CBSE Sample Papers for Class 11 Chemistry Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 5 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

All questions are compulsory.
This question paper contains 37 questions.
Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.
Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.
Questions 28 – 34 in Section C are long-answer I type questions carrying 3 marks each.
Questions 35-37 in Section D are long-answer II type questions carrying 5 marks each.
There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.
Use log tables, if necessary. Use of calculator is not allowed.

Section – A

Question 1.
Which of the following molecules will not show geometrical isomerism? [1]
(A)

(B)

(C)

(D)

OR
Arrange the following alkyl halides in decreasing order of the rate of (3-elimination reaction with alcoholic KOH. H
(a)

(b) CH₃ – CH₂ – Br
(c) CH₃ – CH₂ – CH₂ – Br
(A) A > B > C
(B) C > B > A
(C) B > C > A
(D) A > C > B
Answer:
(D)

Explanation :
It will not show geometrical isomer¬ism due to presence of same group on one carbon atom of the double bond.

Commonly Made Error:
Students get confused and choose wrong options.

Answering Tip:
Understand the concept of geometrical isomerism and remember the conditions which are necessary for a compound to show geometrical isomerism to answer here.

Option (D) is correct.

Explanation:
More the number of β-substituents (alkyl groups), more stable alkene will be formed on β-elimination and more will be the reactivity. Thus, the decreasing order of the rate of β-elimination reaction with alcoholic KOH is : A > C > B.

Commonly Made Error:
Generally, students get confuse and choose wrong options.

Answering Tip:
Understand the concept of beta-elimination and remember how increasing in alkyl groups will affect the stability and reactivity of formed alkene.

Question 2.
Number of angular nodes for 4d orbital is ___________ . [1]
(A) 4
(B) 3
(C) 2
(D) 1
Answer:
(C) 2

Explanation:
Angular node is equal to the azimuthal quantum number (l). That means the quantum number (l) determines the number of angular nodes.
For 4d orbital,
n = 4
l = 2
The number of angular node = 1 = 2
Therefore, the number of angular nodes for 4d orbital is, 2.

Commonly Made Error:
Students confuse angular node with radial node and give wrong answer.

Answering Tip:
Understand the concept of both angular and radial nodes and the way it is found as angular nodes are equal to azimuthal quantum number and radial node = n – 1 – 1.

Question 3.
In PO^3-₄ ion the formal charge on the oxygen atom of P – O bond is [1]
(A) +1
(B) -1
(C) -0.75
(D) + 0.75
Answer:
(B) -1

Explanation:
In PO₄^3- ion the formal charge on the O atom of P – O bond =
(Number of valence electrons in free atom) – (Number of lone pair electrons + 1/2 Number of bonding electrons) = 6 – (6 + 1/2 × 2) = 6 – 7 = -1

Question 4.
In which of the following compounds, an element exhibits two different oxidation states ? [1]
(A) NH₂OH
(B) NH₄NO₃
(C) N₂H₄
(D) N₃H
Answer:
(B) NH₄NO₃

Explanation:
NH₄NO₃ is formed of NH₄⁺ and NO₃^– ions. The oxidation number in these two species is different as shown below:
NH₄ ⁺:
Let the oxidation number of N in NH₄ ⁺ be x.
x + 4(+1) = +1
x = -3
NO₃^–:
Let the oxidation number of N in NO₃^– be y.
y + 3(-2) = -1
y = +5

Commonly Made Error:
Students generally do mistakes while finding the oxidation number.

Answering Tips :

Learn the general oxidation number shown by elements. Some elements have more than one oxidation number, hence, learn these special elements too.
While finding the oxidation number of species, remember that sum of the oxidation number of elements
will be equal to the charge on the species. Keeping this in mind you need to calculate.

Question 5.
A hydrocarbon with carbon-carbon single bond is called [1]
(A) alkene
(B) alkane
(C) alkyne
(D) benzene
OR
In a homologous series, a member differs from another member by the molecular mass of:
(A) 10
(B) 12
(C) 14
(D) 28
Answer:
(B) 12

Explanation:
A hydrocarbon with carbon-carbon single bond is called alkane. It is a saturated hydrocarbon.

