THERMODYNAMICS READ THE INSTRUCTIONS CAREFULLY

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1 TERMOYNMIS RE TE INSTRUTIONS REFULLY. The test is of hours duration.. The maximum marks are 60.. This test consists of 47 questions. 4. For each question in Section, & you will be awarded marks if you have darkened only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. Minus one (-) mark will be awarded for wrong answer 5. For each question in Section you will be awarded marks. No Negative Marking. 6. For each question in Section E (Matrix Match), you will be awarded marks for each row in which you have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this Section. 7. Keep your mobiles switched off during Test in the alls. Section (Single orrect hoice Type) This Section contains 6 multiple choice questions. Each question has four choices ), ), ) and ) out of which ONLY ONE is correct. (Mark only One choice) 6 = 78 Marks. When 0. g of butanol was burnt in a suitable apparatus, the heat evolved was sufficient to raise the temperature of 00 g of water by 5 0. The heat of combustion of butanol in kcal/mole will be (mol mass of butanol = 74): a. 4.8 b. 74 c. 7 d. 70 Sol. eat evolved from 0. g = 00 5 = 000 cal = kcal eat evolved from mole, i.e., 74 g kcal. 0.. ccording to second law of thermodynamics, a process (reaction) is spontaneous, if during the process: a. S universe > 0 b. S universe 0 c. system >0 d. S universe = S system Sol. When a spontaneous (e.g., naturally occurring process) takes place, it is accompanied by increase in total entropy of system and surroundings (S universe ). One mole of an ideal gas at 00 K is expanded isothermally from an initial volume of litre to 0 litres. U for the process is (R=8.4 JK mol ) a. 40 J b. Zero c. 40 J d. 9 litre atom Sol. U = 0, T = 0 ( it is isothermal) and the gas is ideal. If the gas is non ideal or it is some chemical reaction then U Which of the following statements is incorrect? a. The specific heat capacity of substance is greater in the solid state than in liquid state b. The specific heat capacity of a substance is greater in gaseous state than in the liquid state c. The latent heat of vaporization of a substance is greater than that of fusion d. The internal energy of an ideal gas is a function of its temperature Sol. (b) is incorrect statement because heat capacity of liquid is more than gaseous state. r. Sangeeta Khanna Ph. EMISTRY OING IRLE

2 r. Sangeeta Khanna Ph. 5. The entropy change for the following reversible process: mole Sn (, 0 ) mole Sn (, 0 ) = 090 J mol a. -7. JK mol b..8 JK mol c. 7. JK mol d. -.8 JK mol Sol. T = 7 + = 86 K - trans 090Jmol Strans T 86K = 7. J K mol g of carbon was converted into carbon dioxide producing X cal of heat. When 0.8 g of carbon was converted in carbon monoxide Y cal of heat was produced. If X > Y then what will be the heat produced when.86 g of carbon monoxide is converted to carbon dioxide? a. X Y b. 5Y 5 X c. 0 X 0 Y d. Y X Sol. (s) + O (g) O (g); (s) O(g) O(g); X 5X 0.8 Y 5Y 0.8 The required equation O(g) O(g) O (g); = 5 X ( 5Y) = 5 Y 5 X For.8 g.86 (5Y 5X) (Y X) 8 7. Observe the graph and identify the correct statement(s): S vap. Entropy S fusion T T Temperature a. T is melting point, T is boiling point b. T is boiling point, T is melting point c. Sfusion is more than Svap d. T is lower than T Sol. T is melting point at which entropy change is entropy of fusion. T is boiling point at which entropy change is entropy of vaporisation. 8. For a reaction M O(g) M(s) + O (g), 0 kj mol and S 0.07 KJ K- mol at atm. Upto which temperature the reaction would not be spontaneous? a. T > 48 K b. T < 48 K c. T = 00 K d. T = 87 K Sol. For a non-spontaneous reaction G ve G TS r. Sangeeta Khanna Ph. EMISTRY OING IRLE

