2nd PUC – KTBS Solutions

2nd PUC Chemistry Question Bank Chapter 4 Chemical Kinetics

Prasanna — Tue, 14 Jul 2026 10:10:08 +0000

You can Download Chapter 4 Chemical Kinetics Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 4 Chemical Kinetics

2nd PUC Chemistry Chemical Kinetics NCERT Textbook Questions and Answers

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3NO(g) → N₂O (g) Rate = K[NO]²
(ii) H₂O₂ (aq) + 3I^– (aq) + 2H⁺ → 2H₂O(l) +I₃^–
Rate = K[H₂O₂][I^–]
(iii) CH₃CHO (g) → CH₄ (g) + CO(g)
Rate = K [CH₃CHO]^3/2
(iv) C₂H₅Cl (g) → C₂H₄ (g) + HCl (g)
Rate = K [C₂H₅Cl]
Ans:
(i) 2
(ii) 2
(iii) 3/2
(iv) 1

Question 2.
For the reaction:
2A + B → A₂B
the rate = k[A][B]² with k = 2.0 × 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Answer:
Initial rate = K [A] [B]²
= 2.0 × 10^-6(mol^-2 L² s^-1 × 0.1 (mol^-1) × (0.2)²mol²L^-2
= 8 × 10^-9 mol L^-1 s^-1
we know that

\(\frac { 1 }{ 2 }\) [rate of disappearance of A] = rate of disappearance of B
∴ When [A] is reduced by 0.1 – 0.06 = 0.04 M
[B] is reduced by \(\frac { 1 }{ 2 }\) × 0.04 = 0.02 M
∴ [B] remaining = 0.2 -0.02 = 0.18 M
Rate of reaction = K [A] [B]²
= 2.0 × 10^-6 mol ^-2 L²¹
s ^-1 × 0.06 × (0.18)² mol³ L^-3
= 3.89 × 10^-9 Ms^-1

Question 3.
The decomposition of NH₃on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4(H mol^-1L s^-1 ?
Answer:
For zero order reaction rate of reaction = K

∴Rate of production of N₂= k = 2.5 × 10^-4 Ms^-1
Rate of production of H₂ = 3k
= 3 × 2.5 × 10^-4
= 7.5 × 10^-4 Ms^-1

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by
Rate = K [CH₃OCH₃]^3/2
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.
Rate = k (P_CH₃OCH₃)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Answer:
Units rate of reaction = bar min^-1
units of rate constant

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Answer:
(i) Concentration of reactants
(ii) Temperature
(iii) Nature of reactants and products
(iv) Exposure to light (Radiation)
(v) presence of catalysts.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction ‘ affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?
Answer:
(i) Four times
[-rate₁ = K[C_A]²
C_A2 = 2C_A
rate₂ = K [2C_A]² = 4K [C_A]²] = 4 rate₁]

(ii) 1/4 times

Question 7.
What is the effect of temperature on the rate constant of a reaction ? How can this temperature effect on rate constant be represented quantitatively.
Answer:
Increasing the temperature on decreasing the activation energy will result in an increase in the rate of reaction and an exponential increase in the rate constant. On increasing the temperature the fraction of molecules which collide with energy greater than Ea increases and hence the rate constant (exponentially)
K = A ^-ea/RT , quantitative representation of temperature effect on rate constant.

Question 8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s	0	30	60	90
[Ester]/mol L^-1	0.55	0.31	0.17	0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer:
(i)

(ii) For pseudo first order reaction,
Average rate = Rate constant × Average concentration Average rate

Question 9.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer:
(i)

(ii) When concentration of B is tripled, rate of reaction increases by 9 times.
(iii) Rate of reaction increases by 8 times.

Question 10.
In a reaction between A and B, the initial rate reaction (r₀) was measured for different initial concentrations of A and B as given below.

A/ mol L^-1	0.20	0.20	0.40
B/ mol L^-1	0.30	0.10	0.05
r_θmol L^-1S^-5	5.07×10^-5	5.07×10^-5	1.43 x 10^-5

what is the order of the reaction with respect to A and B ?
Answer:
Let the rate law be, r = K [A]^x [B]^y
r₁ = K[0.2]^x[0.3]^y = 5.07 × 10^-5 -(1) .
r₂ = K[0.2]^x [0.1]^y = 5.07 ×10^-5 -(2)
r₃ = K [0.4]^x [0.05]^y = 7.6 × 10^-5 -(3) ,

∴ The order of reaction with respect to A is 0.585 and with respect to B is 0.
∴ rate law = K[A]^0.858 [B]⁰ = K [A]^0.585.

Question 11.
The following results have been obtained during the kinetic studies of the reaction 2A + B → C + D

Experiment	[A]/mol L^-1	[B]/mol L^-1	Initial rate of formation of D/mol L^-1min^-1
I	0.1	0.1	6.0 x 10^-3
II	0.3	0.2	7.2 x 10^-2
III	0.3	0.4	2.88 x 10^-1
IV	0.4	0.1	2.40 x 10^-2

Determine the rate law and the rate constant for the reaction.
Answer:
r = K [A]^x [B]^y
r, = K[0.1]^x [0.1]^y = 6.0 × 10^-3 -(i)
r2 = K [0.3]^x [0.2]^y = 7.2 × 10^-2 -(2)
r3 = K [0.3]^x [0.4]^y = 2.88 × 10^-1 -(3)
r4 = K [0.4]^x [Q.l]^y = 2.4 × 10^-2 -(4)

4^x = 4
⇒ x =1

2^y = 4 ⇒ y = 2.

∴ Rate law, r = K [A]¹ [B]²
Rate constant,

= 6M^-2min^-1

Question 12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment	[A]/mol L^-1	[B]/mol L^-1	Initial rate /mol L^-1min^-1
I	0.1	0.1	2.0 x 10^-2
II	x	0.2	4.0 x 10^-2
III	0.4	0.4	y
IV	z	0.2	2.0 x 10^-2

Answer:
rate law = K [A]
from I 2.0 × 10^-2 = K 0.1
⇒ K = 2.0 × 10^-1min^-1
from II 4 .0 × 10^-2 = 2 .0 × 10^-31 x
⇒ K = 2 .0 × 10^-1 mol L^-1
from III y = 2.0 × 10^-1 × 0.4
⇒ y = 8 × 10^-2 mol 1L
from IV 2.0 × 10^-2 = 2.0 × 10^-1 × z
⇒ z = 0.1 mol L^-1.

Question 13.
Calculate the half – life of a first order reaction from their rate constants given below.
(i) 200 s^-1
(ii) 2 min^-1
(iii) 4 years^-1
Answer:

Question 14.
The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
Answer:
[A]₀ = 100
[A] = 80

= 1845 years
∴The age of the sample is 1845 years.

Question 15.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16 th value?
Answer:
K = 60 s^-1

Question 16.
During nuclear explosion, one of the products is MSr with half-life of 28.1 years. If mg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically ?
Answer:
t^1/2 = 28.1 years

[A]₀ = 2.47 × 1μg
For 10 years

[A] =0.78 × 10^-6g = 0.78μg
For 60 years

The amount of ⁹⁰Sr that will remain in the child’s body after 10 years is 0.78μg and after 60 years is 0.229 μg.

