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

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2 6 Thermodynamics. A gas is allowed to expand in a well insulated container against a constant external pressure of.5 atm from an initial volume of.50 L to a final volume of 4.50 L. The change in internal energy U of the gas in joules will be: (Delhi-07) a J b J c. 500 J d. 505 J. For a given reaction, Δ = 35.5 KJ mol - and ΔS = 83.6 JK mol. The reaction is spontaneous at: (Assume that Δ and ΔS do not vary with temperature) (Delhi-07) a. T > 98 K b. T < 45 K c. T > 45 K d. All temperatures 3. Of the following, the largest value of entropy at 5 0 C and atm is that of: (Gujarat-07) a. C 4 b. c. C 6 d. C 4. Under isothermal and reversible conditions, the term free energy in thermodynamics signifies: (Gujarat-07) a. Expansion work done on the system b. Non-expansion work done by the system c. Expansion work done by the system d. Non expansion work done on the system 5. For a sample of perfect gas when its pressure is changed isothermally from P i to P f, the entropy change is given by: (06 - II) P f a. S = nrt ln P i P i b. S = nrt ln P f P f c. S = nr ln P i P d. i S = nr ln P f 6. The correct thermodynamic conditions for the spontaneous reaction at all temperatures is: (06 - I) a. Δ < 0 and ΔS < 0 b. Δ < 0 and ΔS = 0 c. Δ > 0 and ΔS < 0 d. Δ < 0 and ΔS > 0 7. Consider the following liquid-vapour equilibrium. Liquid Vapour. Which of the following relations is correct? (06-I) dnp v dnp v a. = b. = dt T dt RT dng v dnp v c. = d. = dt RT dt RT 8. The heat of combustion of carbon to CO is kj/mol. The heat released upon formation of 35. g of CO from carbon and oxygen gas is: (05) a kj b. 630 kj c. 3.5 kj d. 35 kj 9. For the reaction, X O 4 (l) XO, ΔU =. kcal, ΔS = 0 cal K at 300 K. ence, ΔG is: (04) a..7 kcal b. 9.3 kcal c. 9.3 kcal d..7 kcal 0. In which of the following reactions, standard reaction entropy change ( S ) is positive and standard Gibb s energy change ( G ) decreases sharply with increasing temperature? (0 Pre) a. C (graphite) + O CO b. C (graphite) + O CO c. CO + O CO d. Mg (s) + O MgO (s)

3 0 NEET 3 Years. The enthalpy of fusion of water is.435 kcal/mol. The molar entropy change for the melting of ice at 0 C: (0 Pre) a cal/mol K b. 0.5 cal /mol K c..04 cal/mol K d cal/mol K. Standard enthalpy of vapourisation vap for water at 00 0 C is kj mol. The internal energy of vapourisation of water at 00 0 C (in kj mol ) is: (0 Pre) a b c d Consider the following process A B Δ(kJ/mol) = B C + D Δ(kJ/mol) = 5 E + A D Δ(kJ/mol) = For B + D E + C, Δ will be: a. 35 kj/mol b. 35 kj/mol c. 55 kj/mol d. 75 kj/mol (0 Mains) 4. Enthalpy change for the reaction, 4 is kj. The dissociation energy of bond is: (0 Pre) a kj b kj c kj d kj 5. If the enthalpy change for the transition of liquid water to steam is 30 kj mol at 7 0 C, the entropy change for the process would be: (0 Pre) a. 00 J mol K b. 0 J mol K c..0 J mol K d. 0. J mol K 6. Which of the following is correct option for free expansion of an ideal gas under adiabatic condition? (0 Pre) a. q = 0, ΔT < 0, w 0 b. q = 0, ΔT 0, w = 0 c. q 0, ΔT = 0, w 0 d. q = 0, ΔT = 0, w = 0 7. The following two reactions are known: Fe O 3 (s) + 3CO Fe(s) + 3CO ; Δ = 6.8 kj FeO(s) + CO Fe(s) + CO ; Δ = 6.5 kj. The value of Δ for the following reaction Fe O 3 (s) + CO FeO(s) + CO is: (00 Mains) a kj b kj c. 0.3 kj d. +6. kj 8. For vaporization of water at atmospheric pressure, the values of Δ and ΔS are kj mol and 08.8 JK mol, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is: (00 Mains) a K b K c K d K 9. