1. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ∆U of the gas in joules will be: (Delhi-2017)a. +505 J b. 1136.25 J c. –500 J d. –505 J

2. For a given reaction, ΔH = 35.5 KJ mol-1 and ΔS = 83.6 JK–1 mol–1. The reaction is spontaneous at: (Assume that ΔH and ΔS do not vary with temperature) (Delhi-2017)a. T > 298 K b. T < 425 Kc. T > 425 K d. All temperatures

3. Of the following, the largest value of entropy at 250C and 1 atm is that of: (Gujarat-2017) a. CH4 b. H2 c. C2H6 d. C2H2

4. Under isothermal and reversible conditions, the term “free energy” in thermodynamics signifies:

(Gujarat-2017)a. Expansion work done on the systemb. Non-expansion work done by the systemc. Expansion work done by the systemd. Non expansion work done on the system

5. For a sample of perfect gas when its pressure is changed isothermally from Pi to Pf , the entropy change is given by: (2016 - II)

a. f

i

PS nRT lnP

∆ =

b. i

f

PS nRT lnP

∆ =

c.

f

i

PS nR lnP

∆ =

d. i

f

PS nR lnP

∆ =

6. The correct thermodynamic conditions for the spontaneous reaction at all temperatures is: (2016 - I)a. ΔH < 0 and ΔS < 0b. ΔH < 0 and ΔS = 0c. ΔH > 0 and ΔS < 0d. ΔH < 0 and ΔS > 0

7. Consider the following liquid-vapour equilibrium.

Liquid Vapour. Which of the following relations is correct? (2016-I)

a. v2 2

d1nP HdT T

−∆= b. v

2

d1nP HdT RT

−∆=

c. v2 2

d1nG HdT RT

∆= d. vd1nP H

dT RT−∆

=

8. The heat of combustion of carbon to CO2 is –393.5 kJ/mol. The heat released upon formation of 35.2 g of CO2 from carbon and oxygen gas is: (2015)a. + 315 kJ b. – 630 kJc. – 3.15 kJ d. – 315 kJ

9. For the reaction, X2O4(l) → 2XO2(g), ΔU = 2.1 kcal, ΔS = 20 cal K–1 at 300 K. Hence, ΔG is: (2014)a. –2.7 kcal b. 9.3 kcal c. –9.3 kcal d. 2.7 kcal

10. 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? (2012 Pre)

a. 12

C (graphite) + 12

O2 (g) → CO2 (g)

b. C (graphite) +

12

O2 (g) → CO(g)

c. CO (g) + 12

O2 (g) → CO2 (g)

d. Mg (s) +

12

O2 (g) → MgO (s)

Thermodynamics6

10 NEET 31 Years 11. The enthalpy of fusion of water is 1.435 kcal/mol.

The molar entropy change for the melting of ice at 0°C: (2012 Pre)a. 0.526 cal/mol K b. 10.52 cal /mol Kc. 21.04 cal/mol K d. 5.260 cal/mol K

12. Standard enthalpy of vapourisation ∆Hvap for water at 1000C is 40.66 kJ mol–1. The internal energy of vapourisation of water at 1000C (in kJ mol–1) is: (2012 Pre)a. +40.66 b. +37.56 c. –43.76 d. +43.76

13. Consider the following process

12

A → B ΔH(kJ/mol) = + 150

3B → 2C + D ΔH(kJ/mol) = – 125

E + A → 2D ΔH(kJ/mol) = + 350

For B + D → E + 2C, ΔH will be: (2011 Mains)

a. –325 kJ/mol b. 325 kJ/molc. 525 kJ/mol d. –175 kJ/mol

14. Enthalpy change for the reaction, 4H(g) → 2H2(g)is –869.6 kJ. The dissociation energy of H – H bond is: (2011 Pre)a. +217.4 kJ b. –434.8 kJc. –869.6 kJ d. +434.8 kJ

15. If the enthalpy change for the transition of liquid water to steam is 30 kJ mol–1 at 270C, the entropy change for the process would be: (2011 Pre)

a. 100 J mol–1 K–1 b. 10 J mol–1 K–1

c. 1.0 J mol–1 K–1 d. 0.1 J mol–1 K–1

16. Which of the following is correct option for free expansion of an ideal gas under adiabatic condition? (2011 Pre)a. q = 0, ΔT < 0, w ≠ 0 b. q = 0, ΔT ≠ 0, w = 0c. q ≠ 0, ΔT = 0, w ≠ 0 d. q = 0, ΔT = 0, w = 0

17. The following two reactions are known:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ΔH = – 26.8 kJ FeO(s) + CO(g) → Fe(s) + CO2(g); ΔH = – 16.5 kJ. The value of ΔH for the following reaction Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g) is: (2010 Mains)a. +10.3 kJ b. – 43.3 kJc. –10.3 kJ d. +6.2 kJ

18. For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol–1 and 108.8 JK–1mol–1, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is: (2010 Mains)a. 273.4 K b. 393.4 K c. 373.4 K d. 293.4 K

19. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be: (2010 Mains)a. Infinite b. 3 Joules c. 9 Joules d. Zero