Question 17.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:

for 99% completion,
if [A]₀ = 100
[A] = 100-99= 1
for 90% completion
[A] = 100-99

Question 18.
A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Answer:
[A]₀ = 100.
[A] = 100 – 30 = 70
t = 40min

Question 19.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-5s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor ?
Answer:
Pre – exponential factor = A = Arrhenius
factor = frequency factor
K = 2.418 × 10^-5s^-1
T = 546 K.
Ea= 179.9 × 10^-3J mol^-1
[By Arrhenius equation] K = \(\text { Ae }^{-\mathrm{E}} / \mathrm{kr}\)

Question 20.
onsider a certain reaction A → Products with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol Lr1. Ans: [A]0 = 1 mol L^-1
Answer:
[A]₀ = 1 mol L^-1
t = 100s
k = 2.0 × 10^-2s^-1

log [A] = -0.8684
⇒ [A] = 10 ^-0.8684 = 0.354 mol L^-1

Question 21.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t^1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
Answer:
t^1/2 = 3 hrs

Question 22.
The decomposition of hydrocarbon follows the equation K = (1.5 × 10¹¹ s^-1)
\(e^{\frac{-28000 K}{T}}\) Calculate E_a
Answer:
According to Arrehenius equation
K = Ae^-Ea/RT
Given K = (4.5 × 10¹¹s^-1)\(e^{\frac{-28000 K}{T}}\)
comparing the two equations

E_a = 28000 × R
= 28000 × 8.314 J mol^-1
= 232.79 KJ mol^-1.

Question 23.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – \(\frac{1.25 \times 10^{4} K}{\mathbf{T}}\)
Calculate E_a for this reaction and at what temperature will Its half-period be 256 minutes?
Answer:
According to Arrehenius equation

E_a = 2.303 R × 1.25 × 10⁴ K
= 2.303 × 8.3l4 JK^-1 mol^-1 × 1.25 × 10⁴ K
= 239.34 KJ mol^-1
What t^1/2 = 356 min

Substituting K in given equation
log(4.51 × 10^-5) = 14.34 – \(\frac{1.25 \times 10^{4}}{\mathrm{T}}\)
T = 669 K.

Question 24.
The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kj mol^-1. At what temperature would k be 1.5 × 10⁴s^-1 ?
Answer:
k₁ = 4.5 × 10³ s^-1
k₂ = 1.5 × 10⁴s^-1
E_a = 60 × 10³ J mol^-1
R = 8.314 JK^-1 mol^-1
T₁ = 283 k
T₂ = ?

Question 25.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:
T₁ = 283 k
T₂ = 313 k
k₂ = 4k

2nd PUC Chemistry Chemical Kinetics Additional Questions and Answers

Question 1.
Name the factors on which the rate of a particular reaction depends. (H. S .B 2001)
Answer:
Concentration, temperature, presence of catalyst and light, surface area of reactants.

Question 2.
Write the expression for rate of reaction in terms of each reactant and product for the reaction.
N₂ + 3H₂ → 2NH₃ (P.S.B 2011, HSB 2000)
Answer:

Question 3.
Give the example of a reaction having fractional order. (PSB 2000)
Answer:
Decomposition of acetal dehyde (order =1.5)

Question 4.
Give an example of a psueudo first order reaction. (AISB, DSB 2004, DSB 2006)
Answer:
Acid catalysed hydrolysis of ethyl acetate

Rate = K [CH3COOC2H5] [H20]°

Question 5.
For an elementary reaction,
2 A + B → 3C
the rate of appearance of C at time ‘t’ is 1.3 × 10⁴ mol^-1 S^-1. Calculate at this time
(i) Rate of the reaction
(ii) Rate of disappearance of A. (CBSE sample paper II 2008)
Answer:
Rate

= 1/3 × 1.3 × 10^-4mol L^-1S^-1
= 4.33 × 10^-5 mol L^-1S^-1

Rate of disappearance of A

2/3 × 1.3 × 10^-4 L^-1S^-1
= 8.66 × 10^-5 mol L^-1S^-1

Question 6.
The rate law for a reaction is found to be
Rate = K [NO^–₂] [I^– ] [H⁺]² How would the rate of reaction change when
(i) Concentration of H⁺ is doubled
(ii) Concentration of I^– is halved
(iii) Concentration of each of NO₂, I^– and H⁺ are tripled ?
Answer:
Suppose initially the concentration are [NO^–₂] = a mol L^-1, [I^– ] = b mol L^– and [H⁺] = c mol L^-1
∴ Rate = K abc2

(i) New [H⁺] = 2c
∴ New rate = 4ab (2c)² = 4abc2
= 4 times

(ii) New [I^– ] = \(\frac { b }{ 2 }\)
New Rate = Ka\(\frac { b }{ 2 }\)c²= \(\frac { 1 }{ 2 }\) Kabc²
ie. rate of reaction is halved

(iii) New [NO^–₂] = 3a, [I^– ] = 3b, [H⁺] = 3c
New rate = K (3a) (3b) (3c)²
= 81K abc² = 81 times.

Question 7.
The decomposition of NH₃ on platinum surface

is zero order with k = 2.5 × 10^-4 mol L^-1 S^-1
What are the rates of production of N₂ and H₂
Answer:
2NH₃ -> N₂ + 3H₂

For zero order reaction, rate = K

= 2.5 × 10^-4 mol L^-1 S^-1
Rate of production of H₂ = \(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\)
= 3 × (2.5 × 10^-4 mol L^-1 S^-1)
= 7.5 × 10^-4 mol ^-1 S^-1

Question 8.
Why does coal not burn by itself in air, but once initiated by a flame, continues to bum ?
Answer:
Activation energy for the combustion reaction is very high and is not available at room temperature. Oh applying flame, to a part of the coal and air in contact with the flame absorb heat which provides the necessary activation energy and combustion starts. The heat liberated further provides activation energy for the combustion to continue.

Question 9.
Following reaction takes place in one step:
2NO(g) + O₂(g) ⇌ 2NO₂(g)

\(\frac{\mathbf{r}_{2}}{\mathbf{r}_{1}}\) = 27 or r₂ = 27r₁ ie rate becomes 27
times
There is no effect on the order of reaction.

Question 10.
Rate constant K of a reaction varies with temperature according to the equation
log K = constant . \(\frac{\mathbf{E} \mathbf{a}}{2.303 \mathbf{R}} \frac{\mathbf{1}}{\mathbf{T}}\)
Answer:
where E_a is the energy of activation for the reaction. When a graph is plotted for
log K versus\(\frac { 1 }{ T }\), a straight line with a slope 6670 K is obtained. Calculate the energy of activation for this reaction. State units (R = 8.314JK^-1 mol^-1)
Answer:
Slope of the lin,e
\(\frac{-E a}{2.303 R}\) =-6670K
Ea = 2.303 × 8.314 (JK^-1 mol^-1) × 6670K
= 127711.4J mol^-1.

2nd PUC Chemistry Question Bank Chapter 13 Amines

Prasanna — Tue, 14 Jul 2026 09:32:53 +0000

You can Download Chapter 13 Amines Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 13 Amines

2nd PUC Chemistry Amines NCERT Textbook Questions

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary .and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH₂)₂NH
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(V) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCH₃
(vii) m-BrC₆H₄NH₂
Answer:

Question 2.
Give one chemical test to distinguish between the following pairs of compounds,

Methylamine and dimethylamine
Secondary and tertiary amines
Ethylamine and aniline
Aniline and benzylamine
Aniline and N-methyl aniline.

Answer:
1. Methylamine and dimethylamine
(1) CH₃NH₂ (Methyl amine; 1° amine) – Carbyl amine test positive
(2) (CH₃)₂ NH (Dimethyl amine; 2° amine) – Doesn’t give this test
Carbylamine test:-

2. 2° and 3° amines:-
(1) 2° amine – Libermann Nitrosomine test positive
(2) 3° amine – Libermann Nitrosomine test negative.
Libermann Nitrosomine test :

3. Ethylamine and aniline:
(1) Aromatic amines gives brilliantly coloured orange dye on diazotisation and coupling with phenol.
(2) Ethylamine do not give a dye on diazotisation and coupling with phenol.
Diazotisation:

4. Aniline and benzylamine:-
(1) Aromatic (1 °) amine gives coloured dye on diazotisation and coupling with phenol.
(2) Other non aromatic 1° amines and which are not an aromatic 1° amine, will not give a dye.
Diazotisation:-
(a)

(b)

5. Aniline and N-methylaniline:-
aniline (1° amine)-Carbylamine test positive
N-methylaniline (2° amine) – Negative
Carbylamine test:-

Question 3.
Account for the following:

1. pK_b of aniline is more than that of methylamine.
Answer:
(i) Aniline is weaker base than methylamine
K_b of aniline is less than K_b of methylamine.
pK_b of aniline is more than pK_b of methylamine.
PK_b = -logK_b
(a) Aniline is weaker base than methylamine because due to resonance, lone pair on N₂ gets delocalised over the ring, so electron pair availability is reduced.