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be: (00 Mains) a. Infinite b. 3 Joules c. 9 Joules d. Zero 0. Match List-I (Equations) with List-II (Type of process) and select the correct option: (00 Mains) List-I Equations List-II Type of processes A. K p > Q (i) Non-spontaneous B. G < RT ln Q (ii) Equilibrium C. K p = Q Spontaneous and (iii) endothermic D. Spontaneous T = (iv) S a. A(i) B(ii) C(iii) D(iv) b. A(iii) B(iv) C(ii) D(i) c. A(iv) B(i) C(ii) D(iii) d. A(ii) B(i) C(iv) D(iii). For an endothermic reaction, energy of activation is E a and enthalpy of reaction is Δ (both of these in kj/mol). Minimum value of E a will be: (00 Pre) a. Equal to zero b. Less than Δ c. Equal to Δ d. More than Δ. Standard entropies of X, Y and XY 3 are 60, 40 and 50 JK mol respectively. For the reaction 3 X + Y XY 3, = 30kJ to be at equilibrium, the temperature should be: (00 Pre) a. 500 K b. 750 K c. 000 K d. 50 K 3. The pressure exerted by 6.0 g of methane gas in a 0.03 m 3 vessel at 9 C is: (Atomic masses : C =.0, =.0 and R = 8.34 JK mol ): (00 Mains) a. 56 Pa b Pa c Pa d Pa

4 Thermodynamics 4. From the following bond energies: bond energy: kj mol C = C bond energy: kj mol C C bond energy: kj mol C bond energy: kj mol Enthalpy for the reaction, C C + - will be: (009) a kjmol b. 0.0 kjmol c kjmol d kjmol 5. The values of Δ and ΔS for the reaction, C (graphite) + CO CO are 70 kj and 70 JK, respectively. This reaction will be spontaneous at: (009) C a. 90 K b. 000 K c. 50 K d. 70 K 6. For the gas phase reaction, PCl5( g) PCl3( g) + Cl( g), Which of the following conditions is correct? (008) a. > 0 and S < 0 b. = 0 and S < 0 c. > 0 and S > 0 d. < 0 and S < 0 7. Bond dissociation enthalpy of, Cl and Cl are 434, 4 and 43 kjmol respectively. Enthalpy of formation of Cl is: (008) a. 45 kjmol b. 93 kjmol c. 45 kjmol d. 93 kjmol 8. Which of the following are not state functions? (008) (I) q + w (II) q (III) w (IV) -TS a. (II) and (III) b. (I) and (IV) c. (II), (III) and (IV) d. (I), (II) and (III) 9. Given that bond energies of and Cl Cl are 430 kj mol and 40 kj mol respectively and f for Cl is 90 kj mol, bond enthalpy of Cl is: (007) a. 380 kj mol b. 45 kj mol c. 45 kj mol d. 90 kj mol C 30. Consider the following reactions: (i) + (aq) + O (aq) = O (l) = X kj mol (ii) + O = O(l), = X kj mol (iii) CO + = CO + O, = X 3 kj mol (iv) C ( g) O ( g) 5 + =CO + O (l), = X 3 kj mol Enthalpy of formation of O (l) is: (007) a. + X 3 kj mol b. X 4 kj mol c. + X kj mol d. X kj mol 3. The enthalpy and entropy change for the reaction: Br (l) + Cl BrCl are 30 kjmol and 05 JK mol respectively. The temperature at which the reaction will be in equilibrium is: (006) a K b. 373 K c. 50 K d. 400 K 3. Identify the correct statement for change of Gibb s energy for a system (ΔG system ) at constant temperature and pressure. (006) a. If ΔG system < 0, the process is not spontaneous b. If ΔG system > 0, the process is spontaneous c. If ΔG system = 0, the system has attained equilibrium d. If ΔG system = 0, then system is still moving in a particular direction 33. Assume each reaction is carried out in a open container. For which reaction will Δ = ΔE? (006) a. CO + O CO b. + Br Br c. C (s) + O + CO d. PCl 5 PCl 3 Cl 34. The enthalpy of hydrogenation of cyclohexene is 9.5 kj mol. If resonance energy of benzene is 50.4 kj mol, its enthalpy of hydrogenation would be: (006) a kj mol b kj mol c. 08. kj mol d kj mol

