CBSE Sample Papers for Class 11 Chemistry Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 1 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 1 type questions carrying 3 marks each.
• Questions 35 – 37 in Section D are long-answer 11 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.
What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL of solution? [1]
(A) 4 mol L^-1
(B) 20 mol L^-1
(C) 0.2 mol L^-1
(D) 2 mol L^-1
OR
One mole of any substance contains 6.022 × 10²³ atoms/molecules. Number of molecules of H₂S0₄ present in 100 mL of 0.02 M H₂SO₄ solution are
(A) 12.044 × 10²⁰ molecules
(B) 6.022 × 10²³ molecules
(C) 1 × 10²³ molecules
(D) 12.044 × 10²³ molecules
Answer:
Option (C) is correct.
Explanation : Weight of NaCl (w) = 5.85 g
Volume of solution (V) = 500 mL
Molecular weight of NaCl = 58.5 g
Molarity = \(\frac{\mathrm{w} \times 1000}{\text { Molecular weight } \times \text { Volume in } \mathrm{mL}}\)
= \(\frac{5.85 \times 1000}{58.5 \times 500}\) = 0.2 mol L^-1

OR
Option (A) is correct.
Explanation :
Number of moles of H₂S0₄ = molarity × volume in mL
= 0.02 × 100 = 2 milli moles
= 2 × 10^-3 mol
Number of molecules of H₂SO₄ = Number of moles × N_A
= 2 × 10^-3 × 6.022 × 10²³
= 12.044 × 10²⁰ molecules

Question 2.
In NO^3- ion, the number of bond pairs and lone pairs of electrons in nitrogen atom are: [1]
(A) 2, 2
(B) 3, 1
(C) 1, 3
(D) 4, 0
Answer:
Option (D) is correct.

Explanation:
The number of valence electrons in nitrogen atom = 5+1 (due to negative charge). One O-atom forms double bond with N-atom and two O-atoms shared with two electrons of N-atom. So, 3 O-atoms shared with 8-electrons of N-atom. Hence, Number of bond pairs (or shared pairs) = 4 and Number of lone pairs = 0

Question 3.
We know that the relationship between K_c and K_p is
K_p = K_c(RT)^Δn
What would be the value of An for the reaction
NH₄Cl(s) ⇌ NH₃(g) + HCl(g) [1]
(A) 1
(B) 0.5
(C) 1.5
(D) 2
Answer:
Option (D) is correct.

Explanation:
Δn = (Number of moles of gaseous products) – (Number of moles of gaseous reactants) = 2 – 0 = 2

Commonly Made Error
Students sometimes forget that only gaseous components are to be taken into account for calculation.

Answering Tip
Remember that An is calculated only for gaseous products.

Question 4.
The largest oxidation number exhibited by an element depends on its outer electronic configuration. In which of the following outer electronic configurations, the element will exhibit highest oxidation number? [1]
(A) 3d¹ 4s²
(B) 3d³ 4s²
(C) 3d⁵ 4s¹
(D) 3d⁵ 4s²
Answer:
Option (D) is correct.

Explanation:
Highest oxidation number of any transition element is the sum of (n-l)d electrons and ns electrons. Therefore, higher the number of electrons in 3d orbitals, greater is the oxidation number. In the given above choices, the oxidation numbers are as below :
(A) 1 + 2 = 3
(B) 3 + 2 = 5
(C) 5 + 1 = 6
(D) 5 + 2 = 7 (1)

Question 5.
The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H⁺ ion to

portion which can be shown as

OR
Which of the following is the correct JUPAC name? [1]
(A) 3-Ethyl-4,4-dirnethylheptane
(B) 4,4-Dirnethyl-3-ethylhepta ne
(C) 5-Ethyl .44-dimethyiheptane
(D) 4,4-Bis(rnethyl)-3-cthylheptane
Answer:
Option (A) is correct.

Explanation:
Since, double bond is a source of electrons and the charge flows from source of more electron density, therefore, n electrons of the double bond attack the proton.

OR

Option (A) is correct.