3 r. Sangeeta Khanna Ph. 0 0 T S 70 T < K 9. What is the free energy change G, when.0 mole of water at 00 and atm pressure is converted into steam at 00 and atm pressure? a. 540 cal b cal c cal d. Zero Sol. t 00, water and steam are at equilibrium. So G = 0 0. In an adiabatic process, which of the following is true? a. q = +w b. q = 0 c. E = q d. PV = 0 Sol. In an adiabatic process, the system is thermally insulated so that there is no exchange of heat. ence q = 0.. The enthalpy of formation O (g), O() and propene are 9.5, and 0.4 kj mol. The enthalpy of isomerisation of cyclopropane to propene is -.0 kj mol. The heat of combustion of cyclopropane at 98 K is a kj mol b kj mol c kj mol d kj mol Sol. + O O (g) = -9.5 kj mol + O O () = kj mol + 6(g) = 0.4 kj mol Pr opene 9 + O O (g) + O () =? = -.0 ( 9.5) ( 85.8) 0.4 = -09. kj mol. The difference between the heat of reaction at constant pressure and at constant volume at 7 for the following reaction is (gr) + O (g) O(g) (R = 8. J mol K ) = Propene 4 = -.0 kj mol a. 490 kj mol b J mol c. +.5 kj mol d..490 kj mol. The enthalpy of atomization of graphite is 75 kj mol and that of hydrogen is 8 kj mol. The enthalpy of formation of methane is -76 kj mol. The enthalpy of formation of the bond from gaseous carbon and atomic hydrogen is a. +48 kj b. +55 kj c. -55 kj d. -48 kj 4. ow much heat is released when a mixture containing 0.0 g N and 0.0 g O reacts by the equation 4N (g) + 5O (g) 4NO (g) + 6 O (g) = -906 J a. -6 J b. 666 J c. 566 J d. J 5. Which of the following salts shall cause maximum cooling when one mole of the salt is dissolved in the same amount of water: (Integral heat of solution at 5 0 is given below for each solute). a. KNO ( = 5.4 KJ) b. Nal ( = 5.5 KJ) c. KO ( = KJ) d. r ( = -8. KJ) r. Sangeeta Khanna Ph. EMISTRY OING IRLE

4 r. Sangeeta Khanna Ph. 6. When.00 mol of benzene is vaporized at a constant pressure of.00 atm and at its normal boiling point of 80.,.9 kj are absorbed and PV for the vaporization process is equal to.90 kj, then a. E =.0 kj and =.9 kj. b. E = 6.8 kj and =.9 kj. c. E =.9 kj and =.0 kj. d. E =.9 kj and = 6.8 kj. 7. The reaction 4g(s) + O (g) g O(s) favours g O at low temperature, but it favours g and O at high temperatures. ow can this be explained in terms of and S? a. is negative and S is negative. b. is negative and S is positive. c. is positive and S is negative. d. is positive and S is positive. 8. In figure () below oxygen molecules represented by unshaded spheres, and chlorine molecules, represented by shaded spheres, are in separate compartments. Figure() shows the equilibrium state of the system after the stopcock separating the two compartments is opened. ssuming that oxygen and the chlorine behave as ideal gases, what are the signs (+, -, or 0) of, S, and G for this process? () () a. = +, S = -, G = + b. = 0, S = +, G = - c. = 0, S = -, G = + d. = -, S = +, G = - 9. Ethyl alcohol is produced by the fermentation of glucose, 6 O 6. 6 O 6 () 5 O() + O (g) = -8 kj Given that the enthalpy of formation is find the enthalpy of formation for 6 O 6. kj/mol for 5 O() and -9.5 kj/mol for O (g), a. -44 kj/mol b. -60 kj/mol c. -75 kj/mol d. -67 kj/mol 0. Given that = - kj for the hydrogenation of acetylene, : (g) + (g) (g) and the following bond dissociation energies, estimate a value for the bond dissociation energy. ond, kj/mol a. 050 kj/mol b. 8 kj/mol c. 807 kj/mol d. 97 kj/mol. For hypothetical reversible reaction (g) (g) (g); -0 kj if standard entropies of, and are 60, 40 and 50 JK mol - respectively. The above reaction will be at equilibrium at: a. 400 K b. 500 K c. 50 K d. 00 K. The amount of energy released when 0 ml of 0.5 M NaO are mixed with 00 ml of 0. M l is x kj. The heat of neutralization (in kj mol - ) is: a. 00 x b. 50 x c. +00 x d x Sol. 0 ml of 0.5 M NaO = = 0 millimoles = 0 milliequivalents 00 ml of 0. M l = = 0 milli equivalents. 0 milliequivalents on neutralization give heat = x kj g eq., i.e., 000 milli eq. will give heat = 00 x kj i.e., 00 kj neut r. Sangeeta Khanna Ph. 4 EMISTRY OING IRLE

