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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), , THERMODYNAMICS (ENERGETICS), 6., LEVEL-V, SINGLE ANSWER QUESTIONS, 1., , The work done by a weightless piston in, causing an expansion V (at constant, temperature), when the opposing pressure,, P is variable, is given by:, (A) W PdV, , 2., , 7., , (B) W = 0, , (C) N 2 g,10atm N 2 g,1atm , (D) Fe 1mol, 400 K Fe 1mol,300K , 8., , constant temperature and pressure:, (A) if G system 0 , the system is still moving in a, particular direction, (B) if G system ve , the process is not, , (B) U = - q, thermally conducting walls, , spontaneous, , (C) U = q - w, closed system., , 4., , (C) if G system ve , the process is spontaneous, (D) if G system 0 , the system has attained, , 9., , 3O 2 H f 16 KJ / mole ., 2O 3 , , 5., , 56, , We can say that, (A) ozone is more stable than oxygen, (B) ozone is less stable than oxygen and ozone, decomposes forming oxygen readily, (C) oxygen is less stable than ozone and oxygen, readily forms ozone, (D) none of the above, Which reaction either endothermic or, exothermic characteristics has the greater, chance of occuring spontaneously?, (A) One in which entropy change is positive, (B)One in which entropy change is negative, (C) One in which free energy change is negative, (D) One in which equilibrium has been established, , Identify the correct statement for change of, Gibb’s energy for a system G system at, , (A) U = Wad ; wall is insulated, , 3., , (A) zero (B) –ve (C) +ve (D) , Change in entropy is negative for:, (A) Bromine (l) Bromine (g), (B) C s O 2 g CO2 g , , (C) W PV, (D) None of these, No heat is absorbed by the system from the, surroundings, but work (w) is done on the, system. What type of wall does the system, have?, , (D) U = q - w, open system, Which one of the following bonds has the, highest average bond energy (kcal/mol), (A) S = S (B) C C (C) C N (D) N N, For the reaction, , S for the reaction,, MgCO3 s MgO s CO 2 g will be, , 10., , equilibrium, Work Done inThermodynamical Process,, 1st Law, Internal Energy & Heat Capacity, The internal energy when a system goes, from state A to B is 40 kJ/mol. If the system, goes from A to B by a reversible path and, returns to state A by an irreversible path., What would be the net change in internal, energy?, (A) 40 kJ, (B) > 40 kJ, (C) < 40 kJ, (D) zero, One, mole, of, an, ideal, gas, , C, , V, , 20 J K 1 mol 1 initially at STP is, , heated at constant volume to twice the initial, temperature. For the process, W and q will, be, (A) W 0;q 5.46 kJ, (B) W 0;q 0, (C) W 5.46 kJ;q 5.46 kJ, Narayana IIT/PMT Academy
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), 20., , Stearic acid [CH 3 CH 2 16 CO2 H ] is a fatty, acid, the part of fat that stores most of the, energy. 1.0g of stearic acid was burnt in a, bomb calo rimeter. The bomb had a heat, capacity of 652 J /0 C . If the temprature of, , 24., , A g AB g ; H 0 50kJ / mol, , If the molar ratio of A2 g to A g is 5:3 in a, set of product gases , then the energy involved in, , C p H 2O 4.18 J / g 0C ], , the decomposition of 1 mole of A 2 B g is:, , A) 39.21 kJ, B) 29.91kJ, C) 108kJ, D) 9.32 kJ, A 0.05 L sample of 0.2 M aqeous, hydrochloric acid is added to 0.05 L of 0.2, M aqeous ammonia in a calorimeter. Heat, capacity of entire calorimeter syatem is 480, J/K. The temprature increase is 1.09 K., , A)48.75 kJ/mol, B) 43.73 kJ/mol, C) 46.25 kJ/mol, D) 64.2 kJ/mol, Calculate the enthalpy for the following, reacttion using the given bond energies (kJ/, mol), , 25., , C H 414; H O 463;, , H Cl 431, C Cl 326;, C O 335, , , HCl aq. NH 3 aq. NH 4Cl aq , A) -52.3 B) -61.1 C)-55.8, , CH 3 Cl g H 2O g , , At 250 C , 1 mole of MgSO4 was dissolved, in water, the heat evolved was found to be, 26., , A) -23 kJ /mol, B) -43 kJ/mol, C) -59 kJ/mol, D) -511kJ/mol, What is the bond enthalpy of Xe F bond, if Ionization energy of Xe =279 kcal/mol, , B.E. F F 38 kcal / mol , electron affinity, , MgSO4 s 7 H 2O l MgSO4 7 H 2O s , is, A) -105 kJ/mol, B) 77.4kJ / mol, C) 105kJ / mol, D) 77.4kJ/mol, The enthalpies of neutralization AOH and a, strong base BOH by HCl are -12250 cal/mol, and -13000cal/mol respectively. When one, mole of HCl is added to a solution containing, 1 mole of AOH and 1 mole of BOH, the, enthalpy change was-12500 cal/ mol. In what, ratio is the acid distributed between AOH, and BOH respectively., A) 2:1, B) 2:3, C) 1:2, D) 3:2, , , , , , , , CH 3 OH g HCl g , , D) -58.2, , 91.2 kJ. One mole of MgSO4 7H 2 O on, dissolution gives a solution of the same, composition accompained by an absorption, of 13.8kJ. H for the reaction, , 58, , undergo, , A2 B g , , r H 0 in kJ/mol for the following reaction, is:, , 23., , can, , A2 g B g ; H 0 40 kJ / mol, , to 39.30 C , how much heat was released, when the stearic acid was burbed ? [Given, , 22., , A2 B g , , decomposition to form two set of products, , 500g water c 4.18 J / g 0 C rose from 25.0, , 21., , Substance, , 27., , of F=85kcal/mol, A) 24kcal/mol, B) 34kcal/mol, C) 8.5 kcal/mol, D) 16.2 kcal/mol, Entropy, Gibbs Energy, 2nd Law, 3rd Law, & Spontaneous Process, In the conversion of lime stone to lime:, CaCO3 s CaO s CO 2 g The values of, –1, Ho and So are 179.1 kJ mol and 160.2, , JK–1 mol–1 respectively at 298 K and 1 bar., Assuming Ho and So remains constant, with temperature, at which, minimum, temperature conversion of lime stone to lime, will be spontaneous:, (A) 1118 K, (B) 1008 K, (C) 1200 K, (D) 845 K, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, MULTI ANSWER QUESTIONS, 44., , In an isothermal ideal gas expansion, A) w 0, , B) U1 U 2, V, , 2, D) q nRT ln V, 1, In an adiabatic process, the work involved, during expansion or compression of an ideal, gas is given by, , C) H1 H 2, 45., , A) nCV T, , B), , nR, T2 T1 , 1, , T2 P1 T1 P2 , , C) nRPext P P, , 1 2, , V2, D) 2.303 RT log V, 1, , 46., , The, , value, , of, , H transition of C graphite C (diamond), is 1.9 kJ/mol at 250 C entropy of graphite, is higher than entropy of diamond. This, implies that, A) C (diamond) is more thermodynamically, stable than C (graphite ) at 250 C, B) C (graphite) is more thermodynamically stable, than C (diamond ) at 250 C, C) diamond will provide more heat on complete, combution at 250 C, , THERMODYNAMICS (ENERGETICS), 48. Which statements are correct?, (A) A spontaneous chemical reaction has, maximum total entropy, (B) The entropy increases in an irreversible, adiabatic process, (C) The entropy decreases in a reversible, adiabatic process, (D) The entropy does not change in a reversible, adiabatic process, 49. The enthalpy of reaction depend upon:, (A) The manner by which the reaction is carried out, (B) Temperature at which the reaction is carried out, (C) Physical state of reactants and products, (D) Whether the reaction is carried out at, constant pressure or at constant volume, 50. Select the correct statements:, (A) State of a system is assumed to be in internal, equilibrium and the temperature and pressure, are uniform throughout the system, (B) Thermal drift ina system with time is more in, Dewar flask than in insulated system., (C) Thermal drift in a system with time is more, in non-insulated system than in in1sulated system., (D) Thermal drift in a system with time is more, in insulated system than in non-insulated system, 51. Which are not correct representation at, equilibrium:, V, , S/ R, 1, (A) V e, 2, , V, , S / RT, 2, (C) V e, 1, , 52., , Gtransition of C diamond , , D), 47., , C graphite , , is -ve, , Select the correct statements:, (A) temperature can be raised in an insulated, system by doing electrical or mechanical work, (B) temperature can be lowered in an insulated, system by doing electrical or mechanical work, (C) temperature cannot be lowered in an, insulated system by doing electrical or mechanical, work, (D) temperature cannot be raised in an insulated, system by doing electrical or mechanical work, , Narayana IIT/PMT Academy, , 53., , (B) K eG, , o, , / RT, , P, , H / RT, 2, (D) P e, 1, , Select the correct statements:, (A) Dissolution of KCl in water shows an, increase in entropy, (B) Acetic acid in benzene shows a decrease in, entropy, (C) The benzene solution containing acetic acid, is more ordered than molecular acetic acid., (D) The solution containing ionised acitic acid is, less ordered than molecular acetic acid., Which of the following expression may be, represents for the spontaneous reaction, (A) G T.P ve, , (B) G T.P. ve, , (C) H P.T ve, , (D) SE.V ve, 61
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THERMODYNAMICS (ENERGETICS), 54. Choose the correct statements:, (A) There is net reduction in energy when a large, number of gas phase K+ and Cl– ions come, together to form a bulk ionic solid, (B) There is a net increase in energy when a, large number og as phase K+ and Cl– ions come, together to form a bulk ionic solid, (C) The net effect of all the ion interactions in, the solid KCl lattice, is attraction, because the, cation or anion is the closest to ions of opposite, charge, (D) The net effect of all the ion interactions in, the solid KCl lattice, is repulsion, because the, anion and cation is the closest to ions of same, charge., 55. Choose the correct statements:, (A) The enthalpy change for the reverse of any, process is the negative of the enthalpy change, for the original process, (B) The enthalpy change for the reverse fo any, process is not the negative of the enthalpy change, for the original process and depends on the path, taken from the initial state of the final state, (C) Generally, the higher the freezing point, the, higher the enthalpy of melting, (D) Generally, the higher the freezing point, the, lower the enthalpy of melting, 56. During the formation of an ion pair, say, between a gaseous K atom and a gaseous, Cl atom, as they approach each other, one, contribution to energy is the energy., (A) Needed to form the K ions from a K atom, in the gas phase, which is the electron affinity of, potassium, (B) Needed to form the K+ ions from a K atom, in the gas phase, which is the ionisation energy, of potassium, (C) Released when the Cl- ion forms from a Cl, atom in the gas phase which is the electron affinity, of chlorine, (D) Needed to form the Cl- ion from a Cl atom, in the gas phase which is the ionisation energy of, chloride, 57. Which of the following statements is/are, correct for bond enthalpies in case of, general diatomic molecules?, 62, , 58., , 59., , JEE ADVANCED, (A) Bond enthalpy increases with the increase, in the number of bonds between the atoms, (B) Bond enthalpy decreases with the increase, in the number of bonds between the atoms, (C) Bond enthalpy increases with the increase, in the size of the neighbouring atoms, (D) Bond enthalpy decreases with the increase, in the size of the neighbouring atoms, Which of the following statements is/are, correct, (A) Absolute value of internal energy cannot be, determined, (B) Absolute value of heat content can be, determined, (C) Absolute value of entropy can be determined, (D) All the three E, H and S are extensive, properties, In the reaction, 2H 2 g O 2 g 2H 2 O l , H x kJ, , 60., , 61., , (A) x kJ is the heat of formation of H2O, (B) x kJ is the heat of reaction, (C) x kJ is the heat of combution of H2, (D) x/2 kJ is the heat of formation of H2O, Predict in which of the following, entropy, increases/decreases:, (A) A liquid crystallizes into a solid., (B) Temperature of a crystalline solid is raised, from 0 K to 115 K., (C) 2NaHCO3(s), , Na2CO3(s) + CO2(g) + H2O(g), (D) H2(g) 2H(g), The vapour pressure of solid benzoic acid, has been found to obey the relationship (in, the neighbourhood of 298 K) as:, ln, , P, b, a, ; where a = 22.88 and b =, o, P, T, , 1.07×104K Which of the following values are, correct at 298 K for the sublimation of, benzoic acid?, (A) G o 32.34 kJ mol1, (B) Ho 88.96kJ mol1, (C) So 190 JK 1 mol1, (D) Ho 88.96kJ mol1, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, 62. Select the incorrect statements about the, plots of ln K vs 1/T, G o, (A) Linear with slope equal to, and, R, , intercept H, (B) Curve, (C) Linear with slope, 63., , 64., , 65., , H o, So, and intercept, R, R, , (D) none of these, A 250 W electric heater raised the, temperature of a calorimeter by 4.22°C in, 55s. When the oxidation of a methanol, sample was carried out in the