(b) In methyl amine due to ‘+I’effect of methyl group, availability of lone pair on N₂ atom is increased So is strong base.

2. Ethylamine is soluble in water whereas aniline is not.
Answer:
(i) Lower aliphatic amines are soluble in water because of their ability to form hydrogen bonds with water.
(ii) But in aniline due to larger hydrocarbon part, it hinders hydrogen bond formation with water.

3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Answer:
Methylamine + H₂O ⇌ Methyl ammonium hydroxide

OH- can precipitate out Fe, AI, Cr etc…
3 [CH₃NH₃]⁺ OH + FeCl₃ → Fe (OH)₃↓+ 3CH₃NH₃Cl
Fe (OH)₃ is unstable, so can be hydrated ferric oxide

4. Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
Answer:
In aniline NH₂ increases e density at o- and p- positions due to +R effect. But when it is nitrated by nitrating mixture, substantial amount of m-nitro aniline is formed.
But in acidic medium (Nitrating mixture – HNO₃ (c) + H₂SO₄ (c)).
NH₂ + H⁺ ⇌ NH₃⁺, NH₃⁺ is e^– withdrawing group and so is m – directing. So gives larger amount of m- nitroaniline.

5. Aniline does not undergo Friedel-Crafts reaction.
Answer:
Friedel-Craft reaction:-

But aniline undergoes salt formation with A1C13 (Lewis acid used as catalyst). So N acquires positive charge and hence acts as strong deactivating group for further reaction.

6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Answer:
In aromatic diazonium salts, due to resonance their is dispersal of positive charge on benzene ring.
But in aliphatic diazonium salts, resonance is not possible, so aliphatic diazonium salts are less stable than aromatic diazonium salts.

7. Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:
Gabriel phthalimide synthesis

Reaction between alkyl halide (R – X) and ammonia gives mixture, of 1°, 2° and 3° amines with tetraalkyl ammonium halide. Mixture is not separable.
To get pure 1° amine, Gabriel phthalimide synthesis is preferred.

Question 4.
Arrange the following:
1. In decreasing order of the pKb values:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and CH₅NH₂
Answer:
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂ NH

2. In increasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
Answer:
C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂ NH

3. In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
Answer:

(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂.
Answer:
C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂

4. In decreasing order of basic strength in gas phase:
C₆H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃,
Answer:
(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃
Note: In GAS PHASE, 3° > 2° > 1° Ammonia [No Hydrogen bending and ‘+I’ effect].

5. In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
Answer:
(CH₃)₂ NH < C₂H₅NH₂ < C₂H₅OH
Note: Stronger H-bonding, more b.p more solubility
Alcohol > Amine In amines on H attached to N takes part in H-bond.

6. In increasing order of solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂.
Answer:
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂
more H-bonding, more solubility

Question 5.
How will you convert:
1. Ethanoic acid into methanamine
Answer:

2. Hexanenitrile into 1-aminopentane
Answer:

3. Methonal to ethanoic acid
Answer:

4. Ethanamine into methanamine
Answer:

5. Ethanoic acid into propanoic acid
Answer:

6. Methanamine into ethanamine
Answer:

7. Nitromethane into dimethylamine
Answer:

8. Propanoic acid into ethanoic acid
Answer:

Question 6.
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Answer:
Hinsberg test:- Benzenesulphonyl chloride (C₆H₅SO₂Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides.
(a) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide.

The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.

(b) In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed.

Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.

(c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride, in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines. However, these days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride.
1° amine – clear solution which on acidification gives insoluble product.
2° amine – insoluble product in solution, unaffected by acid
3° amine – No reaction, dissolves an acidification.

Question 7.
Write short notes on the following:

Carbylamine reaction
Diazotisation
Hofmann’s bromamide reaction
Coupling reaction
Ammonolysis
Acetylation
Gabriel phthalimide synthesis.

Answer:
1. Carbylamine reaction
1° amines (aromatic/aliphatic) On heating with chloroform and alcoholic solution of KOH gives isocyanides/carbylamines producing unpleasant smell.

2° and 3° (aromatic/aliphatic) do not give this reaction.

2. Diazotisation .
Aromatic 1° amines into diazonium salts by action of HNO₂ and dil HCl at ice cold temperature as HNO₂ and product are unstable at higher temperature.

3. Hofmann’s bromamide reaction
Involves heating of an amide with bromine and caustic alkali to yield an amine with one carbon less than original amide.

4. Coupling reaction
Diazonium salt acts as electrophile and brings about substitution in electron rich aromatic ring such as phenol and amines. The result is formation of dye.
(i) with phenol – In basic medium (9 – 10 pH) at low temperature (273 – 278 K)

(ii) with amines – In acidic medium (ph = 4 – 5) at low temperature (273 – 278 K)

5. Ammonolysis – Alkyl halide (1°) react with alcoholic ammonia to form 1°, 2°, 3° amines along with 4°ammonium salts. It is carried out by heating alkyl halide with ale solution of NH3 in sealed tube at 370 K.
If ammonia is used in excess, 1° amine is major product

If alkyl halide is excess, it forms all 1°, 2°, 3°, 4° ammonium salts

6. Acetylation
Aliphatic amines:
Introduction of acetyl group (CH₃CO-) into any molecule.
1° and 2° amines react with acetyl chloride to form amides
But 3° amines has NO H on N, so do not react.

7. Gabriel phthalimide synthesis

Reaction between alkyl halide (R – X) and ammonia gives mixture, of 1°, 2° and 3° amines with tetraalkyl ammonium halide. Mixture is not separable.
To get pure 1° amine, Gabriel phthalimide synthesis is preferred.

Question 8.
Accomplish the following conversions:

Nitrobenzene to benzoic acid
Benzene to m-bromophenol
Benzoic acid to aniline
Aniline to 2,4,6-tribromofluorobenzene
Benzyl chloride to 2-phenylethanamine
Chlorobenzene to p-chloroaniline
Aniline to p-bromoaniline
Benzamide to toluene
Aniline to benzyl alcohol

Answer:
1. Nitrobenzene to benzoic acid

2. Benezene to m-bromophenol

3. Benzoic acid to aniline

4. Aniline to 2,4,6-tribromofluorobenzene

5. Benzyl chloride to 2-phenylethanamine

6. Chlorobenzene to p-choloroaniline

7. Aniline to p-bromoaniline

8. Benzeamide to toluene

9. Aniline to benzeyl alcohal

Question 9.
Give the structures of A,B and C in the following reactions:

Answer:

Answer:

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer:

As ‘C’ is expected as a 1° amine obtained by Hoffmann bromamide degradation of ‘B’ an amide obtained from ‘A’ a carboxylic acid on treatment with NH₃ and heat.
So

Question 11.
Complete the following reactions:
1. C₆H₅NH₂ + CHCI₃ + alc.KOH →
Answer:
C₆H₅NH₂ + CHCl₃ + alc.KOH → C₆H₅NC + 3 KCl + 3H₂O

2. C₆H₅N₂Cl + H₃PO₂ + H₂O →
Answer:

3. C₆H₅NH₂ + H₂SO₄ (conc.) →
Answer:

4. C₆H₅N₂Cl + C₂H₅OH →
Answer:

5. C₆H₅NH₂ + Br₂(aq) →
Answer:

6. C₆H₅N₂ + (CH₃CO)₂ →
Answer:

Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer:
In this reaction, potassium phthalimide gives phthalimide anion which acts as nucleophile and attacks electrophilic carbon of alkyl halide.