Explanation:
In a homologous series, a member differs from another member by -CH2 group, i.e., molecular mass of 12 + 2 = 14.

Question 6.
The frequency, wavelength and velocity of light are related to each other by: [1]
(A) c = λv
(B) λ = cv
(C) v = cλ
(D) c = λ/v
Answer:
(A) c = λv

Explanation:
The relation between the frequency, wavelength and velocity of light is given by c = λv.

Read the passage given below and answer the following questions:
Because ethane exists as a gas at normal temperature and pressure, exposure occurs by inhalation. Concentrations of ethan in natural gas range from 5 to 10%. It is also found in the exhaust of diesel (—1.8%) and gasoline (1.3 – 2.0%) engines. Small amounts of ethane, along with other Q and C4 alkanes and alkenes, have been detected in mined coal samples. Ethane emissions from cigarettes have been measured at 1600 ug per cigarette. Typical background air concentrations in major US cities range from 0.05 to 0.5 ppm. Because it is lighter than air, a major spill would not be expected to migrate and affect adjacent properties or neighbourhoods. It is possible to spill liquid ethane from a refrigerated tank, causing frostbite upon contact with the skin due to rapid evaporation and .. loss of heat.

Answer the questions (7) to (10) given below:

Question 7.
The number of moles of carbon atoms in 3 moles of ethane will be [1]
(A) 6 moles
(B) 12 moles
(C) 3 moles
(D) 1 mole
Answer:
(A) 6 moles

Explanation :
1 mole of ethane (C₂H₆) contains 2 moles of carbon atoms.
So, 3 moles of C₂H₆ will contain = 2 × 3 = 6 mol of carbon atoms.

Question 8.
How many moles of hydrogen atom is present in 3 moles of ethane? [1]
(A) 16 moles
(B) 12 moles
(C) 18 moles
(D) 3 mole
Answer:
(C) 18 moles

Explanation:
1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
So, 3 moles of C₂H₆ will contain = 6 × 3 = 18 mol of hydrogen atoms.

Question 9.
How many molecules of ethane will be there in 3 moles of ethane? [1]
(A) 6.02 × 10²³ molecules
(B) 1.8066 × 10²⁴ molecules
(C) 18.066 × 10²⁴ molecules
(D) 1.8066 × 10²³ molecules
Answer:
(B) 1.8066 × 10²⁴ molecules

Explanation:
1 mole of C₂H₆ contains 6.022 × 10²³ molecules of ethane.
So, 3 moles of C₂H₆ will contain
= 3 × 6.022 × 10²³
= 18.066 × 10²³ molecules
= 1.8066 × 10²⁴ molecules

Question 10.
Can liquid ethane cause frostbite? How? [1]
Answer:
Liquid ethane is highly vaporisable, so can cause frostbite when in contact with skin due to vapourisation and loss of heat from the affected area.

Question 11.
Assertion (A): All isotopes of a given element show the same type of chemical behaviour.
Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom. [1]
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) A is true but R is false.
(D) Both A and R are false.
Answer:
(A) Both A and R are correct and R is the correct explanation of A.

Explanation:
Chemical properties of atoms are controlled by the number of electrons, which are determined by the number of protons in the nucleus. Number of neutrons present in the nucleus have very little effect on the chemical properties of an element.
Therefore, all the isotopes of a given element show same chemical behaviour.

Question 12.
Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.
Reason (R): In crystals, molecules organise in an ordered manner. [1]
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) A is true but R is false.
(D) Both A and R are false.
Answer:
Option (A) is correct.

Explanation:
Entropy of system decreases when the fiquid changes into solid. In crystalline state, molecules are ordered having less randomness.