5 r. Sangeeta Khanna Ph.. Given N (g) + l (g) Nl (g) + l (g); N (g) + (g) N (g) ; (g) + l (g) l (g); The heat of formation of Nl (g) in the terms of, and is: a. f b. f c. f d. None Sol. im : N l Nl Operate Eqn. (ii) Eqn.(i)- Eqn.(iii) We get ( ) ( ). 4. The entropy change involved in the isothermal reversible expansion of 5 moles of an ideal gas from a volume of 0 dm to 00 dm at 00 K is approximately: a. 95 J K b. 9 JK c. 9.5 JK d. 4.5 JK Sol. V S.0 nr log V 00 0 = log = 95 J K 5. alculate the resonance energy of N O from the following data; f of N O = 8 kj mol. ond energies of N N, N = N, O = O and N = O bonds are 946, 48, 498 and 607 kj mol respectively. a. -88 kj mol b. -70 kj mol c. -8 kj mol d. -58 kj mol Sol. N N + / O = O N = N = O f = E of reactants - E of products = [E(N N) + / E(O = O) [E(N = N) + E(N = O)] = ( ) = 70 kj Resonance energy = f (observed) - f (cal.) = 8 70 = -88 kj mol 6. gas allowed to expend reversibly under adiabatic conditions. What is zero for such a process. a. G b. T c. S d. W SETION (ssertion & Reason Type) This Section contains 5 multiple choice questions. Each question has four choices ), ), ) and ) out of which ONLY ONE is correct. (Mark only One choice) 5 = 5 Marks () If both ssertion and Reason are true and Reason is a correct explanation of ssertion. () If both ssertion and Reason are true but Reason is not a correct explanation of ssertion. () If ssertion is true but Reason is false. () If ssertion is false but Reason is true.. ssertion () : Reversible process and a cyclic process are the same type of processes. Reason (R): cyclic and E cyclic are zero. a. () b. () c. () d. (). ssertion () : G and for a process always have the same sign. Reason (R) : G 0 f of r (l) is equal to zero. a. () b. () c. () d. () r. Sangeeta Khanna Ph. 5 EMISTRY OING IRLE

