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      Class 11 CHEMISTRY – JEE

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      • Chemistry
      • Class 11 CHEMISTRY – JEE
      CoursesClass 11ChemistryClass 11 CHEMISTRY – JEE
      • 1. Stoichiometry 1
        13
        • Lecture1.1
          Introduction & POAC 30 min
        • Lecture1.2
          Mole Stoichiometric relationship 29 min
        • Lecture1.3
          Successive reaction & Limiting reagent 25 min
        • Lecture1.4
          Gas Stoichiometry 29 min
        • Lecture1.5
          Important Types of Reactions 25 min
        • Lecture1.6
          Avogadro’s No.1 30 min
        • Lecture1.7
          Mole & Number 28 min
        • Lecture1.8
          Atomic, Molecular Wt 26 min
        • Lecture1.9
          Ionic wt, Avg. At. Wt. 15 min
        • Lecture1.10
          Molar wt. 27 min
        • Lecture1.11
          Molar Volume & Gas Analysis 30 min
        • Lecture1.12
          Gas Analysis 17 min
        • Lecture1.13
          Empirical Formula Determination 26 min
      • 2. Stoichiometry 2
        18
        • Lecture2.1
          Acid Base definition 23 min
        • Lecture2.2
          Acidity & Basicity 32 min
        • Lecture2.3
          Acidic Strength 30 min
        • Lecture2.4
          Acidic Strength 23 min
        • Lecture2.5
          Conjugate Acid-Base pair, Basic Strength 48 min
        • Lecture2.6
          Oxidation & Reduction 50 min
        • Lecture2.7
          Calculation of Oxidation Number 46 min
        • Lecture2.8
          O.A. & R.A., Balancing by Oxidation Number Method 01 hour
        • Lecture2.9
          Balancing by Ion Electron Method. 35 min
        • Lecture2.10
          Eq. Wt. 1 – n factor & Eq. Wt. Concept 47 min
        • Lecture2.11
          Eq. Wt. 2 – Eq. Concept 35 min
        • Lecture2.12
          Volumetric Analysis 43 min
        • Lecture2.13
          Volumetric analysis 44 min
        • Lecture2.14
          Titration – Acid Base Titration 49 min
        • Lecture2.15
          Titration – Acid Base Titration, Indicator 56 min
        • Lecture2.16
          Titration – Redox Titration-8 58 min
        • Lecture2.17
          Titration – Redox Titration, volume Strength of H2O2 50 min
        • Lecture2.18
          Titration – Redox Titration, Iodometry, Oleum, Bleaching Powder 49 min
      • 3. Thermodynamics & Thermochemistry
        19
        • Lecture3.1
          Zeroth Law 55 min
        • Lecture3.2
          1st law – System, Properties, State 40 min
        • Lecture3.3
          1st Law, Process, Internal energy, Work 43 min
        • Lecture3.4
          Work done in Irreversible process, Isobaric Process 49 min
        • Lecture3.5
          Isochoric Process & problems TD 42 min
        • Lecture3.6
          Isothermal irreversible Process, Problems on TD 46 min
        • Lecture3.7
          Adiabatic Process 49 min
        • Lecture3.8
          Problems on TD 44 min
        • Lecture3.9
          Thermochemistry & Enthalpy 38 min
        • Lecture3.10
          Hess’s Law, Kirchhoff’s Law 43 min
        • Lecture3.11
          Enthalpy of Formation, combustion 39 min
        • Lecture3.12
          Enthalpy of Hydrogenation, Hydration, dissolution, lattice energy 40 min
        • Lecture3.13
          Enthapy of Neutralisation, atomisation, Bond Energy 47 min
        • Lecture3.14
          Resonance energy & problems 45 min
        • Lecture3.15
          2nd Law, Entropy-positional 40 min
        • Lecture3.16
          TD Entropy, 3rd Law, Entropy change in a reaction 45 min
        • Lecture3.17
          Gibb’s free energy 43 min
        • Lecture3.18
          Efficiency, engine, pump & Carnot engine 39 min
        • Lecture3.19
          Chapter Notes – Thermodynamics & Thermochemistry
      • 4. Atomic Structure
        22
        • Lecture4.1
          Introduction, Cathode rays & Anode rays 41 min
        • Lecture4.2
          J.J. Thomson Model, Millikan Oil Drop Experiment 38 min
        • Lecture4.3
          Rutherford Experiment 51 min
        • Lecture4.4