Question 13.
How can you separate a mixture of iodine and sodium chloride?
OR
Which technique can be used for purification of iodine that contains traces of NaCl? [1]
Answer:
A mixture of iodine and sodium chloride can be separated by sublimation as iodine undergoes sublimation.

Question 14.
What type of hydrocarbons are present in high octane gasoline?
OR
What do you mean by cracking? [1]
Answer:
Branched chain aliphatic and/or aromatic hydrocarbons.
OR
The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking.

Question 15.
Can we separate two liquids A (b.p. 353 K) and B (b.p. 365 K) present in a mixture by simple distillation? [1]
Answer:
No, because in simple distillation, vapours of both the liquids will be formed simultaneously and will condense together in receiver as the difference between the boiling points is very less. They can be separated by fractional distillation.

Question 16.
Why entropy of steam is more than that of water at its boiling point? [1]
Answer:
Entropy of steam is more than that of water at its boiling point because in steam, the molecules can move freely so there is more disorder, i.e. more entropy.

Question 17.
Can a moving cricket ball have a wave character? Justify your answer. [1]
Answer:
Yes, a moving cricket ball has a wave character. It possesses wave length which is called de Broglie wavelength and is given by the equation.
λ = \(\frac{h}{m v}=\frac{h}{p}\)

Question 18.
Can we use K in place of Na for preparation of Lassaigne’s extract? [1]
OR
How will you separate a mixture of o-nitrophenol and p-nitrophenol ?
Answer:
K is not used normally because it is more reactive than Na. It reacts violently on heating which makes the handling of tube very difficult.
OR
It can be separated by steam distillation, o-nitrophenol being volatile distills over along with water while p-nitrophenol being non-volatile remains in the flask.

Question 19.
Define the term redox titration. [1]
OR
Arrange X, Y and Z in order of increasing reducing power.
Answer:
A titration by which the strength of an oxidant or reductant is determined by titrating it with standard solution of reductant or oxidant using a redox sensitive indicator is called redox titration.

Greater the value of more negative standard reduction potential of the metal, greater will be its reducing power.

Cation	Standard Reduction Potential (V)
X	+ 0.52
Y	-3.03
Z	-1.18

∴ Increasing order of reducing power:
X < Z < Y

Question 20.
Give different isomers of C₄H₁₀ with their IUPAC names.? [1]
Answer:

Section – B

Question 21.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen. [2]
OR
What is the distance of separation between 3^rd and 4^th orbit of H-atom?
Answer:

Commonly Made Error:
Generally, students take n₁ and n₂ values wrongly. They also forget to write the formula sometimes.

Answering Tips

Write working formula followed by substitution of values.
p- Write the answer in standard form with proper units.
Learn the n values for all the series in the hydrogen spectrum.

Distance of 3^rd orbit
d₃ = \(\frac{52.9 \times n^2}{Z}\)
= \(\frac{52.9 \times(3)^2}{1}\)
= 476.1 pm
Distance of 4^th orbit
d₄ = \(\frac{52.9 \times n^2}{Z}\)
= \(\frac{52.9 \times(4)^2}{1}\)
= 846.4 pm (1)
∴ Distance of separation between 3^rd and 4^th orbit of H-atom = Distance of 4^th orbit – Distance of 3^rd orbit from the nucleus
= 846.4 – 476.1
= 370.3 pm = 3.703 Å

Commonly Made Error:
Students often forget to find the difference in distance between the asked orbits.

Answering Tip:
Always read the given problem thoroughly and then writing formula, do the calculations. After calculations done, check once again for the completion of problem as no questions of the problem are missed out.