5 NEET 3 Years 35. The absolute enthalpy of neutralisation of the 4. For the reaction: reaction: MgO(s) + Cl(aq) MgCl (aq) + O(l) will be: (005) a. Less than kj mol b kj mol c. Greater than kj mol l d kj mol 36. Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? (005) a. Exothermic and decreasing disorder b. Endothermic and increasing disorder c. Exothermic and increasing disorder d. Endothermic and decreasing disorder 37. A reaction occurs spontaneously if: (005) a. TΔS < Δ and both Δ and ΔS are +ve b. TΔS > Δ and both Δ and ΔS are +ve c. TΔS = Δ and both Δ and ΔS are +ve d. TΔS > Δ and Δ is + ve and ΔS is ve 38. If the bond energies of, Br Br and Br are 433,9 and 364 kj mol respectively, the 0 for the reaction (004) + Br Br is:- a kj b. + 6 kj c. 03 kj d. 6 kj 39. Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is:- (004) a. S system S surroundings > 0 b. S System > 0 only c. S surroundings > 0 only d. S system + S surroundings > Standard enthalpy and standard entropy changes for the oxidation of ammonia at 98 K are kj mol and 45.6 JK mol, respectively. Standard Gibbs energy change for the same reaction at 98 K is:- (004) a kj mol b kjmol c. 53. kj mol d.. kjmol 4. The work done during the expansion of a gas from a volume of 4 dm 3 to 6 dm 3 against a constant external pressure of 3 atm is (004) a. 608 J b J c. 304 J d. 6 J C O 3CO + 4 O(l) at constant temperature, E is :- (003) a. +RT b. 3RT c. +3RT d. RT 43. The densities of graphite and diamond at 98 K are.5 and 3.3 g cm 3, respectively. If the standard free energy difference ( G ) is equal to 895 J mol, the pressure at which graphite will be transformed into diamond at 98 K is: (003) a Pa b Pa c Pa d Pa 44. What is the entropy change (in J K mol ) when one mole of ice is converted into water at 0 C? (The enthalpy change for the conversion of ice to liquid water is 6.0 KJ mol at 0 C): (003) a. 0.3 b..03 c..98 d Formation of a solution from two components can be considered as:- (i) Pure solvent separated solvent molecules, (ii) Pure solvent separated solvent molecules, (iii) Separated solvent and solute molecules solution, 3 Solution so formed will be ideal if: (003) a. soln = b. soln = + 3 c. soln = 3 d. soln = For which one of the following equations is react equal to p for the product: (003) a. N + O 3 N O 3 b. C 4 + Cl C Cl (l) + Cl c. Xe + F XeF 4 d. CO + O CO 47. The molar heat capacity of water at constant pressure, C, is 75 JK mol. When.0 KJ of heat is supplied to 00 g of water which is free to expand, the increase in temperature of water is: (003) a.. K b..4 K c. 4.8 K d. 6.6 K 48. Unit of entropy is: (00) a. JK mol b. J mol c. J K mol d. JK mol