6 r. Sangeeta Khanna Ph.. ssertion(): In any reversible cyclic process, the net increase in entropy of the system is zero. Reason(R): ecause S universe = S system + S surr. and in case of reversible cyclic process S system is positive, but of surrounding is negative in same amount. That is why S universe = 0 a. () b. () c. () d. () 4. ssertion (): Thermodynamic process means change in at least one of the state variables of the system. Reason (R): Process in open system accompanied by an exchange of matter and energy between the system and the surroundings. a. () b. () c. () d. () 5. ssertion (): There is no change in internal energy in a cyclic process. Reason (R): yclic process is the one in which the system returns to its state after a number of reactions. a. () b. () c. () d. () SETION (Paragraph Type) This Section contains 4 paragraphs. Each of these questions has four choices ), ), ) and ) out of which ONLY ONE is correct. = 9 Marks Passage - mixture of powdered aluminium and powdered iron (III) oxide is called, Thermit and is used in fire bombs and as a source of thermal energy for welding. The reaction of aluminium and iron (III) oxide is very exothermic and the temperature reaches 400. The products of the reaction are iron and aluminium oxide which is a solid under ordinary conditions. l (s) + Fe O (s) Fe (s) + l O (s) = kj [t. Wt. of l = 7, Fe = 56]. ow much thermal energy released when 6.0 g of l reacts with excess of Fe O? a. 586 kj b. 69 kj c..4 0 kj d..7 0 kj. Which of the following expression is used to calculate the standard entropy change for the reaction? a. r S = [S Fe 0 + S 0 O ] [S 0 l l + S 0 O ] b. Fe r S = S 0 O - S 0 l Fe O c. r S = [S Fe 0 + S 0 O ] - S 0 Fe O d. Fe r S = [S 0 Fe + S 0 O ] [S 0 Fe l + S 0 O. The equation which holds good for the above reaction is a. r > r U b. r < r U c. r = r U d. r = r U 4. The correct statement for r is a = [ f Fe 0 + f 0 l O ] [ f 0 l + f ] b = FeO f 0 0 l O f ] Fe ] r. Sangeeta Khanna Ph. 6 EMISTRY OING IRLE Fe O c = [ f Fe 0 ] [ f 0 l ] d = [ f 0 0 ] [ l O f ] Fe O Passage - Fuel cells are electrochemical cells which correct the chemical energy of a fuel directly to electrical energy. Fuel cells differ from dry cells and lead-acid accumulators in having a steady supply of reactants from which to produce an electric current. Fuel cells use a wide variety of fuels including hydrogen, hydrocarbons and alcohols. In the methane-oxygen fuel cell, the following reaction take place. 4(g) + O (g) O (g) + O (l) = -89 kj/mol The standard free energies of formation of 4(g), O (g) and O (l) are 50.8 kj/mol, kj/mol and -7. kj/mol respectively.

7 r. Sangeeta Khanna Ph. 5. The standard free energy change for the reaction is a kj/mol b kj/mol c. -88 kj/mol d kj/mol Sol. G = (-7.) (-50.8) = -88 kj/mol 6. The efficiency of this methane-oxygen fuel cell is a. 65.% b. 96.9% c. 9.8% d. 5.8% 7. The cell potential for this methane-oxygen fuel cell is a V b. +. V c. -. V d V Sol. G = -nfe 0 cell, n = 8 8. The standard internal energy change of cell reaction is a. 89 kj/mol b kj/mol c kj/mol d kj/mol Paragraph Most of the reactions are accompanied by either evolution of heat or absorption of heat. The value of exact enthalpy change depends upon the quantities of the reactants, physical state of the reactants and products, allotropic modification involved, temperature etc. epending upon the nature of reaction, we call the enthalpy changes as enthalpy of combustion, enthalpy of formation, enthalpy of neutralization etc. For many reactions, enthalpy changes cold not be determined experimentally. The problem was solved by the Russian chemist, G.. ess, by putting forward a law known after him a ess s law of constant heat summation. lternatively, the heat of reaction can be calculated from bond energies of different bonds involved. 9. In 4, formation of ( = ) and ( ) is -590 kj/mol and - kj/mole respectively. What is enthalpy change when ethylene polymerizes to form polythene? a kj mol b. + 7 kj mol c. -59 kj mol d. -7 kj mol Sol. n ( ) n In one molecule of =, when one = double bond breaks, three single bonds are formed. owever, in the complete polymer chain, number of single bonds formed is two per = double bond broken. Energy released in forming one mole of - bonds = = 66 kj Net energy released per mole of ethylene = = 7 kj mol i.e., 7 kj - mol Passage 4 Some organic compounds contain double bonds. The simplest example is ethene, 4. Such compounds are said to be unsaturated, they contain less than the maximum amount of hydrogen. Ethene for example can be converted into ethane 6. The reaction of adding hydrogen to a double bond is known as hydrogenation. The heat change in a hydrogenation reaction is the enthalpy of hydrogenation. yclohexene yclohexane enzene 0. alculate the enthalpy of the following reaction: = (g) + (g) (g) The bond energies of,, = and are 44, 47, 65 and 45 kj mol respectively: a. 5 kj mol b. 5 kj mol c. 7 kj mol d. +7 kj mol r. Sangeeta Khanna Ph. 7 EMISTRY OING IRLE

