Category: Chemistry

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Ionisation of Weak Acids

We have already learnt that weak acids are partially dissociated in water and there is an equilibrium between the undissociated acid and its dissociated ions. Consider the ionisation of a weak monobasic acid HA in water.

Applying law of chemical equilibrium, the equilibrium constant K_c is given by the expression

…………. (8.9)

The square brackets, as usual, represent the concentrations of the respective species in moles per litre. In dilute solutions, water is present in large excess and hence, its concentration may be taken as constant say K. Further H₃O⁺ indicates that hydrogen ion is hydrated, for simplicity it may be replaced by H⁺. The above equation may then be written as,

………….. (8.10)

The product of the two constants K_C and K gives another constant. Let it be K_a

…………. (8.11)

The constant K_a is called dissociation constant of the acid. Like other equilibrium constants, K_a also varies only with temperature. Similarly, for a weak base, the dissociation constant can be written as below.

………….. (8.12)

Ostwald’s Dilution Law

Ostwald’s dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c). Degree of dissociation (α) is the fraction of the total number of moles of a substance that dissociates at equilibrium.

We shall derive an expression for ostwald’s law by considering a weak acid, i.e. acetic acid (CH₃COOH). The dissociation of acetic acid can be represented as

CH₃COOH ⇄ H⁺ + CH₃COO^–

The dissociation constant of acetic acid is,

Substituting the equilibrium concentration in equation (8.13)

…………. (8.14)

We know that weak acid dissociates only to a very small extent. Compared to one, α is so small and hence in the denominator (1 – α) ~ 1. The above expression (8.14) now becomes,

………….. (8.15)

Let us consider an acid with K_a value 4 × 10^-4 and calculate the degree of dissociation of that acid at two different concentration 1 × 10^-2M and 1 × 10^-4M using the above expression (8.15) For 1 × 10^-2M,

= 2 × 10^-1
= 0.2

For 1 × 10^-4M

= 2

When the dilution increases by 100 times, (Concentration decreases from 1 × 10^-2M to 1 × 10^-2M), the dissociation increases by 10 times. Thus, we can conclude that, when dilution increases, the degree of dissociation of weak electrolyte also increases. This statement is known as Ostwald’s dilution Law.

The concentration of H⁺ (H₃O⁺) can be caluculated using the K_a value as below.

[H⁺] = αC (Refer table) ………….. (8.16)

Equilibrium molar concentration of [H⁺] is equal to αC

Similarly, for a weak base

K_b = α² and α = \(\sqrt{\frac{K_{b}}{C}}\)
[OH^–] = αC
(or)
[OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}\) ………….. (8.18)

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The pH Scale

We usually deal with acid / base solution in the concentration range 10^-1 to 10^-2M. To express the strength of such low concentrations, Sorensen introduced a logarithmic scale known as the pH scale. The term pH is derived from the French word ‘Purissance de hydrogene’ meaning, the power of hydrogen. pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

pH = – log₁₀[H₃O⁺] …………. (8.5)

The concentration of H₃O⁺ in a solution of known pH can be calculated using the following expression.

[H₃O⁺] = 10^-pH (or) [H₃O⁺] = antilog of (-pH) …………. (8.6)

Similarly, pOH can also be defined as follows

pOH = -log₁₀[OH^–] …………….. (8.7)

As discussed earlier, in neutral solutions, the concentration of [H₃O⁺] as well as [OH⁺] is equal to 1 × 10^-7M at 25°C . The pH of a neutral solution can be calculated by substituting this H₃O⁺ concentration in the expression (8.5)

pH = – log₁₀[H₃O⁺]
= – log₁₀10^-7
= (-7)(-1)log₁₀10 = +7(1) = 7 [∵log₁₀¹⁰ = 1]

Similary, we can calculate the pOH of a neutral solution using the expression (8.7), it is also equal to 7. The negative sign in the expression (8.5) indicates that when the concentration of [H₃O⁺] increases the pH value decreases.