same, calorimeter, the temperature rose from, 22.49°C to 26.77°C. Then,, (A) The amount of heat supplied by the electric, heater is 13.75 kJ, (B) The amount of heat supplied by the electric, heater is 4.6 kJ, (C) The enthalpy of change for the oxidation is, 1.39 kJ, (D) The enthalpy of change for the oxidation is, 13.9 kJ, Choose the correct statements:, (A) The enthalpy of combustion of a fuel per, gram (expressed without a negative sign) is called, its specific enthalpy, (B) The enthalpy of combustion of a fuel per, gram (expressed without a negative sign) is called, its enthalpy density, (C) The enthalpy of combustion of fuel per litre, (expressed without a negative sign) is called its, specific enthalpy, (D) The enthalpy of combustion of fuel per litre, (expressed without a negative sign) is called its, enthalpy density, 200 g of water is contained in a beaker of, mass 150 g at 20°C. The temperature of, water is required to be raised to 80°C. It is, given that the specific heat of water is, 4.184J/(g°C) and of glass is 0.78J/(g°C)., Then, (A) Heat required by water is 50.2 kJ, (B) Heat required by glass is 7.0 kJ, (C) Heat required by glass 9.4 kJ, (D) Total heat required is 57.2 kJ, , Narayana IIT/PMT Academy, , THERMODYNAMICS (ENERGETICS), 66. Which statements are correct?, (A) A spontaneous chemical reaction which starts, far from equilibrium always gives irreversible, process., (B) The entropy increases in an irreversible, adiabatic process, (C) The entropy decreases in reversible adiabatic, process, (D) The entropy does not change in a reversible, adiabatic process, 67. Select the correct statements:, (A) Gibbs energy plays the same role for a, chemical system as the gravitational enrgy plays, for a purely mechanical systems, (B) Gibbs energy is minimum at definite pressure, and temperature for a chemical system at, equilibrium, (C) A positive value of G o does not mean that, the reaction does not occur spontaneously., (D) A negative value of G o means for the, spontaneous conversion of reactants into, products, if both are in standard states, 68. Point out the correct statements:, (A)Oxides of nitrogen are thermodynamically, unstable with respect to decomposition into, elements, (B) A spontaneous process does not necessarily, reduce internal energy or enthalpy of a system, (C) For thermodynamic equilibrium Stotal 0, (D) Entropy can not decrease in an isolated, system, 69. Which of the following statements are, correct?, (A) The entropy of an isolated system increases, in an irreversible process, (B) The entropy of an isolated system remains, unchanged in a reversible process, (C) Entropy can never decrease, (D) DS(system) as well as DS (surroundings), are negative quantities, 70. Which one of is not correct for a cyclic, process as shown in figure?, , 63
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), , , 3, , S, aA bB , cC dD, (0 K ), , (0 K ), , A) r S1 cSC dS D ( aS A bS B ), B) r S 2 aS A bS B , C) r S3 0., , 71., , A) dU 0, B) q w, C)W =314 J, D)W = 31.4 J, Select the correct statements for the, equilibriun under standard conditions, , 74., , D) r S 4 cSC dS D , Which of the following figures given below, shows adiabatic process;, , H 2O( s ) , H 2O( l ) ; S1, H 2O( l ) , H 2O( v ) ; S 2, H 2O( s ) , H 2O( v ) ; S3, , 72., , A) S1 S 2, , B) S 2 S1, , C) S3 S 2, Given, , D) S3 S1, , A) II,III, C) II,IV, 75., , 2 Fe2O3 s 4 Fe s 3O2 g ;, G10 1487kJ mol , , ,, , 6CO g 3O2 g 6CO2 g ;, r G20 1543.2kJ mol 1, , A) r G 0 for reduction of iron oxide by CO is, , N 2 is 1.20, C)the raio of slopes of P-V curves for, , +56.2 kJ mol 1, by CO, , C) Fe2O3 canno t be reduced by CO, spontaneously, , 73., , D) The reduction of Fe2O3 takes part in higher, part of blast furnace, Whic of the following is incorrect for the, change shown below?, r S 1, , aA bB , cC dD, , 64, , (298 K ), , ( 298), , r S 2 , , r S 4, , He, N 2 and O3 are expanded adiabatically, and their expansion curves, between P, and V are plotted under similar, conditions.About the ratio of the slopes,, which one is not correct;, A)the ratio of slopes of P-V curves for He and, O3 is 1.25, B)the ratio of slopes of P-V curves for He and, , correct statement is:, , B) Fe2O3 can be reduced, spontaneously, , B) I,III, D) I,IV, , N 2 and O3 is 1.05., , 76., , D)the slope of He is least steeper and for O3 is, most steeper., Heat of neutralisation of strong acid and, strong base under I atm and 25 C is -13.7, kcal. If standard Gibbs energy change for, dissociation of water to, H and OH is 19.14 kcal,the change is, standard entropy for dissociation of water, in cal K 1 mol 1 is;, A)18.25, B)110.2, C)-18.25, D) none of these., Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, 77. An ideal gas is allowed to expand both, reversily and irreversibly in an isolated, system. If T1 is the initial temperature,which, of the following statement is correct?, , THERMODYNAMICS (ENERGETICS), , 1.19 107 J, 81., , A) (T f )irrev (T f )rev, , to 1000 C. Which is correct, , B) (T f Ti for reversible process but T f Ti for, irreversible process., , A) U during the process is 223.51cal, B) H during the process is 372.56 cal, C) entropy change during the process is, , C) (T f ) rev (T f )irrev, , 1.122 cal k 1 mol 1, , D) T f Ti for both reversible and irreversible, process, One mole of, CH 3COOH undergo, dimerization in vapour state at 127 C, as 2CH 3COOH g , CH 3COOH 2 g , , 79., , 80., , 82., , A Band B C the pressure at A and C, is same. The correct statement(s) is/are, , if, , dimer formation is due to two Hbonds involved in dimer, each of, 33kJ strength and the degree of dimersation, of, acetic acid 98.2%, which is correct, A) S 0 for dimerization is negative, B) S 0 for dimerization is positive, C) S 0 for dimerization is -104.1J/mol, D) S 0 for dimerization is + 1.04J/mol, Work done in expansion of an gas from 4, littre to 6 litre against a constant external, pressure of 2.5 atm was used to heat up 1, mole of water at 293 K. If specific heat of, , 1.9518 104 J, C) work done on the gas is 1.19 107 J, D) Free energy change for the process is Narayana IIT/PMT Academy, , (1,100), A, , B(3,600), C(1,300), , T K, , 83., , A) Work involved in the path AB is zero, B) In the path AB work will be done on the gas, by the surroundings, C) Volume of gas at C= 3 volume of gas at A, D) Volume of gas at B is 16.42 litres, For a process to be spontaneous, A) Gsystem T , P 0, , water is 4.184J g 1 K 1 ,, the final, temperature of water is nearly ?, A)300K B)456K C)278K D)600K, 2 mole of a perfect gas at 27 0 C if is, compressed reversibly and isothermally, from a pressure of, 1.01 105 Nm 2 to 5.05 106 Nm 2 which is, correct among the following, A) work done on the gas is 1.9518 104 J, B) Free energy change for the process is, , D) U , H are same for the process, One mole of an ideal gas is subjected to a, two step reversible process as show in figure., , P(atm ), , 78., , 3, R is, 2, heated at a constant pressure of 1 atm 250 c, , If one mole of an ideal gas with Cv , , B) S system S, , surrounding, , 0, , C) S system S, , surrounding, , 0, , D) Gsystem T , p 0, 84., , The normal boiling point of a liquid ‘X’ is, 400K. Which of the following statement, is true about the process X l X g ?, A) At 400 K and 1 atm pressure G 0, B) At 400K and 2 atm pressure G ve, C) at 400K and 0.1 atm pressure G ve, D) at 410 K and 1 atm pressure G ve, 65
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THERMODYNAMICS (ENERGETICS), 85. Which of the following statements (s) is/are, false?, A) all adibatic process are isoentropic (or, isentropic) processes, , 86., , B) When Gsystem T , p 0; the reaction must, , OH (g) O (g) H(g) are 501.87kJmol1 and, , be exothermic, C) dG VdP SdT is applicable for closed, system, both PV and non-PV work, D) The heat of vaporisation of water at 1000 C, is 40.6 kJ/mol. When 9gm of water, vapour condenses to liquid at 1000 C of 1 atm,, , 423.38kJmol1 . The bond enthalpy of O-H bond, , then Ssystem 54.42 J / K, Form the following data, mark the option, (s) where H is correctly written for the, given reaction. Given, , H aq OH aq H 2O l ;, , H, , solution, , 89., , 0, , S (g) Hf = 222.80 kJ mol–1, (A) –277.49 kJ mol–1 (B) 277.49 kJ mol–1, (C) –349.15 kJ mol–1 (D) 349.15kJmol–1, , of HA g 70.7kJ / mol, , H solution of BOH g 20kJ / mol, H ionization of HA 15kJ / mol and BOH is a, strong base Reaction H r kJ / mol , A) HA aq BOH aq BA aq H 2O -42.3, B) HA g BOH g BA aq H 2O -93.0, C) HA g H aq A aq , , 88., , is, (A) –462.625 kj mol–1 (B) 462.625 kJ mol–1, (C) -713.54 kJ mol–1 (D) 713.54 kJ mol–1, In which of the following , the bond enthalpy, and bond dissociation enthalpy are identical?, (A) H- H bond enthalpy in H2 (g), (B) O-H bond ehthalpy in H2O (g), (C) C- H bond enthalpy in CH4(g), (D) N- H bond enthalpy in NH3 (g), Find the bond ehthalpy of S-S bond from, the following data., H0f = -147.23 kJ mol–1, C2H5 - S-C2H5 (g), 0, C2H5 - S-S-C2H5 (g) Hf = -201.92 kJ mol–1, , H 57.3 kJ, , -55.7, , D) B aq OH aq BOH aq -20.0, COMPREHENSION TYPE QUESTIONS, Comprehension–1, The bond dissociation energy of a diatomic, molecule is also called bond energy. However,, the bond dissociation energy depends upon the, nature of bond and also the molecule in which, the bond is present., The bond energy of N — H bond in NH3 is, equal to one-third of the energy of dissociation, of NH3 because there are three N-H bonds and, those of C — H bond in CH4 is equal to onefourth of the energy of dissociation of CH4., 66, , 87., , JEE ADVANCED, Heat of a reaction = Bond energy of reactants Bond energy of products, The enthalpy changes for the reaction, H 2 O (g) H (g) OH (g) and, , COMPREHENSION -2, Two vessels A and B are connected via a, stopcock. The vessel A is filled with a gas at a, certain pressure and the vessel B is completely, evacuated. The entire assembly is immersed in a, large vat of water and is allowed to come to, thermal equilibrium with the water. The stopcock, is opened and the gas is allowed to expand till, both the vessels are uniformly occupied. After, sometime when the vessel has again come to, thermal equilibrium, temperature of the water is, recorded. The result shows that the temperature, of water after the experiment is the same as that, before., 90. For the expansion referred to above, which, of the following is true?, (A) du = 0 (B) du >0 (C) du < 0 (D) dq > 0, 91. Taking ‘U’ as a function of T and V, under, the given conditions of the experiment., Choose the correct statement., (A) The change in energy of a gas with change, of volume at constant temperature is a positive, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, quantity., (B) The energy of the gas is a function of, temperature only., (C) The change in energy of a gas with change, of volume at constant temperature is a negative, quantity., (D) The result is applicable to both ideal and real, gases., 92. The expansion that occurred is, (A) isothermal reversible expansion, (B) isothermal irreversible expansion, (C) adiabatic reversible expansion, (D) isothermal free expansion, Comprehension–3, The first law of thermodynamics was gives as q, = U + (-w); where q is heat given to a system, and U represents increase in internal energy, and -w is work done by the system. Various, processes such as isothermal, adiabatic, cyclic,, isobaraic and isochoric process in terms of first, law of thermodynamics leads for important, results. The molar heat capacity for 1 mole of, monoatomic gas is 3/2R at constant volume and, 5/2R at constant pressure., 1., Increase in internal energy from state A to state, B is 50 J., 93. Which of the following statements are, correct :, (1) Both work and heat appears at the boundaries, of system., (2) Heat given to a system is given +ve sign., (3) Heat given to a system is equal to increase in, internal energy under isothermal conditions, (4) Heat given to a system is used to increase, internal energy under isochoric conditions, (5) Both work and heat are not state functions, but their sum (q + w) is state function., (A) 1, 2, 4, 5, (B) 1, 3, 4, 5, (C) 1, 2, 3, 4, (D) 2, 3, 4, 5, 94. A system is allowed to move from state A to, B following path ACB by absorbing 80 J of, heat energy. The work done by the system, is 30J. The work done by the system in, reaching state B from A is 10 J through path, ADB which statements are correct:, 1. Increase in internal energy from state A to state, B is 50 J., Narayana IIT/PMT Academy, , THERMODYNAMICS (ENERGETICS), 2. If path ADB is followed to reach state B,, U 50 J, , 95., , 3. If work done by the system in path AB is 20, J, the heat absorbed during path AB = 70J., 4. The value UC - UA is equal to UD - UB., 5. Heat absorbed by the system to reach B from, A through path ADB is 60 J., (A) 1, 5, (B) 1, 3, 5, (C) 1, 2, 3, 5, (D) 1, 4, 5, 1 mole of a monoatomic gas is expanded, through path ABC as shown in figure., , Select the correct statements :, (1) If specific heat of gas are 0.125 and 0.075, cal/g, the mol. wt. of gas = 40., (2) Temperature at point A, B, C are 273, 546, and 273 K respectively., (3) U for the process A to B is 2.27 kJ., (4) U for the process B to C is 3.44 kJ., (5) U for the overall cycle A B C A, is 3.4 kJ., (A) 1, 2, 3, 4, (B) 3, 4, 5, (C) 1, 3, 4, (D) 1, 2, 5, Comprehension–4, Enthalpy of neutralization is defined as the, enthapy change when 1 mole of acid/base is, completely neutralized by base/acid in dilute, solution. For strong acid and strong base, neutralization net chemical change is, H aq OH aq , H 2 O l ; r H 0 55.84kJ / mol, H 0 ionization of aqueous solution of strong acid, and strong base is zero. When a dilute solution, of a weak acid or base is neutralized, the enthapy, 67
