Text Book Back Questions and Answers - Chapter 10 - Surface Chemistry - 12th Chemistry Guide Samacheer Kalvi Solutions

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Surface Chemistry
Textual EvaluationSolved
Multiple Choice Questions
Question 1.
For freudlich isotherm a graph of $\log \frac{x}{m}$ is plotted against $\log P$. The slope of the line and its $y-a x i s$ intercept respectively corresponds to
(a) $1 / n, \mathrm{k}$
(b) $\log 1 / n, \mathrm{k}$
(c) $1 / n, \log \mathrm{k}$
(d) $\log 1 / n, \log \mathrm{k}$
Answer:
(c) $1 / n, \log \mathrm{k}$
$\frac{x}{m}=\mathrm{k} \cdot \mathrm{p}^{1 / \mathrm{n}}$
$\log \left(\frac{x}{m}\right)=\log \mathrm{k}+\frac{1}{n} \log \mathrm{p}$
$\mathrm{y}=\mathrm{c}+\mathrm{mx}$
$\mathrm{m}=\frac{1}{n}$ and $\mathrm{c}=\log \mathrm{k}$
Question 2.
Which of the following is incorrect for physisorption?
(a) reversible
(b) increases with increase in temperature
(c) low heat of adsorption
(d) increases with increase in surface area
Answer:
(b) increases with increase in temperature
Physisorption is an exothermic process. Hence increase in temperature decreases the physisorption.
Question 3.
Which one of the following characteristics are associated with adsorption?
(a) $\Delta \mathrm{G}$ and $\Delta \mathrm{H}$ are negative but $\Delta \mathrm{S}$ is positive
(b) $\Delta \mathrm{G}$ and $\Delta \mathrm{S}$ are negative but $\Delta \mathrm{H}$ is positive
(c) $\Delta \mathrm{G}$ is negative but $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are positive
(d) $\Delta \mathrm{G}$. $\mathrm{AH}$ and $\Delta \mathrm{S}$ all are negative.
Answer:
(d) $\Delta \mathrm{G}, \Delta \mathrm{H}$ and $\Delta \mathrm{S}$ all are negative.
Adsorption leads to decrease in randomness (entropy).i.e. $\Delta \mathrm{S}<0$ for the adsorption to occur, $\Delta \mathrm{G}$ should be - ve. We know that $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ if $\Delta \mathrm{S}$ is - ve, $\mathrm{T} \Delta \mathrm{S}$ is + ve. It means that $\Delta \mathrm{G}$ will become negative only when $\Delta \mathrm{H}$ is - ve and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}$

Question 4.
Fog is colloidal solution of
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas
dispersion medium-gas, dispersed phase-liquid
Question 5.
Assertion: Coagulation power of $\mathrm{Al}^{3+}$ is more than $\mathrm{Na}$.
Reason: greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechuize nile)
Question 6.
Statement: To stop bleeding from an injury, ferric chloride can be applied. Which comment about the statement is justified?
(a) It is not true, ferric chloride is a poison.
(b) It is true, $\mathrm{Fe}^{3+}$ ions coagulate blood which is a negatively charged sol
(c) It is not true; ferric chloride is ionic and gets into the blood stream.
(d) It is true, coagulation takes place because of formation of negatively charged sol with $\mathrm{Cl}^{-}$.
Answer:
(b) It is true, $\mathrm{Fe}^{3+}$ ions coagulate blood which is a negatively charged sol
Question 7.
Hair cream is
(a) gel
(b) emulsion
(c) solid sol
(d) sol.
Answer:
(b) emulsion
Emulsion dispersed phase, Dispersion medium -liquid