For example, if the [H₃O⁺] increases from to 10^-7 to 10^-5M the pH value of the solution decreases from 7 to 5. We know that in acidic solution, [H₃O⁺]>[OH^–], i.e; [H₃O⁺]>10^-7. So, we can conclude that acidic solution should have pH value less than 7 and basic solution should have pH value greater than 7.

Relation Between pH and pOH

A relation between pH and pOH can be established using their following definitions

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Ionisation of Water

We have learnt that when an acidic or a basic substance is dissolved in water, depending upon its nature, it can either donate (or) accept a proton. In addition to that the pure water itself has a little tendency to dissociate. i.e, one water molecule donates a proton to an another water molecule. This is known as auto ionisation of water and it is represented as below.

In the above ionisation, one water molecule acts as an acid while the another water molecule acts as a base. The dissociation constant for the above ionisation is given by the following expression

…………… (8.3)

The concentration of pure liquid water is one. i.e, [H₂O]² = 1
∵ K_w = [H₃O⁺][OH^–] …………. (8.4)

Here, K_w represents the ionic product (ionic product constant) of water.

It was experimentally found that the concentration of H₃O⁺ in pure water is 1 × 10^-7 at 25°C. Since the dissociation of water produces equal number of H₃O⁺ and OH^–, the concentration of OH^– is also equal to 1 × 10^-7 at 25°C.

Therefore, the ionic product of water at 25°C is

K_W = [H₃O]⁺[OH^–] …………. (8.4)
K_W = (1 × 10^-7)(1 × 10^-7)
= 1 × 10^-14.

Like all equilibrium constants, K_w is also a constant at a particular temperature. The dissociation of water is an endothermic reaction. With the increase in temperature, the concentration of H₃O⁺ and OH^– also increases, and hence the ionic product also increases.

In neutral aqueous solution like NaCl solution, the concentration of H₃O⁺ is always equal to the concentration of OH^– whereas in case of an aqueous solution of a substance which may behave as an acid (or) a base, the concentration of H₃O⁺ will not equal to
[OH^–].

We can understand this by considering the aqueous HCl as an example. In addition to the auto ionisation of water, the following equilibrium due to the dissociation of HCl can also exist.

HCl + H₂O ⇄ H₃O⁺ + Cl^–

In this case, in addition to the auto ionisation of water, HCl molecules also produces H₃O⁺ ion by donating
a proton to water and hence [H₃O⁺]>[OH^–]. It means that the aqueous HCl solution is acidic. Similarly, in basic solution such as aqueous NH₃, NaOH etc…. [OH^–]>[H₃O⁺].

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Strength of Acids and Bases

The strength of acids and bases can be determined by the concentration of H₃O⁺ (or) OH^– produced per mole of the substance dissolved in H₂O. Generally we classify the acids/bases either as strong or weak. A strong acid is the one that is almost completely dissociated in water while a weak acid is only partially dissociated in water.

Let us quantitatively define the strength of an acid (HA) by considering the following general equilibrium.

The equilibrium constant for the above ionisation is given by the following expression

………… (8.1)

We can omit the concentration of H₂O in the above expression since it is present in large excess and essentially unchanged.

……………. (8.2)

Here, K_a is called the ionisation constant or dissociation constant of the acid. It measures the strength of an acid. Acids such as HCl, HNO₃ etc… are almost completely ionised and hence they have high K_a value (K_a for HCl at 25°C is 2 × 10⁶).

Acids such as formic acid (K_a = 1.8 × 10^-4 at 25°C), acetic acid (1.8 × 10^-5 at 25°C) etc.. are partially ionised in solution and in such cases, there is an equilibrium between the unionised acid molecules and their dissociated ions. Generally, acids with K_a value greater than ten are considered as strong acids and less than one considered as weak acids.