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THERMODYNAMICS (ENERGETICS), of neutralization is some what less because of, the absorption of heat in the ionization of the, weak acid or base for weak acid /base, H 0 neutralization H 0 ionization r H 0, , H, 96., , If, , , , OH H 2O , enthalpy, , of, , neutralization, , of, , CH 3COOH by HCl is -49.86 kJ /mol then, enthalpy of ionization of CH 3COOH is, , 97., , A) 5.98 kJ/mol, B) 5.98kJ / mol, C) 105.7kJ / mol, D) 10.57kJ/mol, 0, What is H for compleate neutralization, of strong diacidic base A OH 2 by HNO3, , 98., , A) -55.84 kJ, B) -111.68 kJ, C) 55.84kJ / mol, D) -49.86kJ/mol, Under the some conditions how many mL of, , 0.1 M NaOH and 0.05 M H 2 A (strong, diprotic acid) solution should be mixed for a, total volume of 100 mL produce the highest, rise in temprature, A) 25: 75, B) 50:50, C) 75: 25, D) 66. 66:33.33, Comprehension–5, Work done by the system in isothermal reversible, V2, process is : wrev. = -2.303 nRT log V . Also in, 1, , case of adiabatic reversible process work done, , nR, [T2 T1 ] ., -1, During expansion disorder increases and the, increase in disorder is expressed in terms of, by the sytem is given by : wrev. =, , q, change in entropy S rev. . The entropy, T, changes also occurs during transformation of one, H, . Both, T, entropy and enthalpy changes obtained for a, process were taken as a measure of spontaniety, of process but finally it was recommended that, decrease in free energy is responsible for, spontaniety and G H T S ., , state to other and expressed as T , , 68, , 99., , JEE ADVANCED, Which of the following statements are, correct:, (1) The expansion work for a gas into vacuum is, equal to zero., (2) 1 mole of a gas occupying 3 litre volume on, expanding to 15 litre at constant pressure of 1, atm does expansion work 1.215 kJ., (3) The maximum work done during expansion, of 16 g O2 at 300 K from 5 dm3 to 25 dm3 is, 2.01 kJ., (4) The S for S L is almost negligible in, comparison to S for L G ., V, (5) S = 2.303 nR log 2 . (at constant T), V1, , (A) 2, 3, 4, 5, (B) 1, 2, 3, 4, 5, (C) 1, 2, (D) 4, 5, 100. The heat of vaporisation and heat of fusion, of H2O are 540 cal/g and 80 cal/g. The ratio, of, , Svap ., S fusion, , for water is:, , (A) 6.75 (B) 9.23 (C) 4.94 (D) 0.2, 1, , 2Ag(S) + O2 (g) ; attains, 2, equilibrium at temperature . . . K is:, (The H and S for the reaction are 30.5 kJ/, mol and 66 J/ mol/ K), (A) 462.12, (B) 237, (C) 373, (D) 273, 102. A chemical change will definitely be, spontaneous if :, (A) H = -ve, S = -ve and low temperature, (B) H = +ve, S = -ve and high temperature, , 101. Ag2O(S), , (C) H = -ve, S = +ve and any temperature, (D) H = +ve, S = +ve and T S<H, , KEY, LEVEL (V), SINGLE ANSWER QUESTIONS, 1. A, 4. B, , 2. A., 5. C, , 3. D, 6. C, , Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, 7. D, 8. D, 9. D, 10. A, 11. A, 12. B, 13. C, 14. C, 15. C, 16.C, 17.A, 18B, 19.C, 20.A, 21.A, 22.A, 23.A, 24.B, 25.B, 26.B, 27.A, 28.A, 29.D, 30.D, , 31.B, , 34.B, 37.C, 41.C, , 35.B, 38.D, 42.A, , 32.C, , HINTS & SOLUTIONS, , 1., , 36.B, 39.D, 43.B, , 40.A, , 45.(A,B,C), 48.(A, B, D), 50.(A, C), , 2., , note, , that, , opposing pressure is not constant throughout., This is possible for adiabatic wall and thus, U Wadiabatic, , 46.B, , 53(A, C, D), 55.(A, C), 57.(A, D), 59.(B, D), 61.(A, B, C), 63.(A, D), 65.(A, B), 67.(A, B, D), 69.(A, B), , 70.(a,b,c), 71.(b,c,d), 72.B, 73.(d), 74.(a), 75.(d), 76.(b), 77.(a), 78.(c), 79.(a), 80.A,B, 81.(a,b,c), 82(c,d), 83(B,D), 84.(A,B,C), 85.(A,B,C,D) 86.(A,B,C), COMPREHENSION TYPE QUESTONS, 87., (B), 88., (A), 89., (B), 90., (A), 91., (A) 92., (D), 93., (A), 94., (C), 95., (A), 96., (A), 97., (B), 98., (B), 99., (B), 100. (C), 101. (A), 102. (C), , 3., , 4., , 5., 6., , 7., , 8., , (D), B.E. of S = S, C C , C N and N N are, 523, 839, 891 and 941 kJ mol–1 respectively., (B), O3 possesses more energy than O2, (C), If G ve , process is spontaneous., (C), n 1 and thus, entropy increases., (D), The gaseous phase have more entropy and, thus, S is +ve in (A) and (B). Also decrease in, pressure increases disorder and thus, S is +ve, in (C). In (D) the disorder decreases in liquid, state due to decrease in temperature. Thus, S =, –ve., (D), G system ve , the system is spontaneous, , process, G system 0 , the system has attained, equilibrium G system ve , the system is non, spontaneous., 9., 10., , (D), In a cyclic process E 0, (A) W PV P 0 0, , Q CV T 20 273 5460 J 5.46 kJ, 11., , 12., Narayana IIT/PMT Academy, , SINGLE ANSWER QUESTIONS, (A), Wrev PdV or PV ;, , 33.C, , MULTI ANSWER QUESTIONS, 44.(B,C,D), 47.(A, C), 49.(B, C, D), 51.(A, C, D), 52.(A, B, C), 54.(A, C), 56(B, C), 58.(A, D), 60.(A, B , C, D), 62.(A, B), 64.(A, D), 66.(A, B, D), 68.(A, B, C, D), , THERMODYNAMICS (ENERGETICS), LEVEL(V), , (A), DU = 0 ;, -q = w, vapour equilibrium n 3 0 3 than, U 30000 3 2 500 27kcal ., (B), 69
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), At constant volume PV 0 , Q U, 13., (c), , 1, C s O2 g CO g ;, 2, , w Pext V f Vi , , f H y x kJ / mol, , 60 103 40 103 100 103 , 105 , , , , 1000, 1000 , 0.60, , 14., , 17., , (A), 1, H2 g O2 g H2O l , 2, , 105 100 10 3 0.04 10 3 0.1 10 3 , , H 241kJ , , ; w 9994 J, , C6 H10 , , (c), , H 2O l ,323 K , H 2O l ,373K , , U 2, , Eq. (i) + eq. (ii) – eq. (iii) gives, , U, , H 2 O 9, 323 K 3, H 2 O 9, 373 K , , H 241 3800 3920 121kJ, , CV , m H 2O, g 33.314 8.314, 18., , U 2 H 2 ng RT 37.6 ;, , U, , total, , U1 U 2 U 3, , for C 6 H10 H 2 C6 H12, (B), Number, of, moles, , , Cv, m l T U vap Cvap Cvm g T, , , 15., , 75 50, 25 50, 37.6 , ;, 1000, 1000, , Number, , , 42.6 kJ / mol, (c), , n H 25 10 5, , of, , moles, , of, , Ca(OH) 2, , MV 0.01 50, , 50 105, 1000, 1000, , In the process of neutralisation 25×10–5 mole, H+ will be completely neutralised, , , ;, , H 140 25 105 kcal 0.035kcal 35cal, , 19., , Conceptual, , 20., , (a), , (c), , qreaction qbomb qwater, , qreaction C bomb m, , C s O2 g CO2 g ;, , CO g 1/ 2O2 g CO2 g , Equation (1) -(2), , water, , c T, , 652 500 4.18 14.3, , r H1 xkJ / mol .... (1), , 70, , HCl, , n OH 2 50 105 10 3, , 5476kJ / mol, (C – H)6 = 6 × 90 = 540, 620 = 540 + C – C, C – C = 80., , r H 2 y kJ / mol, , of, , MV 0.01 25, , 25 105, 1000, 1000, HCl H Cl, , r H 0 8 394 9 286 250 , , 16., , 17, O 2 g 6CO2 g 5H 2 O l , 2, , ...(ii), H 3800 kJ , C 6 H12 9O2 g 6CO 2 g 6H 2 O l , ...(iii), H 3920 kJ , , U1, , 25J / K mol, , ..(i), , 21., , 39210 J or 39.21kJ, (a) m mole of acid 0.05 0.2 0.01, , r H 0 , , 480 1.09, 52.32 kJ / mol, 0.01 1000, Narayana IIT/PMT Academy
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), , G T,P ve ;, , SE, V ve, , and, , part of blast furnace., 73., , H P.T. ve, 54., 55., 56., 57., 58., , 59., 60., 61., 62., 63., , 64., 65., , (A, C); factual, (A, C); factual, (B, C); defined, (A, D), (A, D) Internal energy depends upon large, no. of factors, U = Etra + Erot + Enuclear + Evib......, The exact calculation is not possible., (B, D); defined, (A, B, C, D); Increase in randomness increases, the entropy., (A, B, C), (A, B), (A, D), Heat supplied by electric heater is = 250 J/sec, x 55 sec = 13.75 kJ, (A, D); defined, (A, B), q msT for water and glass, , (d) r S4 S p S R CSc dS D 0 0, 74., , 75., , (A, B, D); factual, (A, B, D); factual, , 70., , (a,b,c), For a cyclic process dU=0, , for gas 1, (d)Ratio of slope for gas II ; slope for, He, N 2 and O3 are 7/5,5/4 and 4/3, respectively., , 76., )bG H T S H for H 2O H OH si, +13.7 kcal, , S , 77., , 66., 67., , (a) Adiabatic slope are more steeper than isothermal., Slope of adiabatic process= slope of, isothermal process, , H G 13.7, 19.14 1, 1 11.2 cal k0.1102, kcal mol, mol 1, T, 298, , (a) Work done in (-ve) reversible process is, maximum. Thus, in reversible pro, cess T f rev T f irrev, , 78., , q U w , , C., , So l:, , 2CH 3COOH g , CH 3COOH 2 g , 1, , q w, , Also, w= area covered by sphere, , 0, 0.982, 2, , 1-0.982, , 2, (V V (20)2, r 2 2 1 , 1003.14 314 J, 2 , 22, , 71., , (b,c,d) H 2O( s ) has more ordered, arrangement.Also, , S 2 S H, , 2o( v ), , , S 01 S H, , 2o( l ), , , SH, , 2 o( v ), , , SH, , SH, , 2o( l ), , K0, , CH 3COOH 2, 2, CH 3COOH , , Now,, , , , 0.982, 2 0.018, , fo r, H 0, 2 33kJ 66kJ, Thus, G 0 H 0 T S 0, , ;, , 2o( s ), , is maximum and thus S 2 S1, , 2, , 1515.4, , dimerization, , 2.303 RT log K 0 H 0 T S 0, 2.3038.314 400log 1515.4 66103 400 S 0, , 72., , , , 1, , , 2, , G for 2 Fe2O3 600 4 Fe 6CO2 rG rG, , =+1487-1543.2=-56.2 kJmol 1, The reduction occurs spontaneously in lower, 74, , 79., , 24359.2 66000 400 S 0, , 41640.8, 104.102 J / mol, 400, .Ans (A), S 0 , , Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, Solution: Since, work is done against constant, pressure process is irreversible., Given, V (6 4) 2 litre, p = 2.5 atm, , THERMODYNAMICS (ENERGETICS), , ds , , w pext V 2.5 2 5 litre-atm, 5 1.987, 5 1.987 4.187, cal , J 506.31J, 0.0821, 0.0821, C=4.184J/g/Kor4.184J/mol, Now this work is used in heating 1 mole water, , , W n C T, , 80., , Wrev 2.303nRT log10, , T2, , nCp.dT, T, nCp loge 2, T, T1, , S , , , , s 2.303 nC p log10, , T1, , T2, T1, , T 6.723, , 5, 373, 2.303 1 R log10, 1.122 cal k 1 mol 1, 2, 298, , t emperature, , = T1 T 293 6.723 299.723K, Ans (A,B), reversible process:, , nC p .dt, dqrev, ds , T, T, , or, , 506.31 1 4.184 T , , Final, , dqrev nC p .dt, , Also,, , 82, , (c,d), , p1, p2, VB , , At VA , , 1 R 100, 100 R, 1, , 1 R 600, 200 R, 3, , 1.01 105, B VA so expansion of gas takes place, 1.9518 104 Vjoule, 6, 5.05 10, VB 200 0.0821 16.42 L, Since W reversible is a measure of free change, , 2.303 2 8.314 300 log10, , 81., , G Wrev Wmax 1.9518 10 4 J, Ans A,B,C, Cv , , 3, 3, 5, R C p Cv R R R R, 2, 2, 2, , Heat given at constant pressure m.C p .T, or, Now work done in the process PV, , 83, 84, 85, 86, , COMPREHENSION TYPE QUESTIONS, 87., , 5, 5, H or q p 1 R 373 298 or H 1 1.98775 372.56 cal, 2, 2, , nRT2 nRT1 , w p V2 V1 p , , , p , p, , , , pv nRT , , U 223.51 cal, Narayana IIT/PMT Academy, , (B) The enthalpy of dissociation of the O-H, bond depends on the molecular species from, which H-atom is being separated., H 2O(g) , H(g) + OH (g); H0 =, 501.87 kJ mol-1, However to break O-H bond in hydroxyl a, different quantity of heat (423.38 kJ mol-1) is, BE, (O, H), =, 01.87kJmol1 423.38kJmol1, = 462.625, 2, , nR T2 T1 1 1.987 373 298 , 149.225cal, from I law of thermodynamics, U q W 372.56 149.05, , (B,D), (A,B,C), (A,B,C,D), (A,B,C), , 88, , kJ mol-1, (A), In case of diatomic molecules, bond enthalpy, and bond dissociation enthalpy are identical, 75