6 Thermodynamics 49. In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true? (00) a. E = W 0, q = 0 b. E = W = q 0 c. E = 0, W = q 0 d. W = 0 E = q mole of ideal gas at 7 C temperature is expanded reversibly from litre to 0 litre Find entropy change (R = cal/mol K): (00) a. 9. b. 0 c. 4 d eat of combustion 0 for C(s), and C 4 are 94, 68 and 3 Kcal/mol, then 0 for C (s) + C 4 is: (00) a. 7 Kcal b. Kcal c. 70 Kcal d. 85 Kcal 5. Which reaction is not feasible? (00) a. KI + Br KBr + I b. KBr + I KI + Br c. KBr + Cl KCl + Br d. O + F 4F + O 53. Change in enthalpy for reaction O (l) O (l) + O If heat of formation of O (l) and O (l) are 88 & 86 KJ/mol respectively: (00) a. 96 KJ/mol b KJ/mol c KJ/mol d. 948 KJ/mol 54. When mol gas is heated at constant volume, temperature is raised from 98 to 308 K. eat supplied to the gas is 500 J. Then which statement is correct? (00) a. q = w = 500J, U = 0 b. q = U = 500J, w = 0 c. q = w = 500 J, AU = 0 d. U = 0, q = w = 500 J 55. Enthalpy of C4 + O C3O is negative. If enthalpy of combustion of C 4 and C 3 O are x and y respectively. Then which relation is correct? (00) a. x > y b. x < y c. x = y d. x y 56. PbO PbO ; G 98 < 0 SnO SnO ; G 98 > 0 Most probable oxidation state of Pb & Sn will be: (00) a. Pb +4, Sn +4 b. Pb +4, Sn + c. Pb +, Sn + d. Pb +, Sn For the disproportionation of copper Cu + Cu + + Cu, E 0 is:- (Given E 0 for Cu + /Cu is 0.34 V & E 0 for Cu + /Cu + is 0.5 V) (000) a V b V c V d V 58. Cell reaction is spontaneous when: (000) a. ΔG 0 is negative b. ΔG 0 is positive c. ΔE 0 is positive Red d. ΔE0 is negative red 59. Zn + O ZnO ; ΔG 0 = - 66 J Zn + S ZnS ; ΔG 0 = - 93 J S + O SO ; ΔG 0 = J ΔG 0 for the following reaction is: ZnS + 3O ZnO + SO (000) a J b J c J d J 60. At 7 C, latent heat of fusion of a compound is\ 930 J/mol. Entropy change is: (000) a J/mol-K b J/mol-K c J/mol-K d J/mol-K 6. For the reaction C 5 0(l) + 3O CO + 3 O(l) which one is true? (000) a. Δ = ΔE RT b. Δ = ΔE + RT c. Δ = ΔE + RT d. Δ = ΔE RT 6. In an endothermic reaction, the value of is: (999) a. Zero b. Positive c. Negative d. Constant 63. In the reaction 3 S( s) + O( g) SO3( g) + x kcal and SO( g) + O( g) SO3( s) + y kcal the heat of formation of SO is: (999) a. (x + y) b. (x - y) c. (x + y) d. (x - y) 64. At 5 o C and 730 mm pressure, 380 ml of dry oxygen was collected. If the temperature is constant, what volume will the oxygen occupy at 760 mm pressure? (999) a. 365 ml b. g c. 0g d. 0g 3

7 4 NEET 3 Years 65. Identify the correct statement regarding entropy: (998) a. At absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero b. At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve c. At absolute zero of temperature, the entropy of all crystalline substance is to be zero d. At 0 o C, the entropy of a perfectly crystalline substance is taken to be zero 66. One mole of an ideal gas at 300K is expanded isothermally from an initial volume of L to 0L. The E for this process is (R= cal mol K ): (998) a cal b. Zero c. 38. cal d. 9 L atm 67. Given that C + O CO, 0 = x kj CO + O CO 0 = y kj The enthalpy of formation of carbon monoxide will be: (997) y x a. b. x y c. y x d. x y 68. Which of the following is the correct equation? (996) a. U = W + Q b. U = Q W c. W = U + Q d. None of these 69. If enthalpies of formation for C 4, CO and O (l) at 5 0 C and atm pressure are 5, 394 and 86 kj/mol respectively, then enthalpy of combustion of C 4 will be: (995) a kj/mol b. + 4 kj/mol c. 4. kj/mol d. 4 kj/mol 70. A chemical reaction is catalyzed by a catalyst X. ence X (995) a. Reduces enthalpy of the reaction b. Does not affect equilibrium constant of reaction c. Decreases rate constant of the reaction d. Increases activation energy of the reaction. 