20. Match List-I (Equations) with List-II (Type of process) and select the correct option: (2010 Mains)

List-I Equations List-II Type of processes

A. Kp> Q (i) Non-spontaneousB. ∆G° < RT ln Q (ii) Equilibrium

C. Kp = Q (iii) Spontaneous and endothermic

D. T = HS

∆∆

(iv)Spontaneous

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)

21. For an endothermic reaction, energy of activation is Ea and enthalpy of reaction is ΔH (both of these in kJ/mol). Minimum value of Ea will be: (2010 Pre)a. Equal to zero b. Less than ΔHc. Equal to ΔH d. More than ΔH

22. Standard entropies of X2, Y2 and XY3 are 60, 40 and 50 JK–1mol–1 respectively. For the reaction

22 3

1 3X Y XY , H 30kJ2 2

+ ∆ =

to be at

equilibrium, the temperature should be: (2010 Pre)a. 500 K b. 750 K c. 1000 K d. 1250 K

23. The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129°C is: (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 JK–1 mol–1): (2010 Mains)a. 215216 Pa b. 13409 Pac. 41648 Pa d. 31684 Pa

11Thermodynamics

24. From the following bond energies:

H − H bond energy: 431.37 kJ mol–1

C = C bond energy: 606.10 kJ mol–1

C − C bond energy: 336.49 kJ mol–1

C − H bond energy: 410.50 kJ mol–1

Enthalpy for the reaction,

C

H

H

C + H - H

H

H

C

H

H

H C

H

H

H

will be: (2009)a. –243.6 kJmol–1 b. –120.0 kJmol–1

c. –553.0 kJmol–1 d. –1523.6 kJmol–1

25. The values of ΔH and ΔS for the reaction, C(graphite) + CO2 (g) → 2CO(g) are 170 kJ and 170 JK–1, respectively. This reaction will be spontaneous at: (2009)

a. 910 K b. 1000 K c. 510 K d. 710 K

26. For the gas phase reaction, ( ) ( ) ( )5 3 2PCl g PCl g Cl g+ , Which of the

following conditions is correct? (2008)a. ∆H > 0 and ∆S < 0b. ∆H = 0 and ∆S < 0c. ∆H > 0 and ∆S > 0d. ∆H < 0 and ∆S < 0

27. Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJmol–1 respectively. Enthalpy of formation of HCl is: (2008)a. 245 kJmol–1 b. 93 kJmol–1

c. –245 kJmol–1 d. –93 kJmol–1

28. Which of the following are not state functions? (2008)

(I) q + w (II) q

(III) w (IV) H-TSa. (II) and (III) b. (I) and (IV)c. (II), (III) and (IV) d. (I), (II) and (III)

29. Given that bond energies of H – H and Cl – Cl are 430 kJ mol–1 and 240 kJ mol–1 respectively and ∆Hf for HCl is – 90 kJ mol–1, bond enthalpy of HCl is: (2007)a. 380 kJ mol–1 b. 425 kJ mol–1

c. 245 kJ mol–1 d. 290 kJ mol–1

30. Consider the following reactions:

(i) H+(aq) + OH–

(aq) = H2O(l)

∆H = – X1 kJ mol–1

(ii) H2(g) +

12

O2(g) = H2O(l),

∆H = – X2 kJ mol–1

(iii) CO2(g) + H2(g) = CO(g) + H2O,

∆H = – X3 kJ mol–1

(iv) ( ) ( )2 2 25C H g O g2

+ =2CO2 (g) + H2O (l),

∆H = – X3 kJ mol–1

Enthalpy of formation of H2O (l) is: (2007)a. + X3 kJ mol–1 b. – X4 kJ mol–1

c. + X1 kJ mol–1 d. – X2 kJ mol–1

31. The enthalpy and entropy change for the reaction: Br2(l) + Cl2(g) → 2BrCl(g) are 30 kJmol–1 and 105 JK–1 mol–1 respectively. The temperature at which the reaction will be in equilibrium is: (2006)a. 285.7 K b. 373 K c. 250 K d. 400 K

32. Identify the correct statement for change of Gibb’s energy for a system (ΔGsystem) at constant temperature and pressure. (2006)a. If ΔGsystem < 0, the process is not spontaneousb. If ΔGsystem > 0, the process is spontaneousc. If ΔGsystem = 0, the system has attained equilibriumd. If ΔGsystem = 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 ΔH = ΔE? (2006)a. 2CO(g) + O2(g) → 2CO2(g)

b. H2(g) + Br2(g) → 2HBr(g)

c. C(s) + 2H2O(g) → 2H2(g) + CO2(g)

d. PCl5(g) → PCl3(g) → Cl2(g)

34. The enthalpy of hydrogenation of cyclohexene is –119.5 kJ mol–1. If resonance energy of benzene is –150.4 kJ mol–1, its enthalpy of hydrogenation would be: (2006)a. – 358.5 kJ mol–1 b. – 508.9 kJ mol–1

c. – 208.1 ∴kJ mol–1 d. – 269.9 kJ mol–1

12 NEET 31 Years 35. The absolute enthalpy of neutralisation of the

reaction:

MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) will be: (2005)a. Less than - 57.33 kJ mol–1

b. –57.33 kJ mol–1

c. Greater than –57.33 kJ mol–l

d. 57.33 kJ mol–1

36. Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? (2005)a. Exothermic and decreasing disorderb. Endothermic and increasing disorderc. Exothermic and increasing disorderd. Endothermic and decreasing disorder

37. A reaction occurs spontaneously if: (2005)

a. TΔS < ΔH and both ΔH and ΔS are +veb. TΔS > ΔH and both ΔH and ΔS are +vec. TΔS = ΔH and both ΔH and ΔS are +ved. TΔS > ΔH and ΔH is + ve and ΔS is –ve

38. If the bond energies of H – H, Br – Br and H – Br are 433,192 and 364 kJ mol–1 respectively, the ∆H0 for the reaction (2004)

H2(g) + Br2(g) → 2HBr(g) is:-a. + 103 kJ b. + 261 kJc. – 103 kJ d. – 261 kJ

39. Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is:- (2004)a. ∆Ssystem – ∆Ssurroundings > 0

b. ∆SSystem > 0 only

c. ∆Ssurroundings > 0 only

d. ∆Ssystem + ∆Ssurroundings > 0

40. Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are – 382.64 kJ mol–1 and –145.6 JK–1 mol–1, respectively. Standard Gibbs energy change for the same reaction at 298 K is:- (2004)a. – 339.3 kJ mol–1 b. – 439.3 kJmol–1

c. – 523.2 kJ mol–1 d. – 221.1 kJmol–1

41. The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (2004)a. – 608 J b. + 304 J c. – 304 J d. – 6 J

42. For the reaction: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) at constant

temperature, ∆H – ∆E is :- (2003)a. +RT b. –3RT c. +3RT d. –RT

43. The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm–3, respectively. If the standard free energy difference (∆G°) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is: (2003)a. 9.92 × 108 Pa b. 9.92 × 107 Pac. 9.92 × 106 Pa d. 9.92 × 105 Pa

44. What is the entropy change (in J K–1 mol–1) 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–1 at 0°C): (2003)a. 20.13 b. 2.013 c. 2.198 d. 21.98

45. Formation of a solution from two components can be considered as:-

(i) Pure solvent → separated solvent molecules, ∆H1

(ii) Pure solvent → separated solvent molecules, ∆H2

(iii) Separated solvent and solute molecules solution, ∆H3

Solution so formed will be ideal if: (2003)a. ∆Hsoln = ∆H1 + ∆H2 + ∆H3b. ∆Hsoln = ∆H1 + ∆H2 – ∆H3c. ∆Hsoln = ∆H1 – ∆H2 – ∆H3d. ∆Hsoln = ∆H3 – ∆H1 – ∆H2

46. For which one of the following equations is ∆H°react equal to ∆H°p for the product: (2003)a. N2(g) + O3(g) → N2O3(g)b. CH4(g) + 2Cl2(g) → CH2Cl2(l) + 2HCl(g) c. Xe(g) + 2F2(g) → XeF4(g)d. 2CO(g) + O2(g) → 2CO2(g)

47. The molar heat capacity of water at constant pressure, C, is 75 JK–1 mol–1. When 1.0 KJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is: (2003)a. 1.2 K b. 2.4 K c. 4.8 K d. 6.6 K

48. Unit of entropy is: (2002)a. JK–1 mol–1 b. J mol–1 c. J–1K–1 mol–1 d. JK mol–1

13Thermodynamics

49. In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true? (2002)a. ∆E = W ≠ 0, q = 0b. ∆E = W = q ≠ 0c. ∆E = 0, W = q ≠ 0 d. W = 0 ∆E = q ≠ 0

50. 2 mole of ideal gas at 27°C temperature is expanded reversibly from 2 litre to 20 litre Find entropy change (R = 2 cal/mol K): (2002) a. 92.1 b. 0 c. 4 d. 9.2

51. Heat of combustion ∆H0 for C(s), H2(g) and CH4(g) are – 94, – 68 and – 213 Kcal/mol, then ∆H0 for C (s) + 2H2 (g) → CH4 (g) is: (2002)a. – 17 Kcal b. – 111 Kcalc. – 170 Kcal d. – 85 Kcal

52. Which reaction is not feasible? (2002)a. 2KI + Br2 → 2KBr + I2b. 2 KBr + I2 → 2KI + Br2c. 2 KBr + Cl2 → 2KCl + Br2d. 2H2O + 2F2 → 4HF + O2

53. Change in enthalpy for reaction 2H2O2 (l) →2H2O (l) + O2(g) If heat of formation of H2O2(l) and H2O (l) are –188

& – 286 KJ/mol respectively: (2001)a. – 196 KJ/mol b. + 196 KJ/molc. + 948 KJ/mol d. – 948 KJ/mol

54. When 1 mol gas is heated at constant volume, temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct? (2001)a. q = w = 500J, ∆U = 0b. q = ∆U = 500J, w = 0c. q = w = 500 J, AU = 0d. ∆U = 0, q = w = – 500 J

