Chemistry Form 6 Chap 05 New

CHAPTER 5 KINETIC CHEMISTRY

Chemical kinetic ~ study of rate of reaction and the factor which affect them and the mechanism by which the chemical reaction occurs.
Rate of reaction ~ change of concentration of a reactant or a product per unit time
When reactants  products


Re action rate = −

d [reac tan ts] d [products] = (unit = mol dm −3 s −1 ) dt dt

Example 1


aA (aq) + bB (aq) → cC (aq) + dD (g)





In terms of reactant A, rate = – 1 d [A] a dt In terms of reactant B, rate = – 1 d [B] b dt In terms of product C, rate = 1 d [C] c dt In terms of product D, rate = 1 d [D] d dt

Example 2


2 P (aq) + 3 Q (aq) → 5 S (aq) + 4 T (aq)





In terms of reactant P, rate = – 1 d [P] 2 dt In terms of reactant Q, rate = – 1 d [Q] 3 dt In terms of product S, rate = 1 d [S] 5 dt In terms of product T, rate = 1 d [T] 4 dt

Graph of concentration of product against time [product] / mol dm-3

time / s

Graph of concentration of reactant against time [reactant] / mol dm-3

time / s

Important notes From the 2 graphs, both gradients decrease with time (become less steep), showing that the rate of reaction is decreasing with the time taken for the reaction to complete
When gradient = 0, rate of reaction is zero and the reaction has complete
When time, t = 0 ; the gradient (rate of reaction) of the reactant is the initial rate of reaction.


Example 1 A student investigated the rate of reaction between sodium and ethanol. A freshly cut piece of sodium was weighed and added to a large excess of ethanol. The total volume of gas liberated was recorded every minute. The results are tabulated below







time / min

0

total volume of gas / cm3

0

1

2

3

4

5

6

7

8

23.0 36.5 46.0 51.0 54.5 57.0 58.5 58.5

Plot the graph of volume of gas against time Find the overall rate of reaction Calculate the rate of reaction at 4th minute and 6th minute Calculate the initial rate of reaction

Overall rate of reaction = (58.5 – 0.0) / (7 – 0) = 8.36 cm3 / min Rate of reaction in the 4th min = (55 – 47) / (4.8 – 3.0) = 4.44 cm3 / min Rate of reaction in the 6th min = (59 – 55) / (7.0 – 5.0) = 2.00 cm3 / min Initial rate of reaction, instantaneous rate at t = 0 Rate of reaction in 0 min = (35.0 – 0.0) / (1.0 – 0.0) = 35.0 cm3 / min

5. 2 Theory of Reaction Rates


The kinetic molecular theory of gases, gas molecules frequently collide with one another. Therefore, it seems logical to assume that chemical reactions occur as a result of collisions between reacting molecules. In terms of the collision theory of chemical kinetics, then, we expect the rate of a reaction to be directly proportional to the number of molecular collisions per second, or to the frequency of molecular collisions rate of reaction ∞




There are 2 theories, the collision theories and transition state theory, to explain how the rate of reaction change with concentration, temperature or the presence of catalyst

8.2 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 II..IIIIIIIIII energy required to initiate a chemical reaction IIIIIIIIIIIIIIIIIIIIIIIIII. 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 A–B + C  A + B–C is a one step reaction where it can be described as A–B + C  [A---B---C]  A + B–C




In some chemical reaction, it may involve more than 1 steps. Example in the hydrolysis of 2-bromo-2-methylpropane.