Question 22.
Identify the oxidant and reductant in the reaction : [2]
I₂(aq) + 2S₂O^–₃(aq) > 2I(aq) + S₄O₆^2-(aq)
Answer:
I₂(aq) + 2S₂O^–₃(aq) > 2I(aq) + S₄O₆^2-(aq)
For the identification of oxidant and reductant, the oxidation numbers of I and S in each species are calculated as :
Oxidation number of I in I₂ = 0
Oxidation number of S in S₂O₃^2-
= 2(x) + 3(-2) = – 2
2x – 6 = -2
2x = -2 + 6 = 4
x = \(\frac{4}{2}\) = +2
Oxidation number of I in I^-1 = -1
Oxidation number of S in S₄O₆2^–
= 4(x) + 6(-2) = -2
4x – 12 = -2
4x = – 2 + 12 = 10
x = \(\frac{10}{4}\) = +2.5
Now,

Since, in this reaction, the oxidation number of I decreases from -1 to 0. So, it acts as oxidant which oxidises S₂O₃^2-. Whereas, the oxidation number of S increases from +2 to +2.5. So, it acts as reductant which reduces I₂.
Oxidant = I₂
Reductant = S₂O₃^2-

Commonly Made Error:
Students do mistakes while calculating the oxidation number of I and S in species. This mistake confuses them and give wrong answers.

Answering Tips

Explain the relation of oxidation number with oxidant and reductant.
Calculate the oxidation number of I and S in species by substituting correct oxidation number of other species and equating them to the charge of species. Once this part is done correctly, explain the relation of oxidation number with oxidant and reductant.

Question 23.
Convert benzene into: [2]
(a) Toluene
Answer:

(b) BHC
Answer:

Commonly Made Error:
Students do mistakes while writing conversion reactions. They sometime miss reactants, conditions in the reactions.

Answering Tip:
While writing conversion, students must:

Write all the reactants required.
The conditions responsible for conversion.

Question 24.
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. [2]
Answer:
The ethylation reaction of benzene involves the addition of an ethyl group on the benzene ring. Such a reaction is called a Friedel-Craft alkylation reaction. This reaction takes place in the presence of a Lewis acid. Any Lewis acid like anhydrous FeCl₃, SnCl₄ BF₃, etc. can be used during the ethylation of benzene.

Question 25.
Account for the following : [2]
(a) BF₃ molecule has zero dipole moment although B-F bonds are polar.
(b) The structure of NH₃ is pyramidal.
OR
Draw the Lewis dot structure of C0₃^2- ion.
Answer:
(a) In case of BF₃ molecule, the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third i.e.,

M₁ and M₂ cancel each other, hence dipole moment becomes zero.

(b) Structure of NH₃ is pyramidal as it contains one lone pair and three bond pairs. If there was a bond pair instead of a lone pair, the geometry should be tetrahedral. Since, a lone pair is present and the repulsion between lone pair and bond pair causes the angle between the bond pairs to reduce to 107°.

Total no. of valence electrons of C0₃
= 4 + 3 × 6 = 22
Total no. of valence electrons on C0₃^2-
= 22 + 2 = 24
The skeletal structure of C0₃^2- is

In the above structure, octet of oxygen is not complete. Hence, multiple bonding is required between C and O.

Commonly Made Error:
Students show the structure without explaining how the valence electrons are calculated.

Answering Tip:
Initially show the number of valence electrons for the compound. Then show how for an ion, the valence electrons are changed. Then write structure of the ion.

Question 26.
Write IUPAC names of the products obtained by the ozonolysis of the following compounds: [2]
(a) Pent-2-ene
(b) 3,4-Dimethylhept-3-ene
Answer:
(a) Pent-2-ene undergoes ozonolysis as:

The IUPAC name of product (I) is ethanal and product (II) is propanal.

(b) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:

The IUPAC name of product (I) is butan-2-one and product (II) is pentan-2-one.

Commonly Made Errors:

Some students write the reaction till ozonide formation.
Some student write only one product.
Some often do not do the naming correctly.

Answering Tip:
Understand the ozonolysis reaction and learn how ozonide formed and how products ‘ formed. Practice as many reactions as possible : to avoid mistakes.