Question 8 .
Which one of the following is correctly matched?
(a) Emulsion - Smoke
(b) Gel - butter
(c) foam-Mist
(d) whipped cream - sol
Answer:
(b) Gel - butter
Question 9.
The most effective electrolyte for the coagulation of $\mathrm{As}_2 \mathrm{~S}_3$ Soils
(a) $\mathrm{NaCI}$
(b) $\mathrm{Ba}\left(\mathrm{NO}_3\right)_2$
(c) $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
(d) $\mathrm{AI}^{3+}$
Answer:
(d) $\mathrm{AI}^{3+}$
$\mathrm{As}_2 \mathrm{~S}_3$ is a - vely charged colloid. It will be most effectively coagulated by the cation with greater valency.
i.e., $\mathrm{Al}^{3+}$.
Question 10.
Which one of the is not a surfactant?
(a) $\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_{15}-\mathrm{N}-\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{Br}$
(b) $\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_{15}-\mathrm{NH}_2$
(c) $\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_{16}-\mathrm{CH}_2 \mathrm{OSO}_2-\mathrm{Na}^{+}$
(d) $\mathrm{OHC}-\left(\mathrm{CH}_2\right)_{14}-\mathrm{CH}_2-\mathrm{COO}^{-} \mathrm{Na}^{+}$
Answer:
(b) $\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_{15}-\mathrm{NH}_2$
Question 11.
The phenomenon observed when a beam of light is passed through a colloidal solution is
(a) Cataphoresis
(b) Electrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect-scattering of light

Question 12 .
In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using $\mathrm{K}_2 \mathrm{SO}_4$
(i). $\mathrm{Na}_3 \mathrm{PO}_4$
(ii). $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
(iii). and $\mathrm{NaCI}$
(iv). Their coagulating power should be
(a) II $>$ I $>$ IV $>$ III
(b) III $>$ II $>$ I $>$ IV
(c) I $>$ II $>$ III $>$ IV
(d) none of these
Answer:
(b) III $>$ II $>$ I $>$ IV
Question 13
Collodion is a $4 \%$ solution of which one of the following compounds in alcohol - ether mixture?
(a) Nitroglycerine
(b) Cellulose acetate
(c) Glycoldinitrate
(d) Nitrocellulose
Answer:
(a) Nitrocellulose
pyroxylin (nitro cellulose)
Question 14 .
Which one of the following is an example for homogeneous catalysis?
(a) manufacture of ammonia by Haber's process
(b) manufacture of sulphuric acid by contact process
(c) hydrogenation of oil
(a) Hydrolysis of sucrose in presence of all $\mathrm{HCI}$
Answer:
(a) Hydrolysis of sucrose in presence of all $\mathrm{HCl}$
Both reactant and catalyst are in same phase. i.e. (1)
Question 15 .
Match the following.

Answer:
(a) (iv) (i) (ii) (iii)
Question 16.
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of $\mathrm{AS}_2 \mathrm{~S}_3$ are given below
(I) $(\mathrm{NaCI})=52$
(II) $(\mathrm{BaCl})=0.69$
(III) $\left(\mathrm{MgSO}_4\right)=0.22$
The correct order of their coagulating power is
(a) III $>$ II $>$ I
(b) I $>$ II $>$ III
(c) I $>$ III $>$ II
(d) II $>$ III $>$ I
Answer:
(a) III $>$ II $>$ I
coagulating power $\pm \frac{1}{\text { coagulation value }}$
Question 17.
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then
(a) $\Delta \mathrm{H}$ increases
(b) $\Delta \mathrm{S}$ increases
(c) $\Delta \mathrm{G}$ increases
(d) $\Delta \mathrm{S}$ decreases
Answer:
(a) $\Delta \mathrm{S}$ decreases $-\Delta \mathrm{S}$ is -ve

Question 18.
If $\mathrm{x}$ is the amount of adsorbate and $\mathrm{m}$ is the amount of adsorbent, which of the following relations is not related to adsorption process?
(a) $x / m=f(P)$ at constant $T$
(b) $x / m=f(T)$ at constant $P$
(c) $P=f(T)$ at constant $x / m$
(d) $\mathrm{x} / \mathrm{m}=\mathrm{PT}$
Answer:
(d) $\mathrm{x} / \mathrm{m}=\mathrm{mPT}$
Question 19.
On which of the following properties does the coagulating power of an ion depend?
(a) Both magnitude and sign of the charge on the ion.
(b) Size of the ion alone
(c) the magnitude of the charge on the ion alone
(d) the sign of charge on the ion alone.
Answer:
(a) Both magnitude and sign of the charge on the ion.
Question 20.
Match the following.

