chemistry class 11 cbse pdf, Assignments of Chemistry

Download chemistry class 11 cbse pdf and more Assignments Chemistry in PDF only on Docsity! BRILLIANT PUBLIC SCHOOL , SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No. - 330419 XI - Chemistry Chapterwise Topicwise Worksheets with Solution Session : 2014-15 Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-843301 Website: www.brilliantpublicschool.com; E-mail: brilliantpublic@yahoo.com Ph.06226-252314, Mobile: 9431636758, 9931610902 CHEMISTRY (Class 11) Index Chapters page 1. Some Basic concepts of chemistry 01 2. Structure of Atom 16 3. Classification of Elements and Periodicity in Properties 34 4. Chemical Bonding and molecular Structure 46 5. States of Matter: Gases and Liquids 66 6. Thermodynamics 79 7. Equilibrium 93 8. Redox Reactions 114 9. Hydrogen 125 10. S-Block Elements 138 11. Some P-Block Elements 150 12. Organic Chemistry: some basic Principles and Techniques 161 13. Hydrocarbons 182 14. Environmental Chemistry 220 Ans6: Compounds Elements Mixtures Water Silver Tea Carbondioxide Platinum Steel Ans7 The components of a mixture can be separated by physical methods like handpicking, filtrations, crystallization, distillation etc. Ans8: Pure Substances Mixtures Glucose Air Gold Milk Sodium Ans9: Molecules consist of different atoms or same atoms. e.g. molecule of hydrogen contains two atoms of hydrogen where as molecule of water contain two atoms of hydrogen and one of oxygen. Compound is formed when two or more than two different atoms e.g. water carbondioxide, sugar etc. Ans10: The constituents of a compound can not be separated by physical methods. They can only be separate by chemical methods. CBSE TEST PAPER-02 CLASS - XI CHEMISTRY (Basic Concepts of Chemistry) Topic: - Properties of matter and their measurement Marks: 20 1. How are physical properties different from chemical properties? [1] 2. What are the two different system of measurement? [1] 3. Write seven fundamental quantities & their units. [2] 4. What is the difference between mass & weight? How is mass measured in laboratory? [2] 5. How is volume measured in laboratory? Convent 0.5L into ml and 30cm3 to dm3 [2] 6. What is the SI unit of density? [1] 7. Convert 350C to oF & K. [2] 8. What are the reference points in thermometer with Celsius scale? [1] 9. What is the SI unit of volume? What is the other common unit which in not an SI unit of volume. [1] 10. What does the following prefixes stand for – [2] (a) pico (b) nano (c) centi (d) deci CBSE TEST PAPER-02 CLASS - XI CHEMISTRY [ANSWERS] Topic: - Properties of matter and their measurement Ans1: Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance whereas the measurement of chemical properties require a chemical change to occur e.g. colour, odour etc are physical properties and combustion, basicity etc are chemical properties. Ans2: The different system of measurement are English system and the metric system. Ans3: Physical Quantity SI unit 1. Length (l) Metre (m) 2. Mass (m) Kilogram (kg) 3. Time (t) Second (s) 4. Electric Current (I) Ampere (A) 5. Thermodynamic Temperature (T) Kelvin (K) 6. Amount of substance (n) Mole (mol) 7. Luminous Intensity (Iν ) Candela (Cd) Ans4: Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on am object the mass of a substance is determined with the help of an analytical balance in laboratory. Ans5: In the laboratory volume of a liquid can be measured by using graduated cylinder, burette, pipette etc. 1L = 1000 ml 1000cm3 = 1dm3 0.5L = 500 ml 30cm3 = 1 100 0 30× 3dm = 0.03dm3 CBSE TEST PAPER-03 CLASS - XI CHEMISTRY [ANSWERS] Topic: - Uncertainty in measurement Ans1: Precision means the closeness of various measurements for the same quantity. Accuracy is the agreement of a particular value to the true value of the result. Ans2: Significant figures are meaningful digits which are known with certainty. The uncertainty in experimental or the calculated value is indicated by mentioning the number of significant figures. Ans3: (a) 4.01 × 102 – Three (b) 8.256 – Four (c) 100 – One Ans4: Law of definite proportions states that a given compound always contains exactly the same proportion of elements by weight. Ans5: The law of multiple proportions rays that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of other element are in a ratio of small whole numbers. e.g. hydrogen and oxygen can combine to form water as well as hydrogen peroxide. Here, the masses of oxygen (16g & 32g) which combine with a fixed mass of hydrogen (2g) bear a simple ratio i.e., 16:32 = 1:2. Ans6: According to Avogadro’s law, equal volumes of gases at the same temperature and pressure should contain equal number of molecules. Ans7 Postulates of Dalton’s atomic theory – 1. Matter consists of indivisible atoms. 2. All the atoms of a given element have identical properties including atomic mass. Atoms of different element differ in mass. 3. Compounds are formed when atoms of different elements combine in a fixed ratio. 4. Chemical reaction involves reorganization of atoms. These are neither created nor destroyed Ans8: One atomic mass unit (amu) is defined as a mass exactly equal to one – twelfth the mass of one carbon – 12 atom. Ans9: 2 6 (2 12) (6 1) 30C H = × + × = 12 22 11 2 4 3 4 (12 12) (22 1) (11 16) 342 (2 1) 32 (4 16) 98 (1 3) 31 (4 16) 98 C H O H SO H PO = × + × + × = = × + + × = = × + + × = Ans10: When a substance does not contain discrete molecules as their constituent units and have a three dimensional structure, formula mass is used to calculate molecular mass which is sum of all the atomic masses of atom present in the formula. CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (Basic Concepts of Chemistry) Topic: - Mole concept, percentage composition Marks: 20 1. What is the value of one mole? [1] 2. At NTP, what will be the volume of molecules of 236.022 10× H2? [1] 3. Calculate the number of molecules present in 0.5 moles of CO2? [1] 4. Give one example each of a molecule in which empirical formula and molecular formula are (i) same (ii) Different. [2] 5. 1L of a gas at STP weighs 1.97g. What is molecular mass? [1] 6. Write empirical formula of following – [4] CO, Na2CO3, KCl, C6H12, H2O2, H3PO4, Fe2O3, N2O4. 7. Calculate the number of moles in the following masses – [2] (i) 7.85g of Fe (ii) 7.9mg of Ca 8. Vitamin C is essential for the prevention of scurvy. Combustion of 0.2000g of vitamin C gives 0.2998g of CO2 and 0.819g of H2O. What is the empirical formula of vitamin C? [3] CBSE TEST PAPER-05 CLASS - XI CHEMISTRY (Basic Concepts of Chemistry) Topic: - Stoichiometry and Stoichiometric calculations Marks: 20 1. What is stoichiometry? [1] 2. How much potassium chlorate should be heated to produce 2.24L of oxygen at NTP? [2] 3. Write an expression for molarity and molality of a solution. [2] 4. Calculate the weight of lime (CaO) obtained by heating 2000kg of 95% pure lime stone (CaCO3) [2] 5. The substance which gets used up in any reaction is called -------------- [1] 6. What is 1molal solution? [1] 7. 4 litres of water are added to 2L of 6 molar HCl solutions. What is the molarity of resulting solution? [2] 8. What volume of 10M HCl and 3M HCl should be mixed to obtain 1L of 6M HCl solution? [2] CBSE TEST PAPER-05 CLASS - XI CHEMISTRY [ANSWERS] Topic: - Stoichiometry and Stoichiometric calculation Ans1: Stoichimetry deals with the calculations of masses of reactants and products involved in a chemical reactions. Ans2: 2KClO3