7. Standard state Gibb s free energy change for isomerization reaction cis--pentene trans- -pentene is 3.67 kj/mol at 400 K. If more trans- -pentene is added to the reaction vessel, then: (995) a. Equilibrium remains unaffected b. Equilibrium is shifted in the forward direction c. More cis--pentene is formed d. Additional trans--pentene is formed. 7. For a reaction to occur spontaneously: (995) a. must be negative b. S must be negative c. ( T S) must be negative d. ( + T S) must be negative. 73. The average kinetic energy of an ideal gas, per molecule in S.I. units, at 5 0 C will be: (995) a J b J c J d J 74. During isothermal expansion of an ideal gas, its: (99, 94) a. Internal energy increases b. Enthalpy decreases c. Enthalpy remains unaffected d. Enthalpy reduces to zero. 75. Following reaction occurring in an automobile C O 6CO + 8 O. The sign of, S and G would be: (994) a., +, + b. +, +, c. +,, + Wd., +, 76. For the reaction N + 3 N 3, =? (99) a. E + RT b. E RT c. = RT d. E RT 77. If is the change in enthalpy and E, the change in internal energy accompanying a gaseous reaction, then: (990) a. is always greater than E b. < E only if the number of moles of the products is greater than the number of moles of the reactants c. is always less than E d. < E only if the number of moles of products is less than the number of moles of the reactants.

8 Thermodynamics 5 Answer Key d c c b d d b a a b d a d d a d d c d c d b c b b c d a b d a c b c b b b c d a a b none d a c b a b d a b a b b d c a a a a b d a c b a b d b c c d c d b d EXPLANATIONS. (d) NCERT (XI) Ch - 6, Pg. 60 ΔU = q + W For adiabatic process, q = 0 ΔU = W = P.ΔV =.5 atm ( )L =.5 L-atm J = J 505 J. (c) NCERT (XI) Ch - 6, Pg. 78 ΔG = Δ TΔS For spontaneous reaction, ΔG must be negative. For negative value of ΔG, Δ should be greater than TΔS. It is possible when T > 45 K. Δ = 35.5 KJ/mol = TΔS = (45) (83.6) = TΔS > Δ ence, the reaction is spontaneous 3. (c) NCERT (XI) Ch - 6, Pg. 75 The molecule having more number of bonds have largest value of entropy C 6 have large value of entropy. 4. (b) NCERT (XI) Ch - 6, Pg. 78 Under isothermal reversible conditions, the term free-energy in thermodynamics signifies that No expansion work done by the system. ΔG (system) = W non expansion 5. (d) NCERT (XI) Ch 6 Pg. 78 Pi S = nrtln P f For isothermal process T i = T f = 0 Pi S =nrln P 6. (d) NCERT (XI) Ch - 6, Pg. 78 f For any spontaneous process, G = ( ve) & S = (+ve) that is increase in entropy So, at < 0 & S > 0 at all temperatures according the reaction will be spontaneous. 7. (b) NCERT (XI) Ch - 6 dlnp v = (this is clausius clapeyron equation.) dt RT 8. (a) NCERT (XI) Ch - 6, Pg. 65 C (s) + O CO = KJ/mole when heat of combustion is given then r = R - P = 0 - (-393.5) = KJ/mole So heat of formation of CO = 393.5KJ/mole or we can say eat released on formation of 44 gm/o