(CH3)3C–Br + OH–  (CH3)3C–OH + Br– Step 1 : (CH3)3C–Br  (CH3)3C+ + Br– Step 2 : (CH3)3C+ + OH–  (CH3)3C–OH

(CH3)3C–Br (CH3)3C+ (CH3)3C–OH

Rate Laws and Rate Equation









Consider the chemical reaction below where x A (aq) + y B (aq) → z C (aq) + w D (s) The dependence of the reaction rate on the concentration of each reactant is given by an equation called the rate law, where the rate equation can be written as rate = k [A]m[B]n ; k = rate constant m, n = reaction order Rate constant, k is the proportionality constant in the rate equation The larger the value of k ; greater the rate of reaction ; faster the reaction
Value of k remain constant no matter how the concentration of reactants @ products changed
k will only change with temperature or the addition of catalyst







m and n in the rate equation is the order of reaction, where it is the sum of power to which the concentration of the reactants is raised to in the rate equation
The value of m and n can only be determined experimentally.
There’s no relationship between the order of reaction and stoichiometric coefficient, so
m≠ x ; n ≠ y
The overall order of reaction is the summation of the respective order for each reactant involve Overall reaction = m + n
Given the order of reaction with respect to A = 1st order ; B = 2nd order, the overall order = 1 + 2 = 3 The unit of rate constant, k depend on the overall rate of reaction.

The Unit of Rate Constant Overall reaction order

Example

Zero order

rate = k [A]0

mol dm-3 s-1

First order

rate = k [A]

s-1

Second order

rate = k [A][B]

mol-1 dm3 s-1

Third order

Rate = k [A][B]2

mol-2 dm6 s-1

Fifth order

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

mol-4 dm12 s-1

Unit of rate constant, k

First Order of Reaction



rate depends on the concentration of a single reactant raised to the first power. Example : A (aq)  product

d [A] rate = − and dt −

d [ A] = k dt [ A]

rate = k [A ]

integration

so

d[A ] − = k [A] dt

d [A] ∫ [A] = − k ∫ d t → ln [A] = − kt + c

At time, t = 0 ; so c = ln [A]0.

when substitute into the eq.

ln [A]t = – kt + ln [A]0

Graph of rate against concentration [A] Rate / s-1

Rate ∝ [A]

[A] / mol dm-3

Graph of ln [A] against t ln [A] / mol dm-3 s-1 [A]0

gradient = k

time / s

Graph of rate against concentration [A] ln [A] 0 / [A]

gradient = k

Time / s

Half life in 1st order of reaction


In the graph of concentration against time, half life is defined t½ ~ Time required to decrease the concentration by half

[ A]0



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

In 1st order of reaction, the 1 t½ = 2 t ½ = 3 t ½ If the order of reaction is known, half life can also be deduced from the equation based on the order of reaction ln [A]0 - ln [A]t = k t since at t½ ; [A]t = [A]0 / 2 so replace => ln [A]0 - ln [A]0 / 2 = k t ½ ln 2 = k t½ Equation of half-life in 1st order :

t½ = 0.693 / k

Second order of reaction


rate depends on the concentration of a single reactant raised to the power of two or on the concentrations of two different reactants, each raised to the first power
rate = k [A]2 or rate = k [A] [B]

rate = −


d [ A] − = k dt 2 [ A] A]

Integrating the equation will make it become −∫




d [ A] = k [ A] 2 dt

d [ A] = k ∫dt [ A] 2

At time t = 0 and [A] become [A]0

∴ At time t , [ A] = [ A]t

1 [ A]

= kt + C C=

1 [ A]0

1 1 − =kt [ A]t [ A]0

Graph of 2nd order Graph of rate against [A]2 Rate / mol-1 dm3 s-1

Graph of 1/[A] against time 1/[A] / mol-1 dm3

Rate ∝ [A]2

[A]2 / mol2 dm-6

Time / s

Half life in 2nd order by graph When the concentration decrease by half each time
The time taken for each stage is the same, where Half life ; 1st t1/2 ; 2nd t1/2 = 2 ( 1st t1/2 ) 3rd t1/2 = 2 ( 2nd t1/2 ) = 4 ( 1st t1/2 )
Example


[ A]0 10→s [ A]0 → [ A]0 → [ A]0 150s → [ A]0 30 s 70 s 1 2

1 4

1 8

1 16

Equation for half life of 2nd order


By using equation, the half time can be deduced where at t1/2

[ A]0 [ A]t = 2


When [A]t is substituted into the 2nd order equation

1 1 − = k t1 / 2 [ A]0 [ A]0 2


When the equation is derived, the equation is obtained

1 = k t1 / 2 [ A]0

→

t1 / 2

1 = k [ A]o

Zero order of reaction





A reaction is said to be zero order of reaction with respect to a particular reagent if the rate of reaction does not depend on the concentration of reagent Consider the following zero order of reaction : A  products

d [ A] rate = − = k [ A]0 dt


Integrate the equation above will become :