Explanation:
Because, while writing IUPAC name, the alkyl groups are written in alphabetical order. Thus, lower locant 3 is assigned to ethyl. Prefix, di, tri, and tetra are not included in alphabetical order.

Question 6.
Which of the following is a Lewis acid? [1]
(A) AlCl₃
(B) MgCl₂
(C) CaCl₂
(D) BaCl₂
Answer:
Option (A) is correct.

Explanation :
Lewis add are compounds which can take electron from a donor compound. Since, AlCl₃ is electron deficient so act as a Lewis acid.

Read the passage given below and answer the following questions:
Entropy is a measure of degree of randomness. Entropy is directly proportional to temperature. Every system tries to acquire maximum state of randomness or disorder. Entropy is a measure of unavailable energy.
Unavailable energy = Entropy x Temperature
The ratio of Entropy of Vaporization and boiling point of a substance remains almost constant.
Answer the questions (7) to (10) given below:

Question 7.
Which of the following has S = +ve? [1]
(A) H₂(g) + Cl₂(g) → 2HCl (g)
(B) Boiling of egg
(C) Crystallisation of Sugar
(D) Formation of Complex Compound
Answer:
Option (B) is correct.
Explanation .-Boiling of egg leads to denaturation of proteins which has random coil structure. Hence, ΔS = +ve.

Question 8.
The sign of ΔS in the reaction, N₂(g) + O₂(g) → 2NO(g) is: [1]
(A) +ve
(B) – ve
(C) Zero
(D) None of these
Answer:
Option (C) is correct.

Explanation:
It will be spontaneous only if
ΔS = +ve.

Question 9.
Which of the following has S = – ve? [1]
(A) Adsorption
(B) Dissolution of NH₄Cl in water
(C) H₂ → 2H
(D) 2NaHC0₃(s) → Na₂C0₃ + C0₂ + H₂O
Answer:
Option (A) is correct.

Explanation:
In case of adsorption, entropy decreases due to attraction between adsorbent and adsorbate.

Question 10.
The law of thermodynamics, which helps to determine absolute entropy is: [1]
(A) Zeroth law
(B) I law
(C) II law
(D) III law
Answer:
Option (D) is correct.

Explanation:
Nernst invented third law of thermodynamics which helps to determine absolute entropy of various substances.

Question 11.
Assertion (A): Generally, ionisation enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(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) Both A and R are not correct.
(D) A is not correct but R is correct.
Answer:
Option (A) is correct.

Question 12.
Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R): The enthalpies of all elements in their standard state are zero.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Answer:
Option (B) is correct.

Explanation:
Combustion reactions breaks the bonds of organic compound molecules, and the resulting water and carbon dioxide bonds always release more energy than was used to break them originally. That’s why burning organic compounds produces energy and is exothermic.

Question 13.
Cr in ground state has how many unpaired electrons. (Atomic number of Cr = 24)
Answer:
₂₄Cr = 2,8,13,1
= 1s², 2s², 2p⁶, 3s²3p⁶ 3d⁵4s¹
Number of unpaired electrons in Cr in ground state = 6

Question 14.
Arrange the following in order of increasing covalent character is this phenomenon observed : MCI, MBr, MF, MI (where M = alkali metal). [1]
Answer:
The covalent character in the M-X bond depends on the difference in electronegativity (ΔEN) between the two atoms. If ΔEN is small, the bond will have a large amount of covalent character. If ΔEN is large, the bond will have a small amount of covalent character. As we move down the group, the size of the halogens increases and their electronegativity decreases. Hence, the increasing order of covalent character in the M-X bond will be,
MF < MCI < MBr < MI

Question 15.
Which has more energy of electron 4p or 5s?
OR
Nitrogen has correct configuration of 1s², 2s², 2p¹_x, 2p¹_y, 2p¹_z is described by which principle?
Answer:
5s has more energy.
OR
It is described by Hund’s rule of maximum multiplicity.