55. Enthalpy of 4 2 31CH O CH OH2

+ → is negative. If

enthalpy of combustion of CH4 and CH3OH are x and y respectively. Then which relation is correct? (2001)a. x > y b. x < y c. x = y d. x ≥ y

56. PbO2 → PbO ; ∆G298 < 0 SnO2 → SnO ; ∆G298 > 0

Most probable oxidation state of Pb & Sn will be: (2001)

a. Pb+4, Sn+4 b. Pb+4, Sn+2

c. Pb+2, Sn+2 d. Pb+2, Sn+4

57. For the disproportionation of copper 2Cu+ → Cu+2 + Cu, E0 is:- (Given E0 for Cu+2/Cu is 0.34 V & E0 for Cu+2/Cu+ is 0.15 V) (2000)a. 0.49 V b. - 0.19 V c. 0.38 V d. - 0.38 V

58. Cell reaction is spontaneous when: (2000)a. ΔG0 is negative b. ΔG0 is positivec. ΔE0

Red is positive d. ΔE0red is negative

59. 2Zn + O2 → 2ZnO ; ΔG0 = - 616 J

2Zn + S2 → 2ZnS ; ΔG0 = - 293 J

S2 + 2O2 → 2SO2 ; ΔG0 = - 408 J

ΔG0 for the following reaction is:

2ZnS + 3O2 → 2ZnO + 2SO2 (2000)a. - 731 J b. - 1317 J c. - 501 J d. + 731 J

60. At 27° C, latent heat of fusion of a compound is\ 2930 J/mol. Entropy change is: (2000)a. 9.77 J/mol-K b. 10.77 J/mol-Kc. 9.07 J/mol-K d. 0.977 J/mol-K

61. For the reaction C2H50H(l) + 3O2(g) → 2CO2(g) + 3H2O(l) which one is true? (2000)a. ΔH = ΔE – RT b. ΔH = ΔE + RTc. ΔH = ΔE + 2RT d. ΔH = ΔE – 2RT

62. In an endothermic reaction, the value of ∆H is: (1999)a. Zero b. Positivec. Negative d. Constant

63. In the reaction

( ) ( ) ( )

( ) ( ) ( )

2 3

2 2 3

3S s O g SO g 2x kcal2

1and SO g O g SO s y kcal2

+ → +

+ → +

the heat of formation of SO2 is: (1999)a. (x + y) b. (x - y)c. (2x + y) d. (2x - y)

64. At 25oC 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? (1999)a. 365 ml b. 2g c. 10g d. 20g

14 NEET 31 Years 65. Identify the correct statement regarding entropy:

(1998)a. At absolute zero temperature, entropy of a

perfectly crystalline substance is taken to be zerob. At absolute zero of temperature, the entropy of a

perfectly crystalline substance is +vec. At absolute zero of temperature, the entropy of all

crystalline substance is to be zerod. At 0o 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 1L to 10L. The ∆E for this process is (R= 2 cal mol–1 K–1): (1998)a. 163.7 cal b. Zero c. 1381.1 cal d. 9 L atm

67. Given that C + O2 → CO2, ∆H0 = – x kJ

2CO + O2 → 2CO2 ∆H0 = – y kJ

The enthalpy of formation of carbon monoxide will be: (1997)

a. 22−y x b. 2x – y

c. y – 2x d. 22−x y

68. Which of the following is the correct equation? (1996)a. ∆U = ∆W + ∆Q b. ∆U = ∆Q – Wc. ∆W = ∆U + ∆Q d. None of these

69. If enthalpies of formation for C2H4 (g), CO2 (g) and H2O (l) at 250C and 1 atm pressure are 52, – 394 and – 286 kJ/mol respectively, then enthalpy of combustion of C2H4(g) will be: (1995)a. + 141.2 kJ/mol b. + 1412 kJ/molc. – 141.2 kJ/mol d. – 1412 kJ/mol

70. A chemical reaction is catalyzed by a catalyst X. Hence X (1995)a. Reduces enthalpy of the reactionb. Does not affect equilibrium constant of reactionc. Decreases rate constant of the reactiond. Increases activation energy of the reaction.

71. Standard state Gibb’s free energy change for isomerization reaction cis-2-pentene

trans- 2-pentene is –3.67 kJ/mol at 400 K. If more trans-

2-pentene is added to the reaction vessel, then: (1995)a. Equilibrium remains unaffectedb. Equilibrium is shifted in the forward directionc. More cis-2-pentene is formedd. Additional trans-2-pentene is formed.

72. For a reaction to occur spontaneously: (1995)a. ∆H must be negativeb. ∆S must be negativec. (∆H – T∆S) must be negatived. (∆H + T∆S) must be negative.

73. The average kinetic energy of an ideal gas, per molecule in S.I. units, at 250C will be: (1995)a. 6.17 × 10-20 J b. 7.16 × 10-20 Jc. 61.7 × 10-21 J d. 6.17 × 10-21 J

74. During isothermal expansion of an ideal gas, its: (1991, 94)a. Internal energy increasesb. Enthalpy decreasesc. Enthalpy remains unaffectedd. Enthalpy reduces to zero.