Answer:
(d) (iv) (iii) (ii) (i)
Short Answer Questions
Question 1.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:
1. It is reversible
2. It has low heat of adsorption
3. It has weak van der Waals forces of attraction with adsorbent.
4. It increases with increase in pressure.
5. It forms multimolecular layer.
Question 2.
Differentiate physisorption and chemisorption.
Answer:
Chemical adsorption orChemisorption or Activated adsorption
1. It is very slow
2. It is very specific depends on nature of adsorbent and adsorbate.
3. chemical adsorption is fast with increase pressure, it can not alter the amount.
4. When temperature is raised chemisorption first increases and then decreases.
5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate, Heat of adsorption is high i.e., from $40-400 \mathrm{~kJ} / \mathrm{mole}$.
6. Monolayer of the adsorbate is formed.
7. Adsorption occurs at fixed sites called active centres. It depends on surface area .
8. Chemisorption involves the formation of activated complex with appreciable activation energy.
9. Physical adsorption or van der waals adsorption or Physisorptlon
10. It is irreversible
Physical adsorption or van der waals adsorption or physisorption.
1. It is instantaneous
2. It is non-specific
3. In Physisorption. when pressure increases the amount of adsorption increases.
4. Physisorption decreases with increase in temperature.

5. No transfer of electrons
6. Heat of adsorption is low in the order of $4 \mathrm{OkJ} / \mathrm{mole}$.
7. Multilayer of the adsorbate is formed on the adsorbent.
8. It occurs on all sides.
9. Activation energy is insignificant.
10. It is reversible.
Question 3.
In case of chemisorption, why adsorption first increases and then decreases with temperature?
Answer:
1. Chemisorption involves a high activation energy so it is also referred to as activated adsorption.
2. It is found in chemisorption that it first increases and then decreases with increase in temperature. When adsorption is plotted, the graph first increaes and then decreases with temperature.
3. The initial increase illustrates the requirement of activation of the surface for adsorption is due to fact that formation of activated complex requires certain energy. But later it decreacs at high temperature is due to desorption as the kinetic energy of the adsorbate increases (exothermic nature)
Question 4.
Which will be adsorbed more readily on the surface of charcoal and why; $\mathrm{NH}_3$ or $\mathrm{CO}_2$ ?
Answer:
1. The gaseshaving low critical temperature are adsorbed slowly, while gases with high critical temperature are adborbed readily.
2. Among $\mathrm{CO}_2$, and $\mathrm{NH}_3, \mathrm{NH}_3$ will be more readily adsorbed on the surface of the charcoal. This is because the critical temperature of ammonia gas is quite high than the $\mathrm{CO}_2$. Hence, it easily combines with the materials than the $\mathrm{CO}_2$ whether it is solid, liquid or any gases.
Question 5.
Heat of adsorption is greater for chemisorptions than physisorption. Why?
Answer:
Chemisorption has higher heat of adsorption. because in chemisorption the chemical bonds are much stronger. In adsorbed state the adsorbate is hold on the surface of adsorbent by attractive forces (bond). And chemisorption is irreversible one. Therefore, heat of adsorption is greater for chenil sorptions than physisorption. Chemisorption, heat of adsorption range $40-400 \mathrm{~kJ} / \mathrm{mole}$.