2KCl + 3O2 2moles 3moles 2(39 + 35.5 + 3 × 16) 22.4 × 3L = 245g = 67.2L 67.2L of oxygen is produced from 245g of KClO3 2.24L of oxygen is produced from = 245 2.24 67.2 × = 8.17g of KClO3 Ans3: number of moles of solute Molarity Volume of solution in Litres = number of moles of solute Molality Mass of solvent in kg = Ans4: 100kg impure sample has pure CaCO3 = 95 = 95kg ∴200kg impure sample has pure CaCO3 = 95 200 100 × = 190kg CaCO3
CaO + CO2 Since 100kg CaCO3 gives CaO = 56kg 190kg CaCO3 will give CaO = 56 190 100 × = 106.4kg Ans5: The substance that gets used up in any reaction is called limiting reagent. Ans6: one molal solution is solution in which one mole of solute is present in 1000g of solvent. Ans7 Initial volume, V1 = 2L Final volume, V2 = 4L + 2L = 6L Initial molarity, M1 = 6M Final molarity = M2 M1V1 = M2V2 2 2 6 2 6 6 2 2 6 M L M L M L M M L × = × ×= = Thus the resulting solution is 2M HCl. Ans8: Let the required volume of 10M HCl be V liters. Then, the required volume of 3M HCl be (1 – V) Liters. M1V1 + M2V2 = M3V3 10 3 (1 ) 6 1 10 3 3 6 V V V V × + × − = × + − = 7 3 3 0.428 428 . 7 V V L mL = = = = Then the volume of 10M HCl required = 428mL & volume of 3M HCl required = 1000mL – 428mL = 572mL Ans8: The cathode ray discharge tube experiment performed by J.J. Thomson led to the discovery of negatively charged particles called electron. A cathode ray tube consists of two thin pieces of metals called electrodes sealed inside a glass tube with sealed ends. The glass tube is attached to a vacuum pump and the pressure inside the tube is reduced to 0.01mm. When fairly high voltage (10, 000V) is applied across the electrodes, invisible rays are emitted from the cathode called cathode rays. Analysis of this rays led to the discovery electrons. Ans9: The canal ray experiment led to the discovery of – (i) The anode rays, travel in straight line (ii) They are positively charged as they get deflected towards the –ve end when subjected to an electric and magnetic field. (iii) They depend upon the nature of gas present in the cathode tube. (iv) The charge to mass ration (e/m) of the particle is found to depend on the gas from which they originate. (v) They are also material particles The analysis of these proportions led to the discovery of positively charged proton. CBSE TEST PAPER-02 CLASS - XI CHEMISTRY (Structure of Atom) Topic:-Atomic Models Marks: 20 1. Name the scientist who first gave the atomic model. [1] 2. What is an isotope? [1] 3. What are isobars? [1] 4. What are isotones? [1] 5. What is an atomic number? [1] 6. What is a mass number? [1] 7. Find out atomic number, mass number, number of electron and neutron in an element 4 02 0 × ? [2] 8. Give the main features of Thomson’s Model for an atom. [2] 9. Give the drawbacks of J.J. Thomson’s experiment. [1] 10. What did Rutherford conclude from the observations of rayα − scattering experiment? [2] 11. Why Rutherford’s model could not explain the stability of an atom? [1] CBSE TEST PAPER-02 CLASS - XI CHEMISTRY [ANSWERS] Topic: - Atomic Models Ans1: J.J. Thomson, in 1898 first proposed the atomic model called raising-pudding model. Ans2: Atoms of the same elements having same atomic number but different mass number are called isotopes. 1 2 3 1 1 1: , 35 37 12 13 14 , , , 17 17 6 6 6 eg H H and H Cl Cl C C C Ans3: Atoms of different elements which have same mass number but different atomic nos. eg: 14 14 , 6 7 C N 40 40 40 , , 18 19 20 Ar K Ca Ans4: Atoms of different elements which contains the same number of neutron. eg. 14 15 16 , , 6 7 8 C N O Ans5: Atomic number is defined as the number of protons presents in the nucleus of an atom or the number of electron present in a neutral atom of an element. Ans6: Maas number of an element is the number of proton and neutron present in the nucleus of an atom. Ans7: The mass no. of is 40× The atomic no. of is 20× No. of proton is = Z – A = 40 – 20 = 20 CBSE TEST PAPER-03 CLASS - XI CHEMISTRY [ANSWERS] Topic: - Bohr’s Model of an atom Ans1: 400nm to 750nm Ans2: (1) Dual character of the electromagnetic radiations i.e. wave like and particle like properties, and (2) Atomic spectra explained only by assuming quantized electronic energy levels in atoms. Ans3: When electrically charged particles moves under acceleration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the form of wave called electromagnetic waves or radiations. Ans4: Wavelength, c v λ = Substituting 83 10 / secc m= × And 98.7v MHz= 698.7 10 /cyles ses= × ( )61 10 / secMHz cycles=∵ 8 6 3 10 / sec 3.0395 98.7 10 / sec m mλ ×∴ = = × Ans5: The ideal body, which emits and absorbs all frequencies, is called a black body and the radiation emitted by such a body is called black body radiation. Ans6: Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. Ans7: The energy of quantum (E) is directly proportional to the frequency (v) of the radiation. , E v or E hv α = , . hv c or E where v and c velocity and wavelenght λ λ λ = = = = ' 'h Planck’s constant = 346.626 10 .Jh−× Ans8: (i) Frequency, c v λ = We know, 83 10 /c m s= × 0 10 8 8 10 3 3 3000 3000 10 . 3 10 / 3 10 / 3000 10 3 10 10 A m m s m s v m λ − − − = = × × ×∴ = = × × × 8 1 15 1 7 1 10 sec 1 10 sec 1 10 − − − ×= = × × (ii) Energy of the photon E = hv We know, 34 156.625 10 10E −= × × 196.625 10 joules−= × Ans9: Planck was able to explain the distribution of intensity in the radiation from black body as function of frequency or wavelength at different temperature. Ans10: The exact frequency distribution of the emitted radiation from a black body depends only on its temperature. CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (Structure of Atom) Topic:-Photoelectric Effect Marks: 20 1. Define photoelectric effect. [1] 2. How does the intensity of light effect photoelectrons? 3. What is threshold frequency? [1] 4. Name the scientist who demonstrated photoelectric effect experiment. [1] 5. What did Einstein explain about photoelectric effect? [1] 6. What is the relation between kinetic energy and frequency of the photoelectrons? [2] 7. Calculate energy of 2mole of photons of radiation whose frequency is 145 10 .Hz× [1] 8. What is emission and absorption spectra? [2] 9. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum? [2] 10. Spectral lines are regarded as the finger prints of the elements. Why? [2] CBSE TEST PAPER-05 CLASS - XI CHEMISTRY (Structure of Atom) Topic:-Quantum Mechanical Model of the Atom Marks: 20 1. States Heisenberg’s Uncertainty Principle. [1] 2. Give the mathematical expression of uncertainty principle. [2] 3. How would the velocity be effected if the position is known? [1] 4. We don not see a car moving as a wave on the road why? [1] 5. Give the de – Broglie’s relation. [1] 6. Why cannot the motion of an electron around the nucleus be determined accurately? [2] 7. Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length 101 10 .−× [2] 8. Calculate the uncertainty in the velocity of a wagon of mass 4000kg whose position is known accurately of 10m± [1] 9. What is the physical significance of 2ψ up? [1] CBSE TEST PAPER-05 CLASS - XI CHEMISTRY [ANSWERS] Topic: - Quantum Mechanical Model of the Atom Ans1: It states that ‘It is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron. Ans2: Mathematically, it can be given as 4x hx p π∆ × ∆ ≥ ( ), 4 , 4 x x hor x mv hor x U m π π ∆ × ∆ ≥ ∆ × ∆ ≥ Where x∆ is the uncertainty in position and xp∆ ( )xv∆ is the uncertainty in momentum (or velocity) of the particle. Ans3: If the position of the electron is known with high degree of accuracy ( x∆ is small), then the velocity of the electron will be uncertain ( )( )is large. .xV∆ Ans4: According to de Broglie’s relation, 1 . . h i emv m λ λα= the mass of the car is very large and its wavelength ( )λ or wave character is negligible. Therefore, we do not see a car moving like a wave. Ans5: According to de Broglie, every particle in motion is associated with a wavelength and other wave characteristics. He deduced the relation that wavelength ( )λ of a particle in motion is equal to the Planck’s constant (h) divided by the momentum (p) of the particle. i.e. 1h p mv λ = = Where m is the mass, v is the velocity after particles Ans6: Because there is an uncertainty in the velocity of moving electron around the nucleus (Heisenberg’s Uncertainty Principle). Ans7: According to uncertainty Principle . 4 , 4 h x p h or p x π π ∆ ∆ = ∆ = ∆ 34 2 1 10 26 1 6.626 10 , 4 3.143 10 ) 5.27 10 kgm s or p m kgms − − − − ×∆ = × × = × Ans8: Uncertainty in velocity ( )v∆ is given by 34 2 1 3 39 1 4 6.6 10 22 4 4 10 ( 10 ) 7 1.3 10 h v m x kgm s kg m ms π − − − − ∆ ≥ ∆ ×= × × × × ± = × ∴The uncertainty in the velocity of the wagon is 39 11.3 10 ms− −= × Ans9: 2ψ represent probability of finding an electron. Since, l = 3 represents f – orbital, hence the given orbital is a 4f orbital. (iv) Here, n = 4 and l = 2 Since, l = 2 represents d – orbital, hence the given orbital is a 4d – orbital. (v) n = 4 and l = 1 since, l = 1 means it is a p – orbital, hence the given orbital can be designated as – 4p orbital. Ans9. (i) Mn (z = 25), Mn4+ (z = 21) The electronic configuration of Mn4+ to Given by 1s2 2s2 2p6 3s2 3p6 3d3 As the outermost shell 3d has 3 electrons, thus the number of unpaired electrons is 3. (ii) Fe (z = 26), Fe3+ (z = 23) The electronic configuration of Fe3+ is given lay 1s2 2s2 2p6 3s2 3p6 3d5 The number of unpaired electron is 5. (iii) Cr (z = 24), Cr2+ (z = 22) The electronic configuration of Cr2+ is 1s2 2s2 2p6 3s2 3p6 3d4 The number of unpaired electron is 4. (iv) Zn (z = 30), Zn2+ (z = 28) The electronic configuration of Zn2+ is 1s2 2s2 2p6 3s2 3p6 3d10 The number of unpaired electron is 0. CBSE TEST PAPER-01 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Genesis of Periodic classification 1. How many elements are known at present? [1] 2. Who was the first scientist to classify elements according to their properties? [1] 3. What is the basis of triad formation of elements? [1] 4. Stale the modern ‘Periodic law’? [1] 5. Define and state Mendeleev’s periodic law. [1] 6. How did Mendeleev arrange the elements? [2] 7. Name the two elements whose existence and properties were predicted by Mendeleev though they did not exist then. [2] 8. Describe the main features of Mendeleev’s periodic table? [3] CBSE TEST PAPER-01 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Genesis of Periodic classification [ANSWERS] Ans1. There are about 114 elements known at present. Ans2. The German Chemist, Johann Dobereiner in early 1829 was the first to consider the idea of trends among properties of element. Ans3. The middle element of each of the triads had an atomic weight about half way between the atomic weights of the other two. Also the properties of the middle element were in between those of the other two members. Dobereiner’s relationship is known as the haw of triads. Ans4. The physical and chemical properties of the elements are periodic functions of their atomic numbers. Ans5. Mendeleev’s Periodic law states that ‘The properties of the elements are periodic function of their atomic weights’. Ans6. Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. Ans7. Mendeleev predicted not only the existence of gallium and germanium, but also described some of their general physical properties. Ans8. (i) In Mendeleev table, the elements were arranged in vertical columns, and horizontal rows. The vertical columns were called groups and the horizontal rows were called periods. (ii) There were in all eight groups. Group I to VIII. The group numbers were indicated by Roman numerals. Group VIII occupy three triads of the elements each i.e. in all nine elements. (iii) There were seven periods to accommodate more elements the period 4, 5, 6 and 7 were divided into two halves. The first half of the elements were placed in the upper left corner and the second half in the lower right corner of each box. Ans6. p – block elements : The elements in which the last electron enters the p – orbital of their outermost energy level are called p – block elements. It contains elements of group 13,14, 15, 16, 17 and 18 of the periodic table. General electronic configuration of p – block elements is ns2 np1-6. Ans7. d – block elements :- The elements in which the last electron enters the d – orbitals of their last but one energy level constitute d – block elements. There block consists of the elements lying between s and p block starting from 4th period and onwards. They constitute groups 3 to 12 in the periodic table. General electronic configuration is (n – 1) d1-10 ns1-2. Ans8. f – block elements : The elements in which the last electron enters the f – orbital of their atoms are called f – block elements. In these elements the last electron is added to the third to the outermost energy level. These consist of two series of elements placed at the bottom of the periodic table known as Lanthanoid and actinoid series. General electronic configuration is (n-2)f1-14 (n-1)d0-1 ns2. CBSE TEST PAPER-03 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Trends in physical properties of elements 1. Predict the position of the element in the periodic table satisfying the electronic configuration (n-1) d1 ns2 for n=4, [1] 2. How does atomic size change in a group? [1] 3. Why Li and Mg show resemblance in