75. Following reaction occurring in an automobile 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g). The sign of ∆H, ∆S and ∆G would be: (1994)a. –, +, + b. +, +, – c. +, –, + Wd. –, +, –

76. For the reaction N2 + 3H2 2NH3, ∆H = ?

(1991)a. ∆E + 2RT b. ∆E – 2RTc. ∆H = RT d. ∆E – RT

77. If ∆H is the change in enthalpy and ∆E, the change in internal energy accompanying a gaseous reaction, then: (1990)

a. ∆H is always greater than ∆Eb. ∆H < ∆E only if the number of moles of the

products is greater than the number of moles of the reactants

c. ∆H is always less than ∆Ed. ∆H < ∆E only if the number of moles of products

is less than the number of moles of the reactants.

15Thermodynamics

1. (d) NCERT (XI) Ch - 6, Pg. 160

ΔU = q + WFor adiabatic process, q = 0∴ΔU = W ⇒ = – P.ΔV= –2.5 atm × (4.50 – 2.50 )L= – 2.5 × 2 L-atm ⇒ –5 × 101.3 J= – 506.5 J ⇒ –505 J

2. (c) NCERT (XI) Ch - 6, Pg. 178

ΔG = ΔH – TΔS

For spontaneous reaction, ΔG must be negative. For negative value of ΔG, ΔH should be greater than TΔS. It is possible when T > 425 K.

ΔH = 35.5 KJ/mol = 35500

TΔS = (425) (83.6) = 35530

TΔS > ΔH

Hence, the reaction is spontaneous

3. (c) NCERT (XI) Ch - 6, Pg. 175

The molecule having more number of bonds have largest value of entropy

⸫ C2H6 have large value of entropy.

4. (b) NCERT (XI) Ch - 6, Pg. 178

Under isothermal reversible conditions, the term “free-energy” in thermodynamics signifies that “No expansion work done by the system”.

–ΔG(system) = Wnon–expansion

5. (d) NCERT (XI) Ch – 6 Pg. 178

∆S = nRTln i

f

PP

For isothermal process Ti = Tf = 0

i

f

PSP

∴∆ = nRln

6. (d) NCERT (XI) Ch - 6, Pg. 178

For any spontaneous process, ∆G = (–ve) & ∆S = (+ve) that is increase in entropySo, at ∆H < 0 & ∆S > 0 at all temperatures according the reaction will be spontaneous.

7. (b) NCERT (XI) Ch - 6

v2

dlnP HdT RT

−∆= (this is clausius clapeyron equation.)

8. (a) NCERT (XI) Ch - 6, Pg. 165

C(s) + O2 → CO2(g) ∆H = -393.5 KJ/molewhen heat of combustion is given then ∆Hr = HR - HP

= 0 - (-393.5) = 393.5 KJ/moleSo heat of formation of CO2 = 393.5KJ/moleor we can say

Heat released on formation of 44 gm/O2

EXPLANATIONS

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17d c c b d d b a a b d a d d a d d

18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34c d c d b c b b c d a b d a c b c

35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51b b b c d a a b none d a c b a b d a52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68b a b b d c a a a a b d a c b a b

69 70 71 72 73 74 75 76 77d b c c d c d b d

Answer Key

16 NEET 31 Years = 393.5 KJHeat released on formation of 35.2 gm of

2393.5CO 35.2

44= ×

= 314.8 ⇒ 315KJ

9. (a) NCERT (XI) Ch - 6, Pg. 161

According to 1st law of law of thermodynamics

∆H = ∆U + ∆ngRT

∆H = 2.1 + 2 × 2 × 300

= 3.3kcal = 3.3 – 300(20) = –2.7kcalAccording to 3rd law of thermodynamics:

∆G = ∆H – T∆S

10. (b) NCERT (XI) Ch - 6, Pg. 176As C changes from solid to gas, randomness increases. A significant decrease in Gibbs free energy will be observed.

11. (d) NCERT (XI) Ch - 6, Pg. 178

According to 3rd law of thermodynamics:

syst.

3

HST

1.435 10S 5.26 cal / K mol.273

∆∆ =

×∆ = =

12. (a) NCERT (XI) Ch - 6, Pg. 166oVapH∆ for water at 100oC = 40.66 KJ/mole

At 100oC / 373 K

H2O(l) → H2O(g)

∆ng = 1

we know that ∆H = ∆E + ∆ngRT

∆E = ∆H - ∆ng RT

∆E = (40.66 × 103J) - (1 × 8.314 × 373)

∆E = 37599J/mole = 37.56 KJ/mole

13. (d) NCERT (XI) Ch - 6, Pg. 173 – 175

According to the data given – 12

A → B ∆H = +150

3B → 2C + D ∆H = –125

E – A → 2D ∆H = +350

For B + D → E + 2C

∆H = 150 × 2 + (–125) –350

= –175 KJ

14. (d) NCERT (XI) Ch - 6, Pg. 172

2H → 2H2 ∆H = 369.6 KJ

For dissociation enthalpy of H-H 869.6 434.82

= = KJ

15. (a) NCERT (XI) Ch - 6, Pg. 178According to 3rd law of thermodynamics

HS

T∆

∆ =

vap 1 1vap

H 30kJS 100J mol kT 300

− −∆∆ = = =

16. (d) NCERT (XI) Ch – 6 Pg. 160

For free expansion of ideal gas

q = 0, ΔT = 0 & w = 0

17. (d) NCERT (XI) Ch - 6, Pg. 171-173

Fe2O3(s) + 3CO (g) → 2Fe (s) + 3CO2(g)