          Quantum Mechanics, BlackBody Radiation Experiment 41 min
        • Lecture4.5
          Wave 44 min
        • Lecture4.6
          Photoelectric Effect 46 min
        • Lecture4.7
          Problems on Photoelectric Effect 35 min
        • Lecture4.8
          Atomic Structure 44 min
        • Lecture4.9
          Bohr Theory 47 min
        • Lecture4.10
          H – Spectrum 49 min
        • Lecture4.11
          Problems on Bohr’s Theory 40 min
        • Lecture4.12
          Adv. Problems on Bohr Theory & Sommerfeld model 51 min
        • Lecture4.13
          Quantum Mechanical Model for Atomic Structure 47 min
        • Lecture4.14
          Schrodinger wave equation 54 min
        • Lecture4.15
          No. of Orbitals & Quantum no 45 min
        • Lecture4.16
          Orbital Curve, RPD curve, Definition of Node 46 min
        • Lecture4.17
          Calculation of Node, Orbital Picture 43 min
        • Lecture4.18
          Radial Probability curve, MPD, Avg. distance, Screening effect, Zeff 38 min
        • Lecture4.19
          Multielectron system, Electronic configuration 56 min
        • Lecture4.20
          Stability of Elec. Configuration 36 min
        • Lecture4.21
          Chapter Notes – Atomic Structure
        • Lecture4.22
          NCERT Solutions – Atomic Structure
      • 5. Chemical equilibrium
        9
        • Lecture5.1
          Introduction, Eqb constant & Eqb Position 51 min
        • Lecture5.2
          Types of Eqb Constant, Heterogeneous Eqb, Reaction Quotient 45 min
        • Lecture5.3
          Range of Eqb Constant 43 min
        • Lecture5.4
          Problems on Chemical Eqb 41 min
        • Lecture5.5
          Problems on Chemical Eqb 42 min
        • Lecture5.6
          Le-chatelier Principle 42 min
        • Lecture5.7
          Le-Chatelier Principle 35 min
        • Lecture5.8
          Eqb & 2nd Law of TD 26 min
        • Lecture5.9
          NCERT Solutions – equilibrium
      • 6. Ionic Equilibrium
        17
        • Lecture6.1
          Electrolyte, Dissociation of H2O, Nature of Solution 46 min
        • Lecture6.2
          PH scale, Log & Antilog 40 min
        • Lecture6.3
          PH of Strong Acid, Base Solution 51 min
        • Lecture6.4
          PH of Weak Acid, Base solution 41 min
        • Lecture6.5
          PH of mixture of Acids, Bases 46 min
        • Lecture6.6
          PH of Polybasic acids 40 min
        • Lecture6.7
          PH of Salt Solution 1 43 min
        • Lecture6.8
          PH of salt solution 2 52 min
        • Lecture6.9
          Common ion effect, Buffer solution 49 min
        • Lecture6.10
          Buffer Capacity 45 min
        • Lecture6.11
          Titration & PH Curve 1 40 min
        • Lecture6.12
          Titration & PH curve 2 46 min
        • Lecture6.13
          Acid Base indicator 35 min
        • Lecture6.14
          Solubility Equilibrium 47 min
        • Lecture6.15
          Precipitation of Solid, Qualitative analysis of cation 44 min
        • Lecture6.16
          Complex ion equilibrium 23 min
        • Lecture6.17
          Chapter Notes – Equilibrium
      • 7. Introduction & Development of Org. Chemistry
        3
        • Lecture7.1
          Introduction & Development Of Organic Chemistry 44 min
        • Lecture7.2
          Introduction & Syllabus 36 min
        • Lecture7.3
          NCERT Solutions – Org. Chemistry
      • 8. Nomenclature of Org. Compounds
        16
        • Lecture8.1
          Alkane 59 min
        • Lecture8.2
          Alkane 31 min
        • Lecture8.3
          Alkyl Group & Types Of Hydrogen 01 hour
        • Lecture8.4
          Alkene 54 min
        • Lecture8.5
          Alkenyl 32 min
        • Lecture8.6
          Alkyne & Alkenyl 47 min
        • Lecture8.7
          Cycloalkane 43 min
        • Lecture8.8
          Cycloalkene 35 min
        • Lecture8.9
          Bicycloalkane & Spirane 35 min
        • Lecture8.10
          Acid & Aldehyde 45 min
        • Lecture8.11
          Ester & Acid Halides 28 min
        • Lecture8.12
          Amide & Nitrile 28 min
        • Lecture8.13