But due to much less reactivity of aryl halide (nucleophilic substitution reaction), 1° aromatic amines can’t be prepared.

Question 13.
Write the reactions of
1. aromatic and
2. aliphatic primary amines with nitrous acid.
Answer:
1. Aromatic amines with nitrous acid
Aromatic 1° amines react with HNO₂ at 273 – 278 K to form diazonium compound.

2. Aliphatic amines with HNO₂ at 273 – 278 K to give corresponding alcohol, N₂, along with other products.

Question 14.
Give plausible explanation for each of the following:
1. Why are amines less acidic than alcohols of comparable molecular masses?
Answer:

But as ‘O’ is more electronegative than ‘N’. So ‘O’ can accommodate more positive charge and is more stable than ‘N’. So alcohol liberates H+ easily than amines.
So alcohols are more acidic than amines.

2. Why do primary amines have higher boiling point than tertiary amines?
Answer: 1° amines (NH2) has stronger and extensive H – bonding.
3° amines (> N-) has no H, so no H – bonding.
So 1° amines have higher b.p than 3° amines.

3. Why are aliphatic amines stronger bases than aromatic amines?
Answer:

Alkyl amines has positive I effect/electron donating effect, so high electron density on N and stable alkyl ammonium ion formation.
Aromatic amines due to resonance effect and lower stability of aryl ammonium ion than aryl amine, e- density on N is less.
So Alkyl amine is stronger base than aromatic amines.

2nd PUC Amines Intext Questions

Question 15.
Classify as primary, secondary or tertiary amines.
1.

2.

3. (C₂H₅)₂ CHNH₂
4. (C₂H₅)₂ NH
Answer:
1.

→ Primary(As only one of H is replaced by other group, contains – NH₂ group)

2.

→ No H attached to N, so teritary ( As, > N group is teritary)

3. (C₂H₅)₂ CHNH₂ —> Primary (contains – NH₂ group)
4. (C₂H₅)₂NH —> Secondary (contains -NH group)

Question 16.
(i) Write a structure of different isomeric amines corresponding to C₄H₁₁N
(ii) Write IUPAC names of all isomers
(iii) What type of isomerism is exhibited by different pair of amines
Answer:
C₅H₁₁N
Primary amines (-NH₂)

Question 17.
Convert:
(a) Benzene into aniline
Answer:

(b) Benzene into N,N – dimethyl aniline
Answer:

Question 18.
Arrange following in increasing order of their Basic character:
1. C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂
Answer:
C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

2. C₂H₅NH₂, (C₂H₅)NH, (C₂H₅)₃N, C₆H₅NH₂
Answer:
C₆H₅NH₂ < C₂H₅NH₂ < C₂H₅)₃_N < (C₂H₅)₂NH

3. CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Answer:
C₆H₅NH₂ < C₂H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Question 19.
Compliete following acid-base reaction and name the products:
Answer:

Question 20.
Write reactions of final alkylation product of aniline with excess of methyl iodine is presence of solution
Answer:

Question 21.
Write chemical reaction of aniline with benzoyl chloride and write product name.
Answer:

Question 22.
Write structure of isomers of C₃H₉N. Write IUPAC names of isomers which will liberate N₂ gas on treatment with nitrous acid. .
Answer:

Question 23.
Convert
1. 3-methylaniline into 3-nitrotolune
Answer:

2. Aniline into 1,3,5 – tribromobenzene
Answer:

2nd PUC Amines Additional Questions

Question 1.
Complete the reaction.
C₆H₅N₂Cl + CH₃CH₂OH → (Benzene)
Answer:
C₆H₅N₂Cl + CH₃CH₂OH → C₆H₆ +CH₃CHO + N₂ + HCl

Question 2.
Write the IUPAC name of C₆H₆NHCH₃ .
Answer:
N – Methyl aniline.

Question 3.
Which is more soluble in water, NH₃ or methylamine ?
Answer:
NH₃ is more soluble in water due to more H-bonding.

Question 4.
Which of the following amines has largest value of pK_b?
p-Toluidine, Aniline, p- Nitroaniline.
Answer:
Higher the pK_b value, weaker is the base

Presence of -CH₃ group in the ring will increase the basic character as it is an electron releasing group. Presence of -NO₂ group in the ring will decrease the basic character as it an electron withdrawing group. Thus p-nitro aniline will have largest value of pK_b.

Question 5.
State two important uses of aniline. (DSB 2000)
Answer:

In the manufacture of dyes and drugs.
In the preparation of phenyl isocyanide required for the manufacture of polyurethane plastic.

Question 6.
Why do amines behave as nucleophiles ?
Answer:
Amines behave as nucleophiles. For eg: amines react with alkyl halides, acid chlorides etc.

Question 7.
Why are aqueous solution of amines basic in nature ? (DSB 2006)
Answer:
Aqueous solution of amines which are freely soluble in water (like CH₃NH₂, C₂H₅NH₂ etc) are basic in nature. It can be explained as follows,
For eg:-

The aqueous solution of methyl amine will be basic in nature, due to the presence of ions.

Question 8.
Explain why silver chloride is soluble in aqueous solution of methyl amine. (CBSE 1997, 1998,2002,2004)
Answer:
Due to the formation of water soluble complex
AgCl + 2CH₃NH₂ → [Ag(CH₃NH₂)₂]+Cl

Question 9.
Write IUPAC name of. (Delhi CBSE 1999)

Answer:
p-nitro-N, N dimethyl aniline.

Question 10.
For an amine RNH₂, write the expression for K_b to indicate its base strength. (DBS 2000)
Answer:

Question 11.
What does K_b value for an amine stand for? (AISB 2000)
Answer:
K_b value for an amine stands for its basic strength. More the K_b value of an amines more is its basic strength.

Question 12.
Give one example of anbident nucleophile. (CBSE 2001)
Answer:

acts as ambident nucleophiles.

Question 13.
Why do amines acts as nucleophiles ? (AISB 2007)
Answer:
Aliphatic amines act as nucleophiles. This is due to the availability of lone pair of electrons on nitrogen

Aromatic amines do not behave as nucleophiles. This is because the lone pair of electrons on nitrogen in aromatic amines is involved in resonance with benzene ring, and as such not available for nucleophije attack.

Question 14.
Write IUPAC name of. (AISB 2004)

Answer:
4 – Methoxy benzene amine.

Question 15.
How is phenyl aminomethane obtained from phenyl nitrile. (AISB 2004)
Answer:

Question 16.
Why is an alkyl amine more basic than ammonia ? (CBSE Delhi 2009)
Answer:
This is because of the presence of electron releasing alkyl group (positive I effect) which increases the electron density on nitrogen.

Question 17.
Give the IUPAC name of. H₂N — CH₃— CH₂ — CH = CH₂ (CBSE Delhi 2010)
Answer:
But-3-en-l-amine.

2nd PUC Amines Short Answers Questions

Question 1.
How are the following conversions accomplished
(1) Aniline to chlorobenzene
(2) Nitrobenzene to phenol
(3) Aniline to benzoic acid
Answer:
(1)

Question 2.
Account for the following observation:
1. Tertiary amine does not undergo acylation reaction
2. Aniline readily reacts with bromine to give 2,4,6- tribromo aniline (CBSE 2002, 2004)
Answer:
1. Tertiary amine do not undergo acylation reaction because they do not contain a H-atom on the N-atom.
2. This is due to the strong activating effect of the amino group, Halogenation of amines occur very fast and halogens enter at all the three (one P-and two o-positions) positions even in the absence of the catalyst.