ΔH = –26.8 KJ --- (1)

FeO (s) + CO (g) → Fe (s) + CO2 (g)

ΔH = –16.5 KJ --- (2)

For ΔH of Fe2O3(g) + CO(g) → 2FeO(s) + CO2(g)

equation (1) –2 × (2)

= –26.8 – 2 × 16.5 = +6.2 KJ

18. (c) NCERT (XI) Ch - 6, Pg. 179

Gibbs free energy for transformation = 0 HTS

∆=∆

340.63 10T 373.4K108.8

×∆ = =

19. (d) Work done in vacuum = 0 as Pext = 0.

20. (c) NCERT (XI) Ch - 6, Pg. 178–179

a → (iv) b → (i) c → (ii) d → (iii)

Kp > Q ← Reaction moves in forward direction.

[ΔG < RT lnQ] ΔG = (+ve) reaction becomes non – spontaneous

Kp = Q Reaction attains equilibrium

HT HS

∆> = ∆∆

= (+ve) ⇒ endothermic reaction

When ΔH < TΔS reaction is spontaneous.

21. (d) NCERT (XI) Ch - 6, Pg. 115

For endothermic reaction, Ea activation energy

17Thermodynamics

will be more than ∆H (difference between Ef – Eb), activation energy of forward reaction and backward.

22. (b) NCERT (XI) Ch - 6, Pg. 172, 178

∆So = Σ∆Sproducts – Σ∆SReactants

= 50 – (30 + 60)

∆So = –40 J-1 mol-1

∆Go = ∆Ho – T∆So, in equilibrium ∆Go = 0 o 1 1

o 1 1

H 30 10 KJ molT 750KS 40 JK mol

− −

− −

∆ − ×= = ⇒∆ − ×

23. (c) NCERT (XI) Ch - 6, Pg. 161

According to the ideal gas equation:

PV = nRT

=nRTPV

6 8.34 402 41648 Pa19 0.03

×= × =

24. (b) NCERT (XI) Ch - 6, Pg. 172

For the given reaction, enthalphy of reaction can be calculated as

= B.E. (reactant) – B.E. (Product)

= [B.E. (C – C) + B.E.(H – H) + 4 × B.E.(C – H)]

–[B.E. (C – C) + 6 × B.E.(C – H)]

= [606.10 + 431.37 + 4 × 410.50]

–[336.49 + 6 × 410.50]

= 2679.47 – 2799.49 = – 120.02 kJ mol–1

25. (b) NCERT (XI) Ch - 6, Pg. 178For the reaction to be spontaneous, ∆G = –ve, Given, ∆H = 170 kJ = 170 × 103J ∆S = 170 JK–1 mol–1

Applying, ∆G = ∆H – T∆S, the value of ∆G = –ve only when T∆S > ∆H, which is possible only when T = 1110 K.∴ ∆G = 170 × 103 – (1110 × 170) = – 18700J

Thus, reaction is spontaneous at T = 1110 K

26. (c) NCERT (XI) Ch - 6, Pg. 178

( ) ( ) ( )5 3 2PCl g PCl g Cl g+

This is an endothermic reaction ∆H (+ve)

Also the number of moles of product is greater than that of reactant, ∆S > 0 spontaneity increases.

27. (d) NCERT (XI) Ch - 6, Pg. 173-174

According to the chemical equation:

2[H]+ 2[Cl] → 2HCl

∆H for H2 = 434 KJ

∆H for Cl2 = 242 KJ

H2 + Cl2 → 2HCl, ∆H = – 186 KJ

1

f186H of HCl 93KJ mol2

−−∆ = = −

28. (a) NCERT (XI) Ch - 6, Pg. 156-157

Heat 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 = ∆H – T∆S are state functions.

29. (b) NCERT (XI) Ch - 6, Pg. 172

2 21 1HCl H Cl2 2

= +

∆H = ΣB.E.(products) – ΣB.E.(reactants)

( ) ( )1 1430 240 90 670 902 2

= + − − = × +

= 335 + 90 = 425 kJ mol–1

30. (d) NCERT (XI) Ch - 6, Pg. 167

H2(g) + 12

O2(g) = H2O(l) ∆H = –X2 kJ mol–1

This reaction shows the formation of H2O, and the X2 represents the enthalpy of formation of H2O, 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.

31. (a) NCERT (XI) Ch - 6, Pg. 178According to 3rd Law of thermodynamics-

HST

HTS

∆∆ =

∆=∆

330 10 J 285.7K105×

= =

32. (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.