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THERMODYNAMICS (ENERGETICS), because each r efer s t o t he react io n, H2(g) 2H(g); BE (H - H) = 433.93, kJ mol-1, 89., (, B, ), BE, =, , (S, , -, , S), , =, , UB-UA = 50 J, J, , 95., , Hvap (s) H0f, , =, , 0;, , du, , =, , 0, , 98., , 94., , 76, , 1 22.4, 8.314 J =, 0.0821, 40, 4.12, , 3.44, kJ, , 0.075, , = 2268.37 J =, = 2.27 kJ, (b) For max rise in temp; max neutralization of, H and OH required, , u , , 0, V T, , 99., , if we take equal volume, all H ( 5 m-mole), will react with all OH (5m -mole), (B) all are correct, , since it is a free expansion, it is applicable to only, ideal gases., (D), If the external pressure is only infinitesimally, smaller than the pressure of the gas, the, expansion is said to take place reversibly. If,, however, the external pressure is much smaller, than the gas pressure, the expansion occurs, irreversibly. If the external pressure is zero, the, expansion is known as free expansion., (A), q =, +, U, PV, at T = constant, q = PV, (C) ACB = AC + BC, ADB = AD +, B, D, Heat absorbed 80 J, Work done by the, system, =, 10, J, w 10 J, Also, w = -10 J, w = -30, , 100., , (C) Sv , , Since, , 93., , U Cv T, , = 1 22.4 L atm =, , u , , dV 0, V T, , 92., , 2, 273, M, T, =, 273, K, At B, P = 1 atm. V = 44.8 L, T, =, 546, K, C, P = 0.5 atm., V = 44.8 L, T = 273 K, constant, V, , U P V, , u , u , du , dT , dV, T v, V T, , dT, , L, , , M = 40, , At, , Also At, , = 222.80 - {-201.92 - (-147.23)}, = 277.49 kJ mol-1, (A) The gas expands against a zero opposing, pressure. Since dw = -Popp dv, it is obvious that, the work involved in the expansion in zero., du, =, dq, +, dw, Hence, du, =, dq, Since there is no change in temperatuare dq = 0, Hence du = 0, (A), u, =, f(T,, V), , Since, , At A, P = 1 atm. V =, , 0.125 - 0.075 =, , 0, Hvap (s) {H (C 2H5S, H, SC, 2H, 5 )}, f (C, 2H, 5 S C 2H5 ), , 91., , R, M, , 22.4, , 0, f, , 90., , (A) Cp - Cv =, , JEE ADVANCED, q = 50 + 10 = 60, , dV, , 0,, , follows, , that, , H v 540, , ;, 373 373, , LEVEL - VI, MATRIX-MATCHING QUESTIONS, 1., , Match Column - I with Column - II, Column - I, (A) C6 H12 O 6 aq C6 H12 O 6 aq ; H, (B) CaO aq Ca OH 2 aq ; H, (C) C O 2 CO 2 ; H 298K, (D) Cgraphite O 2 CO 2 ; H 298K, Column - II, (p) Exothermic, (q) Endothermic, (r) H is heat of reaction, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, (s) H is heat of formation of CO2, (t) H is standard heat of fomration of CO2, 2., Match Column - I with Column - II, [SR - Rohmbic, SM - Monoclinic], Column - I, Column - II, , 3., , THERMODYNAMICS (ENERGETICS), 5., Column-I (Process for ideal gas), A) Reversible isothermal expansion, B) Reversible adiabatic compression, C) Adiabatic free expnansion, D) Irreversible isothermal compression, Column-II (Entropy change), , (A) f H o O 2 ,g , , (p) Zero, , (B) f H o SR , , (q) Not equal to zero, , p) S surrounding 0, , (C) f H o SM , , (r) Most stable form, , q) S surrounding 0, , (D) f H o PWhite , (s) Less stable form, Column-I, A) Reversible isothermal expansion of an, , r) S surrounding 0, 6., , B) Reversible adiabatic compression of an, C) irreversible adiabatic expansion of an, D) Irreversible isothermal compression of, H 0 an ideal gas, Column-II, , p) H nC p.m T 0 pressure, , V2 , p) w 2.303nRT log V ideal gas, 1, , q) U 0 temprature, r) G V P at constant temperature, , q) PV constant ideal gas, , P2 , s) G nRT ln P ideal gas, 1, Column-I, , nR, r) w 1 T2 T1 ideal gas, 4., , s) S system 0, Column-I, A) heating of an ideal gas at constant, B) Compression of liquid at constant, C) Reversible process for an ideal gas, D) Adiabatic free expnasion of an, Column-II, , 7., , A) O2 g , , Column - I, (a) Van’t Hoff isochore, (b) Kirchhoff’s eq., (c) Clausius Clepeyron eq., (d) Gibbs-Helmholtz eq., Column - II, , B) O3 g , C) Br2 g , D) H 2O l , Column-II, , P2 Hv 1 1 , (p) ln P R T T , 1, 1, 2, , p) f H 0 ve; f S 0 ve, , ( G ) , (q) G H T T , , P, , r) f H 0 ve; f S 0 ve, , q) f H 0 ve; f S 0 ve, , s) f H 0 0 f S 0 0, , d ln K P H 0, , (r), dT, RT 2, , ( G ) , (s) T C p, , P, Narayana IIT/PMT Academy, , 8., , Column-I (Process for ideal gas), A) Reversible isothermal compression, B) Isothermal free expansion, C) Reversible adiabatic expansion, D) Reversible ideal gas expansion, Column-II (Entropy change), 77
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THERMODYNAMICS (ENERGETICS), , 9., , p) S system 0, , (Partial derivative), , q) S system 0, , U , A) T , , V, , p) C p, , H , B) T , , P, , q) CV, , G , C) T , , P, , r) -S, , G , D) P , , T, , s) V, , r) S system 0, s) Information insufficient, Column-I, A) Gsystem T , P 0, D) Gsystem T , P 0, , 10., , Column-II, p) Process is in equilibrium, s) System is unable to do useful work, Column-I, , ASSERTION – REASON TYPE, , A) Gsystem T , P, B) Work done in reversible isothermal, C) G for reversible isothermal expansion, D) S gas for isothermal expansion, Column-II, V , , 2, p) nR ln V , 1, , P , , 2, q) nRT ln P ideal gas expansion, , , , 1, , , , r) nFE of an ideal gas, , 11., , P1 , s) nR ln P of an ideal gas, 2, Column-I, A) Revesible adiabatic compression, B) Reversible vaporisation of liquid, , 13., , 14., , C) 2N g N 2 g , , , D) MgCO3 s MgO s CO2 g , Column-II, , 78, , 15., , p) S, , system 0, , q) S, , system, , r) S, 12., , JEE ADVANCED, (Thermodynamic varible), , 0, , surrounding, , 16., , 0, , s) S surrounding 0, Column-I, , Column-II, , Each question contains STATEMENT – 1, (Assertion) and STATEMENT – 2 (Reason)., Each question has 4 choices (A), (B), (C) and, (D) out of which ONLY ONE is correct., (A) Statement-1 is True, Statement-2 is True;, Statement-2 is a correct explanation for, Statement-1, (B) Statement-1 is True, Statement-2 is True;, Statement-2 NOT a correct explanation for, Statement-1., (C) Statement-1 is True, Statement-2 is False, (D) Statement -1 is False, statement-2 is True, Statement – 1:The heat absorbed during the, isothermal expansion of an ideal gas against, vacuum is zero., Statement – 2: The volume occupied by, molecule of an ideal gas is assume to be zero., Statement – 1: For isothermal reversible, expansion the work done will be the maximum, (Wmax), Statement – 2: The work done for the isothermal, reversible compression is taken as work done, minimum (Wmin), Statement – 1 : Heat of neutralization of HCl, by NaOH is more than that by NH4OH, Statement – 2: NaOH is stronger base than, NH4OH., Statement – 1 : The change in internal energy, and change in enthalpy does not depend upon, path by which change are brought in., Statement – 2: Both are path independent and, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, E and H are state function, 17. Statement – 1 : Entropy change of a process is, different when it is carried out reversibly and, irreversibly.., Statement – 2: Entropy change is the amount, of heat absorbed at a given temperataure., , THERMODYNAMICS (ENERGETICS), neutralization between CH 3COOH and, NaOH is 11.7 k .cal . Calculate heat of, , 23., , INTEGER TYPE QUESTIONS, 18., , 19., , The enthalpy change involved in the, oxidation of glucose is 2880 kJ mol 1 ., Twenty five percent of this energy is, available for muscular work. If 100 kJ of, muscular work is needed to walk one, kilometer, what is the maximum distance, that a person will be able to walk after eating, 125 g of glucose?, Calculate the enthalpy change when, infinitely dilute solutions of CaCl2 and, Na2 CO3 are mixed. H 0f for Ca2+ (aq),, , 24., , A sample of ideal gas 1.4 is heated at, , 25., , constant pressure. If an amount of 85J of, heat is supplied to gas, find U ., The free energy change when 1 mole of, NaCl is dissolved in water at 298 K. is –x, KJ find out value of ‘x’ given–, (a) Lattice energy of NaCl 778 kJ mol 1, (b) Hydration energy of NaCl 775 kJ mol 1, , 26., , occupying a volume of 5 dm3 isothermally, , 129.80, 161.65 and 288.45 kcal mol 1, respectively., An intimate mixture of ferric oxide and, aluminium is used as solid fuel in rockets., Calculate the fuel value per cm3 of the, mixture. Heats of formation and densities, are as follows:, , 27., , until the volume becomes 25 dm3 . (Ignoring, the sign of value), What is the entropy change for the, conversion of one gram of ice to water at, 167 K and one atmospheric pressure?, , 28., , H fusion 6.025 kJ mol 1 , The enthalpy of transition of crystalline, boron to amorphous boron at 15000 C is 0.4, , and, , CaCO 3, , (s), , 1, , H f Al2 O3 399kcalmol ;, H f Fe2 O3 195.92kcalmol1, , Density of Fe2O3 5.2 g / cm3 ;, 21., , Density of Al 2.7 g / cm3, Calculate the entropy change for the, following reversible process:, Tin at 130 C, Tin, 1 mol at 1 atm, 1 mol at 1 atm, , H, 22., , (c) Entropy change at 300 K 40 J mol 1, Calculate the maximum work done in kJ, expanding 16 g of oxygen at 300 K and, , are, , CO23 aq , , 20., , ionization of CH 3COOH ., A gas occupies 2 litre at STP. It is provided, 58.63 joule heat so that its volume becomes, 2.5litre at 1 atm. Calculate change in its, internal energy., , trans, , 29., , 30., , 2288Jmol1 , , Heat of neutralization between HCl and, NaOH is 13.7 k .cal . If heat of, , Narayana IIT/PMT Academy, , 31., , kcal mole 1 . Assuming at w.t of boron 10,, the change in enthalpy of transition 50g, boron from crystalline to amorphous form, is, At 00 C , if enthalpy of fusion of ice is 1365, kcal/mol. The molar entropy, change for, melting of ice at 00 C is, A reaction becomes spontaneous only at, 500K. If H at 500K is 3.0 kJ, the change, in entropy at 500K, How much of the following are extensive, properties?, Resistance, Electromotive force, Heat enthalpy,, 79
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THERMODYNAMICS (ENERGETICS), Dipole moment, Heat capacity,, Specific Heat, Denisity specific volume, 32. How much of the following are intensive, properties?, Vapour pressure , Molarity , Refractive index,, Dielectric constant, Osmotic pressure, Molarity,, Specific gravity, Molar volume, , JEE ADVANCED, A) Lattice energy of NaCl 778 kJ mol 1 ., B) Hydration energy of, NaCl 774.3 kJ mol 1, 38., , SUBJECTIVE TYPE QUESTIONS, 33., , One mole of an ideal gas is heated at, constant pressure from 0 C to 100 C, A)Calculate work done., B) If the gas were expanded isothermally and, reversibly at 0 C from 1 atm to some other, , pressure P1 , what must be the final pressure if, the maximum work is equal t o the wo rk, involved in (a)?, 34.. Water is boiled under a pressure of 1.0, atm.When an electric current of 0.50A from, a 12V. supply bis passed for 300 s through a, resistance in thermal contact with it. it is, found that 0.789 g of water is vaporized., Calculate the molar internal emergy and, enthalpy changes and enthalpy changes at, boiling point(373.15K)., 35. 14g oxygen at 0 C and 10 atm are subjected, to reversible adiabatic expansion to a, pressure of 1 atm.Calculate the work done, in;, A) Litre atm., , C) Entropy change at 298 K=43 J mol 1. ., How much heat is required to change 10 g, ice at 0 C to steam at 100 C ? Heat of fusion, and heat of vaporization for H 2O are, , 39., , c, a, l, /, g, 80 cal / g and 540, respectively.Specific heat of water is 1cal/, g., The vapour pressure of benzene is, 1.53 104 Nm 2 at 303K and 5.2 104 Nm2, at 333K. Calculate the mean latent heat of, evaportion of benzene over this temprature, range, , 40., , Molar heat capacity of CD2O (deutero, formaldehyde) at constant pressure is, 14 cal mol 1 K 1 at 1000K. Calculate the, entropy change associated with cooling of, 3.2g of CD2O vapour from 1000K to 900K, , 41., , 42, , I mole of an ideal gas at 25 C is subjected, to expand reversibly ten times of its initial, volume. Calculate the change in entropy, of expansions., The standard enthalpy and entropy changes, for the reaction in equlibrium for the forward, direction are given as, CO2 ( g ) H 2 ( g ), CO( g ) H 2O( g ) , , B) Calori (Given, Cp / C v 1.4 ), , H , , 14 0.0821141.4 273 , In litre-atm 32 , 1.4 1, , 41.16 kJ mol 1, 300 k, , H , , 4.24 102 kJ mol 1, 300 k, , 36., , Wrev 11.82 litre-atm (or) -288 cal, Calculate the equilibrium consthant for the, reaction given below at 400 K if, , H , 1200 k, , , S 1200k 32.93kJ mol 1, , H 0 77.2 kJ mol 1 and S 0 122 JK 1mol 1, , Caculate K P at each temprature and predict the, direction of reaction at 300K and 1200 K,when, , PCl5 g PCl3 g Cl2 g , 37., , 80, , Calculate the free energy change when I, mole of NaCl is dissoved in water at 298, K. Given., , 32.93 kJ mol 1, , PCO PCO2 PH 2 PH 2O 1 atm at initial state, , 43., , At 300 K, the standard enthalpies of, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, formation of C6H5COOH(s), CO2(g) and, H2O(l) are -408, -393 and -286 kJ/ mol, respectively. Calculate the heat of, combustion of benzoic acid at 300, (i) Constant pressure (ii) constant volume, 44. The heat of combustion of glycogen is about, 476 kJ/mol of carbon. Assume that average, heat loss by an adult male is 150 watt. If we, were