          Alcohol & Sulphonic Acid 37 min
        • Lecture8.14
          Isonitrile, Amine, Nitroalkane, Halo Compounds 39 min
        • Lecture8.15
          Ketone, Anhydride & Ether 34 min
        • Lecture8.16
          Polyfunctional Group Compounds 41 min
      • 9. GOC 1- Hybridisation, Resonance, Aromaticity
        16
        • Lecture9.1
          Concept Of Hybridisation 42 min
        • Lecture9.2
          Sp3, Sp2 Hybridisation 44 min
        • Lecture9.3
          Sp Hybridisation, Relative Study Of Sp3, Sp2, Sp Orbitals 46 min
        • Lecture9.4
          Effect Of Hybridisation On Bond Length, Planar Nature 59 min
        • Lecture9.5
          Concept Of Resonance 39 min
        • Lecture9.6
          Doing Resonance 18 min
        • Lecture9.7
          Resonance Hybrid, Cannonical St. , Resonance Energy 44 min
        • Lecture9.8
          Condition Of Resonance 40 min
        • Lecture9.9
          Writing Cannonical St. 39 min
        • Lecture9.10
          Relative Stability Of Cannonical St. 37 min
        • Lecture9.11
          Resonance Energy 45 min
        • Lecture9.12
          Effect Of Resonance On Bond Length, Enthalpy Of Hydrogenation 43 min
        • Lecture9.13
          Introduction To Aromaticity 43 min
        • Lecture9.14
          Introduction To Aromaticity 39 min
        • Lecture9.15
          Unsaturation Factor 31 min
        • Lecture9.16
          Chapter Notes – GOC General Organic chemistry
      • 10. GOC 2 - Substituent effect
        5
        • Lecture10.1
          Substituent Effect, Hyperconjugation 48 min
        • Lecture10.2
          Substituent Effect, Hyperconjugation 43 min
        • Lecture10.3
          Substituent Effect, Mesomeric Effect 47 min
        • Lecture10.4
          Substituent Effect, Inductive Effect 46 min
        • Lecture10.5
          Substituent Effect, Electromeric Effect, Staric Effect, Relative M & I Effect 41 min
      • 11. GOC 2 - Reactive Intermediate
        6
        • Lecture11.1
          Reactive Intermediate, Carbocation 45 min
        • Lecture11.2
          Reactive Intermediate, Carbocation, Carbonium Ion Rearrangement 42 min
        • Lecture11.3
          Reactive Intermediate, Carbonium Ion Rearrangement 41 min
        • Lecture11.4
          Reactive Intermediate, Carbanion 36 min
        • Lecture11.5
          Reactive Intermediate, Free Radical 47 min
        • Lecture11.6
          Reactive Intermediate, Carbene & Nitrene 42 min
      • 12. GOC 2 - Acid, base, Electrophile, Nucleophile
        3
        • Lecture12.1
          Acid Base, Electrophile Nucleophile 50 min
        • Lecture12.2
          Acid Base, Electrophile Nucleophile 47 min
        • Lecture12.3
          Hard Acid Base, Electrophilic Nucleophilic Strength 40 min
      • 13. Isomerism
        20
        • Lecture13.1
          Structural Isomers 39 min
        • Lecture13.2
          Tautomerism 37 min
        • Lecture13.3
          Stability Of Tautomers 43 min
        • Lecture13.4
          Factors Affecting Stability, Catalysis In Tautomerism 39 min
        • Lecture13.5
          Geometrical Isomerism 41 min
        • Lecture13.6
          E-z Nomenclature, Properties Of G.i. 43 min
        • Lecture13.7
          No. Of G.i., Interconversion Of G.i. 48 min
        • Lecture13.8
          Optical Isomerism & Its Conditions 50 min
        • Lecture13.9
          Different Types Of Projections, R-s Configuration 57 min
        • Lecture13.10
          Relationship Between Optical Isomers 45 min
        • Lecture13.11
          Dissymmetry In A Molecule 44 min
        • Lecture13.12
          Enantiomers, Mesomers, Diastereomers 39 min
        • Lecture13.13
          Special Case Of Optical Isomerism 47 min
        • Lecture13.14
          No. Of Optical Isomers, Stereoisomers 45 min
        • Lecture13.15
          D,l Configuration, Retention & Inversion 36 min
        • Lecture13.16
          Measurement Of Optical Activity 45 min
        • Lecture13.17
          No. Of Isomers 35 min
        • Lecture13.18
          Resolution Of Optical Isomers, Syn, Anti Addition, Elimination. 28 min
