CHEMISTRY PRE-U CHEMISTRY SEM 1 CHAP 5.pdf

PRE-UNIVERSITY CHEMISTRY SEMESTER 1 962 / 1 CHAPTER 5 : KINETIC CHEMISTRY

Past Year Questions Analysis

CHAPTER 5 : KINETIC CHEMISTRY 5.1

Rate of reaction

5.2

Rate law, Order of reactions and rate constants

5.3

The effect of temperature on reaction kinetics

5.4

The role of catalysts in reactions 2007

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2013 Sem 1

2014 Sem 1

Topic P1 5. Kinetic Chemistry

1

P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2

5b

1

1b 2c

3

3

3

6

3

A

1

B, C

A

B, C

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17

5.1 Rate of reaction 1. Chemical kinetics is the area of chemistry concerned with the speeds, or rates, at which a chemical reaction occurs. Kinetics, in this case is referring to the rate of a reaction, which is the change in the concentration of a reactant or a product with time. Generally, rate of reaction is inversely proportional with time, therefore, longer the time taken for a reaction to occur, lower the rate of reaction

1 rate ∝ time 2. Generally in chemical reaction Reactants
Products For example, when A
B (Concentration decrease) (Concentration increase)

b) Consider the following chemical equations : a A (aq) + b B (aq) → c C (aq) + d D (g) Rate of reaction can also be expressed base on the stoichiometry coefficient, where it can be written as Rate of reaction based on A

1 d [A] Rate = − a dt Rate of reaction based on C

1 d [ C] Rate = c dt

Rate of reaction based on B

1 d [B] Rate = − b dt Rate of reaction based on D

1 d [ D] Rate = d dt

As a conclusion,

1 d [A] 1 d [B] 1 d [ C] 1 d [ D] Rate = − = − = = a dt b dt c dt d dt

For example, in a chemical reaction of Ostwald Process, where 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) Rate of reaction based on A

1 d [ NH 3 ] Rate = − 4 dt Rate of reaction based on C

1 d [ NO] Rate = 4 dt

Rate of reaction based on B

1 d [O 2 ] Rate = − 5 dt Rate of reaction based on D

1 d [ H 2 O] Rate = 6 dt

As a conclusion,

1 d [ NH 3 ] 1 d [O 2 ] 1 d [ NO] 1 d [ H 2 O] Rate = − = − = = 4 dt 5 dt 4 dt 5 dt

5.1.1 Theory of Reaction Rates  Collision Theory – use to explain the effects of concentration and temperature on rate of reaction.  Base on 3 main ideas  Molecules must collide to react  Molecules must collide at the right orientation  Molecules must collide at the minimum amount of energy called activation energy.

Molecule must collide to react  Theory – use to explain the effects of concentration and temperature on rate of reaction.  Molecule must collide to react  In order for a reaction to occur, there must be physical interactions that take place, where the molecules are able to collide with each other and form a chemical reaction. Rate of collision is directly influenced by the following factors :  Concentration : As the concentration of particles increased, frequency of collision occur more rapidly. This will increase the chances of effective collision and hence increase the rate of reaction. From the diagram below, we can see that, as the number of particles increase with concentration hence increase the frequency of collision between particles. the frequency of collision increase significantly from (a) < (b) < (c) as the number of particles increase

 Temperature : When temperature increase, particles absorbed the energy supplied and stored in the form of kinetic energy. Particles with higher kinetic energy can move faster, and hence has a higher frequency of collision between particles. Furthermore higher kinetic energy allows more particles to have energy higher than the activation energy, hence increase the rate of reaction.

 Molecule must collide in the right orientation in order to form the right product.  Consider the following reaction occurs. 2 AB
A2 + 2 B.  For the reaction to take place, when the molecules collide, it must collide under the correct way, in order for that particular reaction to happen

Diagram of collision with the correct orientation. Note that, the position of the 2 molecules must be in the correct position. If not, the reaction will not take place, such as the diagram below

 If the molecules collide at a wrong orientation or position is not correct, the reaction will not occur, as illustrated in the 2 diagrams below

 Molecule must possessed certain amount of kinetic energy called activation energy  Even if the 2 conditions above is fulfilled – the molecules collide as in (a) ; it is collided in the right orientation as in (b), does it mean, the reaction will occur?  Note the diagram below Diagram of collision with the correct orientation. Note that, even though it collides with the right orientation, the reaction does not occur. This is due to the collision is too gently, and does not have the enough energy to react. So, this minimum required for the reaction to happened is called activation energy

minimum amount of Activation energy is defined as ……..………………………… energy required to initiate a chemical reaction …………………………………………………………………….  So, in order for a reaction to take place, there must be a certain amount of energy absorbed in order for the molecules which collide at the right orientation to happen.  As discussed during Maxwell-Boltzmann distribution graph, when temperature increase, more particles has energy higher than activation energy, and this will increase the rate of reaction.  Activation is always endothermic, as heat is required in order for molecule to collide effectively and form a new compound.  The energy profile bellow shows the reaction of both endothermic and exothermic reaction 