18 NEET 31 YearsIf ∆G = zero ⇒ system is in equilibrium.

33. (b) ∆H = ∆E + ∆ng RT

if ∆ng = zero then ∆H = ∆E

∆ng for reaction, H2(g) + B2(g) → 2HBr(g) is zero i.e.

∆H = ∆E

34. (c) ∆H = 119.5 KJ/mole

Enthalpy of hydrogenation of benzene

= (3 × ∆H) - (Resonance energy)

= [3 × (-119.5)] - [-150.4]

= -208.1 KJ/mole

35. (b) MgO(s) + 2HCl(aq.) → MgCl2 (aq.) + H2O (l)

Enthalpy of neutralisation will be less than –57.33 J mol L–1 as per the experimental data. However, question is incomplete.

36. (b) For endothermic reaction ∆H is (+ve) & increasing disorderness or randomness increases the spontaneity of the reaction

∆G = ∆H – T∆S

(+ve) – (+ve) = (–ve) negative

So, ∆G is (–ve) negative.

37. (b) For spontaneous reaction ∆H < 0

∆G = ∆H – T∆S

∆G = (–ve) for spontaneous reaction

∆S = (+ve), ∆H = (+ve)

T∆S > ∆H ⇒ ∆G = (–ve)

38. (c) H – H + Br – Br → 2H – Br

433 + 192 2 × 364

= 625 = 728

Energy absorbed Energy released

ΔH0 = 728 – 625 = 103 kJ

here ΔH0 = –103 kJ

(–) represent released Heat

39. (d) 2

1

VS R nV

∆ =

ΔSsys + ΔSsurr > 0.

40. (a) ΔG = ΔH – TΔS

145.6382.64 2981000− − −

= –382.64 + 43.38

= –339.3KJ mol–1

41. (a) Work = –Pext × Volume change

= 3 × 101.32 × (6 – 4)

= 6 × 101.32

= –607.92J ⇒ W = –608J

42. (b) C3H8(g) + 5O2 → 3CO2(g) + 4H2O

Δng = 3 – 6 = –3

ΔH = ΔE + PΔV

or ΔH – ΔE = PΔV

ΔH – ΔE = ΔnRT = –3RT

43. (None) ΔG = –PΔV = work done (w)

312 12V 10 L3.31 2.25

− ∆ = − ×

= 1.71 × 10–3L

ΔG = w = – (–1.71 × 10–3) × P × 101.3J

3

1895p1.71 10 100.3−=

× ×

= 10.93 × 103 atm

= 11.07 × 108pa

No option is correct

44. (d) revq 6000S 21.978T 273

= = = J/K

45. (a) Heat 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, ΔH is +ve & for anhydrous salt, ΔH is –ve

46. (c) For H0reaction = ( ) ( ) ( )o o o

F 4 F F 2H XeF H Xe 2 H F ∆ − ∆ + ∆ Enthalpies of formation & elementary substances Xe & F2 are taken as zero.

47. (b) Molar heat capacity = 75 JK–1mol–1

18g of water = 1 mole = 75JK–1mol–1

1g of water 175 JK18

−=

19Thermodynamics

Q = m.c.Δt = 10 10t 2.4 K

75×

∆ = =

48. (a) Entropy q(s)T

=

= JK-1 mol-1

49. (b) q = ΔE + WAs system is closed q = 0 and work is done on system∴W ≠ 0Temperature and hence internal energy of the system increases.

ΔE ≠ 0

50. (d) Change of entropy

( )2 2V

1 1

dqdsTT VS C In R ....... 1T V

=

∆ = + n

T2 = T1 = 270C = 300K

putting value in eq - (1)

ΔS = 4.605 cal/mol-K.

Entropy change for 2 mol of gas

= 2 × 4.605 = 9.2 cal/K.

51. (a) (i) C (s) + O2 (g) → CO2 (g)

ΔHi = - 68 × 2 k cal mol-1

(ii) 2H2 (g) + O2 (g) → 2H2O

ΔHii = - 682 k cal mol-1

(iii) CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O

ΔHiii = -213 kcalmol-1

(iv) C(s) + 2H2(g) → CH4(g)

ΔHiv = ?

Apply Hess’s law

(i) + (ii) + (iii)

= (- 94 - 68 × 2 + 213) kcal ⇒ -17 kcal

52. (b) 2KBr + I2 → 2KI + Br2

As electronegative character decreases down the group, so option (b) is not feasible.

53. (a) ΔH0f = ΣH0

f(product) – ΣH0f (reactant)

for the given equation:

2H2O2(l) → 2H2O(l) + O2(g)

ΔH0f = 2 × ΔH0

f (H2O) 2 × ΣH0f (H2O2)

= 2 × (-286 KJ mol-1) -2 × (-188) KJ mol-1

= -196 KJ mol–1

54. (b) we know that ΔH = ΔE + ΔPV

ΔH = ΔE + PΔV + VΔP = 0

When ΔV = 0; W = 0; ΔH = ΔE + PΔV

As ΔE = q + W, ΔE = q

In the present problem

ΔH = 500J

ΔH = ΔE = 500J, q = 500J, W = 0

55. (b) 4 2 31CH O CH OH2

+ →

ΔH = x - y; given ΔH = negative

Hence x - y < 0

56. (d) The sign and magnitude of Gibbs energy is a criterion of spontaneity for a process. When ΔG > 0 or +ve, it means Gproduct > Greactant, as ΔG = Gproduct + Greactant the reaction will not take place spontaneously, i.e. the reaction should be spontaneous in reverse direction.