8 r. Sangeeta Khanna Ph. Sol. The given equation is: = + =?.E rec tan ts. E products = 4 ( ).E + ( = ).E + ( ).E ( ).E + 6( ).E = [4(44) ] [ (44)] = ( ) = - 5 kj mol SETION (MORE TN ONE NSWER) This Section contains 4 multiple choice questions. Each question has four choices ), ), ) and ) out of which ONE OR MORE may be correct. 4 = Marks. Entropy increases in case of the reaction: a. N (g) + (g) N (g) b. I (g) (g) + I (g) c. gno (aq) + Nal (aq) gl (s) + NaNO d. ao (s) ao (s) + O (g), Sol. (a) No. of moles decreases. ence, entropy decreases. (b) No of substances increases. ence, entropy increases (c) Reactants give large number of ions in the solution. s the product gl is solid and O is almost unionized, the no. of ions decreases. ence, entropy decreases. (d) s one of the products is gaseous, entropy increases. Which of the following are correct at 98 K? a. element 0 b. element 0 G 0 f c. element 0 d. compound 0 S 0 f, Sol. (a) and (b) Explanation: For an element at 98 K G0 f, 0 0 f (a) and (b) are correct. (c) and (d) are incorrect as at 98 K, any substance has some entropy and also at 98 K any compound has some free energy.. Which of the following affect the heat of reaction? a. Physical states of reactants and products b. llotropic forms of elements c. Temperature d. Reaction carried out at constant pressure or constant volume,,, Sol. (a) O (l) has different than O(g) (b) (diamond) has different c than that of graphite. (c) is correct will change with change in temperature because heat capacity varies with temperature. (d) is correct because w is different at constant pressure or constant volume. When solids and liquids are involved difference is negligible but if gases are involved difference is more. 4. When one mole of a gas is expanded isothermally and reversibly at 00 K from a volume of 0 L to 0 L then: a. The work done in the expansion is.79 kj mol b. hange in enthalpy equal to.79 J mol c. hange in enthalpy equal to zero d. The work done in the expansion is zero, Sol. (a) is correct. V 0 w RT ln log V 0 79 Jmol.79 kj mol (b) is incorrect because = E + PV = E + RT = R = 0 0 = 0 0 f G 0 f r. Sangeeta Khanna Ph. 8 EMISTRY OING IRLE

9 r. Sangeeta Khanna Ph. SETION E (Matrix Type) This Section contains questions. Each question has four choices (,, and ) given in olumn I and five statements (p, q, r, and s) in olumn II. ny given statement in olumn I can have correct matching with one or more statement(s) given in olumn II. For example, if for a given question, statement matches with the statements given in q and r, then for that particular question, against statement, darken the bubbles corresponding to q and r in the ORS. No negative Marking 8 = 6 Marks. Match the following: olumn I olumn II () Enthalpy (P) Intensive property () Temperature (Q) Path function () Free energy (R) Function of E, P, V, () Work (S) State function Sol.() (R) and (S), () (P), (S); () (R) and (S), () (Q). Match the following: olumn I olumn II () Zeroth law of thermodynamics (p) eals with energies () First law of thermodynamics (q) eals with temperature measurements (c) Second law of thermodynamics (r) eals with absolute entropies (d) Third law of thermodynamics (s) eals with spontaneity of a process Sol. q, p, s, r Zeroth law deals with temperature measurements First law deals with energies Second law deals with spontaneity of a process Third law deals with absolute entropies. r. Sangeeta Khanna Ph. 9 EMISTRY OING IRLE

10 NEET 31 Years 11. The enthalpy of fusion of water is kcal/mol. The molar entropy change for the melting of ice at

10 NEET 31 Years 11. The enthalpy of fusion of water is kcal/mol. The molar entropy change for the melting of ice at 6 Thermodynamics. A gas is allowed to expand in a well insulated container against a constant external pressure of.5 atm from an initial volume of.50 L to a final volume of 4.50 L. The change in internal