Question 3.
Give a chemical test to distinguish between aniline and N-methylaniline. (CBSE 2001, 2006)
Answer:
Aniline being a 1° amine gives carbylamine test i.e. on heating with CHCl₃, in presence of alc. KOH, it gives offensive smell of phenyl isocyanide while N-methyl aniline being a 2° amine does not give this test.

Question 4.
Write one chemical reaction each to illustrate the following
1. Hoffmann’s bromamide reaction
2. Gabriel phthalimide synthesis (CBSE 2008)
Answer:

1. Carbylamine reaction
1° amines (aromatic/aliphatic) On heating with chloroform and alcoholic solution of KOH gives isocyanides/carbylamines producing unpleasant smell.

2° and 3° (aromatic/aliphatic) do not give this reaction.

(ii) with amines – In acidic medium (ph = 4 – 5) at low temperature (273 – 278 K)

Question 5.
(a) Arrange the following in an increasing order of basic strength in water:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(b) Arrange the following in increasing order of basic strength in the gas phase
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and CH₃NH₂. (CBSE Delhi 2008)
Answer:
(a) In aqueous solution, basic strength decreases in the order. .
(C₂H₅)NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃
(b) In gaseous phase,(basic strength decreases in the order.
(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃.

Question 6.
Give an example for each describe the following reactions.
1. Hoffmann’s bromamide reaction
2. Coupling reaction
3. Gattermann reaction (CBSE 2009)
Answer:
1.

(ii) with amines – In acidic medium (ph = 4 – 5) at low temperature (273 – 278 K)

2. The reaction of diazonium salt with phenols and aromatic amines to form azo compounds of the general formula Ar – N = N – Ar is called coupling reaction. The coupling with phenol takes place in mildly alkaline medium while with amines it occurs under faintly acidic condition.

3. Gattermann reaction
When benzene diazonium chloride reacts with Halogen acids in presence of Cu powder, aryl halide is formed as.

Question 7.
Complete the following reaction equation ?
(1)

(2) C6HSN2CI + H3PO2 + H2O →
(3) C6H5NH2 + Br2 (aq) 4
Answer:
(1)

(2) and (3) Same as NCRT text book Q.No. 11.

2nd PUC Economics Question Bank Chapter 3 Demand Analysis

Prasanna — Tue, 14 Jul 2026 07:21:44 +0000

You can Download Chapter 3 Demand Analysis Questions and Answers, Notes, 2nd PUC Economics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Economics Question Bank Chapter 3 Demand Analysis

2nd PUC Economics Demand Analysis One Mark Questions and Answers

Question 1.
What is Demand?
Answer:
The concept ‘demand’ refers to the quantity of a good or service that a consumer is willing and able to purchase at various prices, during a period of time. It includes desire for a commodity, ability to pay and willingness to pay.

Question 2.
Qd = f(p) is the demand function. Identify the independent variable in it.
Answer:
In Qd = f(p), the independent variable is ‘p’ (price).

Question 3.
State the Law of Demand.
Answer:
The law can be explained in the following manner: “Other things being equal, a fall in price leads to expansion in demand and a rise in price leads to contraction in demand”.

Question 4.
What is the relationship between price and demand for complementary goods?
Answer:
There is an inverse relationship between price and demand for complementary goods. (When the price of a product becomes more, the demand for its complementary good falls).

Question 5.
What are normal goods?
Answer:
Normal goods are those goods for which the demand increases with rise in income of consumers , and decreases with fall in their income.

Question 6.
Why does the demand curve shift?
Answer:
The shift in demand curve occurs because of changes in all the determinants of demand except price.

Question 7.
Name the method of adding two individual demand curves horizontally.
Answer:
The method of adding two individual demand curves is ‘Horizontal summation’.

Question 8.
Give the meaning of elasticity of demand.
Answer:
Elasticity of demand is generally defined as the responsiveness or sensitiveness of demand to a given change in the price of commodity.

Question 9.
Write the simplified formula of income elasticity of demand.

Question 10.
If change in price and change in expenditure are in the same direction, then what will be the price elasticity of demand?
Answer:
If change in price and change in expenditure are in the same direction, the price elasticity of demand is less than one, i.e., less elastic demand. (P_ed <1).

Question 11.
What is Cross Elasticity of Demand?
Answer:
It may be defined as the proportionate change in the quantity demanded of a particular commodity in response to a change in the price of another related commodity.

Question 12.
What is demand function?
Answer;
The Demand Function refers to a functional relationship between quantities demanded and its determinants. It can be expressed as follows:
Qd = f(P, Pr ,Y, T,………….. ).
Where Qd stands for Quantity demanded, f is function, P is Price of commodity, Pr – price of related/substitutes, Y is income of consumers, T is tastes and preferences of consumers.

2nd PUC Economics Demand Analysis Two Marks Questions and Answers

Question 1.
Mention any four determinants of demand.
Answer:

Price of the product
Price of related goods
Income of consumers and
Tastes and preferences of consumers.

Question 2.
In Qd = 20 – 2p, identify independent variable, dependent variable, constant and co-efficient in it.
Answer:
Qd is dependent variable, P is independent variable, 20 is constant, -2 is coefficient of ‘p’

Question 3.
If Qd = 30 – 2p is the demand function. Suppose the price of onion in the market is Rs.10 per kg. Calculate the quantity demanded.
Answer:
Qd = 30 – 2p; If price is Rs. 10,
Qd = 30 – 2(10)
= 30 – 20
= 10.
The quantity demanded of onion is 10 kgs.

Question 4.
Why does the demand curve slope downwards?
Answer:
The demand curve slopes downwards because of price effect, substitution effect, income effect and operation of law of diminishing marginal utility.

Question 5.
Write the meaning of cross elasticity of demand and its formula.
Answer:
It may be defined as the proportionate change in the quantity demanded of a particular commodity in response to a change in the price of another related commodity

Question 6.
What are complementary goods? Give examples.
Answer:
Complementary goods are those goods which are consumed together or jointly to satisfy human wants. Example, Shoes and socks, vehicles and petrol, bat and ball etc.

Question 7.
Consider the demand for onion. At Rs.10 per kg, demand for onion is 15 kgs. Suppose the price increases to Rs.20 per kg, the demand decreases to 10 kgs. Calculate the price elasticity of demand.
Answer:
Solution:
∆q = 10 – 15 = -5; ∆p = 20 – 10 = 10; then PED is

= -0.33 (the Price Elasticity is 0.33)

Question 8.
There are only two consumers in a market X and Y. Their demand for a good is given below. Calculate the market demand for the goods and draw the market demand curve.

Price	Demand- X	Demand-Y	Market Demand (x + y)
2	16	20	36
4	14	18	32
6	12	15	27
8	10	12	22
10	8	10	¹⁸
12	6	8	14

Question 9.
If there are two consumers in a market and their individual demand functions are Qd₁ = 15 – 2p and Qd₂ = 25 – 3p. Find the Market demand function.
Answer:
The market demand function Qd = Qd₁+ Qd₂
Therefore,
Qd = 15 – 2p + 25 – 3p
Qd = 40 – 5p

When the price is ‘0’, quantity demanded will be
Qd = 40 – 5p
= 40 – 5(0)
=40

When the quantity demanded is ‘0’, the price will be
0 = 40 – 5p
40 – 5p = 0
40 = 5p
P = 40/5
P = 8
So the Market demand is ‘40’ and Market price is Rs.8.

Question 10.
How do the movements along the demand curve occur?
Answer:
The movements along the demand curve occur on the basis of the inverse relationship between price and quantities demanded. When the price is less, demand will be more and when the price is more, demand will be less. So, any changes in price lead to movement on the demand line.