Endothermic process

Exothermic process

Ea

Ea

Theory of transition state  Theory of transition state explained the process that take place during a chemical reaction.  Supposedly the reaction of B–C + A
C + A–B is a one step reaction where it can be described as A + B – C
[A---B---C]
A–B + C The height of the "barrier" in the beginning of the graph is called the activation energy, and the configuration of atoms at the maximum in the potential energy profile is called the transition state, or the activated complex. In another words, A--B--C formed is an activated complex where it usually exist as an intermediate of a chemical reaction, a substance that appear during a reaction but will not form as products

Hydrolysis in 3o haloalkane C(CH3)3Br + OH- →

CH3CH2Br + OH- → CH3CH2OH + BrOccur only in 1 step

Equation C(CH3)3OH + BrOccur in 2 steps Step 1 : Formation carbocation

Process

Step 2 : Nucleophilic attack

Hydrolysis in 1o haloalkane

of

Energy

Energy / kJ

Energ y profile

Reaction coordinate

Rate equati on

Rate = k [C(CH3)3Br]

Reaction coordinate

Rate = k [CH3CH2Br][OH-]

5.2 Rate law, Order of reactions and rate constants 1. From what we had learn so far, we know that concentration is directly proportional to rate of reaction where rate ∝ [concentration] ; rate = k [concentration] The equation bolded is also known as rate equation, (also known as rate law) a way to expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. Rate constant, k, in the other hand, is the proportionality constant of a given chemical reaction. 2. Supposedly if a chemical reaction take placed as shown in the equation below a A (aq) + b B (aq) → c C (aq) + d D (g) The rate equation for the reaction is expressed by k = rate constant [A] and [B] = concentration of A and B x y rate = k [A] [B] x and y = order of reaction with respect to concentration of A and B

rate = k [A]x[B]y

a) Only the concentration of reactants is taken into consideration of rate of reaction, as the speed of the reaction largely depend on the amount and concentration of reactants used in the process. Generally, greater the amount (concentration) of reactant used, greater the rate of reaction b) Rate constant, k, is the proportionality constant of the reaction, in which the value remain constant under constant temperature, regardless of the changes in concentration of the reactants take place. However, rate constant changes with temperature of the reaction

rate = k [A]x[B]y c) order of reaction, x and y, is the power to which the concentration of the reactants (in this case, [A] and [B] respectively) is raised to in the rate equation. i. Note that the order of reaction has no relationship with the stoichiometry coefficient of the chemical equation. Therefore, x ≠ a ; y ≠ b. ii. Order of reaction can only be determined from series of experiments carried out under different concentration. iii. The overall order of reaction can be calculated by summing up the order of reaction for each reactant involved. In this case. Overall order = x + y

d) Once the overall order of reaction is known, the rate constant can be calculated accordingly by substituting the concentration and order of reaction towards the rate of each experiment. Unit of rate constant depend largely on the overall order of reaction. Overall reaction order Zero order First order Second order Third order Fifth order

Example

Unit of rate constant, k

[A]0

(mol dm-3 s-1) = k (mol dm-3)0 k = mol dm-3 s-1

rate = k [A]

(mol dm-3 s-1) = k (mol dm-3)1 k = s-1

rate = k [A][B]

(mol dm-3 s-1) = k (mol dm-3)2 k = mol-1 dm3 s-1

rate = k

Rate = k

[A][B]2

Rate = k [A][B]2[C]2

(mol dm-3 s-1) = k (mol dm-3)3 k = mol-2 dm6 s-1 (mol dm-3 s-1) = k (mol dm-3)5 k = mol-4 dm12 s-1

5.2.1

First order of reaction

1. A first-order reaction is a reaction whose rate depends on the of the equation reactant Integration concentration raised to the first power. Supposedly in a chemical reaction, where A → products Therefore, for this reaction, rate of reaction can be expressed d[A ] rate = − or it can be written as rate = k [A] dt When the rate is substituted between each other, it can be expressed d[A] d [ A] k [ A] = − which can be rearrange as − k dt = dt [ A] d [A ] Integration of the equation ∫ [ A ] = − k ∫ d t → ln [ A ] = − kt + c c is a constant which can be determined when time, t = 0 (initial time), c = ln [A]. Usually, it is indicated as ln [A]0, since it is the initial concentration of the reaction. As a result, the equation that can be applied to a first order reaction is :

ln [A]t = - kt + ln [A]0

Graph in first order reaction

-3

Half life in first order of reaction 3. Order of a reaction can also be determined by using half-life method. Half life, t1/2 is defined as the time required for the concentration of a reactant to decrease to half of its initial 0. concentration, [A]0/2. Flow below shows the decreased of concentration of reactant for each half-life take place

[ A ]0 2 nd t1 / 2 [A ]0 3 rd t1 / 2 [ A ]0 [A ]0   →  →   → 2 4 8 1st t1 / 2

Initial concentration of a reactant is 1.00 mol dm-3 1.00 mol dm-3

1st t1 / 2

  →

0.25 mol dm-3

3rd t1/ 2

0.50 mol dm-3

2nd t1 / 2

 →

 → 0.125 mol dm-3

Half life in first order of reaction (graph)