chemical behaivour? [1] 4. The atomic radius of elements decreases along the period but Neon has highest size among III period element? Why [1] 5. Explain why cations are smaller and anions are larger in radii than their parent atom? [2] 6. Define ionization enthalpy and electron gain enthalpy? [2] 7. How does atomic size change in a group? [2] 8. The size of an atom can be expressed by three radii. Name them. Which of these given the highest, and the lowest value of the atomic radius of an element? [2] 9. Among the elements B, Al, C and Si (a) Which has the highest first ionization enthalpy? (b) Which has the largest atomic radius? [2] 8. Na+ has higher value of ionization enthalpy than Ne, though both have same electronic configuration. [2] CBSE TEST PAPER-03 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Trends in physical properties of elements [ANSWERS] Ans1. (n-1) d1 ns2 = (4-1) d1 4s2 = 3d1 4s2 It lies in fourth period and III B group. Ans2. It increases from top bottom in a group. Ans3. Due to diagonal relationship, since their atomic size, electro negativity and ionisation potential are almost the same. Ans4. Ne is the only element in III period element which has Van der walls radius whereas the rest has covalent radius. And it is known fact that Van der walls radius is always greater than covalent radius. Ans5. The radius of cation is smaller than the parent atom. Cation is formed by the loss of one or more electron from the gaseous atom, but the nuclear charge remains the same. As a result, the nuclear hold on the remaining electrons increases because of the increases in the effective nuclear chanre per electron resulting in decrease in size. Whereas anion is formed by the gain of one or more electrons by the gaseous atom but the nuclear charge is same though the number of electrons has increased. The effective nuclear charge per electron decrease in the anion and the cloud is held less tightly by the nucleus. This causes increase in size. Ans6. Ionization enthalpy – It represents the energy required to remove an electron from an isolated gaseous atom (x) in ground state resulting in the formation of a positive ion. x(g) + Energy → x+ (g) + e- CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Chemical Properties of element [ANSWERS] Ans1. The combining capacity of an element is known as valency. Ans2. In a group, the valency of an element remains constant while in a period it increases from left to right. Ans3. Noble gases on the extreme right are zero valent. Ans4. The tendency of an element to lose electrons decreases in going from left to right in a period. Thus the reactivity of metals goes on decreasing in a period from left right. Ans5. The tendency to lose electrons increases as we go down a group so the reactivity of metals increases down the group. Ans6. The reactivity of non – metals is measured in terms of its tendency to gain electrons to form an ion. The reactivity of non – metals increases from left to right in a period whereas reactivity decreases in a group as we go down the group because the tendency to accept electrons decreases down the group. Ans7. Elements on two extremes of a period easily combines with oxygen to oxides. The normal oxide formed by the element on extreme left is the most basic (eg. Na2O) whereas that formed by the element on extreme right is the most acidic (eg. cl2 O7). Oxides at the centre are however amphoteric (eg. Al2 O3) or neutral (eg. CO). Ans8. Oxides which behave as acids with bases and as a base with an acid are called amphoteric oxide. Ans9. Neutral oxides have no acidic or basic properties. Ans10. Lithium forms covalent bond which is different from its group members because of its anomalous behaviour Li is small in size, large charge / radius ratio and has high electro negativity value. Also it has only 1s2 2s1 orbital for bonding. CBSE TEST PAPER-05 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Miscellaneous Questions 1. What is the general outer electronic configuration of f – block elements? [1] 2. Why do Na and K have similar properties? [1] 3. Arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P. [1] 4. The atomic number of an element is 16. Determine its position in accordance to its electronic configuration. [2] 5. Why are elements at the extreme left and extreme right the most reactive? [2] 6. Why does the ionization enthalpy gradually decreases in a group? [1] 7. Why does electronegativity value increases across a period and decreases down period? [2] 8. How does electronegativity and non – metallic character related to each other? [2] CBSE TEST PAPER-05 CLASS - XI CHEMISTRY (Classification of Elements and Periodicity in Properties) Topic: - Miscellaneous Questions [ANSWERS] Ans1. The general outer electronic configuration of f – block element is (n – 2) f1-14 (n – 1) d0-11 ns2 Ans2. Na and K have similar physical and chemical properties because they have same number of valence electrons. Ans3. P < Si < Be < Mg < Na Ans4. The atomic of the element is 16. The electronic configuration of the element is 1s2 2s2 2p6 3s2 3p4 Thus the element belongs to ‘p-block’ and is placed in third period and 16th group of the periodic table. Ans5. The maximum chemical reactivity at the extreme left (among alkali metals) is exhibited due to the loss of an electron leading to the formation of a cation due to low ionization enthalpy and at the extreme right (among halogens) shown by the gain of an electron forming an anion. Ans6. In a group, the increase in atomic and ionic radii with increase in atomic number generally results in a gradual decrease in ionization enthalpies. Ans7. The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period. The electeonegativity also increases. On the same account electronegativity value decreases with the increase in atomic radii down a group. Ans8. Electronegativity is directly related to the non – metallic character of elements. Electronegativity is inversely related to the metallic properties of elements. Thus the increase in electronegativities across a period is accompanied by an increase in non – metallic properties of elements. Similarly, the decrease in electronegativity down a group is accompanies by a decrease in non – metallic properties of elements. Ans8. Octet rule signifies – (i) It is useful for understanding the structures of most of the organic compounds. (ii) It mainly applies to the second period elements of the periodic table. Ans9. (i) The outer (valence) shell configurations of carbon and oxygen atoms are Carbon : (6) – 1s2 2s2 2p2 Oxygen : (8) – 1s2 2s2 2p4. The valence electrons (4 + 6 = 10) But it does not complete octet, thus multiple bond is exhibited. Thus, (ii) N (2s2 2p3), O (2s2 2p4) 5 + (2 x 6) + 1 = 18 electrons. Thus, Ans10. HNO3 → . . . . . . . . . . C O CBSE TEST PAPER-02 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - Ionic Bonding 1. Define an ionic bonding. [?] 