2nd PUC Economics Demand Analysis Five Marks Questions and Answers

Question 1.
Why does the demand curve slope downwards? Explain.
Answer:
In order to represent the inverse relationship between price and demand, the demand curve must slope downwards. Apart from this basic reason, there are many other factors which make the demand curve to slope downwards. They are as follows:

(a) Operation of the law of Diminishing Marginal Utility:
The law of DMU states that as the consumer acquires larger quantities of any commodity, the additional units of the same product will give him lower utility, and as such he gets a less value for the additional under The law of demand states that in order to induce the consumer to buy more less price must be offered.

(b) Operation of the law of Equi – Marginal Utility:
This law states that utility of the product must be equal to its price in general. As price falls, the equality between the two will be disturbed and in order to re-establish this equality the consumer buys more. Now utility comes to the level of reduced price. Hence, as price falls, a consumer buys more.

(c) Income effect:
A change in demand as a result of change in income is called as Income effect. As price falls, the real income of the consumer increases. With this increased real income (gets more purchasing power with more money in his hands), he buys more.

(d) Substitution effect:
When the price of one product falls, it becomes cheaper when compared to other products ’ for which price remains constant. Hence, a consumer will substitute low priced product to high priced product. The result is that demand for a product rises as price falls.

(e) Price Effect:
When the change in quantities demanded is caused by the change in price, it is called as Price effect. When the price of a product falls, it becomes cheaper and consumer buys more of that and vice versa.
Hence, the demand curve always slopes downwards from left to right.

Question 2.
How does the demand curve shift? Explain with a diagram.
Answer:
(a) Shift in demand curve – Increase and Decrease in Demand:
The increase and decrease in demand are caused by all the determinants of demand except price. When there is change in consumer’s income, tastes and preferences, price of related goods, there may be increase or decrease in demand.

(b) Increase in Demand:
When the income of consumer increases, the demand for the product increases and there will be shift in demand line towards right. For normal goods, the demand, curve shifts to the right. In the diagram given below, D: represents shift in demand line towards right indicating increase in demand.

(c) Decrease in Demand:
When the price of related goods rise, the demand for the product falls and the demand curve shifts towards left. For example, if there is rise in price of petrol, the demand for vehicle decreases, representing backward shift of demand line. In the diagram below, D₂ represents shift in demand towards left indicating decrease in demand.

Question 3.
What is income elasticity of demand? Calculate the income elasticity of demand when income of consumer increases from Rs.10,000 to Rs.12,000 and demand for rice increases from 30 to 40 kgs.
Answer:
Income elasticity of demand may be defined as the ratio or proportionate change in the quantity demanded of a commodity to a given proportionate change in the income. In short, it indicates the extent to which demand changes with a variation in consumers income.

The following formula helps to measure Y_ed

Here, ∆q stands for change in quantity, ∆y is change in Income of consumer, ‘y’ is initial income and ‘q’ is initial quantity.

Solution:
∆q = 40 -30 = 10; ∆y = 12,000 – 10,000 = 2000; then Y_ed is ;

= 1.66 Therefore the Income Elasticity of Demand is greater than one.

Question 4.
What are the factors determining price elasticity of demand?
Answer:
(i) Nature of the Commodity:
In case of comforts and luxuries, demand tends to be elastic because people buy those more only when their prices are low. e.g. TV sets.

In case of necessaries, demand is inelastic because whatever may be the price, people have to buy and use them. e.g. Rice.

(ii) Existence of Substitutes:
If a product has substitutes, demand tends to become elastic because people compare tho prices of substitute goods and cheaper products are purchased, eg. Blades, Soaps.

If a product has no substitutes, demand becomes inelastic because in that case whatever may be the price, people have to buy them. e.g. Onion.

(iii) Durability of the commodity:
If a product is perishable or non durable, demand tends to be inelastic, because people buy them again and again.
If a product is durable, demand tends to be elastic because people buy them occasionally.

(iv) Number of uses of a commodity:
If a product has multiple uses, demand tends to become elastic because, with a fall in price, the same product can be used for many purposes, e.g. Electricity, coal etc.

If a product has only one use, in that case demand becomes inelastic because people have to buy them for a specific single purpose whatever may be the price, e.g. All eatables, seeds, fertilizers, pesticides etc.

(v) Possibility of postponing the use of product:
If there is a possibility to postpone the use of particular product, demand tends to become elastic, e.g. Buying a scooter, motorcycle, TV sets etc. people generally buy these articles when they are cheaper. .

If it is not possible to postpone, demand tends to become inelastic. In this case, whatever may be the price, people have to buy them. e.g. Medicine.

(vi) Level of income of the people:
Generally speaking, demand will be elastic in case of the poor people because even a small change in price will affect the demand for various products.

On the other hand demand will be inelastic in case of rich people because they are ready to spend any amount on buying a product.

(vii) Habits:
If people are not habituated for the use of certain products, then demand tends to be elastic. If products are cheaper, they buy more and if they become costly, they may buy less or may not buy them at all.

When people are habituated for the use of a particular commodity, they do not care for price changes over a certain range, e.g. Cigarettes,, liquor etc. In that case, demand tends to become inelastic.

(viii) Complementary goods:
Goods which are jointly demanded are inelastic in nature, e.g., ink and pens, vehicles and petrol etc. This is because, if people buy one product, they have to buy the supplementary products also without which they cannot make use of the first item.

Demand tends to be elastic in case of independent products, e.g., biscuits, chocolates, ice-creams etc. In this case, consumption or use of a product is not linked to any other products. Hence, they may or may not buy a product.

2nd PUC Economics Demand Analysis Ten Marks Questions and Answers

Question 1.
Explain the law of demand with the help of demand schedule and demand curve.
Answer:
The law can be explained in the following manner: “Other things being equal, a fall in price leads to expansion in demand and a rise in price leads to contraction in demand”.

According to Prof.Samuelson, the law of demand states that ‘People buy more at lower prices and buy less at higher prices, other things remaining the same’.

Demand schedule:
Demand schedule represents the quantities demanded by an individual consumer at different levels of price.

Individual Demand Schedule

Price P (in Rs.)	Demand (Qd) (in Kg)
3	30
4	25
5	20
⁶	15
7	10

In the above individual demand schedule, the consumer is purchasing different quantities at different price levels. At Rs.3 he buys 30 kgs, and at Rs. 4, 25 kgs are bought and so on. As the price increases, the quantities demanded falls.

Demand Curve: The graphical presentation of the demand schedule is called demand curved

In the above diagram, price is measured along y axis and quantities demanded are measured along x axis. The various points on Demand line represent the respective quantities demanded. For example, point ‘c’, quantities demanded is 20 at price Rs.5.

The demand curve slopes downwards from left to right. It shows the rate at which demand changes with respect to change in price. As there is an inverse relationship between price and quantities demanded, the curve is negatively sloped.

Question 2.
Classify the price elasticity of demand and explain them with diagrams.
Answer:
In the words of Prof. Stonier and Hague, “Price elasticity of demand is a technical term used by economists to describe the degree of responsiveness of the demand for a good to a change in its price.”
It is measured by using the following formula.

The rate of change in demand may not always be proportionate to the change in price. A small change in price may lead to very great change in demand. It is called as Elastic Demand. Sometimes even a big change in price may not cause any change in demand. Such a demand is known as Inelastic Demand.

The following formula is used to calculate Price Elasticity of Demand (P_ed):

Here, ∆q stands for change in quantity, ∆p is change in price, ‘p’ is the initial price and ‘q’ is the initial quantity.