Question 27.
(i) When the technique sublimation is used for purification?
(ii) What is the Prussian blue coloured compound formed in the Lassaigne’s test for confirming the presence of nitrogen in the organic compound? [2]
Answer:
(i) This method is used to separate the sublimable compounds from non- sublimable compounds.
(ii) It is Iron (II) hexacyanoferrate(II), Fe₄[Fe(CN)₆]₃.xH₂O. (1)

Commonly Made Error:
Students often write wrong formula of the complex or do not write it at all.

Answering Tip:
Students must write the name of the compound ! with correct valency and correct formula.

Section – C

Question 28.
The reaction 2C + O₂ → 2CO is carried out by taking 24.0 g of carbon and 96.0 g of 02.
Find out
(i) Which reactant is left in excess?
(ii) How much of it is left?
(iii) How many grams of the other reactant should be taken so that nothing is left at the end of the reaction? [3]
Answer:
(i)

24 g of carbon will react with 32 g of O₂ to from 56 g of CO.
So, oxygen is left in excess.
(ii) Amount of oxygen left = 96 – 32 = 64 g
(iii) Mass of other reactant = 96 – 24 = 72 g

Question 29.
25 kg pf N₂ and 6 kg of H₂ are mixed to produce NH₃. [3]
(i) Identify the limiting reagent.
(ii) Calculate the amount of ammonia formed in this reaction.
OR
What weight of iodine is liberated from a solution of potassium iodide when 1 litre of Cl2 gas at 10°C and 750 mm pressure is passed through it ?
Answer:
N₂ + 3H₂ → NH₃
(i) 1 mole N₂ requires 3 moles of H₂
∵ 28 kg N₂ require 6 kg H₂
∴ 25 kg N₂ will require = \(\frac{6}{28}\) × 25 = 5.36 kg H₂
Flence, N₂ is the limiting reagent.

(ii) 1 mole N₂ produces 2 moles NH₃
∵ 28 kg N₂ produce 34 moles NH₃
∴ 25 kg N₂ will produce = \(\frac{34}{28}\) × 25s
= 30.35 kg NH₃

The reaction is 2KI + Cl₂ → 2KCl + I₂
22.4 L at STP 2 × 127 = 254 g
(V₁) volume of Cl₂ gas = 1 L,
V₂ = volume of gas at
(P₁) Pressure = \(\frac{750}{760}\) atm, P₂ = 1 atm V 760 ‘ 2
Temperature (T₁) = 10°C + 273 = 283 K,
T₂ = 273 K
∴ V₂ = \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{\mathrm{T}_2}{\mathrm{P}_2}\)
= \(\frac{750 \times 1 \times 273}{760 \times 283 \times 1} s\) = 0.952 L
∴ Volume of Cl₂ (g) passed at STP = 0.952 L
Now 22.4 L of Cl₂ produces at STP = 254 g of I₂ 0.952 L of Cl₂ at STP will produce
= \(\frac{254}{22.4}\) × 0.952 g of I₂
= 10.795 g of I₂

Question 30.
If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mol of water at 1 bar and 100°C is 41 kj mol^-1. Calculate the internal energy change, when :
(i) 1 mol of water is vaporised at 1 bar pressure and 100°C.
(ii) 1 mol of water is converted into ice. [3]
Answer:
(i) H₂O(1) → H₂O(g)
∆n_(g) = 1 – 0 = 1
∵ ∆H = ∆U + ∆n_(g)RT
or ∆U = ∆H – ∆n_(g)RT
Given, ∆H = 41 kj mol^-1,
T = 100 + 273 = 373 K
∴ ∆U = 41 kj mol^-1 -1 × 8.3 × 10^-3 mol^-1 K^-1 × 373 K
= 37.904 kj mol^-1

(ii) H₂O(I) → H₂O(g)
There is negligible change in volume.
So, p∆V = ∆n_g RT = 0
∴ ∆H = ∆U
or ∆U = 41 kj mol^-1

