen atoms in H₂
  • But they are weak enough to break when the catalysis has been completed
• The reaction takes place between the adsorbed nitrogen and hydrogen atoms which react with each other on the iron surface to form NH₃
• Desorption occurs when the bonds between the NH₃ and iron surface are weakened and eventually broken
• The formed NH₃ diffuses away from the iron surface

Question 47

Iron brings the nitrogen and hydrogen closer together so that they can react and hence increases the rate of reaction

Question 47

Iron brings the nitrogen and hydrogen closer together so that they can react and hence increases the rate of reaction

Answer 47

Question 48

Heterogeneous catalysts are also used in the

Answer 48

• catalytic removal of oxides of nitrogen from the exhaust gases of car engines
• The catalysts speed up the conversion of:
  
  • Nitrogen oxides (NO_y) into harmless nitrogen gas (N₂)
  • Carbon monoxide (CO) into carbon dioxide (CO₂)

Question 49

The catalytic converter has a honeycomb structure containing small beads coated with

Answer 49

platinum, palladium, or rhodium metals which act as heterogeneous catalysts

Question 50

The mode of action of the catalysts is as following

Answer 50

• • Adsorption of the nitrogen oxides and CO onto the catalyst surface
    
    • The weakening of the covalent bonds within nitrogen oxides and CO
    • Formation of new bonds between:
      
      • Adjacent nitrogen atoms to form N₂ molecules
      • CO and oxygen atoms to form CO₂ molecules
    • Desorption of N₂ and CO₂ molecules which eventually diffuse away from the metal surface

Question 51

Homogeneous catalysis often involves

Answer 51

• redox reactions in which the ions involved in catalysis undergo changes in their oxidation number
  
  • As ions of transition metals can change oxidation number they are often good catalysts
• Homogeneous catalysts are used in one step and are reformed in a later step

Question 52

This is a very slow reaction in which the peroxydisulfate (S_2,O₈^2- ) ions oxidise the

Answer 52

• iodide to iodine

S₂O₈^2- (aq) + 2I^- (aq) → 2SO₄^2- (aq) + I₂ (aq)

• Since both the S₂O₈^2- and I^- ions have a negative charge, it will require a lot of energy for the ions to overcome the repulsive forces and collide with each other
• Therefore, Fe³⁺ (aq) ions are used as a homogeneous catalyst
• The catalysis involves two redox reactions:
  
  • First, Fe³⁺ ions are reduced to Fe²⁺ by I^-

2Fe³⁺ (aq) + 2I^- (aq) → 2Fe²⁺ (aq) + I₂ (aq)

• Then, Fe²⁺ is oxidized back to Fe³⁺ by S₂O₈^2-

2Fe²⁺ (aq) + S₂O₈^2- (aq) → 2Fe³⁺ (aq) + 2SO₄^2- (aq)

• By reacting the reactants with a positively charged Fe ion, there are no repulsive forces, and the activation energy is significantly lowered
• The order of the two reactions does not matter
  
  • So, Fe²⁺ can be first oxidised to Fe³⁺ followed by the reduction of Fe³⁺ to Fe²⁺

Question 53

Nitrogen oxides & acid rain

Answer 53

• As fossil fuels contain sulfur, burning the fuels will release sulfur dioxide which oxidises in air to sulfur trioxide, and then dilute sulfuric acid (H₂SO₄) is formed by reaction with water. The result is acidification of rain:

SO₃(g) + H₂O(l) → H₂SO₄(aq)

• Nitrogen oxides can act as catalysts in the formation of acid rain by catalysing the oxidation of SO₂ to SO₃

NO₂(g) + SO₂(g) → SO₃(g) + NO(g)

• The formed NO gets oxidised to regenerate NO₂

NO(g) + ½ O₂(g) → NO₂(g)

• The regenerated NO₂ molecule can again oxidise another SO₂ molecule to SO₃ which will react with rainwater to form H₂SO₄ and so on