Question 16.
Define chemical equilibrium. Give an example. [1]
Answer:
A chemical equilibrium is a system whose chemical composition of the system does not change when dynamic equilibrium is reached.
e.g: N₂(g) + 3H₂(g) → 2NH₃(g)

Question 17.
Calculate the oxidation number of Cr in Cr₂O₇^2-.
OR
What is the oxidation state of K in KO₂?
Answer:
Cr₂O₇^2-
Let, the oxidation number of Cr = x
2(x) + 7(-2) = -2
2x – 14 = – 2
2x = – 2 + 14
x = \(\frac{12}{2}\) = + 6

OR

Let the oxidation number of K = x
x + 2(\(\frac{-1}{2}\)) = 0
x – 1 = 0;
x = + 1.

Commonly Made Error:
Generally, students give incorrect oxidation number values for known elements while ‘ calculating oxidation number. They also miss the charge on the compound, while calculation.

Answering Tip:
Learn the general oxidation number values of elements and also learn the special oxidation number values for oxygen as it varies in many cases. Always remember to write the sum of oxidation number equal to its charge value, if present or to zero if charges are not present in the given compound.

Question 18.
Show the polarisation of carbon-magnesium bond in the following structure.
CH₃—CH₂—CH₂—CH₂—Mg—X [1]
Answer:

Carbon is more electronegative than magnesium.

Question 19.
Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by

OR
If a liquid compound decomposes at its boiling point, which method(s) will you choose for its purification? It is known that the compound is stable at low pressure, steam volatile and insoluble in water. [1]
Answer:
They show metamerism.
OR
Steam distillation can be used for its purification. This method is applied to separate substances which are steam volatile and immiscible with water.

Question 20.
What is Pauli’s exclusion principle? [1]
Answer:
According to this principle, “No two electrons in an atom can have the same set of four quantum numbers.”

Section – B

Question 21.
How is matter classified at macroscopic level ? [2]
Answer:
At the macroscopic level, matter can be classified as mixtures and pure substances. These can be further sub-divided as:

Question 22.
Give the different conformations of ethane with their Newmann Projection formulae. [2]
Answer:
Newman Projection Formulae :

Commonly Made Error:
Generally, students wrongly draw or confuse themselves while drawing both the form of Newmann conformations.

Answering Tip:
Understand the concept in which the Newmann Projection formulae is drawn. Remember hydrogen atoms are nearer in eclipsed form (here for ethane) and hydrogen atoms are far apart in staggered form (here for ethane).

Question 23.
Arrange the following in increasing order of pH. [2]
KNO₃(aq), CH₃COONa(aq), NH₄Cl(aq), C₆H₅COONH₄(aq)
Answer:

Salt	Nature of Aqueous Solution	pH
knO3	Salt of strong acid (HNO3) and strong base (KOH), so, its aqueous solution is neutral.	7
CH3COONa	Salt of weak acid (CH3COOH) and strong base (NaOH), So, its aqueous solutions is basic.	> 7
NH4Cl	Salt of strong acid (HC1) and weak base (NH4OH), so, its aqueous solution is acidic.	< 7
C6H5COONH4	Salt of weak acid (C6H5COOH) and weak base (NH4OH), but NH4OH          is slightly stronger than C6H5COOH. So, its solution is slightly basic.	Slightly > 7

Hence, increasing order of pH of above salts is- NH₄C1 < KN03 < C6H5COONH4 < CH3COONa

Question 24.
Why is sigma bond stronger than a pi bond? Explain. [2]
OR
A compound formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. How many structural isomers possible of the product ?
Answer:
igma bonds are stronger than pi bond because the extent of overlapping in sigma bond is greater than that in pi bond. Sigma bond is formed due to the axial overlap of two orbitals whereas pi bond is formed due to the lateral overlapping of two orbitals. Since, the former is more effective, we can say that sigma bond is stronger than pi bond.

Commonly Made Error
Generally, students do not explain the formation of both the bonds and loose marks.

Answering Tip
It is very important to first explain how the sigma and pi bonds are formed and then explain which one is stronger.