        • Lecture13.19
          Conformational Isomers 51 min
        • Lecture13.20
          Conformers Of Propane, Butane, Cyclohexane & Problems 44 min
      • 14. Reaction Mechanism
        21
        • Lecture14.1
          Introduction, Types Of Organic Reactions 35 min
        • Lecture14.2
          Nucleophilic Substitution Reaction 40 min
        • Lecture14.3
          Sn1 & Sn2 Reaction, Sni Pathway 53 min
        • Lecture14.4
          Reactivity In Sn1 & Sn2 Path 42 min
        • Lecture14.5
          Reactivity In Sn1 & Sn2 Path 36 min
        • Lecture14.6
          Reactivity In Sn1 & Sn2 Path 30 min
        • Lecture14.7
          Reactivity In Sn1 & Sn2 Path 41 min
        • Lecture14.8
          Elimination Reaction 53 min
        • Lecture14.9
          E1 & E2 Reaction, Isotopic Effect 46 min
        • Lecture14.10
          Orientation In Elimination Reaction 45 min
        • Lecture14.11
          Problems On Elimination Reaction 48 min
        • Lecture14.12
          Elimination Vs Substitution 34 min
        • Lecture14.13
          Addition Reaction 51 min
        • Lecture14.14
          Problems On Addition Reaction 46 min
        • Lecture14.15
          Electrophilic Aromatic Substitution Reaction 49 min
        • Lecture14.16
          Orientation In Electrophilic Aromatic Substitution 53 min
        • Lecture14.17
          Reactivity In Electrophilic Aromatic Substitution Reaction 30 min
        • Lecture14.18
          Examples Of Electrophilic Aromatic Substitution Reaction 37 min
        • Lecture14.19
          Examples Of Electrophilic Aromatic Substitution Reaction 37 min
        • Lecture14.20
          Nucleophilic Aromatic Substitution 44 min
        • Lecture14.21
          Benzyne Pathway 27 min
      • 15. Alkane
        7
        • Lecture15.1
          Alkane Preparation 49 min
        • Lecture15.2
          Alkane Preparation & Selective Hydrogenation 31 min
        • Lecture15.3
          Alkane Preparation 40 min
        • Lecture15.4
          Alkane Preparation 38 min
        • Lecture15.5
          Alkane Preparation 32 min
        • Lecture15.6
          Alkane Properties 55 min
        • Lecture15.7
          Alkane Properties & Problems 39 min
      • 16. Alkene
        7
        • Lecture16.1
          Alkene Preparation 45 min
        • Lecture16.2
          Alkene Preparation 36 min
        • Lecture16.3
          Alkene Properties 53 min
        • Lecture16.4
          Alkene Properties 40 min
        • Lecture16.5
          Alkene Properties 42 min
        • Lecture16.6
          Alkene Properties & Ozonolysis 41 min
        • Lecture16.7
          Alkene Properties, Oxidation, Substitution 38 min
      • 17. Alkyl Halides
        4
        • Lecture17.1
          Preparation 38 min
        • Lecture17.2
          Properties 49 min
        • Lecture17.3
          Haloform Reaction 28 min
        • Lecture17.4
          Grignard Reagent 29 min
      • 18. Chemical Bonding
        32
        • Lecture18.1
          Introduction, definition, Concept & Type of Bonding 53 min
        • Lecture18.2
          Ionic Bonding, covalent bonding 50 min
        • Lecture18.3
          Ionic Character in Covalent Bonding, Electronegativity 34 min
        • Lecture18.4
          Dipole Moment 42 min
        • Lecture18.5
          Fajan’s Rule 34 min
        • Lecture18.6
          Model for Covalent Compound, V.B.T. – Lewis St. Model 56 min
        • Lecture18.7
          Lewis Structure Model 45 min
        • Lecture18.8
          Formal Charge 46 min
        • Lecture18.9
          Formal Charge Rule 44 min
        • Lecture18.10
          Resonance 43 min
        • Lecture18.11
          Merits & Demerits of Lewis St. Model 44 min
        • Lecture18.12
          Drawing Lewis St. 30 min
        • Lecture18.13
          VSEPR 1 49 min
        • Lecture18.14
          VSEPR 2 51 min
        • Lecture18.15
          VSEPR 3 51 min
        • Lecture18.16
          VSEPR 4 33 min
        • Lecture18.17
          BackBonding 38 min
        • Lecture18.18
          Bond Angle determination 47 min