2. What changes are observed in atoms undergoing ionic bonding? [2] 3. Mention the factors that influence the formation of an Ionic bond. [2] 4. Which one of the following has the highest bond order? N2, N2+ or N2-. [1] 5. Define bond order. [1] 6. Give reason why H2+ ions are more stable than H2- though they have the same bond order. [2] 7. How would the bond lengths vary in the following species? C2, C2- C22-. [2] 8. What type of bond is formed when atoms have high difference of electornegativity? [1] 9. Out of covalent and hydrogen bonds, which is stronger. [2] 10. Define covalent radius. [2] CBSE TEST PAPER-02 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - Ionic Bonding (ANSWERS) Ans1. An ionic bond (or electrovalent bond) is formed by a complete transfer of one of outer most electrons from the atom of a metal to that of a non – metal. Ans2. Due to the electron transfer the following changes occurs – (i) Both the atoms acquire stable noble gas configuration. (ii) The atom that loses electrons becomes +vely charged called cation whereas that gains electrons becomes –vely charged called anion. (iii) Cation and anion are held together by the coulombic forces of attraction to form an ionic bond. Ans3. Ionic bond formation mainly depends upon three factors – (i) Low ionization energy – elements with low ionization enthalpy have greater tendency to form an ionic bonds. (ii) High electron gain enthalpy – high negative value of electron gain enthalpy favours ionic bond. (iii) Lattice energy – high lattice energy value favours ionic bond formation. Ans4. N2 has the highest bond order. Ans5. Bond order is defined as number of bonds between two atoms in a molecule. Ans6. In H2 - ion, one electron is present in anti bonding orbital due to which destabilizing effect is more and thus the stability is less than that of H2 + ion. Ans7. The order of bond lengths in C2 , C2 - and C2 2- is C2 > C2 - > C2 2-. Ans8. Electrovalent or ionic bond. Ans9. Covalent bond. Ans10. The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation. Ans7. Ans8. Ans9. The dipole moment of the polyatomic molecule depends on individual dipole moments of bonds and also on the spatial arrangement of various bonds in the molecule. Ans10. In BeF2 the dipole moment is zero because the two equal bond dipoles point in opposite directions and cancel the effect of each other. Bond dipoles in Be F2 CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - VSEPR 2 VBT 1. Give the main features of VSEPR Theory. [2] 2. What’s difference between lone pair and bonded pair of electrons? [2] 3. CO2 is linear whereas SO2 is bend – shaped. Give reason. [2] 4. Why does H2O have bent structure? [2] 5. For the molecule, Why is structure (b) more stable than structure (a)? [2] 6. How would you attribute the structure of PH3 molecule using VSEPR model? [2] 7. In SF4 molecule, the lp electrons occupies an equatorial position in the trigonal bipyramidal arrangement to an axial position. Give reason. [2] 8. How is VBT different from Lewis concept? [2] 9. S – orbital does not show any preference for direction. Why? [2] CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - VSEPR 2 VBT [ANSWERS] Ans1. The main postulates of VSEPR theory are as follows : (i) The shape of a molecule depends upon the number of valence shell electron pairs around the central atom. (ii) Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged. (iii) There pairs of electrons tends to occupy such position in space that minimize repulsion and thus maximize distance between them. (iv) The valence shell is taken as a sphere with the electron pairs localizing on the sphere at maximum distance from one another. (v) A multiple bond is treated as it is a single electron pair and two or three electron pairs of a multiple bond is treated as super pair. (vi) When two or more resonance structures can represent a molecule, the VSEPR nodal is applicable to any such structure. Ans2. Lone pair electrons do not take part in bond formation whereas bond pair electrons take part in bond formation. Ans3. In CO2, the bond electron are furtherest away from each other forming1800 angle. Thus, CO2 is linear. . In SO2, the number of bonding pairs is 4 where it has an lone pair of electron which does not participate in bond formation thereby repulsive strain is experienced. Ans4. In water molecule, there are two bonding pairs and two lone pairs of electrons. The shape should have been tetrahedral if there were all bp but two lp are present. Thus the shaped is distorted to an angular shape. Because lp – lp repulsion is more than lp – bp repulsion. H H O CBSE TEST PAPER-05 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - Sigma and pi – bond [ANSWERS] Ans1. A covalent bond formed due to the overlap of orbitals of the two atoms along the line going the two nuclei (orbital axis) is called sigma (σ) bond. Ans2. A covalent bond formed between the two atoms due to the sideways overlap of their p – orbitals is called a pi (π ) bond. Ans3. Orbitals can overlap to a greater extent in a σ - bond due to axial orientation, so σ - bond is strong. Whereas, in a pi – bond sideways overlapping is not to an appreciable extent due to the presence of σ - bond which restricts the distance between the involved atoms. Ans4. In a molecule of ethane, there are 5 σ - bonds (one between C-C , and four between C-H and one π - bond. Ans5. There are 9 σ - bonds (three between C – C and 6 between C – H) and 2 π - bonds. Ans6. pi (π) – bond is always present in molecules containing multiple bond. Ans7. σ - bond can be formed by any of the following types of combinations of atoms orbitals. (a) S – S – overlapping : In this case, there is a lover lap of two half – filled S – orbitals along the inter nuclear axis. (b) S- P overlapping : This type of over lapping occurs between half – filled s-orbitals of one atom and half-filled p-orbitals of another atom. (c) P – P overlapping : This type of overlap takes place between half-filled p-orbitals of the two approaching atoms. Ans8. Covalent bonds are formed due to the overlap of certain orbitals that are oriented favourably in the space. Ans9. According to orbital overlap concept, the formation of a covalent bond between two atoms results by pairing of electrons present in the valence shell having opposite spins. Ans10. The unsymmetrical overlap of orbitals results in zero overlap i-e; between px-s and px- py orbital CBSE TEST PAPER-06 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - Hybridisation 1. Define hybridisation. [1] 2. Give the features of hybridisation. [2] 3. What are the important consolations for hybridisation? [2] 4. Describe the shape of sp, sp2 and sp3 hybrid orbital? [2] 5. State the hybrid orbitals associated with B in BCl3 and C in C2H4 [1] 6. What is the state of hybridization of carbon atoms in diamond and graphite? [1] 7. Ethylene is a planar molecule whereas acetylene is a linear molecule. Give reason. [2] 8. In H2O, H2S, H2Se, H2Te, the bond angle decreases though all have the same bent shape. Why? [2] 9. What type of hybridisation takes place in (i) p in PCL5 and (ii) S in S F6? [1] 10. Out of p-orbital and sp-hybrid orbital which has greater directional character and Why? [2] CBSE TEST PAPER-07 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - Molecular Orbital Theory 1. Define bonding molecular orbital. [1] 2. Define antibonding molecular orbital. [1] 3. Explain diagrammatically the formation of molecular orbital by LCAO. [1] 4. Which one 2-2 2 and O ,O − may exhibit paramagnetism? [1] 5. Why are bonding molecular orbitals more stable than antibonding molecular orbitals? [1] 6. 