Four-
(i) CH₃CH₂CHCl₂
(ii) CH₃CHClCH₂Cl
(iii) CH₃ClCClCH₃
(iv) ClCH₂-CH₂-CH₂Cl

Question 25.
Justify that the reaction :
2Na(s) + H₂(g) → 2NaH(s) is a redox reaction.
Answer:
2Na(s) + H₂(g) → 2NaH(s)
In the above reaction, the compound formed is an ionic compound, which may also be represented as Na⁺ H^–(s). This suggests that one half reaction in this process is :
2Na(s) → 2Na⁺(g) + 2e^– (Oxidation)
and the other half reaction is :
H₂(g) + 2e^– → 2H^–(g) (Reduction)
So, it is evident that sodium is oxidised and hydrogen is reduced in this reaction. So, the complete reaction is a redox reaction.

Commonly Made Error
Generally, students do mistake in writing the half reactions for the equations.

Answering Tip
First understand which element is oxidised and the one reduced in the given redox reaction. Then learn how to write the half reactions with electrons balancing.

Question 26.
What do you understand by the following:
(a) Markovnikov’s rule
(b) Huckel’s rule
OR
Complete the following and name the products: [2]

Answer:
(a) Markovnikov’s Rule:
his rule states that, ‘The negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms, e.g.,

(b) Huckel’s Rule :
This rules states that all planar cyclic conjugated polyenes containing (4n + 2) π electrons where, n = 0,1, 2, ……….. are aromatic in nature.

Commonly Made Error
Some students confuse Markovnikov’s with anti-Markovnikov’s rule.

Answering Tip
Clearly understand the concept of these rules.

OR

Question 27.
The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example. [2]
Answer:
If mass of an object = 1 mg = 10^-6 kg
Then, according to Heisenberg’s uncertainty principle,

Since, the value of Δx.Δv obtained is very small and is insignificant, so, effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles.

Section – C

Question 28.
How would you convert the following compounds into benzene? [3]
(i) Ethyne
(ii) Ethene
(iii) Hexane
Answer:
(i) Benzene from Ethyne:

(ii) Benzene from Ethene :

(iii) Hexane to Benzene :

Commonly Made Error
Generally, students confuse with reactants, reagents, intermediate products, make mistakes in conversion reactions.
Some students get confused in cyclohexane and benzene

Answering Tip
Make a list of conversion reactions in the chapter and learn them. Understand each step in the reaction and the reagents used and their action too. Do not confuse with intermediate products.

Question 29.
Threshold frequency, v0 is the minimum frequency which a photon must possess to eject an electron from a metal. It i§ different for different metals. When a photon of frequency 1.0 × 10¹⁵ s^-1 was allowed to hit a metal surface, an electron having 1.988 × 10^-19 J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
OR
Why was a change in the Bohr Model of atom required? Due to which important development (s), concept of movement of an electron in an orbit was replaced by the concept of probability of finding electron in an orbital? What is the name given to the changed model of atom? [3]
Answer:
Given
v = 1.0 × 10¹⁵ s^-1
K.E. = 1.988 × 10^-19 J
λ = 600 nm = 600 × 10^-9 m
K.E. = h(v – v₀)
K.E. = hv – hv₀

1.988 × 10^-19J = 6.626 × 10^-34 × 1.0 × 10¹⁵ s^-1 – hv₀
hv₀ = 4.638 × 10^-19 J
∴ Threshold frequency (v₀) = \(\frac{4.638 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\)
If λ = 600 nm = 600 × 10^-9 m
Then v = \(\frac{3 \times 10^8 \mathrm{~ms}^{-1}}{600 \times 10^{-9}}\)
= 0.5 × 10¹⁵ s^-1
Sine v < v₀, so, an electron will not be emitted.

Commonly Made Error:
Students often do not convert the units or ! forget to mention unit in the final answer.

Answering Tips:
Check the compatibility of units.