        • Lecture18.19
          Concept of Hybridisation 44 min
        • Lecture18.20
          Sp3, Sp2 Hybridisation 44 min
        • Lecture18.21
          SP hybridisation, Relative study of SP, SP2, SP3 Hybridisation 46 min
        • Lecture18.22
          Hybridsation involving D-orbitals 39 min
        • Lecture18.23
          Hybridsation with D-orbitals, Limitation of Hybridisation 41 min
        • Lecture18.24
          Calculation of Hybridisation of Central Atom, Problems 43 min
        • Lecture18.25
          Merits & demerits of VBT, Introduction to MOT 33 min
        • Lecture18.26
          MO formation, Bond Order 43 min
        • Lecture18.27
          MO with P-orbitals, B2, Magnetic Character 43 min
        • Lecture18.28
          MO of Diatomic Species, Hetroatomic Species 51 min
        • Lecture18.29
          Secondary Bondings 39 min
        • Lecture18.30
          H Bonding 37 min
        • Lecture18.31
          Metallic Bonding 52 min
        • Lecture18.32
          Chapter Notes – Chemical Bonding
      • 19. Periodic Table
        10
        • Lecture19.1
          Development of P.T. 43 min
        • Lecture19.2
          Mandeelev P.T. & Mosley, Modern P.T. 43 min
        • Lecture19.3
          Modern P.T. & Periodic Properties 27 min
        • Lecture19.4
          Atomic Volume & Radius 49 min
        • Lecture19.5
          Atomic Radius, Ionisation Energy 28 min
        • Lecture19.6
          Ionisation Energy 48 min
        • Lecture19.7
          Electron Affinity, Hydration Energy 52 min
        • Lecture19.8
          Electronegativity, Lattice Energy 46 min
        • Lecture19.9
          Oxidising & Reducing Power, Nature of oxides 38 min
        • Lecture19.10
          M.P. & B.P., Density, Bond Energy, Diagonal relationship, Inert Pair Effect 25 min
      • 20. Metallurgy
        7
        • Lecture20.1
          Introduction, Concentration of ore 49 min
        • Lecture20.2
          Roasting, Calcination, smelting 41 min
        • Lecture20.3
          Refining of metal 29 min
        • Lecture20.4
          Pyrometallurgy, electrometallurgy, Hydrometallurgy 32 min
        • Lecture20.5
          Ellingham Diagram 43 min
        • Lecture20.6
          Extraction of Cu & Fe 22 min
        • Lecture20.7
          Extraction of Al & Zn 26 min
      • 21. Hydrogen and its Compounds
        7
        • Lecture21.1
          preparation, properties & Type of Hydrogen 57 min
        • Lecture21.2
          Compounds of Hydrogen, Hydrides, Water, Hydrates 56 min
        • Lecture21.3
          Hardness of Water, H2O2 56 min
        • Lecture21.4
          Problems 48 min
        • Lecture21.5
          Problems 28 min
        • Lecture21.6
          Chapter Notes – Hydrogen and its Compounds
        • Lecture21.7
          NCERT Solutions – Hydrogen
      • 22. S block metals
        8
        • Lecture22.1
          IA 1 – elemental Properties of Alkali metals& its Compounds 57 min
        • Lecture22.2
          IA 2 – Na & its compounds 01 hour
        • Lecture22.3
          IA 3 – Na & its Compounds, Use of Na & K 27 min
        • Lecture22.4
          IIA 1 – Elemental Properties 41 min
        • Lecture22.5
          IIA 2 – Compounds of IIA Metals 53 min
        • Lecture22.6
          IIA 3 – Compounds of Ca 48 min
        • Lecture22.7
          Chapter Notes – S block metals
        • Lecture22.8
          NCERT Solutions – S block metals
      • 23. p block elements
        8
        • Lecture23.1
          Introduction to P – Block & IIIA – elemental properties 51 min
        • Lecture23.2
          IIIA – General properties of compounds & B-compounds 40 min
        • Lecture23.3
          IIIA – Boron compounds, Use of B and Al 35 min
        • Lecture23.4
          IVA – Elemental Properties of C family 46 min
        • Lecture23.5
          IVA – Allotropes of C & compounds of C 01 hour
        • Lecture23.6
          IVA – Compounds of Si 48 min
        • Lecture23.7
          Chapter Notes – p block elements
        • Lecture23.8
          NCERT Solutions – p block elements