9 6 NEET 3 Years = KJ = 75 KJ eat released on formation of 35. gm of 4. (d) NCERT (XI) Ch - 6, Pg CO = = KJ = KJ For dissociation enthalpy of - = = KJ 9. (a) NCERT (XI) Ch - 6, Pg (a) NCERT (XI) Ch - 6, Pg. 78 According to st law of law of thermodynamics According to 3 rd law of thermodynamics S = = U + n g RT T vap 30kJ = Svap = = = 00J mol k T 300 = 3.3kcal = (0) =.7kcal 6. (d) NCERT (XI) Ch 6 Pg. 60 According to 3 rd law of thermodynamics: For free expansion of ideal gas G = T S q = 0, T = 0 & w = 0 0. (b) NCERT (XI) Ch - 6, Pg. 76 As C changes from solid to gas, randomness increases. A significant decrease in Gibbs free energy will be observed. 7. (d) NCERT (XI) Ch - 6, Pg Fe O 3 (s) + 3CO Fe (s) + 3CO. (d) NCERT (XI) Ch - 6, Pg. 78 = 6.8 KJ --- () According to 3 rd law of thermodynamics: FeO (s) + CO Fe (s) + CO = 6.5 KJ --- () Ssyst. = T For of Fe O 3 + CO FeO(s) + CO S = = 5.6 cal / K mol. equation () () 73 = = +6. KJ. (a) NCERT (XI) Ch - 6, Pg (c) NCERT (XI) Ch - 6, Pg. 79 o Vap for water at 00 o C = KJ/mole Gibbs free energy for transformation = 0 At 00 o C / 373 K T = S O (l) O n g = T = = 373.4K 08.8 we know that = E + n g RT 9. (d) Work done in vacuum = 0 as P ext = 0. E = - n g RT 0. (c) NCERT (XI) Ch - 6, Pg E = ( J) - ( ) a (iv) b (i) c (ii) d (iii) E = 37599J/mole = KJ/mole K p > Q Reaction moves in forward direction. 3. (d) NCERT (XI) Ch - 6, Pg According to the data given [ G < RT lnq] G = (+ve) reaction becomes non spontaneous A B = +50 K p = Q Reaction attains equilibrium 3B C + D = 5 T> = = (+ve) endothermic reaction S E A D = +350 When < T S reaction is spontaneous. For B + D E + C = 50 + ( 5) 350. (d) NCERT (XI) Ch - 6, Pg. 5 For endothermic reaction, E a activation energy

10 Thermodynamics will be more than (difference between E f E b ), activation energy of forward reaction and backward.. (b) NCERT (XI) Ch - 6, Pg. 7, 78 S o = Σ S products Σ S Reactants = 50 ( ) S o = 40 J - mol - G o = o T S o, in equilibrium G o = 0 o 30 0 KJ mol T = = 750K o S 40 JK mol 3. (c) NCERT (XI) Ch - 6, Pg. 6 According to the ideal gas equation: PV = nrt P = nrt = = 4648 Pa V (b) NCERT (XI) Ch - 6, Pg. 7 For the given reaction, enthalphy of reaction can be calculated as = B.E. (reactant) B.E. (Product) = [B.E. (C C) + B.E. ( ) + 4 B.E. (C ) ] [B.E. (C C) + 6 B.E. (C ) ] = [ ] [ ] = = 0.0 kj mol 5. (b) NCERT (XI) Ch - 6, Pg. 78 For the reaction to be spontaneous, G = ve, Given, = 70 kj = J S = 70 JK mol Applying, G = T S, the value of G = ve only when T S >, which is possible only when T = 0 K. G = (0 70) = 8700J Thus, reaction is spontaneous at T = 0 K 6. (c) NCERT (XI) Ch - 6, Pg. 78 PCl g PCl g + Cl g ( ) ( ) ( ) 5 3 This is an endothermic reaction (+ve) Also the number of moles of product is greater than that of reactant, S > 0 spontaneity increases. 7. (d) NCERT (XI) Ch - 6, Pg According to the chemical equation: []+ [Cl] Cl for = 434 KJ for Cl = 4 KJ + Cl Cl, = 86 KJ 86 f of Cl = = 93KJ mol 8. (a) NCERT (XI) Ch - 6, Pg eat given to a system or by the system (Q) & work done (W) on the system or by the system are path functions, as they depend on the initial and final position. U internal energy = Q + W G = T S are state functions. 9. (b) NCERT (XI) Ch - 6, Pg. 7 Cl = + Cl = ΣB.E. (products) ΣB.E. (reactants) = ( ) ( 90) = = = 45 kj mol 30. (d) NCERT (XI) Ch - 6, Pg O = O(l) = X kj mol This reaction shows the formation of O, and the X represents the enthalpy of formation of O, according to the definition, it suggests that the enthalpy of formation is the heat evolved or absorbed when one mole of substance is formed from its constituent atoms. 3. (a) NCERT (XI) Ch - 6, Pg. 78 According to 3 rd Law of thermodynamics- S = T T = S J = = 85.7 K (c) Criteria for spontaneity of a process in terms of G is as following. If G = ( )ve Process is spontaneous If G = (+)ve Process does not occour in forward direction spontaniously but may occour spontaniously in backward direction. 7