Let us consider the dissociation of HCl in aqueous solution,

As discussed earlier, due to the complete dissociation, the equilibrium lies almost 100% to the right. i.e., the Cl^– ion has only a negligible tendency to accept a proton form H₃O⁺. It means that the conjugate base of a strong acid is a weak base and vice versa. The following table illustrates the relative strength of conjugate acid – base pairs.

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Acids and Bases

The term ‘acid’ is derived from the latin word ‘acidus’ meaning sour. We have already learnt in earlier classes that acid tastes sour, turns the blue litmus to red and reacts with metals such as zinc and produces hydrogen gas. Similarly base tastes bitter and turns the red litmus to blue.

These classical concepts are not adequate to explain the complete behaviour of acids and bases. So, the scientists developed the acid – base concept based on their behaviour.

Let us, learn the concept developed by scientists Arrhenius, Bronsted and Lowry and Lewis to describe the properties of acids and bases.

Arrhenius Concept

One of the earliest theories about acids and bases was proposed by swedish chemist Svante Arrhenius. According to him, an acid is a substance that dissociates to give hydrogen ions in water. For example, HCl, H₂SO₄ etc., are acids. Their dissociation in aqueous solution is expressed as

The H⁺ ion in aqueous solution is highly hydrated and usually represented as H₃O⁺, the simplest hydrate
of proton [H(H₂O)]⁺. We use both H⁺ and H₃O⁺ to mean the same.

Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example, substances like NaOH, Ca(OH)₂ etc., are bases.

Limitations of Arrhenius Concept

• Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents such as acetone, Tetrahydrofuran etc.
• This theory does not account for the basicity of the substances like ammonia (NH₃) which do not possess hydroxyl group.

Lowry – Bronsted Theory (Proton Theory)

In 1923, Lowry and Bronsted suggested a more general definition of acids and bases. According to their concept, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance. In other words, an acid is a proton donor and a base is a proton acceptor.

When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCl behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as

HCl + H₂O ⇄ H₃O⁺ + Cl^–

When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as

H₂O + NH₃ ⇄ NH⁺₄ + OH^–

Let us consider the reverse reaction in the following equilibrium

H₃O⁺ donates a proton to Cl^– to form HCl i.e., the products also behave as acid and base. In general,
Lowry – Bronsted (acid – base) reaction is represented as

Acid₁ + Base₂ ⇄ Acid₂ + Base₁

The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs.

HCl and Cl^–, H₂O and H₃O⁺ are two conjugate acid – base pairs, i.e; Cl^– is the conjugate base of the acid HCl. (or) HCl is conjugate acid of Cl^–. Similarly H₃O⁺ is the conjugate acid of H₂O.

Limitations of Lowry – Bronsted Theory

(i) Substances like BF₃, AlCl₃ etc., that do not donate protons are known to behave as acids.

Lewis Concept

In 1923, Gilbert. N. Lewis proposed a more generalised concept of acids and bases. He considered the electron pair to define a species as an acid (or) a base. According to him, an acid is a species that accepts an electron pair while base is a species that donates an electron pair. We call such species as Lewis acids and bases.

A Lewis acid is a positive ion (or) an electron deficient molecule and a Lewis base is a anion (or) neutral molecule with at least one lone pair of electrons.

Les us consider the reaction between Boron tri flouride and ammonia.

Here, boron has a vacant 2p orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond. We have already learnt that in coordination compounds, the Ligands act as a Lewis base and the central metal atom or ion that accepts a pair of electrons from the ligand behaves as a Lewis acid.

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Factors Affecting the Reaction Rate

The rate of a reaction is affected by the following factors.

1. Nature and State of the Reactant
2. Concentration of the Reactant
3. Surface Area of the Reactant
4. Temperature of the Reaction
5. Presence of a Catalyst

Nature and State of the Reactant:

We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.

Let us compare the following two reactions that you carried out in volumetric analysis.