Question 6.
In a coagulation experiment $10 \mathrm{~mL}$ of a colloid (X) is mixed with distilled water and $0.1 \mathrm{M}$ solution of an electrolyte $\mathrm{AB}$ so that the volume is $20 \mathrm{~mL}$. It was found that all solutions containing more than $6.6 \mathrm{~mL}$ of $\mathrm{AB}$ coagulate with in 5 minutes. What is the flocculation values of $\mathrm{AB}$ for sol (X)?
Answer:
A minimum of $6.6 \mathrm{~mL}$ of $\mathrm{AB}$ is required to coagulate the sol. The moles of $\mathrm{AB}$ in the sol is $\frac{6.6 \times 0.01}{20}=0.033 \mathrm{moles}$
This means that a minimum of 0.033 moles or $0.0033 \times 1000=3.3$ milli moles are required for coagulating one litre of sol. Flocculation value of $\mathrm{AB}$ for $\mathrm{X}=3.3$
Question 7.
Peptising agent is added to convert precipitate into colloidal solution. Explain with an example.
Answer:
1. Ions either positive or negative of peptizing agent (electrolyte) are adsorbed on the particles of precipitate. They repel and hit each other and break the particles of the precipitate into colloidal size.
2. For example, when we add a small volume of very dilute hydrochloric acid solution peptising agent to a fresh precipitate of a silver chloride, it leads to formation of silver chloride colloidal solution,

Question 8.
What happens when a colloidal sol of $\mathrm{Fe}(\mathrm{OH})_3$ and $\mathrm{As}_2 \mathrm{~S}_3$ are mixed?
Answer:
On mixing $\mathrm{Fe}(\mathrm{OH})_3$ positive sol and $\mathrm{As}_2 \mathrm{~S}_3$ negative sol, mutual coagulation occurs which causes precipitation. When these sol got mixed with each other, due to $\mathrm{Fe}^{3+}$ and $\mathrm{S}^{2-}$ ions neutralisation of charges will happen and precipitate will be formed.
$
\mathrm{Fe}(\mathrm{OH})_3+\mathrm{As}_2 \mathrm{~S}_3 \rightarrow \mathrm{Fe}_2 \mathrm{~S}_3+\mathrm{As}(\mathrm{OH})_3
$

Question 9.
What is the difference between a sol and a gel?
Answer:
Sol
1. The liquid state of a colloidal solution is called sol.
2. The sol does not have a definite structure.
3. The dispersion medium of the sol may be water.
4. The sol can be converted to gel by cooling The sol can be easily dehydrated.
5. The viscosity of the sol is very low.
6. Sol is categorized into lyophobic and lyophilic sols.
7. Example: Blood
Gel
1. The solid or semi solid state of a colloidal solution is called gel.
2. The gel possesses honey comb like structure.
3. The dispersion medium of gel will be hydrated colloid particles.
4. The gel can be converted sol by heating.
5. The gel cannot be dehydrated.
6. The viscosity of the gel is very high.
7. There is no such classification of gel.
8. Example: Fruit jelly, cooked gelatin jelly.
Question 10.
Why are lyophillic colloidal sols are more stable than lyophoblic colloidal sol?
Answer:
1. A lyophilic colloidal sols are stable due to the charge and the hydration of sol particles.
2. Lyophilic sols are more stable than lyophobilc sols because they are highly hydrated in the solution. And since more is the hydration more will be its stability.
3. Lyophilic sols are stabilized by electrostatic charge and hydration where as lyophobile sols are only stabilized by charge, so they easily gets coagulated and requires a stabilising agent. Hence, lyophilic sols
are more stable than the lyophobilc sols.

Question 11.
Addition of Alum purifies water. Why?
Answer:
Purification of drinking water is activated by coagulation of suspended impurities in water using alums containing $\mathrm{Al}<$ sup $>3+$. That is why we are adding to purify water.
Question 12.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
Factors which influence the adsorption of a gas on a solid is as follows:
1. Nature of the gas:
Easily liquifiable gases such as $\mathrm{NH}_3, \mathrm{HCl}$ etc are adsorbed to a great extent in comparison to gases such as $\mathrm{H}_2, \mathrm{O}_2$ etc. This is because van der Waal's forces are stronger is easily liquifiable gases.
2. Surface area of the solid:
The greater the surface area of the adsorbant, the greater is the adsorption of gas on the solid surface.
3. Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.
4. Effect of temperature:
Adsorption is an exothermic process. Thus in accordance with Le - Chatelier's principle, the magnitude of adsorption decreases with an increase in temperature.
Question 13.
What are enzymes? Write a brief note on the mechanism of enzyme catalysis.
Answer:
Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action.

Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell. Mechanism of enzyme catalysis: Mechanism of enzyme catalysed reaction is known as lock and key mechanism.
1. Enzymes arc highly specific in their action.
2. These specificity is due to the pressure of active sites. The shape of active site of any given enzyme is like cavity such that only a specific substrate can fit into it.

In the same way a key fit into lock. The specific binding needs to the formation of an enzyme substrate complex which accounts for high specificity of enzyme catalysed reactions.
3. Once the proper orientation is attained the substrate molecules reacts to form the product in two steps.
4. Since product molecule do not have any affinity for the enzyme they leave the enzyme surface making room for fresh substrate.

step 1: Formation of enzyme - substrate complex

Step 2: Dissociation of enzyme - substrate complex to form product

The rate of the formation of product depends upon concentration of ES.

Question 14.
What do ou mean by activity and selectivity of catalyst?
Answer:
1. Activity of Catalyst:
The activity ofcatalyst is its ability to increase the rate ofa particular reaction, Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants in the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.
2. Selectivity of the catalyst:
The ability of the catalyst to dircct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between $\mathrm{H}_2$ and $\mathrm{CO}$.
(a) $\quad \mathrm{CO}_{(\mathrm{g})}+3 \mathrm{H}_{2(\mathrm{~g})} \stackrel{\mathrm{Ni}}{\longrightarrow} \mathrm{CH}_{4(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$
(b) $\quad \mathrm{CO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \stackrel{\mathrm{Cu} / \mathrm{ZnO}}{\underset{\mathrm{CrO}_3}{\longrightarrow}} \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{g})}$
(c) $\quad \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2(\mathrm{~g})} \stackrel{\mathrm{Cu}}{\longrightarrow} \mathrm{HCHO}_{(\mathrm{g})}$
Question 15.
Describe some feature of catalysis $b$ Zeoliles.
Answer:
1. Zeolites are microporous, crystalline, hydrated aluminosilicates, made of silicon and aluminium tetrahedra.
2. There are about 50 natural zeolites and 150 synthetic zeolites. As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra negative charge. To balance the negative charge, there are extra framework cations for example, $\mathrm{H}^{+}$or $\mathrm{Na}^{+}$ions.
3. Zeolites earring protons are used as solid acids, catalysis and they are extensively used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel,etc.
4. Zeolites earring Na ions are used as basic catalysis.
5. One of the most important applications of zeolites is their shape selectivity. In zeolites, the active sites namely protons are lying inside their pores. So, reactions occur only inside the pores of zeolites.

Question 16.
Give three uses of emulsions.
Answer:
1. The cleansing action of soap is due to emulsions.
2. It is used in the preparation of vanishing cream.
3. It is used in the preparation of cold liver oil.
Question 17.
Why does bleeding stop by rubbing moist alum
Answer:
Blood is a colloidal sol. When we nib the injured part with moist alum then coagulation of blood takes place. Hence main reason is coagulation, which stops the bleeding. Therefore bleeding stop by rubbing moist alum.
Question 18.
Why is desorption important for a substance to act as good catalyst?
Answer:
Desorption is important for a substance to act as a good catalyst, so that after the reaction, the products found on the surface separate out (desorbed) to create free surface again for other reactant molecules to approach the surface and react. If desorption does not occur then other reactants are left with no space on the catalysts surface for adsorption and reaction will stop.
Question 19.
Comment on the statement: Colloid is not a substance but it is a state of substance.
Answer:
The statement is true. Because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example. $\mathrm{NaCl}$ in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves
like a crystalloid while concentrated solutions behaves as a colloid. It is the size of the particles which matters. That is the state in which the substance exists. If the size of the particles lies in the range $1 \mathrm{~nm}$ to $100 \mathrm{~nm}$, it is in the colloidal state.
Question 20.
Explain any one method for coagulation

Answer:
The flocculation and setting down of the sol particles is called coagulation. Various method of coagulation are given below:

1. Addition of electrolytes
2. Electrophoresis
3. Mining oppositively charged sols
4. Boiling.
Addition of electrolytes
A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order
$
\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^{+} \text {, Similarly }\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-3}>\mathrm{SO}_4^{-2}>\mathrm{Cl}^{-}
$
The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles / lit) required to cause precipitation of a sol in 2hours. This value is called flocculation value. The smaller the flocculation value greater will be precipitation.
Question 21.
Write a note on electro osmosis
Answer:
Electro osmosis:
A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the sol particles. This movement of dispersion medium under the influence of electric potential is called electro osmosis.