Write working formula followed by substitution of values.
Do not forget to mention unit with the answer.
OR
A change in the Bohr’s model of atom was required due to following reasons:
(i) An electron is regarded as charged particle moving in well defined circular orbits around the nucleus. An orbit can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible on the basis of Heisenberg uncertainty principle.
(ii) The wave character of the electron was not considered in Bohr model.
Due to these drawbacks in Bohr’s model, concept of movement of an electron in an orbit was replaced by the concept of probability of finding the electron in an orbital due to de-Broglie concept of dual behaviour of electron and Heisenberg uncertainty principle. The changed model of atom is called quantum mechanical.

Question 30.
Write seven fundamental quantities and their units. [3]
Answer:

Fundamental Quantity	SI Unit
1. Length	Metre
2. Mass	Kilogram
3. Time	Second
4. Electric current	Ampere
5. Thermodynamic temperature	Kelvin
6. Amount of substance	Mole
7. Luminous intensity	Candela

Question 31.
BeF₂ is linear while SF₂ is angular though both are triatomic. Explain. [3]
Answer:
BeF₂: 4Be = 2, 2
= 1s², 2s²
Be has 2 valence electrons. These two electrons are used to form bonds with two fluorine atoms. So, it contains two bond pairs but no lone pair. The structure of BeF, is shown as below:
F—Be—F
So, its shape is linear.
SF₂: ₁₆S = 2, 8, 6
= 1s², 2s²2p⁶, 3s²3p⁴
S has 6 valence electrons. Out of these, two valence electrons are used to form bonds with two F-atoms. So, it contains two bond pairs and two lone pairs. The structure of SF² is shown as below :

So, its shape is angular due to lp-lp repulsion

Question 32.
State as to why
(a) C0₂ is more soluble in aqueous NaOH than in water.
(b) A solution of ammonium chloride in water shows pH less than 7.
(c) Aluminium chloride is a Lewis acid.
Answer:
(a) CO₂ when dissolved in water dissociates as per the following equilibrium reaction :
C0₂ + H₂O ⇌ HCO₃^– + H⁺
On adding an alkali such as NaOH, it will react with the H+ ions, thus reducing their concentration. The equilibrium will therefore try to raise the concentration of H⁺ ions to their former level, and it does this by dissolving more C0₂.

(b) AlCl₃ when dissolved in water shows pH less than 7 because it gives weak aluminium hydroxide and strong HCl in water.
AlCl₃ + 3H₂O ⇌ Al(OH)₃ + 3HCl

(c) A substance is said to be Lewis acid if it can accept a pair of electron from other compound in a chemical reaction. AlCl₃ has an electron-deficient aluminium atom. It has only six electrons in its valence shell. So, it readily accepts electrons from other atoms, in an attempt to get a full valence shell of eight electrons. That’s why it generally behaves as a Lewis acid.

Question 33.
The radius of the first Bohr orbit of hydrogen atom is 0.529 A, calculate the radii of
(i) the third orbit of He⁺ ion.
(ii) the second orbit of Li²⁺ ion.
OR
Calculate the wavelength of 1000 kg rocket moving with a velocity of 3000 km/hr. [3]
Answer:
Radius of nth Bohr orbit
r_n = \(\frac{0.529 n^2}{Z}\)
For H-atom, Z = 1,
For first orbit n = 1

(i) For He+ion, Z = 2, third orbit, n = 3
r₃ = \(\frac{0.529 \times(3)^2}{2}\)
= \(\frac{0.529 \times 9}{2}\)
= 2.380 A

(ii) For Li²⁺ ion, Z = 3
Second orbit, n = 2
r₂ = \(\frac{0.529 \times(2)^2}{3}\)
= \(\frac{0.529 \times 4}{3}\)
= 0.7053 Å

Commonly Made Error
Generally, students do mistake in taking the Z and n value.

Answering Tip
Be clear with the Z (atomic number values) for the elements given. Remember that n value depends on which orbit, as for first orbit n = 1, for second n = 2 and soon.