Classification of Price Elasticity of Demand:
On the basis of the degree of price elasticity for different goods, we classify P_ed as follows:

1. Perfectly Elastic Demand:
In this case, a very small change in price leads to an infinite change in demand. The demand curve is a horizontal curve and parallel to x axis. The numerical co-efficient of perfectly elastic demanded is infinity (ED = ∞).

2. Perfectly Inelastic Demand:
In this case, whatever may be the change in price, quantity demanded will remain perfectly constant. The demand curve is a vertical straight line and parallel to Y axis. Quantity demanded would be 10 units, irrespective of price change from Rs. 10.00 to Rs.2.00. Hence, the numerical co-efficient of perfectly inelastic demand is zero. ED = 0.

3.Relatively Elastic Demand: More elastic demand.
In this case, a slight change in price leads to more than proportionate change in demand. One can notice here that a change in demand is more than that of change in price. Hence, the elasticity is greater than one. For e.g., price falls by 3% and demand rises by 9%. Hence, the numerical co-efficient of demand is greater than one.

Here, the percentage change in quantities demanded will be more than percentage change in price, i.e., ∆q > ∆p. M M₁ > P P₁.

4. Relatively Inelastic Demand: Less elastic demand. In this case, a large change in price, say 8% price fall, leads to less than proportionate change in demand: say 4% rise in demand. One can notice here that change in demand is less than that of change in price. This can be represented by a steeper demand curve. Here, elasticity is less than one. (P_ed <1)

Here, the percentage change in quantities demanded will be less than percentage change in price, i.e., ∆q > ∆p, M M₁> P P₁

5. Unitary Elastic Demand:
In this case, proportionate change in price leads to Equal proportionate change in demand. For e.g., 5% fall in price leads to exactly 5% increase in demand. Hence, elasticity is equal to unity. It is possible to come across unitary elastic demand, but it is a rare phenomenon.

Here, the percentage change in quantities demanded will be equal to percentage change in price. i.e., ∆q > ∆p, M M ₁ > P P_1.

Out of the five different degrees, the first two are theoretical and the last one is a rare possibility. Hence, in all our general discussions, we make reference only to two terms- relatively classic demand and relatively inelastic demand.

2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry

Prasanna — Tue, 14 Jul 2026 06:24:41 +0000

You can Download Chapter 3 Electrochemistry Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry

2nd PUC Chemistry Electrochemistry NCERT Textbook Questions and Answers

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Answer:
Mg, Al, Zn, Fe, Cu.

Question 2.
Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V,
Hg²⁺/Hg = 0.79 V
Mg²⁺/Mg = -2,37 V, Cr³⁺/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.
Answer:
The lower the reduction potential, the higher is the reducing power. Hence, the reducing power of the given metals increases inthe following order.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn(s)+2Ag⁺(aq) —> Zn²⁺(aq)+2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:
The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn²⁺ (aq) || Ag⁺ (aq) | Ag(s)
(i) Zn electrode (anode) is negatively charged
(ii) Tons are carriers of current in the cell and in the external circuit, current from silver to Zinc.
(iii) The reaction taking place at the anode is given by,
Zn(s) -H → Zn²⁺(aq) + 2e^–
The reaction taking place at the cathode is given
Ag⁺+ e^– → Ag(s)

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd
(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Calculate the ArG9and equilibrium constant of the reactions.
Answer:
(i) For the given reaction, the Nemst equation can be given as:

∴ E^θ = 1.104V
We know that,
Δ_rG^θ = -nFE^θ
= -2 × 96487× 1.04
= – 213043.296J
= -213.04KJ

Question 5.
Write the Nernst equation and em! of the
following cells at 298 K:
(I) Mg(s)|Mg⁺²(O.OO1M) ||Cu⁺²(0.0001M)|Cu(s) .
(ii) Fe(s)|Fe⁺²(O.OO1 M)||H⁺(1M)H₂(g) (1bar)|Pt(s)
(iii) Sn(s) |Sn²⁺(O.050 M)||H⁺(0.020M|H2(g) (1 bar)|Pt(s)
(iv) Pt(s)|Br₂(l)|Br^–(O.O1O M)||H⁺(O.030 M)| H₂(g) (1 bar)|Pt(s).
Answer:
(i) E^θCr³⁺ /Cr = 0.74V
E^θCd²⁺ / Cd = – 0.40V
The galvanic cell of the reaction IC depicted as :
Cr(s)|Cr³⁺ (aq) || (Cd²⁺ (aq) Cd(s)
Now, the standard cell potential is
E^θ_cell = E^θ_R– E^θL
= -40 – (-0.74)
= +0.34V
∆_rG^θ = —nFE^θ_cell
In the given equation,
n = 6
F = 96487 C mol^-1
E^θ_cell = + 0.34V
Then, ∆_rG^θ =-6 × 96487 mol^-1 × 0.34V
= -196833.48 CV mol^-1
= -196833.48 J mol^-1
= -196.83 KJ mol^-1
Again
∆_rG^θ = -RT In K
=>∆_rG^θ =-2.303RT log K

(ii) E_θ Fe³⁺ /Fe²⁺ = 0.77V
E_θ Ag^–/Ag = 0.80V
(ii) For the given reaction, the Nernst equation can be given as:

(iii) Far the givën reaction, the Nernst equation can be given as:

0.14 – 0.0295 × log 125
= 0.14-0.062
= 0.078 V
= 0.08 V (approx)

(iv) For the given reaction , the nernst equation can be given as:

= – 1.09 – 0.02955 × log ( 1.11× 10⁷)
= – 1.09 – 0.02955(0.0453 + 7)
= -1.09 – 0.208 =-1.298 V.

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag₂O(s) + H₂O(l) → Zn²+ (aq) + 2Ag(s) + 2OH-(aq)
Determine ∆_r G^θ and E^θ for the reaction.
Answer:
The galvanic cell of the reaction is depicted as:
Fe²⁺ (aq) | Fe³⁺ (aq) || Ag⁺ (aq) | Ag(s)
Now, the standard cell potential is
E^θ_cell = E^θ_R – El^θ
= 0.80 – 0.77 ‘
= 0.03 V
Here, n = 1
Then, ∆_r G^θ = – nFE^θ_cell
= -1 × 96487 C mol^-1 × 0.03V
= – 2894.61 J mol^-1
= – 2.89 KJ mol^-1
∆_r G^θ = 2.303 RT In K

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte.
Discuss their variation with concentration.
Answer:
Conductivity of a solution is defined as the conductance of a solution 1 cm in length and area of cross section cm2.1 is represented by K.

Conducti vity always decreases with a decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area area of cross-section A and distance of unit length.

Molar conductivity increases with a decrease in concentration. This is because the total volume of the solution containing one mole of the electrolyte increases on dilution.

Question 8.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.
Answer:

Question 9.
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1.
Answer:
Cell constant = conductivity × Resistance
= 0.146 × 10^-3S C^m-1 × 1500 Ω = 0.219 cm^-1

Question 10.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and thfe results are given below:

Calculate ∆_m for all concentrations and draw a plot between ∆_m and c^1/2. Find the value of ∆⁰_m
Answer:
K = 7.896 × 10^-5 S cm^-1
M = 0.00241

Question 11.
How much charge is required for the following reductions?
(i) 1 mol of Al³⁺ to Al
(ii) 1 mol of Cu²⁺ to Cu
(iii) 1 mol of MnO^4- to Mn²⁺
Answer:
Al³⁺ + 3e → A1
charge required = 3F
(ii) Cu²⁺ + 2e → Cu
charge required = 2F
(iii) MnO^4- + 8H⁺ + Se^– → Mn²⁺ + H₂O
charge required = 5F

Question 12.
How much electricity in terms of Faraday ¡s required to produce
(j) 20.Ogat Ca from molten CaCl₂
(ii) 40.0 g of Al from Almólten Al₂O₃
Answer:
(i) Ca²⁺2 + 2e → Ca
2F can produce I mole (=40 g) Ca
∴ To produce 20 g Ca requires, \(\frac{2 \mathrm{F} \times 20}{40}\) = 1F