Equation of Half life in first order of reaction c) Using the equation of the first order of reaction, we can also calculate the half-life of the reaction, and also rate constant, k, if either one of these information is given. Derived from the equation of first order of reaction : Since after t1/2 ; concentration of the initial reactant decreased [A]0 by half. Therefore,

[ A ]t =

2

substituting into the equation,

[A]0 ln = − kt1 / 2 + ln [A ]0 2

[ A ]0 [ A ]0 Rearranging equation ln [ A ]0 − ln = kt 1 / 2 or ln [ A ] = kt 1 / 2 0 2 2 or simply ln 2 = k t1/2

or

t1/ 2

0.693 = k

5.2.2

Second order of reaction

1. A second-order reaction is a reaction whose rate depends on the concentration of one reactant raised to the second power or on the concentrations of two different reactants, each raised to the first power. A + B → products Rate equation for a second order reaction can be either : rate = k [A]2 or rate = k [B]2 or rate = k [A][B]

d[A] rate = − dt or it can be written as

rate = k [A]2 When the rate is substituted between each other, it can be expressed as d [A] d[A] 2 k [A] = − which can be rearrange as k dt = − dt [A] 2 d [A ] 1 Integration of the equation

−∫

[A ]

2

= k ∫dt →

[A ]

= kt + c

c is a constant which can be determined when time, t = 0, c = 1/[A]. Usually, it is indicated as 1/[A]0, since it is the initial concentration of the reaction:

1 1 = kt + [ A ]t [ A ]0

Graph in second order reaction

Graph of half life in second order reaction

The graph above clearly shows how does the half life occur for a second order of reaction, where times taken for each stage of half life : 1 1 1 [A]0 1  → [ A ]    → [ A ]    → [A]0 0 0 2 4 8 min 2 min 4 min

For a second order of reaction : Since 1st t1/2 = 1 min 2nd t1/2 = 2 x 1st t1/2 = 2 (1 min) = 2 min 3rd t1/2 = 2 x 2nd t1/2 = 2 (2 min) = 4 min As for 4th t1/2 = 2 x 3rd t1/2 = 2 (4 min) = 8 min b) Similar to first order of reaction, the equation of the half life in second order of reaction can be derived from the second order equation. At t1/2 ; [ A] = [ A]0 t 2 1 1

[ A]0 2

−

[ A]0

= k t1 / 2

1 = k t1 / 2 [ A]0

→

t1 / 2

1 = k [ A]o

Zero order reaction 1. First- and second-order reactions are the most common reaction types. Reactions whose order is zero are rare. For a zero-order reaction, where A → products rate equation : rate = k [A]0 rate of reaction can be expressed using the equations below,

d[A] rate = − or it can be written as dt

rate = k

When the rate is substituted between each other, it can be expressed as d[A] k= − which can be rearrange as − k t = d[A] dt Integration of equation ∫ d [ A ] = − k ∫ d t → [ A ] = − kt + c

[A]t = - kt + [A]0

Graph Zero order reaction

3. For a zero order of reaction, since the graph of concentration against time is linear, therefore, it is impractical to determine the half-life of zero order using practical methods. In fact, the linear shape of the graph itself has proven that the reaction is zero order with respect to the reactant involved. a) Though, the half-life of zero order reaction can be calculated using equation method, where the equation is derived from the zero order equation During half life, t1/2, occur ; [ A] = [ A]0 t

when [A]t is substitute in equation

[A]0 [A]0 − = k t1 / 2 2

⇒

2

t1 / 2

[A]0 = 2k

1. A + B
C The reaction above is first order with respect to A and zero order with respect to B. the initial concentration of A and B are 1.50 mol dm-3 and 2.00 mol dm-3 respectively & the initial reaction rate is 6.70 x 10-5 mol dm-3 s-1. rate = k [A]1 [B]0 @ rate = k [A] rate constant, k = rate / [A] k = 6.70 x 10-5 mol dm-3 s-1/ 1.50 mol dm-3 k = 4.47 x 10-5 s-1 c) Using 1st order equation, [A]t at 1 hour is calculated ln [A]t = – k t + ln [A]0 => ln [A]t = - 4.47 x 10-5 x 3600 + ln 1.50 [A]t = 1.28 mol dm-3 Then use rate = k [A] ; rate = 4.47 x 10-5 x 1.28 rate = 5.71 x 10-5 mol dm-3

a) b)

2. A dilute solution of hydrogen peroxide can be used to bleach hair. It decomposes slowly in aqueous solution according to the following equation: 2 H2O2 (aq)
2 H2O (l) + O2(g) A solution with an original concentration of 3.0 mol dm-3 was placed in a bottle contaminated with transition metal ions, which act as catalysts for the decomposition. The rate of decomposition was measured by withdrawing 10 cm3 portions at various times and titrating with acidified 0.10 mol dm-3 KMnO4 (aq). (5 mol of peroxide react with 2 moles of KMnO4.) The following results were obtained: Time / min V of 0.10 mol dm-3 KMnO4 / cm3

0

5

10

15

20

25

30

30.0

23.4

18.3

14.2

11.1

8.7

6.8

Plot a graph of volume of KMnO4 against time. Based on the graph sketch, determine the order of reaction with respect to H2O2, express the rate equation for the reaction and calculate the rate constant of the reaction.