2He does not exist. Explain in terms of LCAO. [2] 7. Define bond order. [1] 8. Define hydrogen bonding [1] 9. What are the types of H-bonding? Which of them is stronger? [1] 10. 3NH has higher boiling point than 3PH . Give reason. [1] CBSE TEST PAPER-07 CLASS - XI CHEMISTRY (Chemical Bonding and Molecular Structure) Topic: - Molecular Orbital Theory [ANSWERS] Ans1. The molecular orbital formed by the addition of atomic orbitals is called bonding molecular orbital. A Bσ = Ψ − Ψ Ans2. The molecular orbital formed by the subtraction of atomic orbitals is called antibonding molecular orbital. A Bσ + = Ψ − Ψ . Ans3. Pg 122 (Pant I) – Fig . 4.19 NCERT. Ans4. 2O would exhibit paramagnetism because it contains one unpaired electron in its Mo configuration. Ans5. Bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital. Ans6. The electronic configuration of helium atom is 21s . Each helium atom contains 2 electrons, therefore, in 2He molecule there would be 4 electrons. These electrons will accommodated in +1 snd 1s sσ σ molecular orbitals leading to electronic configuration : 2 2 2 2 : ( 1s) ( 1 ) 1 order of He (2 2) 0 2 He s Bond is σ σ + − = 2He molecule is there unstable and does not exist. Ans7. Bond order (b.o) is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals i.e; Bond order (b.o) = 1 2 a ( - N )bN If ,b aN N> molecule is stable and If ,b aN N< molecule is unstable. Ans8. Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Ans9. (i) Inter-molecular H-bonding (ii) Intra molecular H-bonding. Inter molecular H-bonding is stronger than intra- molecular H-bonding. Ans10. In 3NH , there is hydrogen bonding whereas in PH3 there is no hydrogen bonding. CBSE TEST PAPER-02 CLASS - XI CHEMISTRY (States of Matter: Gases and Liquids) Topic: - The Gaseous State & The Gas Laws 1. Define Boyle’s law. [1] 2. Why helium and hydrogen gases not liquefied at room temperature by applying very high pressure? [1] 3. At what temperature will the volume of a gas at 00 c double itself, pressure remaining constant? [2] 4. How is the pressure of a given sample of a gas related to temperature at volume? [1] 5. Define absolute zero temperature. [1] 6. 50 cm3 of hydrogen gas enclosed in a vessel maintained under a pressure of 1400 Torr, is allowed to expand to 125 cm3 under constant temperature conditions. What would be its pressure? [2] 7. State the law depicting the volume-temperature relationship. [2] 8. State Avogadro’s Law. Is the converse of Avogadro’s law true? [2] CBSE TEST PAPER-02 CLASS - XI CHEMISTRY (States of Matter: Gases and Liquids) Topic: - The Gaseous State & the Gas Laws [ANSWERS] Ans1. At constant temperature, the pressure of a fixed amount (i.e; number of moles n) of gas varies inversely with its volume. Mathematically, 1 p v ∝ (at constant T and n) => 1 1 = kp v Or, Ans2. Because their critical temperature is lower than room temperature and gases cannot be liquefied above the critical temperature even by applying very high pressure. Ans3. Let the volume of the gas at 00C = Vml. Thus, V1 = Vml V2 = 2Vml T1 = o + 273 T2 = ? = 273k By applying charles law 1 2 1 2 V V T T = = > 2 2 273 V V T = = > T2 = 2 273 546 Vx k V = T2 = 546 - 273 = 2730C Ans4. Pressure is directly proportional to the temperature , i.e; P∝ T. 1 = Kpv Ans5. The lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume is called Absolute zero. Ans6. Since, temperature is constant, we have PV = constant ∴ P1 V1 = P2 V2 P1 = 1400 ; P2 = ? V1 = 50cm3 ; V2 = 125 cm3 ∴ P2 = 1 1 2 1400 50 560 0 125 PV x T rr V = = The final pressure of the gas after expansion would be 560 Torr. Ans7. The law is known Charle’s law. “Pressure remaining constant the volume of a given mass of a gas increases or decreases by 1/273 of its volume at 00C for every one degree centigrade or fall in temperature. Mathematically, o = V + t273 o t V V = (1+ ) 273o t V Where Vt is the volume of the gas at t0C and Vo is its volume at 00C. Ans8. Avogadro’s Hypothesis: This law was given by Avogadro in 1811. According to this law, “Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.” Volume = a constant X Number of Moles (Temperature and pressure constant). The converse of Avogadro’s law is also true. Equal number of molecules of all gases occupy equal volume’s under the same conditions of temperature and pressure. It follows that one gram molecular mass of any gas (containing 6.023x1023 molecules) will occupy the same volume under the same conditions. The volume occupied by one gram molecular mass of any gas at 00C and 760 mm of Hg is 2.4 dm3 (liters) is called the gram molecular volume or simply molar volume. o 273 = V ( ) 273)t t V + Ans4. Molar mass of acetylene (C2 H2) M = (2x12+2x1) g/mol = 26 g/mol Mass of acetylene, m = 50g. Temperature, T = (500C + 273) = 323k Pressure, p = 740mm Hg = 740 0.9737 atm 760 mm = Pv = nRT = m RT M v = 5 0.082 323 = 26 0.9737 mRT x x pM x = 5.23Lv Ans5. Pressure exerted by saturated water vapors is called aqueous tension. Ans6. At S. T. P, R = 8.20578 x 10-2 L at m k-1 mol-1. Ans7. The ratio pv/RT is equal to the number of moles of an ideal gas in the sample. This number should be constant for all pressure, volume and temperature conditions. If the value of this ratio changes with increasing pressure, the gas sample is not behaving ideally. Ans8. Kinetic energy of a molecule is directly proportional to temperature and independent of mass so both the molecules will have the same kinetic energy. CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (States of Matter: Gases and Liquids) Topic: - Behaviour of Real Gases 1. Write Van der waal’s equation for n moles of a gas. [1] 2. Out of NH3 and N2, which will have (i) larger value of ‘a’ and (ii) larger value of ‘b’? [1] 3. What property of molecules of real gases is indicated by van der waal’s constant ‘a’? [1] 4. Under what conditions do real gases tend to show ideal gas behaivour? [1] 5. How are Van der waal’s constants ‘a’ and ‘b’ related to the tendency to liquefy? [1] 6. Mention the two assumptions of kinetic theory of gases that do not hold good. [2] 7. When does a gas show ideal behaivour in terms of volume? [1] 8. Define Boyle point. [1] 9. Calculate the pressure