OR

Given m = 1000kg
v = 3000km/hr
= \(\frac{3000 \times 1000}{60 \times 60}\) m/s
833.33 m/s

According to de-Broglie equation,
λ = \(\frac{h}{m v}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{1000 \times 833.33 \mathrm{~m} / \mathrm{s}}\)
= 7.951 × 10^-40 m

Commonly Made Error
Students often forget to convert the units into SI unit.

Answering Tip
Always check the units properly.

Question 34.
Differentiate between the following (with examples): [3]
(i) Open and closed system.
(ii) Adiabatic and Isothermal process.
(iii) State function and path function.
Answer:
(i) Open and closed system.

Open System	Closed System
It can exchange matter as well as energy with its surroundings.	It does not exchange matter with it surroundings but car exchange energy in that form of heat.
For example: Reaction mixture For exopen beaker.	For example: Hot water in a closed container.

(ii) Adiabatic and Isothermal process.

Adiabatic Process	isothermal Process
In this process, no heat can flow in or out of the system. i.e., dq = 0	In this process, temperature remains constant. i.e., dT = 0

(iii) State function and path function.

State Function	Path Function
The variables which depend upon the initial and final states of a system are called state functions not on the path through which the final state reaches. For example: Internal energy Pressure, Volume, Enthalpy etc.	The variables which depend upon the path followed are called path functions to reach the final state. For example: Heat, Work etc.

Section – D

Question 35.
(a) Differentiate the terms Bond Dissociation Enthalpy and Bond Enthalpy. [5]
(b) The heat of formation of methane is -17.9 kcal mol x. If the heats of atomization of carbon and hydrogen are 170.9 and 52.1 kcal moL1 respectively. Calculate the C-H bond enthalpy in methane.
OR
(i) Define the following:
(a) Standard enthalpy of formation (Δ_fH°).
(b) Standard enthalpy of combustion (Δ_fH°).
(ii) The combustion of 1 mole benzene takes place at 298 K and 1 atm. After combustion C0₂(g) and H₂O(1) are produced and 3267-0 kj of heat is liberated. Calculate standard enthalpy of formation (Δ_fH°) of benzene. Standard enthalpy of formation of C0₂(g) and H₂O(l) are – 393-5 kj/mol and – 285-83 kj mol^-1 respectively.
Answer:
(a)

Bond Dissociation Enthalpy	Bond Enthalpy
It is the enthalpy change accompanying the breaking of one mole of covalent bond of a gaseous covalent compound to give products in the gas phase.	It is the enthalpy change related to the formation of bond or breaking of bond.
The value will differ from one bond to another.	The value will differ from one bond to another.

(b) Given that
C(s) + 2H₂(g) → CH₄(g); Δ_fH⁰
= -17.9 kcal mol^-1
Enthalpy change in reactants :
Heat of atomisation of 1 mole C = 170.9 kcal
Heat of atomisation of 4 moles of H = 4 × 52.1 kcal
= 208.4 kcal

Enthalpy change in products :
Heat of formation of 4 moles of C-H bonds = 4 × x kcal
where x = enthalpy of formation of C-H bonds in kcal/mole
Since, the algebraic sum of all the heat changes is equal to the heat of formation of the above equation, we have :
170.9 + 208.4 + 4x = – 17.9
379.3 + 4x = – 17.9
4x = – 17.9 – 379.3
4x = – 397.2 397.2
x = \(-\frac{397.2}{4}\) = – 99.3 kcal/mole
∴ Bond enthalpy in methane = 99.3 kcal/mole

OR

(i) (a) Enthalpy change accompanying the formation of 1 mole of a compound from its constituent elements, all substances being in their standard states (1 bar pressure and 298 K).
(b) It is the enthalpy change accompanying the complete combustion of one mole of a substance in excess of oxygen or air.
(ii) Reaction for the formation of benzene, 6C(graphite,s) + 3H₂(g) → C₆H₆(l), D_fH° = ?…(i)
The enthalpy of combustion of 1 mol of benzene is:
RC₆H₆(l) + \(\frac{15}{2}\)O₂(g) → O₂(g) + 3H₂O(l); Δ_cH°
= -3267 kJ mol^-1 …(ii)