Question 31.
(a) A certain buffer solution contains equal concentration of X^– and HX. The K_b for X^– is 10¹⁰. What is the pH of the buffer?
(b) Assuming that the buffer in blood is CO_2- HCO_3-, calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH, 7.4 (K_a of H₂CO₃ = 4.5 × 10^-7). [3]
Answer:
(a) Given, K_b = 10-1°
PK_b = – log [K_b] = – log [10^-10]
= 10 log 10 = 10
∵ PK_a + pK_b = 14
or PK_a = 14 – pK_b
= 14 – 10 = 4
For acidic buffer,
pH = pK_a + log\(\frac{\text { [salt] }}{\text { [acid] }}\)
∵ [X^–] = [HX]
∴ pH = pK_a = 4

(b) Given
pH = 7.4
K_a = 4.5 × 10^-7
CO₂ + H₂O ⇌ H⁺ + HCO_3-
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_3^{-}\right]}{\left[\mathrm{CO}_2\right]}\) …….. (i)
Now pH = – log [H⁺]
or [H⁺] = 10^-PH
= 10^-7.4
= 4.0 × 10-⁸
So, from equation (i),

Question 32.
Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br
OR
Write the Lewis dot structure of CO molecule.
Answer:

Commonly Made Errors:

Students only show lewis structure.
Sometimes they miss to write the steps to reach the final answer.

Answering Tip:
When a Lewis dot structure is asked to write for an atom, then writing that atom’s electronic configuration is very important. From which valence electrons of that atom will be known. These details should be given.

Step 1 : Count the total number of valence electrons of carbon and oxygen atoms. The outer (valence) shell configurations of carbon and oxygen atoms are: 2s² 2p² and 2s² 2p⁴, respectively. The valence electrons available are 4 + 6 = 10.

Step 2 : The skeletal structure of CO is written as: CO

Step 3 : Draw a single bond (one shared electron pair) between C and O and complete the octet on O, the remaining two electrons are the lone pair on C.

This does not complete the octet on carbon and therefore we have to resort to multiple bonding (i.e. triple bond) between C and O atoms. This satisfies the,octet rule conditions for both atoms.

Question 33.
(a) Define the equilibrium constant.
(b) For the general reaction:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Derive the relationship between K_p and K_c. [3]
Answer:
(a) Equilibrium constant: It is the ratio of the rate constants of forward reaction to that of backward reaction.

(b) For the general reaction,
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Equilibrium constant,

Commonly Made Error:
Some students miss certain steps in the derivation.

Answering Tip:
The relation between K_p and K_c must be clearly written. Do not miss any step while derivation.

Question 34.
(i) The energy associated with the first orbit in the hydrogen atom is – 2.18 × 10^-18 J atom^-1. What is the energy
associated with the fourth orbit?
(ii) Calculate the radius of Bohr’s third orbit for hydrogen atom. [3]
Answer:
(i) E = – 2.18 × 10^-18 × \(\frac{Z^2}{n^2}\) J atom^-1
For H-atom, Z = 1
Value of n for fourth orbit = 4
∴ E= – 2.18 × 10^-18\(\frac{(1)^2}{(4)^2}\)
= \(\frac{-2.18 \times 10^{-18}}{16}\)
= – 0.136 × 10^-18
= – 1.36 × 10^-19 J

(ii) Radius of third orbit for H-atom
r_n = \(\frac{52.9 n^2}{Z}\)
r₃ = \(\frac{52.9 \times(3)^2}{1}\)
= 52.9 × 9
= 476.1 pm

Section – D

Question 35.
(a) What is the free expansion? Determine work done in case of free expansion of an ideal gas.
(b) 4.0 mol of ideal gas at 2 atm and 25°C expands isothermally to 2 times of its original volume against the external pressure of 1 atm. Calculate work done. If the same gas expands isothermally in a reversible manner,
OR
(a) What is reversible process in thermodynamics ?
(b) Name the thermodynamic process for which
(i) q = 0
(ii) ΔU = 0
(iii) ΔV = 0
(iv) Δp = 0
(c) Water decomposes by absorbing 286.2 kj of electrical energy per mole. When H₂ and O₂ combine to form one mole of H₂O, 286.2 kj of heat is produced, Which thermodynamic law is proved? Write its statement, [5]
Answer:
(a) When a gas expands under vacuum, i.e., no external pressure work on it (p_ex = 0), its expansion is called free expansion.
∵ P_ex = 0
W = 0 in case of free expansion of an ideal gas [∵ W = – p_ex (V₂ – V₁) = 0]
It means no work is done.