1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO₄
2. Redox reaction between oxalic acid and KMnO₄

The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺. In fact heating is required between KMnO₄ and Oxolate ion and is carried out at around 60°C.

The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine.

Let us consider another example that you carried out in inorganic qualitative analysis of lead salts. If you mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantaneously, whereas if you mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.

Concentration of the Reactants:

The rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of collision theory of reaction rates. According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.

Effect of Surface Area of the Reactant:

In heterogeneous reactions, the surface areas of the solid reactants play an important role in deciding the rate. For a given mass of a reactant, when the particle size decreases surface area increases. Increase in surface area of reactant leads to more collisions per litre per second, and hence the rate of reaction is increased. For example, powdered calcium carbonate reacts much faster with dilute HCl than with the same mass of CaCO₃ as marble.

Effect of Presence of Catalyst:

So far we have learnt, that rate of reaction can be increased to certain extent by increasing the concentration, temperature and surface area of the reactant. However significant changes in the reaction can be brought out by the addition of a substance called catalyst.

A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated at the end of the reaction. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.

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Arrhenius Equation – The Effect of Temperature on Reaction Rate

Generally, the rate of a reaction increase with increasing temperature. However, there are very few exceptions. The magnitude of this increase in rate is different for different reactions. As a rough rule, for many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10°C.

A large number of reactions are known which do not take place at room temperature but occur readily at higher temperatures. Example: Reaction between H₂ and O₂ to form H₂O takes place only when an electric spark is passed.

Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to e^-(Ea/RT) and he proposed a relation between the rate constant and temperature.

k = Ae^-(Ea/RT) …………. (1)

Where A the frequency factor,
R the gas constant,
E_a the activation energy of the reaction and,

T the absolute temperature (in K)

The frequency factor (A) is related to the frequency of collisions (number of collisions per second) between the reactant molecules. The factor A does not vary significantly with temperature and hence it may be taken as a constant.

E_a is the activation energy of the reaction, which Arrhenius considered as the minimum energy that a molecule must have to posses to react.

Taking logarithm on both side of the equation (1)

y = c + mx

The above equation is of the form of a straight line y = mx + c.

A plot of ln k Vs 1/T gives a straight line with a negative slope – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\) line with a negative slope –\(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\). If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy as follows.

At temperature T = T₁; the rate constant k = k₁

This equation can be used to caluculate E_a from rate constants K₁ and k₂ at temperatures
T₁ and T₂.

Example 1

The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s^-1 respectively. Caluculate the value of activation energy.
Solution:
According to Arrhenius equation

E_a = log(2) × 2.303 × 8.314 JK^-1mol^-1 × 400K
E_a = 2.305 J mol^-1

Example 2

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)(\(\frac{1}{T}\)) Where E_a is the activation energy. When a graph is plotted for log K Vs \(\frac{1}{T}\) a straight line with a slope of – 4000k is obtained. Caluculate the activation energy.
Solution:
log k = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)(\(\frac{1}{T}\))
y = c + mx
m = – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)
E_a = – 2.303 R m
E_a = – 2.303 × 8.314 JK^-1 mol^-1 × (-4000K)
E_a = 76, 589 J mol^-1
E_a = 76.589 kJ mol^-1

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Collision Theory

Collision Theory was proposed independently by Max Trautz in 1916 and William Lewis in 1918. This theory is based on the kinetic theory of gases. According to this theory, chemical reactions occur as a result of collisions between the reacting molecules. Let us understand this theory by considering the following reaction.

A₂(g) + B₂(g) → 2AB(g)

Fig 7.5 Progress of the Reaction

If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.

Rate ∝ number of molecules colliding per litre per second (collision rate)

The number of collisions is directly proportional to the concentration of both A₂ and B₂.

Collision rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]

Where, Z is a constant.

The collision rate in gases can be calculated from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately 10⁹ collisions per second, i.e., 1 collision in 10^-9 second.