        NCERT Solutions – equilibrium

        7.1. A liquid is in equilibrium with its vapours in a sealed container at a fixed temperature. The volume of the container is suddenly increased, (i) What is the initial effect of the change on the vapour pressure? (ii) How do the rates of evaporation and condensation change initially? (iii) What happens when equilibrium is restored finally and what will be the final vapour pressure?

        Answer: 

        (i) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.

        (ii) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

        (iii) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

        7.2. What is Kc for the following reaction in state of equilibrium?
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-1

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-2
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-3

        7.3.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-4

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-5

        7.4. Write the expression for the equilibrium constant for each of the following reactions
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-6

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-7

        7.5. Find the value of Kc for each of the following equilibria from the value of K
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-8

        Answer:

         
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-9

        7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K. NO(g)+O3 —–>NO2(g) + O2(g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction ?

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-10

        7.7. Explain why pure liquids and solids can be ignored while writing the value of equilibrium constants.

        Answer: 

        This is because molar concentration of a pine solid or liquid is independent of the amount present.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-11
        Since density of pure liquid or solid is fixed and molar mass is also fixed. Therefore molar concentration are constant.

        7.8. Reaction between nitrogen and oxygen takes place as follows:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-12
        If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc – 2.0 x 10-37, determine the composition of the equilibrium mixture.

        Answer: 

        Let x moles of N2(g) take part in the reaction. According to the equation, x/2 moles of O2 (g) will react to form x moles of N2O(g). The molar concentration per litre of different species before the reaction and at the equilibrium point is:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-13
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-14
        The value of equilibrium constant (2.0 x 10-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, is extremely small and can be omitted as far as the reactants are concerned.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-15

        7.9. Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction given below:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-16
        When 0.087 mole of NO and 0.0437 mole of Br2 are mixed in a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Determine the compositions of the equilibrium mixture.

        Answer: 

        The balanced chemical equation for the reaction is:
        According to the equation, 2 moles of NO (g) react with 1 mole of Br2 (g) to form 2 moles of NOBr (g). The composition of the equilibrium mixture can be calculated as follows:
        No. of moles of NOBr (g) formed at equilibrium = 0.0518 mol (given)
        No. of moles of NO (g) taking part in reaction = 0.0518 mol
        No. of moles of NO (g) left at equilibrium = 0.087 – 0.0518 = 0.0352 mol
        No. of moles of Br2 (g) taking part in reaction = 1/2 x 0.0518 = 0.0259 mol
        No. of moles of Br2 (g) left at equilibrium = 0.0437 – 0.0259 = 0.0178 mol
        The initial molar concentration and equilibrium molar concentration of different species may be represented as:
        2NO (g) + Br2(g) ——————> 2NOBr(g)
        Initial moles 0.087 0.0437 0
        Moles at eqm. point: 0.0352 0.0178 0.0518

        7.10.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-17

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-18

        7.11. A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-19

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-20

        7.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-21
        Is this reaction at equilibrium? If not, what is the direction of net reaction?

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-22

         

        7.13. The equilibrium constant expression for a gas reaction is,
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-23
        Write the balanced chemical equation corresponding to this expression.

        Answer:

        Balanced chemical equation for the reaction is 4
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-24

        7.14. If l mole of H20 and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-25
        Calculate the equilibrium constant for the reaction.

        Answer: 

        Number of moles of water originally present = 1 mol
        Percentage of water reacted =40%
        Number of moles of water reacted = 1 x 40/100 = 0.4 mol
        Number of moles of water left = (1 – 0.4) = 0.6 mole According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
        Thus, the molar cone, per litre of the reactants and products before the reaction and at the equilibrium point are as follows:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-26

        7.15. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-27

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-28

        7.16. K =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6  when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-29

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-30

        7.17. The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as:
        CH3COOH(l) + C2H5OH(l)——-> CH3COOC2H5(l) + H2O(l)
        (i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction.
        (ii) At 293 K, if one starts with 1.000 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
        (iii) Starting mth 0.50 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-31
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-32
        Since Qc is less than Kc this means that the equilibrium has not been reached. The reactants are still taking part in the reaction to form the products.

        7.18. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl5  was found to be 0.5 x 10-1 mol L-1. If Kc is 8.3 x 10-3 what are the concentrations of PCl3  and Cl2  at equilibrium?