to assume that all the heat comes from, oxidation of glycogen, how many units of, glycogen (1mole carbon per unit) must be, oxidised per day to provide for this heat, loss?, 45., , Calculate H 0f for chloride ion from the, following data:, 1, 1, H 2 ( g ) Cl2 ( g ) , HCl( g ) ;, 2, 2, , H 0f 92.4 kJ, , , HCl (g) + nH2O , H ( aq ) Cl( aq ) ;, , 46., , 47., , 48., , 49., , H 0f H (aq ) 0.0 kJ, H 0 74.8 kJ, A constant pressure calorimeter consists of, an insulated beaker of mass 92 g made up, of glass with heat capacity 0.75 J/K/g. The, beaker contains 100 mL of 1 M HCl of, 22.6°C to which 100 mL 1M NaOH at 23.4°C, is added. The final temperature after the, reaction is complete is 29.3°C. What is ΔH, per mole for this neutralization reaction ?, Assume that the heat capacities of all, solutions are equal to that of same volumes, of water., The standard enthalpy of combustion at, 25°C of hydrogen, cyclohexene (C6H10) and, cyclohexane (C6H12) are –241, – 3800 and, –3920 kJ/mole respectively. Calculate the, heat of hydrogenation of cyclohexene., Calculate enthalpy change of the following, reaction:, H2C = CH2(g) + H2(g) , H3C CH3(g). The bond energy of C – H, C – C, C =, C, H – H are 414, 347, 615 and 435 kJ/ mol, respectively., For the reaction, N2(g) + 3H2(g) , , , Narayana IIT/PMT Academy, , THERMODYNAMICS (ENERGETICS), 2NH3(g); H = -95.4 kJ and S = -198.3 J/, K, ., Calculate the maximum temperature at, which the reaction will proceed in forward, direction., 50.. Assume that for a domestic hot water supply, 150 kg of water per day must be heated from, 10°C to 65°C and gaseous fuel propane, C3H8 is used for this purpose. What moles, & volume of propane ( in litre at STP ) would, have to be used for heating domestic water., ΔH for combustion of propane is –2050 kJ/, mol & specific heat of water is 4.184 ×10–3, kJ/g., 51. A gas expands from 3dm3 to 5dm3 against a, constant pressure of 3atm. The work done, during expansion is used to heat 10 mole of, water of temperature 290 K. Calculate final, temperature of water. Specific heat of water, = 4.184 J/ g/K., 52. Diborane is a potential rocket fuel which, undergoes combustion according to the, reaction,, B O (s) + 3H O(g), B2H6(g) + 3O2(g) , 2 3, 2, From the following data, calculate the enthalpy, change for the combustion of diborane :, 3, , , B O (s) ;, i) 2 B(s) + O2(g) , 2 3, 2, , ΔH=-1273 kJ/ mol, 1, , , H O(l) ;, ii) H2(g) + O2(g) , 2, 2, , ΔH = -286 kJ/ mol, H O(g) ;, iii) H2O(l) , 2, , ΔH = 44 kJ/ mol, B H (g) ;, iv) 2B(s) + 3H2(g) , 2 6, , ΔH = 36 kJ/ mol, PREVIOUS IIT QUESTIONS, MULTI ANSWER QUESTIONS, 53., , The difference between heats of reaction at, constant pressure and constant volume for, the reaction:, 81
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THERMODYNAMICS (ENERGETICS), 2C6 H 6 l 15O 2 g 12CO2 g 6H 2 O l , , 54., , 55., , at 25°C in kJ is, [IIT JEE 1991], (A) –7.43 (B) 3.72 (C) –3.72 (D) +7.43, For an endothermic reaction where H, represents the enthalpy of the reaction in, kJ/ mole, the minimum value for the energy, of activation will be [IIT JEE 1992], (A) less than H, (B) zero, (C) more than H (D) equal to H, For which change H E ?, [IIT JEE, 1995], (A) H 2 I 2 2HI, (B) HCl NaOH NaCl, , 60., , 61., , (C) C s O 2 g CO2 g , 56., , 57., , 58, , (D) N 2 3H 2 2NH 3, Molar heat capacity of water in equilibrium, with ice at constant pressure is[IIT JEE, 1997], (A) zero, (B) infinty , –1, –1, (C) 40.54 kJ Mol (D)75.48 kJ–1Mol–1, Standard molar enthalpy of formation of CO2, is equal to, [IIT JEE 1997], (A) zero, (B) the standard molar enthalpy of combustion, of gaseous carbon, (C) the sum of standard molar enthalpies of, formation of CO and O2, (D) the standard molar enthalpy of combustion, of carbon (graphite), The Ho f for CO2(g) and H2O(g) are –393.5,, –110.5 and –241.8 kJ mol–1 respectively. The, standard enthalpy change (in kJ) for reaction, CO2 g H 2 g CO 2 g H 2 O g is, , 59., , 82, , [IIT JEE 2000], (A) 524.1, (B) 41.2, (C) –262.5, (D) –41.2, In thermodynamics, a process is called, reversible when, [IIT JEE 2001], (A) surroundings and system change into each, other, (B) there is no boundary between system and, surroundings, (C) the surroundings are always in equilibrium, with the system, , 62., , JEE ADVANCED, (D) the system change into the surroundings, spontaneously, Which one of the following statements is, false?, [IIT JEE 2001], (A) Work is a state function, (B) Temperature is a state function, (C) Change in the state is completely defined, when the initial and final states are specified, (D) Work appears at the boundary of the system, One mole of a non-ideal gas undergoes a, change of state (2.0 atm, 3, 0 L,95(K) (4.0, atm, 5.0 L, 245 K) with a change in internal, energy, U 30.0L atm is [IIT JEE 2002], (A) 40.0 (B) 42.3 (C) 44.0, (D) not defined, because pressure is not constant, Which of the reaction defines Hof ?, [IIT JEE 2003], (A) C diamond O2 g CO 2 g , (B), , 1, 1, H 2 g F2 g HF g , 2, 2, , (C) N 2 g 3H 2 g 2NH 3 g , 1, 2, , (D) CO g O 2 g CO 2 g , 63., , Two moles of an ideal gas is expanded, isothermally and reversibly from 1 litre to, 10 litre at 300 K.The enthalpy change (in, kJ) for the process is, [IIT JEE, 2004], (A) 11.4 kJ, (B) –11.4 kJ, (C) 0 kJ, (D) 4.8 kJ, 64. The enthalpy of vapourisation of liquid is 30, kJ mol–1 and entropy of vapourisation is 75, kJ mol–1 K. The boiling point of the liquid at, 1 atm is [IIT JEE 2004], (A) 250 K, (B) 400 K, (C) 450 K, (D) 600 K, 65.. The value of log10 K for a reaction A B, is (Given Hor 298K 54.07 kJ mol1 ,, Sor 298K 10JK 1 mol1 and, , R =8.314 JK–1 mol–1,, 2.303×8.314×298=5705) [IIT JEE 2007], (A) 5, (B) 10 (C) 95 (D) 100, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, 66. For the process, H 2 O l 1bar,373K H 2 O g 1bar,373K ,, , 67., , the correct set of thermodynamic parameters is, [IIT JEE 2007], (A) G 0, S ve, (B) G 0, S ve, (C) G ve, S 0, (D) G ve, S ve, Which of the following equations correctly, represents the standard heat of formation, , THERMODYNAMICS (ENERGETICS), in kJ/mol is, (A) + 43.93, (B) -43.93, (C) +527.61, (D) -527.61, 71. Identify the state function among the, following :, (1993), (A) q, (B) q-w (C) q / w (D) q + w, 72. For which change H E, (1995), (A) H 2 I 2 2 HI, (B) HCI NaOH NaCI, (C) C s O2 g CO2 g , , H of methane? (1992), 0, , f, , , CH 4 g , (A) C diamond 2 H 2 g , , , 73., , , CH 4 l , (B) C graphite 2 H 2 g , , , CH 4 g , (C) C graphite 2H 2 g , , , CH 4 g , (D) C graphite 4 H g , , , 68., , The products of combustion of an aliphatic, thiol (RSH) at 298 K are: (1992), (A) CO2 g , H 2o g and So2 g , , 74., , (B) CO2 g , H 2o l and So2 g , , 70., , 1, (i) H 2 g O 2 g H 2 O l ,, 2, , E , (B) p 0, , T, , E , (C) T 0, , p, , (D) All of these, , The rusting of iron takes place as follows:, (2005), , Fe 2 2e Fe s ; E 0 0.44V, , (D) CO2 g , H 2o l and So2 l , At a given temperature, the energy of, activation of two reactions is same if :, (A) the specific rate constant for the two reactions, is the same, (1993), (B) the temperature coefficient for the specific, rate constant for the two reactions is the same, (C) H of the reactions is the same but not zero, (D) H of the two reactions is zero, The enthalpy of vaporisation of liquid water, using the data :, (1993), , E , (A) T 0, , v, , 1, 2 H 2e O2 H 2 O l ; E 0 1.23V, 2, , (C) CO2 l , H 2o l and So2 g , , 69., , (D) N 2 3H 2 2 NH 3, Which of the following is correct for an ideal, gas?, , Calculate G 0 for the net process, A) 322kJ mol 1, 75., , C) 152 kJ mol 1, D) 76 kJ mol 1, The direct conversion of A to B is difficult ,, hence it is carried out by the f o l l o w i n g, show path, (2006), C, , A, , D, , B, , H 285.77kJ / mol, , Given, , 1, (ii) H 2 g O 2 g H 2 O g ,, 2, , S AC 50e.u.,, , H 241.84kJ / mol, Narayana IIT/PMT Academy, , B) 161kJ mol 1, , S C D 30e.u.,, S B D 20e.u, , :, 83
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THERMODYNAMICS (ENERGETICS), Where e.u is the entropy unit, then S A B is, , 76., , 77., , 78., , 79., , A) +60e.u, B) +100e.u, C) -60e.u, D) -100e,u, In themodynamics, a process is called, reversible when, (2001), A) the surrounding and system change into each, other, B) there is no boundary between sysatem and, surroundings, C)the surroundings are always in equilibrium with, the system, D) the system changes into the surroundings, spontaneously, Which of the following statements is false, A) Work is a state function, B) Temprature is a state function, C) Change in the state is completely defined whe, the initial and fianl states are, specified, D) Work appers at the boundary of the system, The species which by definition has zero, standard molar enthalpy of formation, at 298 K is, A) Br2 g , , B) Cl2 g , , C) H 2O g , , D) CH 4 g , , 81., , 82., , 83., , JEE ADVANCED, (A) Enthalpy, (B) Temperature, (C) Volume, (D) Refractive index, The following is (are) endothermic reaction, (s), [IIT JEE 1999], (A) Combustion of methane, (B) Decomposition of water, (C) Dehydrogenation of ethane to ethylene, (D) Conversion of graphite to diamond, Among the following which are /is state, function(s), (2009), A) Internal Energy, B) Irreversible Expansion Work, C) Reversible Expansion work, D) Molar Enthalpy, The reversible expansion of an ideal gas, under adiabatic and isothermal conditions is, shown in the figure. Which of the following, statement(s) is (are) correct?, (2011), , P, , Using the data provided, calculate the, , V, , multiple bond energy kJ mol 1 of a C C, , A) T1 T2, , bond in C2 H 2 . That energy is (take the, , C) wisothermal wadiabatic, , bond energy of a C H bond as 350, kJ mol 1 ) (2012), , 84., , 2C s 2C g , H 1410 kJmol 1, , H 2 g 2H g , H 330 kJmol 1, A) 1165 B) 837, , C) 865, , D) 815, , B) T3 T1, , D) U isothermal U adiabatic, For an ideal gas, consider only P-V work in, going from an initial state X to the, final, state Z . The final state Z can be reached by, either of the two paths shown in the figure., Which of the following choice (s) is(are), correct ?[Take S as change in entropy, and as work done [2012], , MULTI ANSWER QUESTIONS, 80., 84, , Identify the intensive quantities from the, following: [IIT JEE 1993], Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, , THERMODYNAMICS (ENERGETICS), Column-I, A) CO2 s CO2 g , , Y, , X, , B) CaCO3 s CaO s CO2 g , Z, , C) 2H . H2 g , D) P white , solid P r e d , solid , , V litre , , Column-II, p) phase transition, q) allotropic change, r) H is positive, s) S is positive, t) S is negative, , A) S x z S x y S y z, B) Wx z Wx y Wy z, C) Wx y z Wx y, D) Sx yz Sx y, , ASSERTION & REASON, , MATRIX MATCHING, 85., , One mole of a monatomic ideal gas is taken, through a cycle ABCDA as shown in the PV, diagram. Column-II gives the characterstics, involved in the cycle. Match them with each of, the processes given in Coloumn-I (2011), , (A), (B), (C), (D), 87., , Column-I, A) Process A B, , 86., , Column-II, p) Internal energy, decreases, B) Process B C q) Internal energy, increases, C) Process C D r) Heat is lost, D) Process D A s) Heat is gained, t) Work is done on, the gas, Match the transformations in Column-I with, approprite options in Coumn-II, , Narayana IIT/PMT Academy, , 88., , 89., , Direction: This section contains reasoning type, questions. Each question has 4 choice (A), (B),, (C) & (D), out of which ONLY ONE is correct., Statement 1 is true; statement 2 is true; statement, 2 is a correct explanation for statement 1., Statement 1 is true; statement 2 is true; statement, 2 is NOT a correct explanation for statement 1., Statement 1 is true; statement 2 is false, Statement 1 is false ; statement 2 is true, Statement 1: The endothermic reactions are, favoured at lower temperature and the, exothermic reactions are favoured at higher, temperature., Statement 2: When a system in equilibrium is, distrubed by changing the temperature, it will tend, to adjust itself so as to overcome the effect of, change. [IIT JEE 1991], Statement 1: The heat absorbed during the, isothermal expansion of an ideal gas against, vacuum is zero., Statement 2: The volume occupied by the, molecules of an ideal gas is zero. [IIT JEE, 2000], Statement 1: For every chemical reaction at, equilibrium, standard Gibbs energy of, reaction is zero., Statement 2: At constant temperature and, pressure, chemical reactions are spontaneous in, the direction of decreasing Gibbs energy., [IIT JEE 2008], 85