Thus, if every collision resulted in reaction, the reaction would be complete in 10^-9 second. In actual practice this does not happen. It implies that all collisions are not effective to lead to the reaction. In order to react, the colliding molecules must possess a minimum energy called activation energy. The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.

Fraction of effective collisions (f) is given by the following expression

f = e^-Ea/RT

To understand the magnitude of collision factor (f), Let us calculate the collision factor (f) for a reaction having activation energy of 100 kJ mol^-1 at 300K.

Thus, out of 10¹⁸ collisions only four collisions are sufficiently energetic to convert reactants to products. This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.

The figure 7.6 illustrates the importance of proper alignment of molecules which leads to reaction. The fraction of effective collisions (f) having proper orientation is given by the steric factor p.

⇒ Rate = p × f × collision rate
i.e., Rate = p × e^-Ea/RT × Z[A₂][B₂] ………….. (1)
As per the rate law,
Rate = k[A₂][B₂] ………….. (2)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
k = p Z e^-Ea/RT

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Half Life Period of a Reaction

The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value. For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.

The rate constant for a first order reaction is given by

Let us calculate the half life period for a zero order reaction.

Hence, in contrast to the half life of a first order reaction, the half life of a zero order reaction is directly proportional to the initial concentration of the reactant.

Example 1

A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989; log 10 = 1)
Solution:
For a first order reaction
k = \(\frac{2.303}{t}\)log[latex]\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}[/latex] …………. (1)
Let [A_°] = 100M
When t = t_90%; [A]=10M (given that t_90% = 8hours)
t = t_90%; [A]=20M

Find the value of k using the given data

Substitute the value of k in equation (2)

t_80% = \(\frac{2.303}{2.303/8hours}\) log(5)
t_80% = 8hours × 0.6989
t_80% = 5.59hours

Example 2

The half life of a first order reaction x → products is 6.932 × 10⁴s at 500k. What percentage of x would be decomposed on heating at 500K for 100 min.(e^0.06=1.06).
Solution:
Given t_1/2 = 0.6932 × 10⁴s
\(\frac{\left[\mathrm{A}_{0}\right]-[\mathrm{A}]}{\left[\mathrm{A}_{0}\right]}\) × 100
We know that
For a first order reaction, t_1/2 = \(\frac{0.6932}{k}\)

Example 3

Show that case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.
Solution:
Let [A_°] = 100
When t = t_99.9%; [A] = (100 – 99.9) = 0.1

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The Integrated Rate Equation

We have just learnt that the rate of change of concentration of the reactant is directly proportional to that of concentration of the reactant. For a general reaction, A → product. The rate law is Rate = \(\frac{-d[A]}{dt}\) = k[A]^x

Where k is the rate constant, and x is the order of the reaction. The above equation is a differential equation, \(\frac{-d[A]}{dt}\), so it gives rate at any instant. However, using the above expression, we cannot answer questions such as how long will it take for a specific concentration of A to be used up in the reaction? What will be the concentration of reactant after a time ‘t’?. To answer such questions, we need the integrated form of the above rate law which contains time as a variable.

Integrated Rate Law for a First Order Reaction

A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following first order reaction,

A → Product

Rate law can be expressed as
Rate = k[A]¹
Where, k is the fist order rate constant.
\(\frac{-d[A]}{dt}\) = k[A]¹
⇒ \(\frac{-d[A]}{[A]}\) = k dt …………… (1)

Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A₀] to [A] at the later time.