        Answer: 

        Let the initial molar concentration of PCl5 per litre = x mol
        Molar concentration of PCl5 at equilibrium = 0.05 mol
        .’. Moles of PCl5 decomposed = (x – 0.05) mol
        Moles of PCl3 formed = (x – 0.05) mol
        Moles of Cl2 formed = (x – 0.05) mol
        The molar conc./litre of reactants and products before the reaction and at the equilibrium point are:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-33

        7.19. One of the reactions that takes place in producing steel from iron ore is the reduction of iron
        (II) oxide by carbon monoxide to give iron metal and C02
        FeO(s) + CO(g) ———>Fe(s) + C02(g) ; Kp = 0.265 atm at 1050 K
        What are the equilibrium partial pressures of CO and C02 at 1050 K if the initial pressures are: PCO = 1.4 atm and PCO2 = 0.80 atm?

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-34
        Since Qp>Kp(0.265), this means that the reaction will move in the backward direction to attain the equilibrium. Therefore, partial pressure of C02 will decrease while that of CO will increase so that the equilibrium may be attained again. Let p atm be the decrease in the partial pressure of C02. Therefore, the partial pressure of CO will increase by the same magnitude i.e., p atm.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-35
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-36

        7.20.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-37

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-38

        7.21. Bromine monochloride (BrCl ) decomposes into bromine and chlorine and reaches the equilibrium:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-39
        The value of Kc is 32 at 500 K. If initially pure BrCl  is present at a concentration of 3.3 x10-3mol L-1what is its molar concentration in the mixture at equilibrium?

        Answer: 

        Let x moles of BrCl decompose in order to attain the equilibrium. The initial molar concentration and the molar concentration at equilibrium point of different species may be represented as follows:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-40

        7.22. At 1127 K and 1 atmosphere pressure, a gaseous mixture of CO and C02 in equilibrium with solid carbon has 90.55% CO by mass.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-41
        Calculate Kc for the reaction at the above temperature.

        Answer:

         
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-42

        7.23. Calculate (a) ∆G– and (b) the equilibrium constant for the formation of N02 from NO and 02 at 298 K
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-43

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-44

        7.24. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure bp increasing the volume?
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-45

        Answer:

        (i) Pressure will increase in the forward reaction and number of moles of products will increase.
        (ii) Pressure will increase in backward reaction and number of moles of products will decrease.
        (iii) The change in pressure will have no effect on the equilibrium constant and there will be no change in the number of moles.

        7.25. Which of the following reactions will get affected by increase in pressure ? Also mention whether the change will cause the reaction to go to the right or left direction.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-46

        Answer: 

        Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different (np ≠ nr) (gaseous). With the exception of the reaction (1); all the reamaining five reactions will get affected by increasing the pressure. In general,

        • The reaction will go to the left if np> nr.
        • The reaction will go to the right if nr > np .
          Keeping this in mind,

        (i) Increase in pressure will not affect equilibrium because np = nr = 3.
        (ii) Increase in pressure will favour backward reaction because np (2) > nr (1)
        (iii) Increase in pressure will favour backward reaction because np (10) > nr (9)
        (iv) Increase in pressure will favour forward reaction because np (1) < nr (2)
        (v) Increase in pressure will favour backward reaction because np (2) > nr(1)
        (vi) Increase in pressure will favour backward reaction because np (1) > nr (0).

        7.26. The equilibrium constant for the following reaction is 1.6 x 105at 1024 K.
        Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-47

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-48
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-49

        7.27.Hydrogen gas is obtained from the natural gas by partial oxidation with steam as per following endothermic reaction:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-50
        Write the expression for Kpfor the above reaction
        How will the value of Kp and composition of equilibrium mixture be affected by:
        (i) increasing the pressure, (ii) increasing the temperature, (iii) using a catalyst?

        Answer: 

        The expression for Kp for the reaction is:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-51
        (i) By increasing the pressure, the number of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of reactants will be formed and the value of Kp will decrease.
        (ii) If the temperature is increased, according to Le Chatelier’s principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of Kp will increase.
        (iii) The addition of catalyst will not change the equilibrium since it influences both the forward and the backward reactions to the same extent. But it will be attained more quickly.

        7.28. What is the effect of:
        (i) addition of H2 (ii) addition of CH3OH
        (iii) removal of CO (iv) removal of CH3OH
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-52

        Answer: 

        (i) Equilibrium will be shifted in the forward direction.
        (ii) Equilibrium will be shifted in the backward direction.
        (iii) Equilibrium will be shifted in the backward direction.
        (iv) Equilibrium will be shifted in the forward direction.