The enthalpy of formation of 1 mol of C02(g)
C(graphite,s) + 02(g) → C02(g); Δ_fH⁰
= – 393.5 kj mol^-1 ………..(iii)

The enthalpy of formation of 1 mol of H20()) is :
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); Δ_fH°
= -285.83 kj mol^-1…(iv)

Multiplying eq. (iii) by 6 and eq. (iv) by 3, we get
6C(graphites) + 60₂(g) → 6C0₂(g), Δ_fH°
= -2361 kj mol^-1

3H₂(g) + \(\frac{3}{2}\)O₂(g) → 3H₂O(l); Δ_fH°
= – 857.49 kj mol^-1

On adding above equations,
6C(graphites) + 3H₂(g) + \(\frac{15}{2}\)O₂(g) → 6C0₂(g) + 3H₂O(1); Δ_fH° = – 3218.49 kj mol^-1 … (v)

On reversing eq. (ii)
6C0₂(g) + 3H₂O(l) → C₆H₆(l) + \(\frac{15}{2}\)O₂(g);
Δ_fH° = 3267 kj mol^-1……………. (vi)

On adding equation (v) and (vi), we get
6C(graphites) + 3H₂(g) → C₆H₆(l); Δ_fH° = 48.51 kj mol^-1
∴ Standard enthalpy of formation of benzene
(Δ_fH°) = 48.51 kj mol^-1

Question 36.
An organic compound contains 69% carbon and 4.8% hydrogen, the remaining is being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. [5]
OR
Give the IUPAC names ol tlw following compounds:

Answer:
Percentage of carbon in organic compound = 69 %
That is, 100 g of organic compound contains 69 g of carbon.
0.2 g of organic compound will contain
= \(\frac{69 \times 0.2}{100}\) = 0.138 g of C
Molecular mass of carbon dioxide, CO₂ = 44 g
That is, 12 g of carbon is contained in 44 g of CO₂
Therefore, 0.138 g of carbon will be contained in
= \(\frac{44 \times 0.138}{12}\) = 0.506g of CO2

Thus, 0.506 g of CO2 will be produced on complete combustion of 0.2 g of organic compound.
Percentage of hydrogen in organic compound is 4.8.
i.e., 100 g of organic compound contains 4.8 g of hydrogen.

Therefore, 0.2 g of organic compound will contain
= \(\frac{4.8 \times 0.2}{100}\) =00096g of H
It is known that molecular mass of water (H20) is 18g.

Thus, 2 g of hydrogen is present in 18 g of water.
0.0096 g of hydrogen will be contained in
= \(\frac{18 \times 0.0096}{2}\) = 0.0864 g of water
Thus, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic compound.

Commonly Made Error
Student often leave the problem after calculating only amount of hydrogen and carbon present. They miss the last part of question.

Answering Tips
Students must know how to calculate molecular 1 masses of common compounds. w They must read the problem carefully to ; understand what is being asked.

OR

(i) 1-phenyl propane
(ii) 2-methyl-l-cyanobutane
(iii) 2, 5-Dimethylheptane
(iv) 3-Bromo- 3-chloroheptane
(v) 3-Chloropropanal

Commonly Made Error
Some student write the incorrect IUPAC name.

Answering Tip:
Remember general rules for writing IUPAC name.

Question 37.
What is resonance? What are the conditions for writing the resonating structures? [5]
Answer:
Resonance: When a molecule cannot be represented by a single structure but its characteristic properties can be described by two or more than two canonical structures, then actual structure is said to be a resonance hybrid of these structures. This phenomenon is known as resonance.

Conditions for Writing Resonating Structures:
The following are essential conditions for writing resonating structures :

• The contributing structures should have same atomic positions. They should differ only in electronic arrangements.
• These structures should have same number of unpaired electrons.
• These structures should have nearly same energy.
• These structures should be written such that negative charge is present on an electronegative atom and positive charge is present on an electropositive atom.
• In contributing structures, like charges should not reside on adjacent atoms.