11 8 NEET 3 Years If G = zero system is in equilibrium. 33. (b) = E + n g RT ΔS sys + ΔS surr > (a) ΔG = Δ TΔS if n g = zero then = E n g for reaction, + B Br is zero i.e. 000 = E = = 339.3KJ mol 34. (c) = 9.5 KJ/mole 4. (a) Work = P ext Volume change = (6 4) = = 607.9J W = 608J Enthalpy of hydrogenation of benzene = (3 ) - (Resonance energy) = [3 (-9.5)] - [-50.4] = -08. KJ/mole 35. (b) MgO(s) + Cl(aq.) MgCl (aq.) + O (l) Enthalpy of neutralisation will be less than J mol L as per the experimental data. owever, question is incomplete. 36. (b) For endothermic reaction is (+ve) & increasing disorderness or randomness increases the spontaneity of the reaction G = T S (+ve) (+ve) = ( ve) negative So, G is ( ve) negative. 37. (b) For spontaneous reaction < 0 G = T S G = ( ve) for spontaneous reaction S = (+ve), = (+ve) T S > G = ( ve) 38. (c) + Br Br Br = 65 = 78 Energy absorbed Energy released Δ 0 = = 03 kj here Δ 0 = 03 kj ( ) represent released eat V 39. (d) S= Rn V 4. (b) C O 3CO + 4 O Δn g = 3 6 = 3 Δ = ΔE + PΔV or Δ ΔE = PΔV Δ ΔE = ΔnRT = 3RT 43. (None) ΔG = PΔV = work done (w) V = L = L ΔG = w = ( ) P 0.3J p 895 = = atm = pa No option is correct q 44. (d) rev 6000 S = = =.978 J/K T (a) eat of solution is defined as the amount of heat evolved or absorbed when one mole of the substance is dissolved in excess of the solvent. For hydrates salt & for salt which do not form hydrates, Δ is +ve & for anhydrous salt, Δ is ve 46. (c) For 0 = o( ) o( ) o( ) reaction F XeF4 F Xe + F F Enthalpies of formation & elementary substances Xe & F are taken as zero. 47. (b) Molar heat capacity = 75 JK mol 8g of water = mole = 75JK mol 75 g of water JK = 8

12 Thermodynamics 0 0 Q = m.c.δt = t = =.4 K 75 q 48. (a) Entropy (s) = T = JK - mol (b) q = ΔE + W As system is closed q = 0 and work is done on system W 0 Temperature and hence internal energy of the system increases. ΔE (d) Change of entropy dq ds = T T V S = CV In + R n...( ) T V T = T = 7 0 C = 300K putting value in eq - () ΔS = cal/mol-k. Entropy change for mol of gas = = 9. cal/k. 5. (a) (i) C (s) + O CO Δ i = - 68 k cal mol - (ii) + O O Δ ii = - 68 k cal mol - (iii) C 4 + O CO + O Δ iii = -3 kcalmol - (iv) C(s) + C 4 Δ iv =? Apply ess s law (i) + (ii) + (iii) = ( ) kcal -7 kcal 5. (b) KBr + I KI + Br As electronegative character decreases down the group, so option (b) is not feasible. 