30

25

20

15

10

5

0 0

5

10

15

20

25

30

From the graph, since first t1/2 is 13.5 min, while second t1/2 is also 13.5 min, therefore, the reaction is first order with respect to H2O2. The rate equation can be expressed as : rate = k [H2O2] rate constant can be calculated using :

t1/ 2

0.693 0.693 = ; k= k 13.5

k = 0.0513 min

−1

3. Substance B decomposed according to the following equation 2B→C+2D The concentrations of B were determined at certain interval time and the graph of 1/[B] against time was obtained 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0

50

100

150

200

250

300

a) Determine the order of reaction with respect to B. Second order of reaction FFFFFFFFFFFFFFFFFFFFFFFFFFFFFF b) From the graph, determine i. the initial concentration At t = 0 ; 1/[B] = 1.0 ; [B] = 1.0 mol dm-3 ii.

rate constant of reaction Using any two point to form tangent Rate constant, k = 5.6 – 3.2 / 200 – 100 = 0.024 mol-1 dm3 s-1

c) Write the rate equation for the reaction Rate = k [B]2

4. Trichloromethane, CHCl3 reacts with sodium hydroxide solution as represented by the following equation: 2 CHCl3 + 7 OH– → CO + HCOO– + 6 Cl– + 4 H2O The reaction is first order with respect to each reactant. a) Write the rate equation for the reaction above. Rate = k [CHCl3] [OH–] FFFFFFFFFFFFFFFFFFFFFFFFFFFFFF b) Determine the rate of production of chloride ions at 28 °C when the rate of loss of trichloromethane is 2.0 x 10–5 mol s–1.

1 d[CHCl3 ] 1 rate = − ; so, rate = − × 2.0 × 10− 5 2 dt 2 rate = 1.0 × 10− 5 − 1 d[Cl − ] 1 d [ Cl ] −5 rate = ; so, 1.0 × 10 = − × 6 dt 6 dt d[Cl − ] = 6.0 × 10− 5 dt

c) If the rate of reaction is r when the concentrations of both trichloromethane and sodium hydroxide are both 2.0 mol dm–3, what is the rate of reaction in terms of r when half of the hydroxide ion is reacted?

2 CHCl3 + 7 OH-
products Initial conc. 2 2 Final conc. 2 – (2/7 x 1) = 12 /7 2 – (2 x ½) = 1 Rate, r = k [CHCl3][OH–] ; r = k [12/7 CHCl3] [1 OH–] r = 12 / 7 k [CHCl3] [OH–] So, rate is 12 / 7 r d) Sketch a graph of the rate of reaction against the concentration of trichloromethane if sodium hydroxide is in excess such that the hydroxide ion concentration remains practically constant in the reaction mixture. rate

[CHCl3]

5.2.4 Determination of order of reaction via experiment 1. Consider the reaction between oxygen and nitrogen monoxide, a key step in the formation of acid rain and in the industrial production of nitric acid . O2 (g) + 2 NO (g) → 2 NO2 (g) The rate equation, expressed in general form, is rate = k [O2]x [NO]y Note that the order of reaction cannot be determined directly from the stoichiometry of the reaction. To find out the orders of reactant with respect to each O2 and NO, we run series of experiments, starting each one with a different set of reactant concentrations and obtaining an initial rate in each case.

Exp 1 2 3 4

Initial concentration of reactant O2 / mol dm-3 NO / mol dm-3 1.10 x 10-2 2.50 x 10-2 2.20 x 10-2 2.50 x 10-2 1.10 x 10-2 5.00 x 10-2 3.30 x 10-2 7.50 x 10-2

Initial rate (mol dm-3 s-1) 2.40 x 10-3 4.80 x 10-3 9.60 x 10-3 x

From each experiment, the rate equations are expressed individually, where Experiment 1 : 2.40 x 10-3 = k (1.10 x 10-2)x (2.50 x 10-2)y Experiment 2 : 4.80 x 10-3 = k (2.20 x 10-2)x (2.50 x 10-2)y Experiment 3 : 9.60 x 10-3 = k (1.10 x 10-2)x (5.00 x 10-2)y Experiment 4 : x = k (3.30 x 10-2)x (7.50 x 10-2)y

Comparing Experiment 2 to Experiment 1 : 4.80 x 10-3 = k (2.20 x 10-2)x (2.50 x 10-2)y 2.40 x 10-3 = k (1.10 x 10-2)x (2.50 x 10-2)y 2 = (2)x Order of reaction with respect to O2 ; x = 1 Comparing Experiment 3 to Experiment 1 : 9.60 x 10-3 = k (1.10 x 10-2)x (5.00 x 10-2)y 2.40 x 10-3 = k (1.10 x 10-2)x (2.50 x 10-2)y 4 = (2)y Order of reaction with respect to NO ; y = 2 From the order of reaction deduced, the rate equation is rate = k [O2][NO]2. The overall order of reaction = 1 + 2 = 3