SnO2 → SnO; ΔG > 0

+ 4(more favourable) + 2

ΔG < 0 or –ve, the reaction or change occurs spontaneously, PbO2 → PbO; ΔG < 0

+4 +2

57. (c) For the reaction 2Cu+ → Cu2+ + Cu

The cathode is Cu+/Cu and anode is Cu+/Cu2+

Given, Cu2+ + 2e– → Cu; 01E = 0.34 V ......(1)

Cu2+ + e– → Cu+; 02E = 0.15 V .......(2)

Cu+ + e– → Cu; 03E = ? .......(3)

0 01 1

0 0 02 3 3

0 01 2

03

03

Now G nFE 2 0.34 FG 1 0.15 F, G 1 E F

Again G G 0.68F0.15F E F

E 0.68 0.15 0.53V

∆ = − = − × ×

∆ = − × × ∆ = − × ×

∆ = ∆ ⇒ −

= − − ×

⇒ = − =

( ) ( )0 0 0 2cell cathode anodeAs, E E Cu / Cu E Cu / Cu+ + += −

= 0.53 – 0.15 = 0.38V

58. (a) For a cell reaction to be spontaneous ∆Go should be negative. As ∆Go = –nFEo, So the value will be negative only when Eo is positive. n = number of electrons involved, F = Value of Faraday.

20 NEET 31 Years 59. (a) For the reactions,

2'Zns → 2Zn + S2; ∆Go = + 293J .....(1)2Zn + O2 → 2ZnO; ∆Go = – 616J .....(2)

S2 + 2O2 → 2SO2; ∆Go = – 408J .....(3)

The ∆Go for the reaction,

2ZnS + 3O2 → 2ZnO + 2SO2 can be obtained by adding eq. (1), (2) and (3)

So, ∆Go = 293 – 616 – 408 = – 731J

60. (a) Q 2930S 9.77J / mol KT 300

∆ = = =

61. (a) ∆H = ∆E + P∆Valso PV = nRT (ideal gas equation) or P∆V = ∆nRT∆n = change in number of gaseous moles ∴ ∆H = ∆E + ∆nRT = ∆ng 2–3 ⇒ –1 ∆H = ∆E – RT

62. (b) For endothermic rxn ER < EP

So, ∆H = EP – ER = +ve

63. (d) ( ) ( ) ( ) ( )2 33S s O g SO g 2x ..... 12

+ +

( ) ( ) ( )3 2 21SO s SO g O y ..... 22

+ −

On adding eq (1) & (2) we get ( ) ( ) ( ) ( )2 2S s O g SO g 2x y kcal.+ + −

64. (a) NCERT (XI) Ch - 6, Pg. 161According to Boyle’s law:P1V1 = P2V2

2

730 380V 365760×

= =

= 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 = 0

67. (a) C(s) + O2(g) → CO2(g); ∆H = x kJ …..(i)

( ) 2( ) 2( )1CO O CO ; kJ2 2

+ → ∆ = −g g gyH …..(ii)

By subtracting equation (ii) from (i) we get, 2 kJ

2 2− ∆ = − − − =

y y xH x

68. (b) This is the mathematical relation of first law of thermodynamics. Here ∆U = Change in internal energy; ∆Q = Heat absorbed by the system and W = Work done by the system.

69. (d) C2H2 + 3O2 → 2 CO2 + 2 H2O

∆H = ∆Hproducts – ∆Hreac

= 2 × (–394) + 2 × (–286) – (52 + 0)

= –1412 kJ/mol

70. (b) Since a catalyst affects equally on both forward and backward reactions, therefore it does not affect equilibrium constant of reaction.

71. (c) If more trans-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the constant K, concentration of cis-2-pentene will also increase. Therefore more cis-2-pentene will be formed.

72. (c) For a reaction to be spontaneous, ∆G (Gibbs free energy change) must be negative. ∆G = ∆H – T∆S ∆H = change in enthalpy, ∆S = change in entropy.

73. (d) NCERT (XI) Ch - 6Temperature (T) = 250C = 298 K.Therefore K.E. per molecule

( )23

3 3 8.314 2982 2 6.02 10

× ×= =

× ×RTn

= 6.17 × 10–21 J 74. (c) During isothermal expansion of an ideal gas, ∆T

= 0. Now we know H = E + PV ⸪ ∆H = ∆E + ∆(PV)

= ∆E + ∆ (nRT) ∴ ∆H = ∆E + nR∆T; ∆T = 0 + 0 = 0

75. (d) (i) The given reaction is a combustion reaction, therefore ∆H is less than 0. Hence, ∆H 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) ∆ng = 2 – 4 = –2, ∆H = ∆E – 2RT

77. (d) If np < nr; ∆ng = np – nr = –ve.Hence, ∆H < ∆E.