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THERMODYNAMICS (ENERGETICS), 90. Statement 1: There is a natural asymmetry, between converting work to heat and, converting heat to work., Statement 2: No process is possible in which, the sole result is the absorption of heat from a, reservoir and its complete conversion into work., [IIT JEE 2008], , JEE ADVANCED, TRUE / FALSE, 96. First law of thermodynamics is not adequate, in predicting the direction of a process., [IIT JEE 1982], 97. Heat capacity of a diatomic gas is higher, than that of a monoatomic gas. [IIT JEE, 1985], SUBJECTIVE, , INTEGER TYPE QUESTION, 98., 91., , One mole of an ideal gas is taken from a to, b along two paths denoted by the solid and, the dashed lines as shown in the graph below., If the work done along the solid line path is, Ws and that along the doted line path is Wd, then the integer closest to the ratio Wd / Ws, is, , An athlete is given 100 g of glucose, (C6H12O6) of energy equivalent to 1560 kJ., He utilizes 50 per cent of this gained energy, in the event, in order to avoid storage of, energy in the body, calculate the weight of, water he would need to perspire. The, enthalpy of evaporation of water is 44 kJ/, mole. [IIT JEE 1989], 99. The standard enthalpy of combustion at 25°C, of hydrogen, cyclohexene (C 6 H 10 ) and, cyclohexane (C6H12) are –241, –3800 and –, 3920 kJ mole respectively. Calculate the, heat of hydrogenation of cyclohexene., [IIT JEE 1989], 100. Using the data (all value are in kcal mol–1 at, 25°C) given below, calculate the bond energy, of C –C and C – H bonds., H ocombustion ethane 372.0, , 92., , Ans : 2, IIT-JEE 2010, For a liquid the vapour pressure is given by, 400, 10, T, Vapour pressure of the liquid is 10 x mm Hg. The, value of x will be----log10 P , , 93., , 94., , 95., , 86, , FILL IN THE BLANKS, A system is said to be ............... if it can, neither exchange matter nor energy with the, surroundings., [IIT JEE 1997], The heatcontent of the products is more than, that of the reactants in an ............... reaction., [IIT JEE 1993], Enthalpy is an ..................... property. [ I IT, JEE 1997], , ;, , H ocombustion propane 530.0 ;, Hocombustion graphite C g 172.0 ; Bond, , energy of H–H = 104.0;, H of of H2O (l) = 68.0, H of of CO2 (g) = 94.0, [IIT JEE 1990], 101. A gas mixture of 3.67 litres of ethylene and, methane on complete combustion at 25°C, produces 6.11 litres of CO2. Find out the, amount of heat evolved on burning one litre, of the gas mixture. The heats of combustion, of ethylene and methane are 1423 and 891, kJ mol–1at 25°C., [IIT, JEE 1991], 102. Determine the enthalpy change of the, reaction. C3 H 8 g C 2 H8 g CH 4 g at, 25°C, using the given heat of combustion, value under standard conditions:, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, Compound, H2(g), CH4(g) C2H6(g), C(graphite), H o kJ / mol –285.8 –890.0, , –1560.0 –, , 393.5, The standard heat of formation of C3H8(g) is –, 103.8 kJ/mol, [IIT JEE 1992], 103. In order to get maximum calorific output, a, burner should have an optimum fuel to, oxygen ratio which corresponds to 3 times, as much oxygen as is required theoretically, for complete combustion of the fuel. A burner, which has less adjusted for metahne as fule, ( with x litre/hour of CH4 and 6x litre/hour of, O2) is to be readjusted for butane, C4H10. In, order to get the same calorific output, what, should be the rate of supply of butane and, oxygen? Assume that losses due to, incomplete, etc. Are the same for both fuels, and the gases behave ideally. (Heats of, combustion: CH4 = 809 kJ /mol; C4H10 =, 2878 kJ/mol) [IIT JEE 1993], 104. The polymerisation of ethylene to linear, polyethlene is represented by the reaction, , THERMODYNAMICS (ENERGETICS), 120 g of glucose?, [IIT JEE 1997], 107. Compute the heat of formation of liquid, methyl alcohol in kilojoules per mole, using, the following data. Heat of vapourisation of, liquid methyl alcohol = 38 kJ/mol. Heat of, formation of gaseous atoms from the, elements in their standard states; H = 218, kJ/, mol;, C, = 715, kJ/mol;, O = 49 kJ/mol. Average bond energies:, C – H = 415 kJ mol, C –O = 365 kJ/mol , O –, H =463 kJ/ mol [IIT JEE 1997], 108. Anhydrous AlCl3 is covalent. From the data, given below, predict whether it would remain, covalent or become ionic in aqueous solution., (Ionisation energy for Al = 5137 kJ mol–1;, H hydration for Al3+ = –4665 kJ mol–1; H hydration, , 109., , nCH 2 CH 2 CH 2 CH 2 n where n has a, , large integral value. Given that the average, enthalpies of bond dissociation for C = C and, C – C at 298K are + 590 and +331 kj mol–1, respectively, calculate the enthalpy of, polymerisation per mole of ethylene at 298, K., [IIT JEE, 1994], 105. The standard molar enthalpies of formation, of cyclohexane (l) and benzene (l) at 25°C, are –156 and +49 kJ mol–1 respectively. The, standard enthalpy of hydrogenation of, cyclohexene (l) at 25° is –119 kJ mol–1. Use, these data to estimate the magnitude of the, resonance, energy, of, benzene., [IIT JEE 1997], 106. The enthalpy change involved in the, oxidation of glucose is –2880 kJ mol–1 ., Twenty five per cent of this energy is, available for muscular work. If 100 kJ of, muscular work is needed to walk one, kilometre, what is the maximum distance, that a person will be able to walk after eating, Narayana IIT/PMT Academy, , 110., 111., , 112., , for Cl– = –381 kJ mol–1) [IIT JEE 1997], From the following data, calculate the, enthalpy change for the combustion of, cyclopropane at 298 K. The enthalpy of, formation of CO2(g), H2O(l) and propene (g), a r e – 3 9 3 . 5 , – 2 8 5 . 8 a n d, 20.42 kJ mol–1 respectively. The enthalpy, of isomerisation of cyclopropane to propene, is –33.0 kJ mol–1., [IIT JEE 1999], Formation of SF6(g), S(g) and F(g) are : –, 1100, 275 and 80 kJ mol–1 respectively., A sample of argon at 1 atm pressure and, 27°C expands reversibly and adiabatically, from 1.25 dm3 to 2.50 dm3. Calculate the, enthalpy change in this process. CV..m for, argon is 12.48 JK–1 mol–1. [IIT JEE 2000], Show, that, the, reaction, 1, CO g O 2 g CO 2 g , 2, , at 300 K, is, , spontaneous and exothermic and, when the, standard entropy change is –0.094 kJ mol–1, K–1. The standard Gibbs free energies of, formation for CO2 and CO are –394.4 and, –137.2 kJ mol–1, respectively., 113. Diborane is a potential rocker fuel which, undergoes combustion according to the, reaction, B2 H 6 g 3O2 B2 O3 s 3H 2 O g . From, , the following data, calculate the enthalpy, change for the combustion of diborane., 87
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THERMODYNAMICS (ENERGETICS), 118. In, , 3, 2B s O2 g B2 O3 s ;, 2, , N 2 O 4 g 2NO 2 g . When 5 moles of, , H 1273kJ mol1, 1, H 2 s O 2 g H 2 l ; H 286kJ mol1, 2, H 2 O l H 2 O g ; H 44 kJ mol, , 1, , 2B s 3H 2 g B2 H 6 g ; H 36 kJ mol, , 1, , [IIT JEE 2000], 114. When 1-pentyne (A) is treated with 4 N, alcoholic KOH at 175°C, it is converted, slowly into an equilibrium mixture of 1.3%, 1-pentyne (A), 95.2% 2-pentyne (B) and, 3.5% of 1, 2-pentyne (C). The equilibrium, was maintained at 175°C. Calculate G o for, the following equilibria: B A ; G1o ? ;, B C ; G o2 ?, , From the calculate value of G1o and G o2, indicate the order of stability of (A), (B) and, (C). Write a resaonable reaction mechanism, showing all intermediates leading to (A), (B), and (C)., [IIT JEE 2001], 115. For the reaction, 2CO O2 2CO 2 ;, H 560kJ . Two moles of CO and one, , mole of O2 are taken in a container of volume, 1 L. They completely from two moles of CO2,, the gases deviate appreciablyfrom ideal, behaviour. If the pressure in the vessel, changes from 70 to 40 atm, find the, magnitude (absolute value) of U at 500 K., (1 atm = 0.1 kJ), [IIT JEE 2006], 116. CV value of He is always 3R/2 but CV value, of H2 is 3R/2 at low temperature and 5R/2, at moderate temperature and more than 5R/, 2 at higher temperature. Explain in two to, three, lines., [IIT JEE 2003], 117. An insulated container contains 1 mol of a, liquid, molar volume 100 ml, at 1 bar. When, liquid is steeply pressed to 100 bar, volume, decreases to 99 ml. Find H and U for, the process., 88, , the, , JEE ADVANCED, following equilibrium, , each is taken and the temperature is kept, at 298 K, the total pressure was found to, be 20 bar. Given, G of N 2 O 4 100 kJ ; G of NO 2 50 kJ ., , (i) Find G of the reaction at 298 K. (ii), Find the direction of the reaction., [IIT JEE 2004], LEVEL-VI, Matrix-Match Type, 1.), (A) - (q, r), (B) - (p, r),, (C) - (p, r, s), (D) - (p, r, t, u), 2.), (A) - (p, r), (B) - (p, r),(C) - (q, s),, (D) - (q, r), 3.) A-P,S;, B-Q,R;; C-R ;, D-S, 4)., (A) - (r), (B) - (s), (C) - (p), (D) - (q), 5) A-Q B-P,S C-P D-R, 6) A-P, B-R C-Q,S, D-Q,S, 7) A-S, B-R,, C-P,, D-Q, 8) A-Q, B-P, C-R D-S, 9) A-P,S B-R, C-Q,S, D-Q,S, 10) A-R B-Q, C-Q, D-P,S, 11) A-S B-P,R, C-Q D-P,R, 12) A-Q, B-P, C-R,, D-S, Assertion – Reason Type, 13. (B), 14., (B), 15., (A), 16. (A), 17., (D), Integer Type, 18)5 km., , 19)3 kcal., , 20)4 kcal ., 21) 8 J K 1 mol 1, 22)2, 23)8 Joule., 24)5J., 25) 9, 26)2 kJ., 27).2 28) 2, 29) 5, 30) 6 31) 3 32) 7, SUBJECTIVE TYPE QUESTIONS, 33. 0.694atn, 34. molar internal enegry change= 37.5 kj/, mol,molar enthaply of vapourization = 40.6kj/, mol, 36. K C 1.958 10 4, 37. -9.1555 kj, 38. 7200 cal, 39. 31.1kj, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, , THERMODYNAMICS (ENERGETICS), , 40. -0.1555cal/deg, 41. 19.15j/k/mol, 42., K P 8.94 10 4 ,at 1200K. K P =0.77, 43. -3199.75 kj, 44. 27.22units, 45., 46., 47., 48., 49., 50., 51., 52., , heat of formation for Cl aq = -167.2 kj, 57 kj, 121 kj / mole, 125 kj, T > 481, 3.77 102 liter, 290.81 K, -2035 kj / mole, , PREVIOUS IIT QUESTIONS, MULTI ANSWE TYPE QUESTIONS, 53. (A), 54. (C), 55. (D), 5 6 ., (B), 57.(D), 58.(B), 59.(C), 60.(A), 61.(C), 62.(B), 63. (C) 64. (B) 65. (B) 66. (A) 67. C, 68. B 69. A 70. A 71. D 72. D, 73. B 74. A 75. A 76. C 77. A, 78. B 79. D, MULTIPLE CHOICE QUESTIONS MORE, THAN ONE CORRECT ANSWER, 80., (B,D) 81., (B, C,D), 82., A,D, 83. A,C,D, 84. A,C, MATRIX MATCHING, 85. A-R,T,P, B-P,R, C-S,Q, DT,R, 86.A-P,R,S, B-R,S, C-T, DQ,T, ASSERTION & REASON, 87., , (D), 88., (A), INTEGER TYPE, 91.2, 92. 9, FILL IN THE BLANKS, 93., Isolated, Endothermic, , (B), , TRUE/ FALSE, Narayana IIT/PMT Academy, , 89., , (D), , 96., True, 97., True, SUBJECTIVE PROBLEMS, 98., 319.1 g, 99., –121 kJ, 100. C–C=82 kcal / C–H = 99 kcal, 101. 5557 kJ, 102. –55.7 kJ/mol, 103. 5.48(X) litre O2, 104. –72 kJ moles–, 105. –152 kJ, 106. 4.8 km, 107. –275 kJ mol–1, 108. IONIC, 109. 2091.32 kJ mol–1, 110. 309.16 kJ mol–1, 111. –115.96 joules, 112. G o is negative; H is negative, 113. –2035 kJ, 114., G1o 15.992 kJ mol1 , Go2 12.312 kJ mol1 ,, H O , B > C < A, , 115., 117., 118., , 557 kj / mol, U 100 bar ml; H 9900 bar ml, , 5.0705×103 kJ mol–1; reverse direction, LEVEL (VI), , 18., Sol:, , INTEGER TYPE QUESTIONS, Ans 5km, Energy available fo r muscular work, 2880 25, 720 kJ / mol ., 100, Energy available for muscular work from, , , , 125 gm glucose =, , 90., , 720, 125 500KJ, 180, 500, , 95., , 94., Extensive, , 19., , Distance travelled = 100 5km ., Ans. 3 kcal., , Sol:, , On mixing CaCl2 aq. and Na2 CO3, CaCl2 Na2 CO3 , CaCO3 2 NaCl, Solutions are very dilute and thus, 100 %, dissociation occurs., 89