(b) Given
n = 4 moles
p = 2 atm
T = 25 + 273 = 298 K
P_ex = 1 atm
∴ W = -p_ex(V_f – V_i.)
Now, initial volume (V_i) = \(\frac{n \mathrm{RT}}{p}\)
= \(\frac{4 \times 0.082 \times 298}{2}\)
= 48.87 L
∵ Volume becomes 2 times of its original volume.
Final volume (Vf) = 48.87 × 2 = 97.74 L
W = – 1 (97.74 – 48.87)
= -1 (48.87)
= -48.87 J
For isothermal reversible expansion of ideal gas,
W_rev, = -2.303 nRT log\(\frac{V_f}{V_i}\)
= – 2.303 × 4 × 8.314 × 298 log\(\left(\frac{97.74}{48.87}\right)\)
= – 2.303 × 4 × 8.314 × 298 × 0.3010
= – 6869.84 J

(a) Reversible Process: It is defined as a system where change takes place infinitesimally slow and the direction of which at any point can be reversed by infinitesimal change in the state of the system.

(b)
(i) For q = 0, heat is constant, so, the process is adiabatic process.
(ii) For ΔU = 0, internal energy is constant which takes place at constant temperature, so, the process is isothermal process.
(iii) For ΔV = 0, volume is constant, so, the process is isochoric process.
(iv) For Δp = 0, pressure is constant, so, the process is isobaric process.

(c) H₂O(g) >H₂(g) + 1/2O₂(g);
Δ_r\(\mathrm{H}^{\Theta}\) = 286.2 kJ mol^-1
H₂(g) + 1/2O₂(g) > H₂O(S);
Δ_r\(\mathrm{H}^{\Theta}\) = – 286.2 kJ mol^-1
It is referred to as first law of thermodynamics. According to this law, “Energy can neither be created nor be destroyed but can be converted from one form to another”.

Question 36.
Derive the derivation of De-Broglie relationship. [5]
Answer:
The relationship may be derived by combining the mass-energy relationship proposed by Max Planck and Einstein.
According to Planck, photon of light having energy E is associated with a wave of frequency u as:
E = hυ …(i)
According to Einstein, mass and energy are related as:
E = mc² …(ii)
where c is the velocity of light
Combining the above two relations in equation (i) and (ii), we get:
hυ = mc²
Now, since

The equation is valid for a photon, de Broglie suggested that on substituting the mass of the particle m and its velocity V in place of velocity of light c, the equation can also be applied to material particles.
Thus, the wavelength of material particles, λ is :
λ = \(\frac{h}{m v}\)
This equation is known as de Broglie equation,
∵ Momentum of particle, p = mv
∴ λ = \(\frac{h}{p}\)
where p stands for the momentum (mv) of the particle. Since h is constant,
∴ λ ∝ \(\frac{1}{\text { Momentum }}\)
It means that the wavelength of a particle in motion is inversely proportional to its momentum.
This equation is known as de Broglie relation.

Question 37.
Give the I.U.P.A.C. names of the following compounds: [5]

Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation,
(a) C₆H₅OH
(b) C₆H₅NO₂
(c) CH₃CH=CHCHO
(d) C₆H₅-CHO
(e) C₆H₅CH₂
Answer:
(i) 2-Bromo-4-methylpentane-3-one
(ii) 4-Methyl-2-nitropentane-3-one
(iii) 2-Ethoxy-4-methoxypentane-3-one
(iv) 2-Methoxy, 4-methylhexane-3-one
(v) 2-Methylpropanoic acid