Using any experiment, rate constant can be calculated. For example, in experiment 1 2.40 x 10-3 = k (1.10 x 10-2) (2.50 x 10-2)2 k = 349 mol-2 dm6 s-1. Once the order of reaction and the rate constant were determined, we can predict the rate of reaction under any concentration of reactants used. For example, in Experiment 4, where rate = k [O2][NO]2 ; rate = 349 (3.30 x 10-2) (7.50 x 10-2)2 rate = 6.48 x 10-2 mol dm-3 s-1

2. Sometimes, using a combination of graphical methods and initial rate methods, the order of reaction can be found individually. For example, in hydrolysis of bromoethane with potassium hydroxide at different concentrations. CH3CH2Br (l) + KOH (aq) → CH3CH2OH (l) + KBr (aq) The following results were obtained from two experiments on such a hydrolysis. In each experiment, the overall [KOH(aq)] remained virtually constant at the value given at the top of the column. time /min

[CH3CH2Br] / mol dm-3 when [KOH] = 0.10 mol dm-3

[CH3CH2Br]/mol dm-3 when [KOH-] = 0.15 mol dm-3

0

0.0100

0.0100

40

0.0079

0.0070

80

0.0062

0.0050

120

0.0049

0.0034

160

0.0038

0.0025

200

0.0030

0.0017

240

0.0024

0.0012

0.01

0.008

0.006

0.004

0.002

0 0

50

100

150

200

250

a) From the reaction, the rate equation can be written as rate = k [CH3CH2Br]x [KOH]y Therefore, to determine the order of reaction with respect to CH3CH2Br, the half-life method is applied, since the graph of concentration of CH3CH2Br against time are plotted accordingly, while the order of reaction with respect to KOH can be obtained using initial rate methods, where the initial rate for both [CH3CH2Br] under different concentration of KOH can be found hence calculated. b) By the mean of half-life method, the first t1/2, second t1/2 and third t1/2 of the concentration of CH3CH2Br under the concentration of KOH 0.15 mol dm-3 occurred at 80 s, 160 s and 240 s. Since the 1st t1/2 = 2nd t1/2 = 3rd t1/2 = 80 s, therefore the order of reaction with respect to CH3CH2Br is first order of reaction.

d) From both methods applied, the rate equation can be written as rate = k [CH3CH2Br] [KOH] The overall order of reaction is 1 + 1 = 2 e) Using any experiment above, rate constant of the reaction can be calculated. In experiment 2 : 1.00 x 10-4 = k (0.0100) (0.15) k = 0.0667 mol-1 dm3 s-1

5.2.5

Reaction Mechanisms

1. Chemical reactions may occur in one way reaction or a reversible reaction. Example of a one way reaction, is the formation of nitrogen dioxide via the reaction of nitrogen monoxide and oxygen gas 2 NO (g) + O2 (g) 2 NO2 (g) (symbolised one way reaction) While the example of a reversible reaction as in production of ammonia from nitrogen gas and hydrogen gas, which is largely used in industrial process via Haber process. N2 (g) + 3 H2 (g) 2 NH3 (g) (symbolised reversible reaction)

2. In a one way reaction, process may be taken in multiple steps, and therefore irreversible back to the reactants, as they might involve in steps that required higher activation energies. For example, in the reaction stated above where 2 NO (g) + O2 (g) 2 NO2 (g) The steps (or simply mechanism) for the reaction of nitrogen monoxide and oxygen to form nitrogen dioxide can be described below. Step 1

:

2 NO slow  → N 2 O 2

Step 2

:

N 2 O 2 + O 2 fast → 2 NO 2

Overall :

2 NO + O2

2 NO2

a) In the process of the formation of nitrogen dioxide, N2O2 is formed temporary, but it will not exist as a product in the end of reaction. Therefore, N2O2 is also known as intermediate, a substance that appear in the mechanism of the reaction but not in the overall balanced equation. b) For each step of reaction, the rate equation can be described accordingly Mechanism

Step 1 Step 2

Reaction Equation

Rate equation

2 NO slow  → N 2 O 2

rate = k [NO]2

N 2 O 2 + O 2 fast → 2 NO 2

rate = k [N2O2] [O2]

From the rate equation proposed for each step in the series of mechanism proposed the order of reaction can be determined straight forward by the stoichiometry coefficient. From the mechanism equation in step 1, order of reaction is second order with respect to NO, while in step 2, order of reaction is first order with respect to each N2O2 and O2 respectively.

c) Note that in the series of step, above the arrow is written with the word "slow" and "fast", which can be interpreted as a slow step of reaction, and a fast step of reaction. In determining the order with respect of each reactant involved, we can make use the mechanism to determine the order of each reactant, since slow step is the rate determining step. Therefore, the rate equation that can represent the overall equation 2 NO + O2 2 NO2 rate = k [NO]2.