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), Ca2 aq. 2Cl aq. 2Na aq. CO32 aq. , , , 0.5 1.987 4.184, joule 50.63 J, 0.0821, Work is carried out at constant P and thus, irreversible., From I law of thermodynamics, , q U W, , , CaCO3 , , 2 Na aq. 2Cl aq. or, Ca 2 aq. CO32 aq. , CaCO3 s , 0, 0, 0, 0, , H HPr0 oducts HRe, actants or H Hf CaCO3 Hf ca2 Hf CO32 , , H0, , of, , a, , , , compo und, , 0, H Formation, 288.45 129.80 161.65 3 k cal, , 20., Sol:, , 58.63 U 50.63, , 24., , Ans. 4 kcal., The required equation is:, , U 8 joule, , Ans., Sol:, Cv , , 5J., Gas is diatomic as 1.4 , thus,, , 5, 7, R and C p R, 2, 2, , Given, H 85 J at constant pressure, , 2 Al Fe2 O3 , Al2 O3 2 Fe, H ?, , H H f products H f reac tan ts , , T , , Also,, , H f Al2O3 2H f Fe 2H f Al H f Fe2O3 , , Now, W nRT n 2 , , U H W, 85 80, , 399 195.92 203.08 ., At. mass of aluminium = 27, Mol. mass of, , Fe2O3 160, Vo lume, , U 5 J ., of, , reactants, , 25., , 160 2 27, , 50.77 cm3 ;, 5.2, 2.7, , Ans., Sol:, Strans , , 22., 23., , Ans., Sol:, Ans., Sol:, , l, , 3 kJ mol 1 3000 J mol 1, S dissolution 40 J mol 1 ; G 9 kJ ., , 8 J K 1 mol 1 ., , Gdissolution H T S 3000 300 40 9000 J, , H trans 2288, , 8 J K 1 mol 1 ., T, 286, 2., Q 13.7 11.7 2, 8 Joule., Work, do ne, , P dV 1 2.5 2.0 , 0.5litre atm, 90, , Ans. 9 kJ ., S, o, , :, , Hdissolution Hionisation H hydration 778 775, , 203.08, 4 kcal ., Fuel value per cm3 , 50.77, , 21., , 40, 80 J, n, , q p H U W , , Also,, , 399 2 0 2 0 195.92 , , 85 140 2 40, , , nC p 7 n R, n, , 26., , Ans., Sol:, , 2 kJ., Reversible work is maximum work., , , , 2.303 , , V , w 2.303 nRT log10 2 , V1 , 16, 25, 8.314 300 log, 32, 5, , Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, SUBJECTIVE TYPE QUESTIONS, 33., , THERMODYNAMICS (ENERGETICS), \, , nR, Work done in adiabatic expansion 1 T2 T1 , , Solution (a) Work involved in heating of gas, , 36., , Wa P.V P (V2 V1 ), 1, 9, 8, ., , 7, , =, , -, , H 0 77.2 mol 1, , S 0 122 JK 1mol 1, , 1, 2.303X1X9.87X 273log to P P1 0.694 atm, 1, , 34., , Solution, , T 400k, , G 0 H 0 T S 0 ;, G 0 77200 400 122 28400 J, , Solution: The vaporization occurs at consstant, pressure therefore the enthalpy change is, equal to the work done by the heater;, , Also, we have, , G 0 2.303RT log10 K c, , where, kc is equilibrium constant, 28400 2.303 8.314 400 log10 kc, , ', , H 0.50 X 12 X 300, , 1800 J 1.8kJ, , Molar enthalpy of vaporization H , , H ', H, , mole ofH 2 O n H 2 o, , kc 1.958 10 4, , 37., , Solution;, 3.7 kJ mol 1 3700 J mol 1, H dissolution H ( ionisation ) H ( hydration ) 778 774.3, , 1.8, 40.6 kJ mol 1, 0.798 , , , 18 , , , , Sdisolution 43 J mol 1, G d is o lu tio n H T S 3 7 0 0 2 9 8 X 4 3 9 1 1 4 J, , o,sA, lH, , U PV U ng RT U RT, , 38., , ( H 2 o(l ) H 2 o( g ), n g 1), , G 9.114kJ, S, o, , 39., , 10 80 10 1100 10 540 7200cal, Sol;-Using Clausius-Clapeyron equation;, , H R T 4 0 .6 8 .3 1 4 X 1 0 3 X 3 7 3 .1 5 3 7 .5 kJ m o l 1, , S, , o, , l, , u, , t, , p1 10atm at T 273 k for, , p2 1atm at T T2 k for, , i, , o, , n, , :, , 2.303log, , 14, moleO2, 32, , H v , , 14, moleO2, 32, , H v , , T .P1 cons tan t, , , 1, , or, , T1, p, 1 log 2, log, T2, p1, , 273, 1, or 1.4 log T 1 1.4 log 10, 2, , Narayana IIT/PMT Academy, , T2 141.4 K, , P2 H v T2 T1 , , P1, R, T1T2, , 2.303RT1T2, P, log 2, (T2 T1 ), P1, , We have, , Fo r adiabatic expansion we have, , T1 P2 , , T2 P1 , , :, , H total H fusion H heating H vaporization, , U molar int ernal energy change, , 35., , l, , 40., , 2 .3 0 3 8 .3 1 4 3 0 3 3 3 3, 5 .2 1 0 4, lo g to, (333 300 ), 1 .5 3 1 0 4, , 31.1 103 J 31.1 kJ, the, H for, CD2O mS T , , colling, , of, , 3.2, 14 100 140 cal, 32, , Cooling is exothermic H 140cal, Now,, 91
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), S , , 41., , H, 140, , 0.1555 cal deg 1, T, 900, , 2.303nRT log to, , , H 0 3201kJ, Also H U ng RT, , Sol;-Wehave , S q1 , T, T, Given, n=1, R=8. 314J, T =298K,, , 44., , U (0.5) 8.314 103 300, , U 3201 1.2471 = -3199.75 kJ, Total energy required in the day, , V1 V ,V2 10V, , , , S 2.3031 8.314logto, , 42., , , , , , 0, 2, G300, 28.44kJ, K 41.16 300 4.24 10, , a, , n, , Units of glycogen required =, , G, , , , 45., , , , Also, G 0 2.303 RT log K p, , H 0 74.8 kJ . . . . .(iii), Add. eq. (ii) and (iii), , Similarly , at 1200K, K P 0.77, Given,, 7C(s) + 3H2(g) + O2(g) , , , 1, 1, H 2 ( g ) Cl2 ( g ) nH 2O , H ( aq ) Cl ( aq ) ;, 2, 2, , C6H5COOH(s);, , H 0 408kJ . . (i), C(s) + O2(g) , CO2(g), , H 0 167.2 kJ . . . . (iv), Subtract eq. (i) from (iv), , H 0 393 kJ . . .(ii), 1, H2(g) + O2 (g) , H2O(l), 2, , , , 1, Cl2 ( g ) aq. e , Cl ( aq .) ;, 2, , H 167.2 kJ, , 0, , 15, O 2 ( g ) , C6H5COOH(s) +, , 2, 7CO2(g) + 3H2O(l), , 1, H 2 ( g ) aq. , H ( aq ) e;, 2, , +, HCl(g) + nH2O(l) , H (aq) + Cl (aq);, , 4, , H 286 kJ . . .(iii), 15, O2 ( g ) , C6H5COOH(s) +, 7CO2(g), 2, + 3H2O(l) ;, H ? . . . .. . . . (iv), By (ii) 7 + (iii) 3 - (i), , Given,, , 1, 1, H 2 ( g ) Cl2 ( g ) , HCl ( g );, 2, 2, H 0 92.4kJ . . . . (ii), , Thus, at 300 K, reaction will process in forward direction and 1000K back ward direction, , 43., , 12960, 27.22, 476, , H 0 0 . . . . (i), , 32.93 1200 2.96 10 2.59kJ, , K p 8.94 10, , (1 watt = J/, , units., , d, 2, , × 24× 60× 60, kJ, 1000, , sec) = 12960 kJ, , 10V, 19.15JK 1mol 1, V, , Solution: At equilibrium G 0 H 0 T S 0, , 0, 1200 k, , = 3201 =, , , , V2, V1, , Heat of formation for Claq 167.2 kJ, 46., , Initial average temperature of the acid and base, =, , 22.6 23.4, 23.0C, 2, , Rise in temperature = (29.3 – 23.0) = 6.3°C, Total heat produced = ( 92 × 0.75 + 200 × 4.184) ×, 6.3, = (905.8) × 6.3 = 5706.54 J, , H 0 [393 7 286 3 408], 92, , Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, Enthalpy of neutralisation = –, , THERMODYNAMICS (ENERGETICS), 50., , 5706.54, 1000 1 = – 57065.4 J = – 57 kJ, 100, , 47., , The required reaction is, C6 H10 H 2 , C6 H12 ,, Cyclohexene, , Cyclohexane, , H1 ? . . . . ., , (i), , Given, , Heat taken up by water = m SΔT, = 150 × 1000 × 4.184 × 10–3 × 55 = 34518 kJ, 2050 kJ heat is provided by 1 mole C3H8, 34518, kJ heat is provided by = 34518/2050, , = 16.83 mole of C3H8, Volume of C3H8 at NTP = 16.83 × 22.4 litre, = 3.77 × 102 litre, 51., Work is done aganist constant P, V 5 3 2dm3 2 litre; P 3 atm., , 1, H 2 O 2 , H 2O,, 2, H 2 241 kJ/mole, , C6 H10 , , W = P.V 3 2 litre atm., , . . . . . (ii), , 17, O 2 , 6CO 2 5H 2 O,, 2, , H 3 = – 3800 kJ/mole, , . . . . . (iii), C6H12 + 9O2 , 6CO2 + 6H2O,, H 4 = – 3920 kJ/mole . . . . . (iv), add (ii) and (iii), C6H10 + H2 + 9O2 , 6CO2 + 6H2O, , 52., , H 5 = 404 kJ/mole, , Substract (iv) from (v), C6H10 + H2 , C6O12, H = 121 kJ/mole, 48., , 49., , H reaction = Bond energy data for the formation, of bond + Bond energy data for the dissociation, of bond, = –[1(C – C) + 6(C – H)] + [1(C = C) + 4(C, – H) + 1(H – H)], = –347 – 2 414 + 615 + 435, = – 125 kJ, Enthalpy change for the reaction = –125 kJ, G = H - T S, For a reaction to be spontaneous, G = -ve, H T S ve or H T S, , 6 4.184 1.987, J 607.57 J, 0.0821, Now this work is used up in heating water, W = n C T, 607.57 = 10 4.184 18 T, T 0.81, Final temperature = T1 + T = 290 +, 0.81 = 290.81 K, The concerned chemical reaction is, B O + 3H O , H, B2H6(g) + 3O2(g) , 2 3(s), 2 (g), =?, The enthalpy change can be calculated in the, following way., , , H [H B2 O3 (s ) 3H H2O (g) ] H B2 H6 (s) ;, , ( H 0f of O2 0), H H 2 O( g ) can be obtained by adding H H 2 O( l ) and, H H 2 O(g ) ,, , 53., , or, , . H E nRT = 12–15 = –3, , 3, , H, 95.4 10, T or, T or 481.0 > T, S, 198.3, Thus, if temperature of system is lesser than, 481 K, the reaction would be spontaneous. At, 481 K, the reaction will be in equilibrium. An, increase in temperature above 481 k will develop, non-spontaneity for the reaction., , i.e. – 286 + 44 = –242 kJ mol–1, –1, H = [–1273 + 3 ´ –242] – 36 kJ mol, = –1273 – 726 – 36, = –2035 kJ mol–1, n = no. of moles of gaseous products – no., of moles of gaseous reactants, H E 3 8.314 298 7.43kJ, , 54., , 55., , Eactivation = (Energy intermediate complex) –, Average energy of reactants), The energy of intermediate complex generally, lies above the energy of products., Hence (C) is correct option., H n RT, , For H n 0 Products, Narayana IIT/PMT Academy, , 93
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THERMODYNAMICS (ENERGETICS), where n =no. of moles of gaseous products, – no. of moles of gaseous reactants., For (A), (B) and (C), n 0, Hence (D) is correct option., 56., , H2O H2O, l , , JEE ADVANCED, ds , , 65., , Hence (B) is correct option., For the equilibrium, A B, G o H o TSo, , s , , G o 2.303RT log10 K, , H H1 H, CP 2, , , (infinity), T, 0, [ T 0 , because two states (liquid and solid), , 57., , dq rev, 30 103, 75 , T 400 K, T, T, , 64., , of water are in equilibrium], Hence (B) is correct option., for, CO 2, is, H of, , (K is, , equilibrium constant), 2.303RT log10 K Ho TSo, 2.303RT log10 K TSo Ho, , given, , by, , lo g 1 0 K , , T S o H o 2 9 8 1 0 5 4 .0 7 1 0 0 0, , 1 0, 2 .3 0 3 R T, 2 .3 0 3 8 .3 1 4 2 9 8, , C graphite O2 g CO 2 g , , 58., , 59., , 60., , Hence (B) is correct option., Hence (D) is correct option., 66., The change given is occuring at the boiling point, of the liquid, where, at a given pressure and, 1, the liquid-vapour system virtually, H H P H R 1 1 0 . 5 2 4 1 . 8 3 9 3 . 5 k J m o l temperature, remains at equilibrium and hence G 0 . Also, = 41.2 kJ mol–1, due to absorption of heat as latent heat of, Hence (B) is correct option, vaporisation, or due to change from liquid to, Since the driving and opposite forces are equal, gaseous state where randomness has also, in case of a reversible process so in such a, increased, S 0 ., process, the surroundings are always in, Hence (A) is correct option, equilibrium with the system., 67., c), Hence (C) is correct option, Natural isotopic form of carbon is graphite and, Work done depends upon the path adopted, that hydrogen is H 2 . The required, , so it is not a state function., Hence (A) is correct option, , equation is C s 2 H 2 g CH 4 g , , 61., , , , , , , , , , HH2 H1 E2 P2V2 E1P1V1 E2 E1 P2V2 P1V1, , 62., , 63., , 94, , =30 + 4×5–2×3= 44 L atm, Hence (C) is correct option., Is not correct because C(graphite) and not, C(diamond) is the standard state., is correct because in it one mole of HD in its, standard state is formed from its elements in their, standard states., is incorrect because in it 2 moles of NH3 are, formed., is incorrect as it involves CO (not an element)., Hence (B) is correct option., H n.C P .T ;, Since T 0, Hence, H 0, Hence (C) is correct option., , 68., 69., 70., , 71., , b)Conceptual, a)Conceptual, a), Enthalpy of vaporisation of liquid water =, 285.77 241.84 43.93kJ mol 1, d), , U q w, U State function U 2 U1 , 72., , d), , ng, , fo r, , the, , reaction,, , N 2 g 3 H 2 g 2 NH 3 g , is not, equal to, 73., 74., , zero hence, , H U n g RT, H U ;, b)Conceptual, a), , Fe s Fe 2 2e , , E 0 0.44V, , Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, , THERMODYNAMICS (ENERGETICS), In adiabatic process expansion occurs by us, ing internal energy hence it decreases while in, isothermal process temperature remains con, stant that’s why no change in internal energ, , 1, 2 H 2e O2 H 2O l E 0 1.23V, 2, ________________________, , 84., 1, Fe s 2 H O2 Fe2 H 2O, 2, , E 0 0.44 12.31.67V, , 76., 77., 78., , 50 30 20 60eu, c)Conceptual, a)Conceptual, b)Conceptual, , 79., , (i) 2C s H2 g H C C H g , H 225 kJmol 1, , 87., , 88., , (ii) 2C s 2C g , H 1410 kJmol 1, (iii) H 2 g 2 H g , H 330kJmol 1, From equation (i):, , , 225 2 H, 1 BEH H 2 BEC H 1 BEC C , C s C g , , , , 89., , 83., , WX Y Z WX Y (Work done in Y Z is, zero as it is an isochoric process), In this case statement-1 is wrong because, endothermic reactions are favoured at high, temperature and exothermic reacts arefavoured, at lower temperature in accordance with Le, Chatelier’s principle., Statement-2 is correct as it is in accordance, with Le Chatelier’s principle., Hence (D) is correct option., By first law of thermodynamics dq= dE+ dW., Under isothermal condition for ideal gas dW =, 0 as volume occupied bythe molecules of ideal, gas is zero. Also (dE)T = 0 as for ideal gas there, is no change in internal energy at constant T, due to no force of attraction between the, molecules., dq = 0+0 =0, Hence, (B) is correct option, At equilibrium G 0 , but standard Gibb’ss, energy G o of a reaction mayor may not be, , 225 1410 330 700 BEC C , , zero. For reaction to be spontaneous G, (Gibbs energy) should be more negative, i.e., G 0 ., Hence, (D) is correct option, Second law of thermodynamics states that total, heat can never be converted into equivalent, amount of work., Hence, (A) is correct optionxz., , 225 1040 BEC C, , 81., , [Entropy(S) is a state, , 225 1410 1 330 2 350 1 BEC C , , 225 1740 700 BEC C, , 80., , Y z, , function, hence additive], , G 0 nFE 0 2 96500 1.67 322kJ, 75., a), , S A B S AC SC D S B D, , S x z S x y S, , BEC C 1040 225 815 kJ mol 1, Intensive properties are those which are, independent of the mass of the system. Hence, (B, D) is correct option., All combustion reactions are accompained by, evolution of heat i.e., they are exothermic. Hence, (B, C, D) is correct option., , T1 T2 because process is isothermal, work done in adiabatic process is less than in, isothermal because area covered by, isothermal curve is more then the area cov, ered by the adiabatic curve, , Narayana IIT/PMT Academy, , 90., , 92., , 93., , 94., , log10 P , , 400, 400, 10 , 10 9 ;, T, 400, , P 109 mm Hg x=9, Isolated; when the boundary is both sealed and, insulated, no interaction is possible with the, surroundings., Endothermic; Since HP > HR or, H P H R H ve, 95