∫ d [ A] = − k ∫ d t



⇒

d [ A] − =k dt

⇒

[ A] = − kt + c

At time, t = 0, the concentration is consider as [A]0  c = [A]0 As a result, the overall equation for zero order of reaction [A]0 – [A]t = k t

a) Graph of rate against concentration

rate

concentration

b) Graph of concentration against time

Concentration

time

Half life for zero order of reaction


During half life, t1/2, occur when [ A] = [ A]0 t 2




[ A]0 [ A]0 = k t1 / 2 ⇒ t1 / 2 = So substitute in equation[ A]0 − 2 2k




Since the concentration is independent with the rate, so the half life of zero order of reaction cannot be determined graphically

8.5

Finding the rate of reaction

The order of reaction and the rate constant can be determined by using :
The initial rate method • The half-life method
The reaction rate method • Linear plot [1] The initial rate method to determine the order of reaction
A series of experiments are carried out, where the concentration of the reactant in each experiment is altered. As a result, the initial rate of each experiment may be different and the order or each reactant can be measured. AC
Rate equation is written as rate, x = k [A]





If doubling the concentration of [A] has no effect on the rate zero of reaction, the reaction is IIIIIII. with respect to A. This is because




If doubling the concentration of [A] doubled the rate of first reaction, the reaction is IIIII.... with respect to A. This is because




If doubling the concentration of [A] increased the rate of second reaction by a factor of four, the reaction is IIIIII. with respect to A. This is because

[2] Using successive half-lives to determine the order of reaction Through the graph of concentration of reactant against time, the order of reaction can be determined (a) For a first order reaction, the half-life is independent of the initial concentration. Thus 1st half life t1/2’ = 2nd half life, t1/2’’ = 3rd half life, t1/2’’’ (b) For a second order reactions, the half-life is inversely proportional to the initial concentration. Thus 1st half life t1/2’ = 2 x 1st half life (2nd half life) = 4 x 1st half life (3rd half life) (c) For zero order of reaction, the graph obtained should be a linear graph. (negative gradient)


Example 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)

Example

b) 1st t ½ = 22 min 2nd t ½ = 46 – 22 = 24 min Since 1st t ½ = 2nd t ½ So reaction is first order with respect to N2O5 c) From the t ½ ; Average = 22 + 24 / 2 = 23 min t ½ = 0.693 / k k = 0.693 / 23 min = 0.0301 min-1

Example 3 • The rate equation for the reaction ; P2 (g) + 2 X2 (g)  C (g) is rate = k [P2]2 [X2]. At a constant temperature, the rate of reaction is x. How does the rate of reaction change if : a) Only partial pressure of P2 is doubled. rate = k [2P2]2 [X2] ; rate = k 4 [P2]2 [X2] so, rate = 4 x b) The partial pressure of P2 and X2 are doubled rate = k [2P2]2 [2X2] ; rate = k 4[P2]2 2[X2] so, rate = 8 x c) The partial pressure of P2 is halved and X2 is tripled rate = k [P2/2]2 [3X2] ; rate = k ¼ [P2]2 3 [X2] so, rate = ¾ x d) The volume of the vessel is halved when V is halved, P is doubled rate = k [2P2]2 [2X2] ; rate = k 4[P2]2 2[X2] so, rate = 8 x e) The volume of vessel is doubled when V is doubled, P is halved rate = k [½ P2]2 [½ X2] ; rate = k ¼ [P2]2 ½ [X2] so, rate = 1/8 x

4.

The initial rate for the reaction, in a series of experiments NH4+ (aq) + NO2- (aq)  N2 (g) + 2 H2O (l) with different concentrations were recorded in the Table below.