5.3

The effect of temperature on reaction kinetics

1. Temperature often has a major effect on reaction rate. In general, increasing the temperature of a reaction increases the average speed of particles and therefore their frequency of collisions. a) Arrhenius proposed that every reaction has an energy threshold that the colliding molecules must exceed in order to react. This minimum collision energy is the activation energy (EA), the energy required to activate the molecules into a state from which reactant bonds can change into product bonds. b) Many of the chemical reactions near room temperature approximately double their rates with a 100C rise in temperature.

2. The effect of temperature towards the rate of reaction can be further explained using Maxwell-Boltzmann distribution graph.

From the graph obtained, particles at 800 K have move collision energy compared to particles at 300 K. This is due to, as the kinetic energy increased, particles moved faster and more particles collides more frequently hence increased the collision energy. Therefore, more particles have energy greater than activation energy, and results higher rate of reaction.

3. Generally, the rate of reaction increased with temperature as it affect the rate by increased the rate constant of the reaction. The dependence of the rate constant of a reaction on temperature can be expressed by using Arrhenius equation.

k = Ae

EA − RT

k = rate constant A = Arrhenius constant T = temperature EA = activation energy R = gas constant (8.31 J mol-1 K-1)

Arrhenius equation shows that the rate constant is directly proportional to A and, therefore, to the collision frequency. In addition, because of the minus sign associated with the exponent EA/RT, the rate constant decreases with increasing activation energy and increases with increasing temperature

Derivation of Arrhenius Equation. a) Arrhenius equation – expressed as natural logarithm of both sides, where the equation can be expressed as :-

EA ln k = ln A − ; RT

EA 1 rearrange : ln k = − + ln A R T

From the rearranged equation, if a graph of ln k against 1/T is plotted, a negative gradient linear line may be obtained, where the gradient, m = EA / R. Therefore, using this method, the activation energy of a reaction can be calculated

Example : The table below shows how the rate constant, k, varies with the temperature for the reaction between H2 (g) and I2 (g)

H2 (g) + I2 (g)
2 HI (g) Temperature (K) Rate constant, k (mol-1 dm3 s-1)

556

575

647

700

791

3.52 x 10-7 1.22 x 10-6 8.59 x 10-5 1.16 x 10-3 3.90 x 10-2

By plotting a suitable graph, determine the activation energy for the reaction. Solution : A table of 1/T and ln k is first calculated and placed accordingly.

1 / T (K-1) ln k

0.00180

0.00174

0.00155

0.00143

0.00126

-14.9

-13.6

-9.36

-6.76

-3.24

− 13.6 − (−9.36) From the graph , gradient = = − 22316 0.00174 − 0.00155 EA 8.31 E A = + 185, 444 @ + 185 kJ mol −1

− 22316 = −

0.0 0.00100 0.00110 0.00120 0.00130 0.00140 0.00150 0.00160 0.00170 0.00180 0.00190

-2.0

-4.0

-6.0

-8.0

-10.0

-12.0

-14.0

-16.0

b) An equation relating the rate constants k1 and k2 at temperatures T1 and T2 can be used to calculate the activation energy or to find the rate constant at another temperature if the activation energy is known. E E At T1 ln k1 = − A + ln A ; At T2 ln k 2 = − A + ln A RT1 RT2 Subtracting both equation at two different temperatures above

ln k1 − ln k 2

 EA   EA  =  − + ln A  −  − + ln A   RT1   RT2 

k1 EA  1 1   ln = − k2 R  T2 T1 

;

Example 2 : The rate constant of a first-order reaction is 3.46 x 10-2 s-1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? Solution :

k1 EA  1 1   ln = − k2 R  T2 T1 

3.46 × 10 ln k2

−2

k2

50.2 × 10  1 1  = −   8.31 298   350 3

=

0.703 s-1

5.4

The role of catalysts in reactions

1. A catalyst is a substance which alter the rate of reaction without changing its chemical composition. Therefore the chemical formula will remain the same after the reaction occur. In most of the chemical reactions, a catalyst is usually added to speed up the reaction (increase the rate of reaction), however, the quantity of catalyst used was only in a small amount, since large amount of catalyst does not significantly increase the rate of reaction. 2. A catalyst works by providing an alternative pathway for a chemical reaction to take place, which required a lower activation energy. However, it will not affect the enthalpy change (∆H) of a chemical reaction.

Endothermic process Energy / kJ

Exothermic process Energy / kJ

Reaction coordinate

Reaction coordinate

3. As proposed by Arrhenius' equation,

k = Ae

EA − RT

when the activation energy, EA, of a chemical reaction decreased, the rate constant of the reaction increased, therefore increased the rate of reaction. However a catalyst do not initiate the reaction, rather it accelerate the reaction that is already occurring

5.4.1

Autocatalysis

1. If the product of a reaction itself acts as a catalyst for the reaction, the product is also known as autocatalyst. For example, the reaction between manganate (VII) ions, MnO4- and ethanedioate ions, C2O42- in the presence of sulphuric acid, H2SO4 : 2 MnO4- + 5 C2O42- + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2O Table below shows the observation from the first drop of KMnO4 is added slowly to until the end of reaction No of drop of KMnO4 First two drops

Observation Purple colour of KMnO4 decolourised slowly

Following drops

Purple colour of KMnO4 decolourised more rapidly.