53. (a) Δ 0 f = Σ0 f (product) Σ0 f (reactant) for the given equation: O (l) O(l) + O Δ 0 = f Δ0 ( O) f Σ0 ( O ) f = (-86 KJ mol - ) - (-88) KJ mol - = -96 KJ mol 54. (b) we know that Δ = ΔE + ΔPV Δ = ΔE + PΔV + VΔP = 0 When ΔV = 0; W = 0; Δ = ΔE + PΔV As ΔE = q + W, ΔE = q In the present problem Δ = 500J Δ = ΔE = 500J, q = 500J, W = (b) C4 + O C3O Δ = x - y; given Δ = negative ence x - y < (d) The sign and magnitude of Gibbs energy is a criterion of spontaneity for a process. When ΔG > 0 or +ve, it means G product > G reactant, as ΔG = G product + G reactant the reaction will not take place spontaneously, i.e. the reaction should be spontaneous in reverse direction. SnO SnO; ΔG > 0 + 4(more favourable) + ΔG < 0 or ve, the reaction or change occurs spontaneously, PbO PbO; ΔG < (c) For the reaction Cu + Cu + + Cu The cathode is Cu + /Cu and anode is Cu + /Cu + Given, Cu + + e 0 Cu; E = 0.34 V...() Cu + + e Cu + 0 ; E = 0.5 V...() Cu + + e 0 Cu; E 3 =?...(3) Now G = nfe = 0.34 F G = 0.5 F, G = E F Again G G 0.68F = 0 = 0.5F E3 F 0 E3 = = 0.53V cell = cathode anode ( ) ( ) As, E E Cu / Cu E Cu / Cu = = 0.38V 58. (a) For a cell reaction to be spontaneous G o should be negative. As G o = nfe o, So the value will be negative only when E o is positive. n = number of electrons involved, F = Value of Faraday. 9

13 0 NEET 3 Years 59. (a) For the reactions, 'Zns Zn + S ; G o = + 93J...() Zn + O ZnO; G o = 66J...() S + O SO ; G o = 408J...(3) The G o for the reaction, ZnS + 3O ZnO + SO can be obtained by adding eq. (), () and (3) So, G o = = 73J Q (a) S = = = 9.77J / mol K T (a) = E + P V also PV = nrt (ideal gas equation) or P V = nrt n = change in number of gaseous moles = E + nrt = n g 3 = E RT 6. (b) For endothermic rxn E R < E P So, = E P E R = +ve 3 S s + O g SO3 g + x... SO3( s) SO( g) + O y...( ) On adding eq () & () we get 63. (d) ( ) ( ) ( ) ( ) ( ) + ( ) ( ) + ( ) S s O g SO g x y kcal. 64. (a) NCERT (XI) Ch - 6, Pg. 6 According to Boyle s law: P V = P V V = = = 365 ml 65. (c) At absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero. It is called third law of thermodynamics 66. (b) Isothermal means temperature is constant. At constant temperature, change in internal energy ( E) remain constant. So, E = (a) C (s) + O CO ; = x kj..(i) y CO( g) + O( g) CO ( g) ; = kj..(ii) By subtracting equation (ii) from (i) we get, y y x = x = kj 68. (b) This is the mathematical relation of first law of thermodynamics. ere U = Change in internal energy; Q = eat absorbed by the system and W = Work done by the system. 69. (d) C + 3O CO + O = products reac = ( 394) + ( 86) (5 + 0) = 4 kj/mol 70. (b) Since a catalyst affects equally on both forward and backward reactions, therefore it does not affect equilibrium constant of reaction. 7. (c) If more trans--pentene is added, then its concentration in right hand side will increase. But in order to maintain the constant K, concentration of cis--pentene will also increase. Therefore more cis--pentene will be formed. 7. (c) For a reaction to be spontaneous, G (Gibbs free energy change) must be negative. G = T S = change in enthalpy, S = change in entropy. 73. (d) NCERT (XI) Ch - 6 Temperature (T) = 5 0 C = 98 K. Therefore K.E. per molecule = 3RT n = = J 3 ( ) 74. (c) During isothermal expansion of an ideal gas, T = 0. Now we know = E + PV = E + (PV) = E + (nrt) = E + nr T; T = = (d) (i) The given reaction is a combustion reaction, therefore is less than 0. ence, is negative. (ii) Since there is increase in the number of moles, therefore S is positive (iii) Since reaction is spontaneous, therefore G is negative. 76. (b) n g = 4 =, = E RT 77. (d) If n p < n r ; n g = n p n r = ve. ence, < E.

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