Experiment 1 2 3 4

Initial concentration (mol dm-3) NH4+ NO20.12 0.10 0.18 0.10 0.12 0.050 0.48 X

Initial rate of reaction (mol dm-3 s-1) 3.6 x 10-6 5.4 x 10-6 1.8 x 10-6 7.2 x 10-6

a) Determine the order of reaction with respect to rate = k [NH4+]a [NO2-]b Experiment 1 : 3.6 x 10-6 = k (0.12)a (0.10)b Experiment 2 : 5.4 x 10-6 = k (0.18)a (0.10)b Experiment 3 : 1.8 x 10-6 = k (0.12)a (0.050)b i) NH4+ ii) NO2Exp 2 : 5.4 x 10-6 = k (0.18)a (0.10)b Exp 1 : 3.6 x 10-6 = k (0.12)a (0.10)b Exp 1 : 3.6 x 10-6 = k (0.12)a (0.10)b Exp 3 : 1.8 x 10-6 = k (0.12)a (0.050)b 5.4 x 10-6 = (0.18)a 3.6 x 10-6 = (0.10)b 3.6 x 10-6 = (0.12)a 1.8 x 10-6 = (0.050) b a =1 b=1

b) Write the rate equation and determine the rate constant of the reaction rate = k [NH4+] [NO2-] Use any Exp :- Exp 1 : 3.6 x 10-6 = k (0.12) (0.10) k = 3.0 x 10-4 mol-1 dm3 s-1 c) Determine the value of X in experiment 4 Exp 4 : 7.2 x 10-6 = 3.0 x 10-4 (0.48) (x) x = 0.050 mol dm-3

5(a) i. A bromoalkane, RBr, is hydrolysed by aqueous sodium hydroxide. R–Br (l) + NaOH (aq)  R–OH + NaBr (b) The following results were obtained from two experiments on such a hydrolysis. In each experiment, the overall [NaOH(aq)] remained virtually constant at the value given time /min

[RBr]/mol dm-3 when [OH-] = 0.10 mol dm-

[RBr]/mol dm-3 when [OH-] = 0.15 mol dm-3

3

0 40 80 120 160 200 240

i.

0.0100 0.0079 0.0062 0.0049 0.0038 0.0030 0.0024

0.0100 0.0070 0.0049 0.0034 0.0024 0.0017 0.0012

Plot these data on suitable axes and use your graphs to determine the following. [4] ii. Use the half-life method to deduce the order of reaction with respect to the bromoalkane. [2] iii. Use the initial rates method to deduce the order of reaction with respect to sodium hydroxide. [5] iv. Construct a rate equation for the reaction and use it to calculate a value for the rate constant. [3]

8.6

The effect of temperature and Arrhenius Equation




When temperature increase, the rate of reaction will also increase IIIIII




One of the reason is because when temperature increase, higher kinetic energy, thus can the molecules possessed III... faster move IIIIIII This will increase the frequency of effective collision causing the rate of reaction to increase. IIIIIIIII This can be further explained using Maxwell – Boltzmann distribution graph. Note the following detail of the graph.

Low energy




proportional to the total The area below the curve is IIIIII. number of molecules involved. Since the total number of molecules is the same at both temperatures, the area must same be the III.




The distribution of molecular speeds(energies) are not symmetrical. At a higher temperature, the peak of the graph, which represents the average molecular speed, moves to the right while the height of peak gets IIIII. lowered This means III there’s a wider spread of speeds (energies) at higher temperatures and the most probable speed increased as the temperature increased.




As a conclusion, as temperature increase, less molecule average speed travels at IIIIIIIIII while more molecules travel higher speed at IIIIIIIIII

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)

 The Arrhenius constant, A, represents the frequency of effective collision between the particles that occur each second. E −

 The term e

A

RT

is measuring the fraction of effective

collision that was resulted in a reaction




If the equation is integrated k = A e

−

EA RT

ln k = – Ea / RT + ln A


When compare to a linear equation => y = m x + c where y = ln k ; x = 1 / T ; m = – Ea / R ; c = ln A




So, when plot a graph of ln k against 1 / T, the activation energy of the reaction can be determined.