Last few drops before reaction end

Purple colour of KMnO4 decoloursied less rapidly and eventually become slower until it no longer decolourised.

5.8 Application of catalysis in industries
Catalysis can be categorised into 2 types, namely
heterogeneous catalyst • homogeneous catalyst 5.8.1 Heterogeneous catalyst
A heterogeneous catalyst is a catalyst which has different phase with reactants. Usually it was between a solid catalyst that is used to catalyse between a gaseous or liquid reactants. We shall study 4 specific examples of heterogeneous catalyst, which are Haber Process, Ostwald Process, Contact Process and Catalytic converter use in automobile exhaust.

5.8.1.1 Haber Process
Ammonia is an extremely valuable inorganic substance used in the fertilizer industry, the manufacture of explosives, and many other applications
The main ingredients use to synthesis ammonia are nitrogen (which can be obtained through fractional distillation of liquefied air) and hydrogen (which can be obtained either from syn gas [C + H2O] or petroleum refining process)


In heterogeneous catalysis, the surface of the solid catalyst is usually the site of the

reaction. The initial step in the Haber process involves the dissociation of N2 and H2 on the metal surface.
Although the dissociated species are not truly free atoms because they are bonded to the metal surface, they are highly reactive.
The highly reactive N and H atoms combine rapidly at high temperatures to produce NH3 molecules

5.8.1.2 Ostwald Process
Nitric acid is one of the most important inorganic acids. It is used in the production of fertilizers, dyes, drugs, and explosives. The major industrial method of producing nitric acid is the Ostwald process. The starting materials, ammonia and molecular oxygen, are heated in the presence of a platinum-rhodium catalyst to 8500C Step 1
This step is the crucial step as it will determine the yield of nitric acid formed. The rest of the steps do not require catalysis and will occur at high temperature
Step 2 :The nitric oxide readily oxidizes (without catalysis) to nitrogen dioxide: 2 NO(g) + O2 (g) → 2 NO2 (g)
Step 3 :When dissolved in water, NO2 forms both nitrous acid and nitric acid: 2 NO2 (g) + H2O (l) → HNO2 (aq) + HNO3 (aq)
On heating, nitrous acid (HNO2) is converted to nitric acid as follows: 3 HNO2 (aq) → HNO3 (aq) + H2O (l) + 2 NO (g)
The NO generated can be recycled to produce NO2 in the second step

5.8.1.3 Contact Process
Sulphuric acid is one of the most widely use inorganic acids. Contact process is still preferable, even today, to synthesise high concentration of sulphuric acid. The following are steps in manufacturing sulphuric acid, starting from heating sulphur with oxygen. Step 1 : S (g) + O2 (g) → SO2 (g)
After sulphur dioxide is formed and filtered, it was further oxidised to form sulphur trioxide, using vanadium (V) oxide, V2O5, as catalyst. This step is crucial as it will influence the amount of H2SO4 formed. Step 2
V2O5 catalyst serve as the active site and provide an alternative solution for

the formation of SO3.
Alternative Step 1 : 2 SO2 + 4V5+ + 2 O2- → 2 SO3 + 4V4+ (Oxidation of SO2 into SO3 by V5+)
Alternative Step 2 : 4 V4+ + O2 → 4 V5+ + 2 O2- (Oxidation of V4+ back into V5+ by oxygen - catalyst regenerate)


Hot sulphur trioxide passes through the heat exchanger and is

dissolved in concentrated H2SO4 to form oleum Step 3 : H2SO4 (l) + SO3 (g) → H2S2O7 (l)
Oleum is reacted with water to form concentrated H2SO4 Step 4 : H2S2O7 (l) + H2O (l) → 2 H2SO4 (l)
The yield of sulphuric acid solution formed is around 30 - 40%. The unreacted suphur dioxide and sulphur trioxide is then further treated in another recycle chamber named as DCDA (Double Contact Double Absorption) Chamber. Through this method, the yield will be maximised to nearly 99.8% of H2SO4

5.8.1.4 Catalytic converter
At high temperatures inside a running car’s engine, nitrogen and oxygen gases react to form nitric oxide
N2 (g) + O2 (g) → 2 NO (g) *Note that this reaction occur only when the engine is very hot. This is due to nitrogen gas, NΞN has a short and strong triple bond, with a high bond energy. So, high amount of heat is required to break the bond. Another phenomenon which caused the same reaction are when air is surrounded by lightning.
When released into the atmosphere, NO rapidly combines with O2 to form NO2. Nitrogen dioxide and other gases emitted by an automobile, such as carbon monoxide (CO) and various unburned hydrocarbons (CxHy), make automobile exhaust a major source of air pollution
To overcome these problem, most of cars nowadays are equipped with catalytic converter, which contain platinum-rhodium catalyst and copper (II) oxide + chromium (III) oxide as co-catalyst