(ii) Al³⁺ + 3e → Al
3F can produce 1 mole (= 27g) Al
∴ To produce 40 g A1 requires
\(\frac{3 \mathrm{F} \times 40}{27}\) = 4.44F

Question 13.
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H₂O to O₂
(ii) 1 mol of FeO to Fe₂O₃.
Answer:
(i) 2H₂O → 4H⁺ + O₂ + 4e
2F of electricity is required for oxidation of 1 mole of H₂O
(ii) Fe²⁺ → Fe³⁺ + e
IF of electricity is required for oxidation of 1 mole FeO

Question 14.
A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer:
Ni²⁺ + 2e^– → Ni
2F (2 × 96500 C) can produce 58.7 g of Ni
Q = It = 5 × 20 × 60 =6000C

Question 15.
Three electrolytic cells A,B,C containing = 0.439 g of Zn solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Answer:

i.e. 108 g of Ag is deposited by 96487 C
Therefore, 1 .45g of Ag is deposited by

Given,
Current = 1.5A
\(\frac{1295.43}{1.5}\) S
∴ Time = 863.6S
= 864 S
= 14.40 min
Again,

i.e. 2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit = 0.426g of Cu

i.e. 2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1295.43 C of charge will deposit = 0.439 g of Zn

Question 16.
Predict the products of electrolysis in
each of the following:
(1) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.
Answer:
(i) At cathode:
The following reduction reactions compete to take place at the cathode
Ag⁺(aq) + e^– → Ag(s); E^θ = 0.80 V
H⁺ (aq) + e^– → \(\frac { 1 }{ 2 }\) H₂ (g); E^θ = 0.00V
The reaction with a higher value of E_θ takes place of the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:
The Ag anode is attacked by NO₃^– ions. Therefore, the silver electrode at the anode dissolves in the solution to from Ag⁺.

(ii) At cathode: Same as above
At anode: Anode is not attackable and hence OH^– ions have lower discharge potential than NO₃^– ions and OH^– ions react to give O₂
OH^– → OH + e^–
4OH → 2H₂O + O₂ (g)
(iii) H₂SO₄ → 2H⁺ + SO^2-₄
HO₂ ⇌ H⁺ + OH^–

At cathode:
2H⁺+ 2e^– → H₂
At anode: 4OH^– → 2H₂O + O₂ + 4e^–
i. e., H₂ will be liberated at cathode and O₂ at anode.

(iv) CuCl₂ → Cu²⁺+2Cl^–

At Cathode: Cu²⁺ ions will be reduced in preference to H⁺ ions
Cu²⁺ + 2e → Cu
At anode: Cl’ ions will be oxidised in preference to OH^– ions.
2Cl^– → Cl₂ + 2e^–
i.e., Cu will be deposited on the cathode and
Cl₂ will be liherated at the anode.

2nd PUC Chemistry Electrochemistry Additional Questions and Answers

Question 1.
A solution of sodium chloride is a better
conductor of electricity at a temperature of 50°C than at room temperature. Why?
Answer:
A solution of NaCl shows greater conduction of electricity at a temperature of 50°C than at room temperature because the ionic mobility of a strong electrolyte such as NaCl increases with an increase in temperature.

Question 2.
Give the relationship between molar conductivity and specific conductivity.
Answer:
Molar conductivity and specific conductivity are related to each other by the given equation.

Where,
Δ_m = Molar conductivity
K = Specific conductivity
C = Molar concentration

Question 3.
Why is it not possible to measure single electrode potential?
Answer:
The process of oxidation or reduction cannot take place alone. However, electrode potential is a relative tendency and can be measured with respect to a reference electrode such as standard hydrogen electrode.

Question 4.
Why is the rusting of iron faster in saline water than in pure water?
Answer:
Strong electrolytes such as sodium chloride are present in saline water. The ions produced from NaCl help in the reduction of oxygen to form water. Hence, the rusting of iron is faster in saline water than in pure water.

Question 5.
What happens Δ⁰_m for weak electrolytes obtained by using Kohlrausch law if the migration of ions is increased to three
Answer:
Δ⁰_m for weak electrolytes obtained by using Kohlrausch law is independent of the migration of ions.

Question 6.
Define molar conductivity of a solution.
Answer:
The molar conductivity of a solution at a given concentration is the conductance of volume ‘V’ of a solution containing 1 mole of _ the electrolytic kept between two electrodes with cross sectional area ‘A’ and distance of unit length.
Or,
Δ_m = \(\frac{\Delta}{1}\) K
Now,
1 = 1 and Δ = V (volume containing 1 mole of the electrolyte)
∴ Δ_m = KV

Question 7.
What are the factors that affect the conductivity of an ionic (electrolytic) solution?
Answer:
The conductivity of an ionic (electrolytic) solution depends upon the following factors.

Temperature
Concentration of electrolyte
Nature of the electrolyte added
Nature of solvent and its viscosity
Size of the ions produced and their solvation.

Question 8.
The electrolysis of a salt solution of a metal was carried out by passing a current of 4A for 45 minutes. This resulted in the deposition of 2.977g of the metal. If the atomic mass of the metal is 106.4 g mol^-1, then calculate the charge present in the metal cation.
Answer:
Let the charge on the metal cation be h i.e. the metal cation is M⁺
Accordingly,
M^h+ + 4e^– → M
Therefore, a current of h × 96500 coulomb will deposit 106.4 g of metal quantity of charge passed = 10800 coulombs
Now, 10800 coulombs will deposit

Hence, the charge present on the metal cation is + 4.

Question 9.
(a) At infinite dilution, the ionic conductance of Ba²⁺ and Cl is 121 and 76 ohm1 cm respectively. What will be the equivalent
conductance of BaCl₂ (in ohm^-1 cm²) at infinite dilution?
(b) What effect does concentration have on the molar conductivity of a strong electrolyte?
Answer:
(a) The molar conductivity of barium chloride is given by the following equation:
Δ⁰_m(BaCl₂) = Δ⁰Ba²⁺ + 2 Δ⁰_Cl-
= 127 + 2 × 76 = 279Ω^-1 cm² mol^-1
Δ⁰ eq = \(\frac{279}{2}\) Ω^-1cm²Eq^-1
= 139.5 Ω^-1cm² Eq^-1
[∴ Eq.wt. of BaCl2 = \(\frac{1}{2}\) × mol.wt]
Hence the equivalent conductance of BaCl₂ at infinite dilution is 139.5 Ω^{-1^{cm² eq^-1}}

(b) The molar conductivity of a strong electrolyte decreases with the square root of concentration in linear fashion, as shown below.

Question 10.
(a) The standard reduction potentials of Fe³⁺ | Fe²⁺ and I^–³ | I^–are 0.77 V and 0.54 V respectively for the reaction 2Fe³⁺ + 3I^– ⇌ 2Fe²⁺ + I^–₃. Calculate the value of equilibrium constant.
(b) How much charge is required 1 mole of Cu²⁺ to Cu
Answer:
(a) The cell may be represented as

Hence, the equilibrium constant for the given cell is 6.025 × 10⁷ (b) When copper is reduced, the following reaction takes place Cu²⁺ + 2e^–→ Cu
Hence, the quantity of charge required for the reduct ion of 1 mole of Cu²⁺= 2F
= 2 × 96500 C = 193000C

2nd PUC Kannada Workbook Answers Chapter 11 Gade Mathu Vistarane

Prasanna — Tue, 14 Jul 2026 05:49:18 +0000

You can Download 2nd PUC Kannada Workbook Answers Pallava Chapter 11 Gade Mathu Vistarane, 2nd PUC Kannada Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

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