1/T / K-1(103)

Ea ( y1 − y 2 ) Gradient = − = R ( x1 − x 2 )

ln k

Temperature (K)

556

575

647

700

791

Rate constant, k (mol-1 dm3 s-1)

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

1 / T (1/K)

0.00180

0.00174

0.00155

0.00143

0.00126

ln k

-14.90

-13.60

-9.36

-6.76

-3.24

Derivation of Arrhenius Law Given 2 different temperatures with different rate constant, k. Activation Energy can also be calculated by the derived equation out from Arrhenius Equation At temperature T1 At temperature T2




EA 1 ln k1 = − + ln A R T1

EA 1 ln k 2 = − + ln A R T2

When minus between 2 temperatures

k1 EA  1 1  ln =−  −  k2 R  T1 T2 

8.7

Reaction Mechanism

In chemical reaction, some take place in a single step but most reactions take place in a sequence of steps.
Using rate of reaction, the reaction mechanisms can be deduce. Reaction mechanism is the sequence of bond making and bond breaking steps that occur during the conversion from reactants to products
Consider the following equation 2 G (aq) + J (s) → L (aq) + Q (g)
The above reaction takes series of elementary steps as shown in the mechanism below :
Step 1 : G (aq) + J (s) → Z (aq) + Q (g) [slow]
Step 2 : Z (aq) + G (aq) → L (aq) [fast]
Overall : 2 G (aq) + J (s) → L (aq) + Q (g)





In each elementary steps, stated [slow] and [fast] indicating how fast the reaction occur. If the elementary is a slow step, then this step is the rate determining step. This step determined the rate of the entire multi-step process. So, as in the example above, the rate equation for


2 G (aq) + J (aq) → L (aq) +

rate = k [G] [J]

Q (g)




The molecularity of each elementary steps can also be deduce.




[Molecularity ~ number of species taking part]




unimolecular bimolecular These steps may be IIIIIIIII , IIIIIIIIII trimolecular or IIIIIII








unimolecular steps involves one reactant only, in the A IIIIIIII elementary step A  B+C rate = k[A] bimolecular steps involves two reactants, in the A IIIIIIII elementary step D+E  F+G rate = k[D] [E] trimolecular A IIIIIIIII steps involves three reactants, in the elementary step H + I + J  K+L rate = k[H] [I] [J]

8.8

Effect of Catalyst

A catalyst is a substance which alter the rate of reaction without changing its chemical composition
It will remain chemically unchanged at the end of the reaction but may physically changes its phase.
In most of the reactions, a catalyst is added to speed up the reaction (increase the rate of reaction)
Catalyst are usually added in a small amount (large amount of catalyst does not significantly increase the rate of reaction, therefore it is unnecessary)
Catalyst does not affect the enthalpy change (∆H) of reaction as it will only lower the activation energy (EA) of the substance.
Catalyst generally increase the rate constant (proved by Arrhenius equation)
Catalyst do not initiate the reaction, rather it accelerate the reaction that is occur


Catalyst works by providing an alternative reaction pathway of a lower activation energy so that more molecules will have energy higher than the activation energy.
Basically, a catalyst is used to lower the activation energy. Provided the energy profile for a chemical reaction which is exothermic, as illustrated in the diagram below


Endothermic reaction Ea for uncatalysed reaction Ea for catalysed reaction

Exothermic reaction Ea for uncatalysed reaction Ea for catalysed reaction

Autocatalysis If a product of a reaction acts as a catalyst for the reaction, the product is called as autocatalyst.
The reaction between manganate (VII) ions, MnO4- and ethanedioate ions, C2O42- in the presence of sulphuric acid, H2SO4 :
2 MnO4- (aq) + 5 C2O42- (aq) + 16 H+ (aq) → 2 Mn2+ (aq) + 10 CO2 (g) + 8 H2O (l)
Observation :
1st drop of manganate (VII) solution => pink colour persist for some time before discharge
2nd drop of manganate (VII) solution => decolourisation become more rapid


Graph of concentration of reactant against time

Graph of concentration of product against time

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.