– ln[A] – (- In[A₀]) = k (t-0)
– ln[A] + In[A₀] = kt
ln(\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) = kt ……….. (2)

This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303. 2.303 log (\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) = kt
k = \(\frac{2.303}{t}\)log(\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) ………… (3)
Equation (2) can be written in the form y = mx + c as below
ln[A₀]-ln[A] = kt
ln[A] = ln[A₀]-kt
⇒ y = c + mx

If we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated. Examples for the first order reaction

(i) Decomposition of Dinitrogen Pentoxide

N₂O₅(g) → 2NO₂(g) + \(\frac{1}{2}\)O₂(g)

(ii) Decomposition of Sulphurylchloride

SO₂Cl₂(l) → SO₂(g) + Cl₂(g)

(iii) Decomposition of the H₂O₂ in aqueous solution; H₂O₂ → H₂O(l) + \(\frac{1}{2}\)O₂(g)

(iv) Isomerisation of Cyclopropane to Propene.

Pseudo First Order Reaction:

Kinetic study of a higher order reaction is difficult to follow, for example, in a study of a second order reaction involving two different reactants; the simultaneous measurement of change in the concentration of both the reactants is very difficult.

To overcome such difficulties, A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,

Rate = k [CH₃COOCH₃] [H₂O]

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e.,concentration of water remains almost a constant.

Now, we can define k [H₂O] = k’; Therefore the above rate equation becomes
Rate = k'[CH₃COOCH₃]
Thus it follows first order kinetics.

Integrated Rate law for a Zero Order Reaction:

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us consider the following hypothetical zero order reaction.

A → product
The rate law can be written as,
Rate = k[A]°
\(\frac{-d[A]}{dt}\) = k(1) (∴[A]° = 1)
⇒ -d[A] = k dt

Integrate the above equation between the limits of [A_°] at zero time and [A] at some later time ‘t’,

Equation (2) is in the form of a straight line y = mx + c
i.e., [A] = – kt + [A_°]
⇒ y = c + mx

A plot of [A] Vs time gives a straight line with a slope of – k and y – intercept of [A_°].

Fig 7.4: A plot of [A] Vs time for a zero order reaction A → product with initial concentration of [A] = 0.5M and k = 1.5 × 10^-2mol^-1L^-1min^-1

Examples for a Zero Order Reaction:

1. Photochemical reaction between H₂ and I₂

2. Decomposition of N₂O on hot Platinum Surface

N₂O(g) ⇄ N₂(g) + \(\frac{1}{2}\)O₂(g)

3. Iodination of Acetone in Acid Medium is Zero Order With Respect to Iodine.

Rate = k [CH₃COCH₃][H⁺]

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Molecularity

Kinetic studies involve not only measurement of a rate of reaction but also proposal of a reasonable reaction mechanism. Each and every single step in a reaction mechanism is called an elementary reaction.

An elementary step is characterized by its molecularity. The total number of reactant species that are involved in an elementary step is called molecularity of that particular step. Let us recall the hydrolysis of t butyl bromide studied in XI standard. Since the rate determining elementary step involves only t-butyl bromide, the reaction is called a Unimolecular Nucleophilic substitution (S_N¹) reaction.

Let us understand the elementary reactions by considering another reaction, the decomposition of hydrogen peroxide catalysed by I^–.

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

It is experimentally found that the reaction is first order with respect to both H₂O₂ and I^–, which indicates
that I^– is also involved in the reaction. The mechanism involves the following steps.

Step: 1

H₂O₂(aq) + I^–(aq) → H₂O(l) + Ol^–(aq)

Step: 2

H₂O₂(aq) + OI^–(aq) → H₂O(l) + I^–(aq) + O₂(g)

Overall Reaction is

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

These two reactions are elementary reactions. Adding equ (1) and (2) gives the overall reaction. Step 1 is the rate determining step, since it involves both H₂O₂ and I^–, the overall reaction is bimolecular.

Differences Between Order and Molecularity:

Order of a Reaction	Molecularity of a Reaction
1. It is the sum of the powers of concentration terms involved in the experimentally determined rate law.	1. It is the total number of reactant species that are involved in an elementary step.
2. It can be zero (or) fractional (or) integer	2. It is always a whole number, cannot be zero or a fractional number.
3. It is assigned for a overall reaction	3. It is assigned for each elementary step of mechanism.