exerted by one mole of CO2 at 273 k if the Van der waal’s constant a = 3.592 dm6 at m mol-1. Assume that the volume occupied by CO2 molecules is negligible. [2] 10. What is the value of compressibility factor Z, of a gas when (i) pressure is low, (ii) pressure is high, (iii) at intermediate pressure. [1] CBSE TEST PAPER-04 CLASS - XI CHEMISTRY (States of Matter: Gases and Liquids) Topic: - Behaviour of Real Gases [ANSWERS] Ans1. ( ) 2 2 an p V nb nRT V   + − =    is Van der waal’s equation for n moles of a gas. Ans2. (i) NH3 will have larger value of ‘a’ because of hydrogen bonding. (ii) N2 should have large value ‘b’ because of larger molecular size. Ans3. Intermolecular attraction Ans4. When the pressure of the gas is very low and the temperature is very high. Ans5. The Van der waal’s constant ‘a’ is a measure of intermolecular attractions. Therefore, the value of ‘a’ reflects the tendency of the gas to liquefy. The gas having larger value of ‘a’ will liquefy more easily. Ans6. The two assumptions of the kinetic theory that do not hold good are – (i) There is no force of attraction between the molecules of a gas. (ii) Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas. Ans7. The gases show ideal behaviors when the volume of the occupied is large so that the volume of the molecules can be neglected in comparison to it. Ans8. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point. CBSE TEST PAPER-05 CLASS - XI CHEMISTRY (States of Matter: Gases and Liquids) Topic: - Liquid state [ANSWERS] Ans1. If the pressure is 1 bar then the boiling point is called standard boiling point of the liquid. Ans2. The energy required to increase the surface area of the liquid by one unit is defined as surface energy. Ans3. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid S.I unit is expressed as Nm-1. Ans4. Surface tension decreases as the temperature is raised. Ans5. Viscosity of liquids decreases as the temperature rises because at high temperatures molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers. Ans6. The pressure of the air is directly proportional to the temperature. Since the temp is higher in summer than in winter, the pressure of the air in the tube of the tyre is likely to be quite high as compared to winter. Ans7. Glycerol has three hydrogen bonds which results in an extensive hydrogen bonding. That is why glycerol is highly viscous. Ans8. (i) = M D V The volume increases, with the increase of temperature. Therefore, density decreases with the rise of temperature. (ii) As the temperature of a liquid is increased, the vapors pressure increases. Ans9. The spheres would start moving faster randomly and colliding with each other. Ans10. A liquid boils when its vapors pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, causes a rise in the boiling temperature of the Liquid. CBSE TEST PAPER 01 CLASS XI CHEMISTRY (Thermodynamics) Topic : Thermodynamic Terms 1. Define a system. [1] 2. Define surroundings. [1] 3. State the first law of thermodynamics. [1] 4. What kind of system is the coffee held in a cup? [1] 5. Give an example of an isolated system. [1] 6. Name the different types of the system. [1] 7. What will happen to internal energy if work is done by the system? [1] 8. From thermodynamic point of view, to which system the animals and plants belong? [1] 9. How may the state of thermodynamic system be defined? [1] 10. Change in internal energy is a state function while work is not, why? [2] CBSE TEST PAPER 01 CLASS XI CHEMISTRY (Thermodynamics) Topic : Thermodynamic Terms [ANSWERS] Ans 1. A system in thermodynamics refers to that part of the universe in which observations are made. Ans 2. The rest of the universe which might be in a position to exchange energy and matter with the system is called its surroundings. Ans 3. The first law of thermodynamics stales that ‘the energy of an isolated system is constant’. Ans 4. Coffee held in a cup is an open system because it can exchange matter (water vapors) and energy (heat) with the surroundings. Ans 5. Coffee held in a thermos flask is an isolated system because it can neither exchange energy nor matter with the surroundings. Ans 6. There are three types of system – (i) Open system (ii) Closed system (iii) Isolated system. Ans 7. The internal energy of the system will decrease if work is done by the system. Ans 8. Open system. Ans 9. The state of thermodynamic system may be defined by specifying values of state variables like temperature, pressure, volume. Ans 10. The change in internal energy during a process depends only upon the initial and final state of the system. Therefore it is a state function. But the wonk is related the path followed. Therefore, it is not a state function. Ans 5. The difference between H∆ and U∆ is not usually significant for systems consisting of only solids and / or liquids because they do not suffer any significant volume changes upon heating. Ans 6. For gases the volume change is appreciable. let VA be the total volume of gaseous reactants, and VB be the total volume of gaseous product. nA be the number of moles of the reactant and nB be the number of moles of the product, Then at constant pressure and temperature, p VA = nA RT p VB = nB RT or p VB - pVA = (nB - nA) RT or p ( ) = RT g V n∆ ∆ where ( ) B Agn n n∆ = and is equal to the difference between the number of moles of gaseous products and gaseous reactants. Substituting the value of p V∆ awe get. ( ) g H U n RT∆ = ∆ + ∆ H=qp ∴ ∆ (heat change under constant pressure) U qv∆ = (heat change under constant volume) ∴ for gaseous system. Ans 7. Extensive property is a property whose value depends on the quantity or size of matter present in the system. Intensive property is a property which do not depend upon the quantity or size of matter present. Ans 8. (i) For isothermal irreversible change Q = -w = pex (vf - vi) (ii) For isothermal reversible change q = - w = nRT ln f i v v = 2.303 nRT log f i v v ( ) g qp qv n RT= + ∆ CBSE TEST PAPER 03 CLASS XI CHEMISTRY (Thermodynamics) Topic : Heat Capacity 1. Define Heat capacity [1] 2. Define specific heat. [1] 3. Give the mathematical expression of heat capacity. [1] 4. It has been found that 221.4J is needed to heat 30g of ethanol from 150C to 180C. calculate (a) specific heat capacity, and (b) molar heat capacity of ethanol. [3] 5. Show that for an ideal gas Cp- Cv = R [2] 6. Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2 R. [2] 7. A 1.25g sample of octane (C18 H18) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 294.05 to 300.78K. If heat capacity of the calorimeter is 8.93 KJ/K. find the heat transferred to calorimeter. [2] CBSE TEST PAPER 03 CLASS XI CHEMISTRY (Thermodynamics) Topic : Heat Capacity [ANSWERS] Ans 1. The heat capacity for one mole of the substance is the quantity of heat needed fo raise the temperature of one mole by one degree Celsius. Ans 2. Specific heat /specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin). Ans 3. The mathematical expression of heat capacity q = c x m x T∆ = C T∆ where c = specific heat m = mass T∆ = temperature change. Ans 4. (a) Specific heat capacity absorbed by the substance of the substance x Rise in temp. Heat Mass C = ( ) -1 0 1 0 0 221.4 221.4 = Jg 30 x 3 30 18 15 j c g c c − − = -1 0 12.46 g J c− Since 10C is equal to 1k, the specific heat capacity of ethanol = 2.46Jg-1 0c-1. (b) Molar heat capacity, Cm = specific heat x molar mass. Therefore, Cm (ethanol) = 2.46 x 46 = 113.2 Jmol-1 0c-1 The molar heat capacity of ethanol is 113.2 J mol-1 0c-1. Ans 5. When a gas is heated under constant pressure, the heat is required for raising the temperature of the gas and also for doing mechanical work against the external pressure during expansion. CBSE TEST PAPER 04 CLASS XI CHEMISTRY (Thermodynamics) Topic : Enthalpy Change of a Reaction [ANSWERS] Ans 1. The enthalpy change accompanying a reaction is called the reaction enthalpy ( ).rH∆ Ans 2. The standard enthalpy of reaction is the enthalpy change for a reaction is the enthalpy change for a reaction when all the participating substances are in their standard states. Ans 3. ( ) ( tan ) oo o f fH H products H reac ts∆ = ∆ − ∆∑ ∑ = [2 X ∆ Hfo Fe2O3(s) ] – [4 ∆ Hf oFe (s) + 3 ∆ Hf oO2(g)] = 2(-824.2kJ) – [ 4 x o + 3 x o ] = 1648.4kJ− Ans 4. C2H4 (g) + 302(g) → 2CO2(g) + 2H2O (g) 2( ) = -393.7kJfH CO∆ 2( ) = -241.8kJfH H O∆ 2 4( ) = +52.3kJfH C H∆ = oo f fHreaction H products H∆ ∆ − ∆∑ ∑ reactants = [2 x OfH∆ (CO2) + 2 x 2( ) O fH H O∆ ] – ( )2 4 f 2 3 x H ( )O OfH C H O ∆ + ∆  = 2 x[(-393.7)m+2x (-241.8)] – [(523.0) + 0)] fot elementary substance =0OfH ∴∆  = [-787.4 – 483.6 ] -53.3 = - 1323.3 kJ. Ans 5. Examples of reactions driven by enthalpy change: The process which is highly exothermic, i.e. enthalpy change is negative and has large value but entropy change is negative is said to be driven by enthalpy change, eg. (i) ( ) ( ) ( )2 2 2 1 2 H g O g H O l+ → -1285.8 J molOfH k∆ = − (ii) ( ) ( ) ( )2 2 33 2N g H g NH g+ → -192 J mol .OH k∆ = − Ans 6. No, the heats released in the two reactions are not equal. The heat released in any reaction depends upon the reactants, products and their physical states. Here in reaction (i), the water produced is in the gaseous state whereas in reaction (ii) liquid is formed. As we know, that when water vapors condensed to from water, heat equal to the latent heat of vaporization is released. Thus, more heat is released in reaction (ii). Ans 7. A chemical reaction involves the breaking of bonds in reactants and formation of new bonds in products. The heat of reaction (enthalpy change) depends on the values of the heat needed to break the bond formation .Thus (Heat of reaction = (Heat needed to break the bonds in reactants – Heat liberated to from bonds in products). OH∆ = Bond energy in (to break the bonds) X Bond energy out (to form the bonds) = Bond energy of reactants – Bond energy of products. Ans 8. (i) The standard enthalpy of formation of CO2 is -393.5 kJ per mole of CO2. That is ( ) -12 , 393.5 mol .OfH CO g kJ∆ = − (ii) The standard enthalpy of combustion of carbon is - 393.5 kJ per mole of carbon i.e. -1( ) 393.5 kJ mol .OH comb c∆ = − CBSE TEST PAPER 05 CLASS XI CHEMISTRY (Thermodynamics) Topic : Spontaneity 1. Define spontaneous process. [1] 2. Define non-spontaneres process. [1] 3. What is the sign of enthalpy of formation of a highly stable compound? [1] 4. Predict the sign of S∆ for the following reaction CaCo3 (s) 5 2 ( ).CaO CO g ∆→ + [1] 5. Two ideal gases under same pressure and temperature are allowed to mix in an isolated system – what will be sign of entropy change? [1] 6. Predict the sign of the entropy change for each of the following changes of state. (a) ( ) ( )Hg l Hg g→ (b) 3 3( ) ( )AgNO S AgNO aq→ (c) I2(g) 2 ( )I S→ (d) ( ) ( )C graphite C diamond→ [2] 7. Explain how is enthalpy related to spontaneity of a reaction? [2] 8. The 2 2 and S for 2Ag O(s) 4Ag(s) +O ( )H g∆ ∆ → are given + 61.17kJ mol-1 and + 132 Jk-1mol-1 respectively. Above what temperature will the reaction be spontaneous? [2] CBSE TEST PAPER 01 CLASS XI CHEMISTRY (Equilibrium) Topic : Equilibrium in physical processes 1. Define dynamic equilibrium. [1] 2. Name the three group into which chemical equilibrium can be classified. [3] 3. What is physical equilibrium? Give an example. [1] 4. What is meant by the statement ‘Equilibrium is dynamic in nature’? [1] 5. On what factor does the boiling point of the liquid depends? [1] 6. State Henry’s law. [1] 7. What happens to the boiling point of water at high altitude? [1] 8. On which factor does the concentration of solute in a saturated solution depends? [1] 9. Mention the general characteristics of equilibria involving physical processes. [2] 10. What conclusion is drawn from the following – Solid ⇌ Liquid H2O(s) ⇌ H2O (l) [1] CBSE TEST PAPER 01 CLASS XI CHEMISTRY (Equilibrium) Topic : Equilibrium in physical processes [ANSWERS] Ans 1. When the reactants in a closed vessel at a particular temperature react to give products, the concentrations of the reactants keep on decreasing, while those of products keep on increasing for sometime after which there is no change in the concentrations of either the reactants or products. This stage of the system is the dynamic equilibrium. Ans 2. Chemical equilibrium can be classified into three groups – (i) The reaction that proceeds nearly to completion and only negligible concentrations of the reactants are left. (ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage. (iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium. Ans 3. Physical equilibrium is an equilibrium between two different physical states of same substance e.g. H2 O(s) ⇌H2O(l) Ans 4. At equilibrium, reaction does not stop rather it still continues, the equilibrium is dynamic in nature. It appears to stop because rate of forward reaction is equal to the rate of backward reaction. Ans 5. Boiling point depends on the atmospheric pressure. Ans 6. The mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the gas above the solvent. Ans 7. Boiling point of water depends on the altitude of the place. At high altitude the boiling point decreases. Ans 8. The concentration of solute in a saturated solution depends upon the temperature. Sugar (soln.) ⇌ sugar (solid). Ans 9. (a) For solid ⇌ liquid equilibrium, there is only one temperature at 1 atm at which two phases can co-exist. If there is no exchange of heat with the surroundings, the mass of the two phases remain constant. (b) For liquid ⇌ vapors equilibrium, the vapors pressure is constant at a given temperature. (c) For dissolution of solids in liquids, the solubility is constant at a given temperature. (d) For dissolution of gases in liquids, the concentration of a gas in liquid is proportional to pressure of the gas over the liquid. Ans 10. Melting point is fixed at constant pressure.