Question 22.
Write a note on catalytic poison
Answer:
Catalytic poison:
Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons. For example, In the reaction, $2 \mathrm{SO}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{SO}_3$ with a Pt catalyst, the poison is $\mathrm{AS}_2 \mathrm{O}_3$.
i.e., $\mathrm{AS}_2 \mathrm{O}_3$ destroys the activity of pt. $\mathrm{AS}_2 \mathrm{O}_3$ blocks the activity of the catalyst. So, the activity is lost.
Question 23.
Explain intermediate compound formation theory of catalysis with an example.
Answer:
The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation. in homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.
Consider the reactions:
$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}$
$\mathrm{A}+\mathrm{C} \rightarrow \mathrm{AC}$ (intermediate)
$C$ is the catalyst
$
\mathrm{AC}+\mathrm{B} \rightarrow \mathrm{AB}+\mathrm{C} \ldots \ldots \ldots \ldots(3)
$
Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.
Example:
The mechanIsm of Fridel crafts reaction is given below
$
\mathrm{C}_6 \mathrm{H}_5+\mathrm{CH}_3 \mathrm{Cl} \stackrel{\text { anhydrous } \mathrm{AlCl}_3}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+\mathrm{HCl}
$
The action of catalyst is explained as follows .
$\mathrm{CH}_3 \mathrm{Cl}+\mathrm{AlCl}_3 \rightarrow\left[\mathrm{CH}_3\right]^{+}\left[\mathrm{AlCl}_4\right]^{-}$
It is an intermediate
$
\mathrm{C}_6 \mathrm{H}_6+\left[\mathrm{CH}_3\right]^{+}\left[\mathrm{AlCl}_4\right]^{-} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+\mathrm{AlCl}_3+\mathrm{HCl}
$
This theory describes,
1. The specificity of a catalyst.
2. The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations
1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.
Question 24 .
What is the difference between homogenous and hetrogenous catalysis?
Answer:
Hornogenous Catalysis:
1. In a catalysed reaction the reactants, products and catalyst are present in the same phase.
2. For example.
$
2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\mathrm{NO}_{(\mathrm{g})}}{\longrightarrow} 2 \mathrm{SO}_{3(\mathrm{~g})}
$
Hence $\mathrm{NO}$ act as catalyst.
3. Homogeneous catalysis explained by intermediate compound formation theory.
Heterogeneous Catalysis:
1. In a reaction, the catalyst is present in a different phase. i.e., catalyst is not present in the same phase as that of reactants and products.

2. For example.
$
2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\mathrm{Pt}_{(\mathrm{s})}}{\longrightarrow} 2 \mathrm{SO}_{3(\mathrm{~g})}
$
Hence $\mathrm{Pt}(\mathrm{s})$ act as catalyst.
3. Hetenogeneous catalysis explained by adsorption theory.
Question 25.
Describe adsorption theory of catalysis.
Answer:
Adsorption theory:
Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also he called as contaçt catalysis.
According to this theory, the reactants arc adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product. The various steps involved in a heterogeneous catalysed rcacton arc given as follows:
1. Reactant molecules diffuse from bulk to the catalyst surface.
2. The reactant molecules are adsorbed on the surface of the catalyst.
3. The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
4. The product molecules are desorbed.
5. The product diffuse away from the surface of the catalyst.
Advantages of adsorption theory:
The adsorption theory explains the following .
1. Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
3. A promoter or activator increases the number of active centres on the surfaces.