        7.29. At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl5) is 8.3 x 10-3 . if decomposition proceeds as:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-53
        (a) Write an expression for Kc for the reaction
        (b) What is the value of Kc for the reverse reaction at the same temperature.
        (c) What would be the effect on Kc if
        (i) More of PCl5is added (ii) Temperature is increased.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-54
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-55
        (c) (i) By adding more of PCl5, value of Kc will remain constant because there is no change in temperature.
        (ii) By increasing the temperature, the forward reaction will.be favoured since it is endothermic in nature. Therefore, the value of equilibrium constant will increase.

        7.30. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2 In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-56
        If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that PCO = PH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp = 0.1 at 400°C.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-57

        7.31. Predict which of the following will have appreciable concentration of reactants and products:
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-58

        Answer: 

        Following conclusions can be drawn from the values of Kc .
        (a) Since the value of Kc is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
        (b) Since the value of Kc is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
        (c) Since the value of Kc is 1.8, this means that both the products and reactants have appreciable concentration.

        7.32.The value of Kc for the reaction 302(g) —>203(g) is 2.0 x 10-50 at 25°C. If equilibrium concentration of 02 in air at 25°C is 1.6 x 10-2, what is the concentration of O3?

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-59
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-60

        7.33.
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-61

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-62

        7.34. What is meant by conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CH–, HClO4 , OH–, CO32-, S2-

        Answer: 

        An acid-base pair which differs by a proton only (HA———> A– + H+) is known as conjugate acid-base pair.
        Conjugate acid:HCN,H20,HCO3–,HS–.
        Conjugate base:NO2–,ClO4–,O2–

        7.35. Which of the following are Lewis Acids?
        H2O,BF3, H+ and NH4+,

        Answer: 

        BF3, H+ ions are Lewis acids.

        7.36. What will be the conjugate bases for the Bronsted acids?HF, H2S04 and H2C03?

        Answer: 

        Conjugate bases: F–, HSO–4 , HCO–3.

        7.37. Write the conjugate acids for the following Bronsted bases:
        NH2, NH3 and HCOO–

        Answer:

        NH3, NH4+ and HCOOH

        7.38.The species H20, HCO3–, HSO4– and NH3 can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-63

        7.39. Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid/Lewis base?
        (a) OH– ions (b) F– (c) H+ (d) BCl3

        Answer: 

        (a) OH– ions can demate an electron pair and act as Lewis base.
        (b) F– ions can donate an electron pair and act’as Lewis base.
        (c) H+ ions can accept an electron pair and act as Lewis acid.
        (d) BCl3 can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.

        7.40. The concentration of hydrogen ions in a sample of soft drink is 3.8 x 10-3 M. What is the pH value?

        Answer: 

        pH = – log [H+] = – log (3.8 x 10-3) = – log 3.8 + 3 = 3 – 0.5798 = 2.4202 = 2.42

        7.41. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

        Answer:

        pH = – log [H+] or log [H+] = – pH = – 3.76 = 4.24
        .-. [H+] = Antilog 4.24 = 1.738 x  10-4 = 1.74 x 10-4 M

        7.42. The ionization constant of HF, HCOOH and HCN at 298 K are is 6.8 x 10-4 , 1.8 x 10-4 and 4.8 x 10-9 respectively, Calculate the ionization constant of the corresponding conjugate base.

        Answer:

        For F– , Kb =Kw/Ka= 10-14/(6.8 x 10-4) = 1.47 x 10-11 = 1.5 x 10-11 .
        For HCOO-, Kb = 10-14/(1.8 x 10-4) = 5.6 x 10-11 
        For CN–, Kb= 10-14/(4.8 X 10-9) = 2.08 x 10-6

        7.43. The ionization constant of phenol is 1.0 x 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-64

        7.44. The-first ionization constant of H2S is 9.1 x 10-8. Calculate the concentration of HS– ions in its 0.1 M solution and how will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 x 10-13, calculate the concentration of S2-under both conditions.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-65
        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-66

        7.45. The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-67

        7.46. It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its PKa.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-68

        7.47. Assuming complete dissociation, calculate the pH of the following solutions:
        (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-69

        7.48. Calculate the pH of the following solutions:
        (a) 2g ofTlOH dissolved in water to give 2 litre of the solution
        (b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
        (c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
        (d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-70

        7.49. The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the PKa of bromoacetic acid.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-71

        7.50. The pH of0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.

        Answer:

        ncert-solutions-for-class-11-chemistry-chapter-7-equilibrium-72

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