An efficient catalytic converter serves two purposes: It oxidizes CO and

unburned hydrocarbons to CO2 and H2O, and it reduces NO and NO2 to N2 and O2. Oxidation : CO (g) + 1/2 O2 (g) → CO2 (g) Oxidation : CxHy + (x + y/4) O2 (g) → x CO2 (g) + y/2 H2O (g) Reduction : NOx (g) → 1/2 N2 (g) + x/2 O2 (g)
The suitable catalyst use is platinum / rhodium (use to oxidise CO and CxHy) based catalyst doped with copper (II) oxide or chromium (III) oxide (use to reduce NOx). Because the catalyst serve these 3 purposes, sometime it is also referred as three-way catalyst

5.8.2 Homogeneous catalyst.
In homogeneous catalysis the reactants and catalyst are dispersed in a single phase.
Acid and base catalyses are the most important types of homogeneous catalysis in liquid solutions
For example, in the hydrolysis of ester


The rate equation of the reaction above is written as
However, this reaction can be catalysed by the addition of hydrogen ion (H+)

from an acidic substance for example, hydrochloric acid or sulphuric acid, where now, the rate equation can be written as rate = k [CH3COOCH2CH3] [H+]
By this, the rate of hydrolysis of ester can be increased by the addition of acidic substance, which may caused faster reaction, without increasing the concentration of ester.


Other than this, homogeneous catalyst can also be exemplified by the

reaction between disulphate ion, (S2O82-) and iodide ion, (I-) where iron (III) ion here can be added as catalyst to increase the rate of reaction. Equation : S2O82- (aq) + 2 I- (aq)
I2 (aq) + 2 SO42- (aq)
The reaction can be catalysed using aqueous iron (III) ion. Step 1 : oxidation of iodide ion by iron (III) ion [catalysed] 2 Fe3+ (aq) + 2 I- (aq)
2 Fe2+ (aq) + I2 (aq)

Step 2 : Oxidation of iron (II) ion back to iron (III) ion by persulphate ion. 2 Fe2+ (aq) + S2O82- (aq)
2 Fe3+ (aq) + 2 SO42- (aq)


Homogeneous catalysis can also take place in the gas phase. This can be

exemplified by the reaction from SO2 to SO3, which is one of the major pollutant in air. SO2 in the air are mainly released by the fumes of volcanic activities. However, recent papers had reported that SO2 emission could also be mainly contributed by the combustion of diesel oil, mining activities (sulphide based ores), and through various chemical industries processes such as Contact Process.
SO2 can be oxidised in air under the presence of nitrogen dioxide according to the equation below
2 SO2 (g) + 2 NO2 (g) → 2 SO3 (g) + 2 NO (g)
2 NO (g) + O2 (g) → 2 NO2 (g)
Overall : 2 SO2 (g) + O2 (g)
2 SO3 (g)
SO3 produced are hygroscopic, hence they react easily with water droplet or rain to form corrosive acidic rain.

5.9 Enzymes as Biological Catalysts
Chemical reactions that occur in our bodies are speeded up by enzymes, which act as biological cataysts.
Enzymes are the largest and most highly specialised class of proteins and are produced by living cells from amino acids. Its molecular weight varies from 12,000 to over 1 million.
Enzymes work under mild conditions and often give 100% yields and may speed a reaction by 106 or 1012 times.
Some enzymes require the presence of metal ions as cofactors, and these are called metalloenzymes. Many but not all metalloenzymes contain a transition element

Type of Enzyme

Metalloenzyme

Function

Arginase

Mn2+

Urea formation

Carboxypeptidase

Zn2+

Digestion of proteins

Ferredoxin

Fe2+ or Fe3+

Glutamic mutase

Co

Nitrogenase

Fe and Mo

Nitrogen fixation

Tyrosinase

Cu+ or Cu2+

Skin pigmentation

Photosynthesis Metabolism of amino acids


Compared to inorganic catalysts, enzymes are specific'in their actions.

Each enzyme catalyses only one type of reaction whereas platinum catalyses several reactions.
In the lock-and-key theory, the active site of the enzyme conforms exactly to the substance molecule. This specificity results from the fact that enzymes are formed from L-amino acids and therefore the active sites are asymmetrical.


Factors affecting enzyme activity:
Temperature - Most enzymes have their highest activity at temperatures from 35°C to 45°C. Above this range, the enzymes start to denature and the reaction rate decreases. Above 80°C, enzymes are permanently denatured.
pH - The structure and geometry of an enzyme's active site changes when the pH of the surrounding medium changes. For example, trypsin (which is active in the small intestine) has its maximum activity at pH 8 whereas pepsin (which is active in the stomach) has an optimum pH of 1.5.
Solvents and salt concentrations can also change the structure of a

protein and the activity levels of enzymes.
The activity of enzymes can be inhibited by heavy metals as mercury, lead and silver. These metals are toxic because they bind irreversibly with free sulphydryl (-SH) functional groups on enzymes, which are then not available to bind with the necessary cofactor.
Compounds in nerve gases combine with the hydroxyl (-OH) functional group on enzymes and cause the enzymes to lose their ability to catalyse a reaction. This is why animals poisoned by nerve gas become paralysed.