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THERMODYNAMICS (ENERGETICS), 95., Extensive; A property which depends upon the, quantity of matter present in the system is called, an extensive property, TRUE/ FALSE, 96., (It only tells that if a process occurs the heat, gained by one end would be exactly equal to, lost by the other. It does not predict the, direction)., 97., In case of polyatomic gases, a part of energy, supplied is used in increasing the internal energy, of the system, and thus some additinal energy, is needed to raise the temperature of the gs, through 1°C. it means that the heat capacity, (amount of heat required to raise the, temperature of the system by 1°C) is higher in, case of polyatomic gases as compared to that, for monoatomic gases., 98., 100 g glucose = 1560 kJ, Energy, utilised, left, in, bo dy, , , 50, 1560 780 kJ, 100, , Hence energy to be given out = 780 kJ, Enthalpy of evaporation of water = 44 kJ / mol, = 44kJ /18 g of water, Thus 44 kJ energy is given by water = 18 g, 1 kJ energy is given by water = 18/44 g, 730 kJ energy will be given by water, , , 99., , 18, 780 319.1g, 44, , Hence amount water to be perspired to avid, storage of energy = 319.1 g, In this case the required reaction is, , JEE ADVANCED, ; H 3920 kJ, To get the desired eqaution add (i) and 9ii) and, substract (iii).Then, H 241 3800 3920 121kJ / mole, , 100., , (i) H1 2 C C 8 C H 3Csg 4 H H, (ii), , From the given data, we can write following, equations, 1, H2 O2 H 2O, 2, ; H 241kJ, , (iii), , (ii), , 17, O2 6CO2 5H 2, 2, , ; H 3800 kJ, (iii), 96, , C6 H12 9O 2 6CO 2 6H 2 O, , H1 2x 8y 3 1724 104 , , and, (iv), , H 2 x 6y 2 172 3 104, , Given, (v), , C O 2 CO 2, , ;, , H 94.0 kcal, , (vi), , 1, H2 O2 H 2O, 2, , ;, , H 68.0 kcal, , (vii), , 7, C2 H 6 O2 2CO 2 3H 2 O ;, 2, H 372.0 kcal, , (viii), , C 2 H8 5O 2 3CO 2 4H 2 O, , ;, , H 530.0 kcal, , If we multiply (v) by 2 and (vi) by 3 and then, from the sum of these new equation substract, (vii), we get (ix), (ix), ;, 2C 3H 2 C 2 H 6, , (i), , C6 H10 , , H 2 1 C C 6 C H 2C s g 3 H H , , Let the bond energy of C – C and C – H bonds, be x kcal and y kcal respectively. Then, we, have, , C 6 H10 H 2 C6 H12, H f ?, , For formation of C3H8, 3C 4H 2 C3 H 8 ,, H1 ?, For formation of C2H6, 2C 3H 2 C 2 H 6 ,, H 2 ?, Thus, , H 2 20.0 kcal, , Again 3–(v) +4×(vi) (viii) gives;, (x), , 3C 4H 2 C3 H 8, , ;, , H1 20.0 kcal, , Solving equation (iii), (iv), (ix) and (x) we get, x + 6y = 676, 2x + 8y = 956, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, or, x = 82 kcal and y = 99 kcal, hence bond energy of C–C bond = 82 kcal, and bond energy of C–H bond = 99 kcal, 101. Combustion of C2H4 and CH4 takes place as, , (iv) C s O2 CO2 ; H 4 393.5kJ, (v) 3C s 4H 2 C3 H8 ; H 4 103.8kJ, The required equation is, C3 H 8 g H 2 g C 2 H 6 g CH 4 g H ?, , C 2 H 4 3O2 2CO2 2H 2 O, 1vol, , THERMODYNAMICS (ENERGETICS), , 2 vol, , We can get the desired equation using the, manipulations given below, [3×(iv) + 5×(i)] – [(v) + (iii) + (ii)], , CH 4 2O2 CO 2 2H 2 O, 1vol, , 1vol, , Let the volume of CH4 in the mixture be x litre., Then the volume of C2H6 in mixture = (3.67 –, x) litre, Volume of CO2 produced by x L of CH4 = x L, Volume of CO2 produced by (3.67–x) L of, C2H4 = 2(3.67–x) L, Total volume of CO2 produced = x +, , 2(3.67–x) L, or, 6.11 = x +2(3.67–x) L x = 12.3 L, Thus, volume of CH4 in mixture = 1.23 L, So volume of C2H4 is mixture = ( 3.67–1.23), L = 2.44 L, Volume of CH 4/ litre of mixture, , , H 3H 4 5H1 H 5 H 3 H 2 , , = [3×(–393.5) + 5× (–285.8)] – [–103.8 –, 1560–590] = –55.7 kJ mole, 103. Combustion of CH4 and C4H10 takes place as, CH 4 2O2 CO 2 2H 2 O, Initial volume(in litre), , combustion , , Volume of C2H4/ litre of mixture, , Since volume of 1 mole of a gas at NTP = 22.4, , , , 809 X , kJ energy is obtained, V, , , , 809 X V, litre C4H10 = 0.281 (X), V 2878, , L, So heat evolved due to combustion of 0.665 L, , (i), , 0.665 1423, 42.32 kJ, 22.40, , litre C4H10, Thus, butane sullpied for the same calorific, output = 0.281 (X) litre, , Hence total heat evolved = 13.32 + (–42.25), = –55.57 kJ, Total heat evolved = 5557 kJ, Negative sign indicates evolution of heat., From the given data, we can get following, equations., , , , H 3 1560 kJ, Narayana IIT/PMT Academy, , 13, O 2 4CO 2 5H 2 O ;, 2, , Volume of O2 required = 3 volume of O2 for, combustion of C4H10, , 1, H 2 O 2 H 2 O ; H1 285.8kJ, 2, , 7, C2 H 6 O2 2CO 2 3H 2 O, 2, , C4 H10 , , H 2878kJ / mol, , 13, volumeof C4 H10, 2, 13, 3 0.281 X 5.48 X litreO 2, 2, , (ii) CH 4 2O 2 CO 2 2H 2 O ; H 2 890 kJ, (iii), , 808 X , kJ, V, , 2878 kJ energy is obtained by 1 mole, , of V litre C4H10, , 2.44, , 0.665L, 3.67, , 102., , 6X, , ; H 809kJmol1, Let the temperature be T and assume volume, of 1 mle of a gas is V litre at this condition., V litre of 1 mole of CH4 gives energy, , on combustion = 809 kJ, X litre of CH4 gives energy on, , , 1.23, 0.335L, 3.67, , of C2H4 , , X, , 3, , ;, 104., , The polymersiation reaction, is nCH 2 CH 2 CH 2 CH 2 n, 97
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THERMODYNAMICS (ENERGETICS), In this process, one double bond (C= C), breaks and two –CO2 groups are linked with, single bonds thus forming three single bonds, (two single bonds are formed when each CH2, group of ethylene (CH2 = CH2) links with, another CH2 - group of another ethylene, molecule)., Therefore, in polymerisation reaction one C =, C is replaced by two C – C bonds or one, mole of C = C bonds are replaced by 2 moles, of C – C bonds, Energy Released = Energy due to formation of, 2 single bonds., = (2×331) kJ = 662 kJ, Energy needed to dissociate one mole of C =, C bonds = 590 kJ, H pot or enthalpy of polymerisation = (590–, , 105., , 662) kJ = –72 kJ moles, C 6 H10 H 2 C6 H12 ; H 119kJ, (involves breaking up of three double bond and, additional of three H2 molecule), C6 H 6 3H 2 C 6 H12 ;, H 3 119 357 kJ, , (involves breaking up of three double bond and, additional of three H2 molecule), Also given 6C 6H 2 C 6 H12 l ; H 156, We have C6 H 6 3H 2 C 6 H12 l ;, , 107., , JEE ADVANCED, The concerned thermochemical reaction is, H, |, C g 4H g O g H C O H, ;, |, H, H ?, , 1, , , H f H Cs Cg 2H H H H O O , 2, , , , 3H CH H CO H O H H Vap CH3OH , , 108., , = [715+2 ×436+249] –[3×415 +365 +463, +38] = –275 kJ mol–1, Total hydration energy of Al3+ and 3 Cl– ions of, AlCl3, , H, , Hydration, , × Hydration energy of Cl–, = [–4665 + 3 ×(–381)] kJ mol–1 = –5808 kJ, mol–1, This amount of energy exceeds the energy, needed for the ionisation of Al to Al3+ (i.e., 5808, > 5137). Because of this AlCl3 becomes less ionic, in aqueos solution., In aqueous solution AlCl3 exists in ionic form, as [Al(H2O)6]3+ and 3 Cl–, 3, , AlCl3 6H 2 O Al H 2 O 6 3Cl, AlCl3 aq AlCl3 aq , , H 357, , 6C 3H 2 C6 H 6 ; H 201kJ, , 106., , Therefore, resonance energy = 49– 201, = – 152 kJ, Energy available for muscular work by 1 mole, of glucose , , 2880 25, 720 kJ, 100, , Thus, 180 g (1 mole ) of glucose (C6H12O6), supplies energy = 720 kJ, 720, 120 480 kJ, Will supply 120 g glucose , 180, , Distance covered by 100 kJ energy = 1km or, 100 kJ is need to walk 1 km, Distance covered by 480 kJ energy, , , 98, , 1, 480 4.8km, 100, , = Hydration energy of Al3+ + 3, , ; H ?, = (Energy released during, H, hydration) – (Energy used during hydration), = (–4665)–(3×381) + 5137 =, –1, –671 kJ mol, Thus, formation of ions will take place., 109. Following equation can be obtained from the, available data, (i), , C s O 2 g CO2 g ;, , H 393.5kJ, , (ii), , 1, H2 g O2 g H2O l ;, 2, , H 2858kJ, , (iii), , 3C s 3H 2 g C 2 H 6 g ;, Narayana IIT/PMT Academy
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JEE MAINS, ADVANCED, JEE, - CW - VOL - I, , THERMODYNAMICS (ENERGETICS), , H 20.42 kJ, , (iv), , CH2, , CH2, , CH2, , (g) C3 H 6 g ;, , The desired equation is, CH2, , n, , CH2, , CH2, CH2, , CH2, , , , 1, , To get the desired equation compute as follows, [3×(i)+3×(ii)] + [(vi)–(iii)], , 112., , = –394.4–[–137.2 + 0] = –257.2 kJ mol–1, Since G o is negative so the reaction is feasible, i.e., spontaneous,, Again, G o H o TS, , Given : SF6 g S g 6F g ;, H ?, , 257.2 H o 300 0.094 H o, 285.4 kJ mol 1, , From the available data, we can write the, following equation, (i), (ii), , S g S g ; H 275.0 kJ, , (iii), , 1, F2 g F g ; H 80.0 kJ, 2, , To get the required equation carry out the, following computation [6×(iii) +(ii)]–(i), i.e., H 6 80 2745 1100.0 1855kJ, , Now, in SF6 we find S–F bonds, therefore the, bond energy, Thus average bond energy for S – F bond, , 111., , CV+R=(12.48+8.314) 20.8], = –115.96 J, Following reaction takes place, , o, o, o, We know, G G Pr oduct G Re ac tan t , , H 2091.32 kJ mol1, , S s 3F2 g SF6 ; H 1100.0 kJ, , H 0.05 20.8 188.5 300 [Cp =, , 1, CO g O 2 CO 2 g , 2, , (g), , 9, O2 g 3CO2 g 3H 2 O l ;, 2, , 110., , PV, 1 1.25, , 0.05, RT 0.082 300, , Now, we know that H n.C p .T, , 9, (g) O2 g 3CO2 g 3H 2 O l ;, 2, , H 2091.32 kJ mol, , 300 8.31 2.50, , ln, T2 12.48 1.25, , Solving the above equation, we get T2 = 188.5, K, Number of moles of argon gas,, , H 33.0 kJ, , CH2, , ln, , or, , 113., , Since the value of H is negative so the reaction, is exothermic., The concerned chemical reaction is, ;, H ?, To get this multiply given equation (ii) by 3 and, add to equation (i) to get equation (v), now multiply given equation (iii) by 3 and add, to equation (v) obtain in above step to get, equation (vi)., Now substract equation (iv) from equation (vi)., 3, 3, 2B s O 2 g 3H 2 g O 2 g , 2, 2, B2 O3 s 3H 2 O l , , H 1273 286 3 2131.0 kJ ..(v), , 1855, , 309.16 kJ mol1, 6, , 3H 2 O l 3H 2 O g , , For adiabatic expansion of a gas, we have, , H 132.0kJ, , ln, , T1 R V2, , ln, T2 CV V1, , Adding,, 2B s 3H 2 g 3O 2 g B2 O3 s 3H 2 O, , H 1999.0kJ, Narayana IIT/PMT Academy, , 99
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JEE ADVANCED, , THERMODYNAMICS (ENERGETICS), Substracting given equation (iv)) from (vi), we, get, , 3R, , considered and f = 3. CV , 2, At moderate temperature both translational and, rotational motions are considered., f = 3+2 (3-translational and 2 rotational), , B2 H 6 g 3O 2 g B2 O3 s 3H 2 O g , , H 2035.0 kJ, , 114., , Calculation of G, G, , o, , Product , 2.303RT log, Reactant , , CV , , At still high temperature translational, rotational, and vibrational motions are considered., f= 3+2+2 (3-translational, 2-rotational,, 2-vibrational), , For the equilibrium reaction B A , we get, and for the reaction B C , we get, 1.3 , , G1o 2.303 8.314 448log, , 95.2, , , 1, 15.992 kJmol, , CV , , 117., , 3.5 , , G o2 2.303 8.314 448log, , 95.2 , , 12.312 kJmol1, , Here, P1 = 1 bar P2 = 100 bar ;, V1 = 100 ml , V2 = 99 ml, For adiabatic process, q 0 ,, , Again, for the reaction A C, , thermodynamics), , 3.5 , , G o3 2.303 8.314 448log, 1.3 , , 1, 3.688kJmol, , q P V2 V1 , , B C , G o2 12.312 kJ mol1, , (first law of, , [W= P(V2–V1)], = 0–100(99–100) = 100 bar mL, Also, H U PV , U P2 V2 P1V1 ;, , 100 99 100 100 1 9900bar mL, , A C , G o3 3.688kJ mol1, , (i)Standard Gibbs free energy change for, the reaction,, , Thus, the correct order of stability B > C > A, , N 2 O 4 g 2NO 2 g , , H U VP, , V 0 , U H VP, , = –560 – [0.1 (40–70)×0.1] = –557 kJ, So, the magnitude is 557 kJ mol–1, 116. In case of helium (monoatomic gas) we have, only three degrees os freedom which correspond, to three translaational motion so the total heat, capacity will increase. The contributors by, vibrational motion is not appreciable at low, temperature but increases from 0 to R when, temperature increases., C V f, , R, , where f is the degree of freedom., 2, , At low temperature only translational motion is, 100, , 7R, 2, , U W, Since, U q W, , Fromthe above calculations, we have, B A , G1o 15.992 kJ mol1, , 115., , 5R, 2, , 118., , G o 2.303RT log K p 0 ; KP = 1, , Initially, PN O PNO 10 bar, 2, , 4, , Reaction quotient, , 2, , P , , , 2, , NO2, , PN2O2, , , , 100, 10, 10, , G o 2G of NO2 G of N 2O 4 100 100 0, , Initial Gibbs free enrgy of the above reaction,, G G o 2.303RTlog QP, G 0 2.303 8.314 298log10, 5.0705 103 kJ mol 1, , (ii) Since initial Gibbs free energy change of, the reaction is positive, so the reverse reaction, will take place., Narayana IIT/PMT Academy
