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***

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 Fe**³⁺** 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

REACTION QUOTIENT

Answer 53

o At any time t, the system can be sampled to determine the amounts of reactants and products present. o Q tells us whether the system is at equilibrium, or in what direction the reaction must proceed in order to reach equilibrium: o Qc < Kc more products must form o Qc > Kc more reactants must form o Qc = Kc at equilibrium (no net change)

Question 54

SUMMARY – EFFECT OF DISTURBANCES

Answer 54