CLS Aipmt 14 15 XI Che Study Package 2 SET 2 Chapter 6

Chapter

6

Thermodynamics Solutions SECTION - A Objective Type Questions 1.

Tea placed in thermos flask is an example of (1) Open system

(2)

Close system

(3) Isolated system

(4)

It can't act as system

Sol. Answer (3) A thermos flask does not allow exchange of energy and matter. Hence, it is an isolated system. 2.

Gaseous system is placed with pressure P1, volume V1 and temperature T1, it has undergone thermodynamic changes where temperature is remaining constant, it is (1) Adiabatic process

(2)

Isothermal process

(3)

Isobaric process

(4)

Isochoric process

Sol. Answer (2) A system which undergoes change such that temperature remains constant. Such a change is called isothermal process. 3.

The respective examples of extensive and intensive properties are (1) Enthalpy, Entropy

(2)

Entropy, Enthalpy

(3) Entropy, Temperature

(4)

Temperature, Entropy

Sol. Answer (3) Entropy is an extensive property (mass dependent). Temperature is intensive since it is mass independent. 4.

A thermally isolated, gaseous system can exchange energy with the surroundings. The mode of energy may be (1) Heat

(2)

Work

(3)

Heat and radiation

(4)

Internal energy

Sol. Answer (2) Since the system is thermally isolated, energy can only be transferred through a non-thermal mode i.e. work. 5.

Which of the following is a state function? (1) q

(2)

w

(3)

q+w

(4)

All of these

Sol. Answer (3) q (heat) and w (work done) are both path functions. q + w = U, which is change in internal energy, which is a state function. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

44 6.

Thermodynamics

Solution of Assignment (Set-2)

For the reaction PCl5(g)  PCl3(g) + Cl2(g) (1) H = E

(2)

H > E

(3)

H < E

(4)

Can’t predicted

(4)

E = s

Sol. Answer (2) PCl5 (g)  PCl 3 (g)  Cl2 (g)

ng = 2 – 1 = 1  H = U + ng RT = U + RT or, H > U.

(∵ RT is positive)

where U = E i.e., change in internal energy. 7.

If ‘r’ is the work done on the system and ‘s’ is heat evolved by the system then, (1) E = r + s

(3)

E = r

(1) H – E = (b – d) RT

(2)

H – E = (c – b) RT

(3) H – E = (a + b) – (c + d) RT

(4)

H – E = (a – d) RT

(2)

E = r – s

Sol. Answer (2) According to 1st law of thermodynamics, U = q + w w = +r

(∵ work is done on the system)

q = –s

(∵ heat is evolved out of system)

 U = r – s 8.

For the reaction, aA(s) + bB(g)   dD(s) + cC(g). Then

Sol. Answer (2) For the reaction, ng = (c – b)

[∵ rest are solid substances]

We know, H = U + ng RT or, H = U + (c – b) RT or, H – U = (c – b) RT 9.

A system absorbs 10 kJ of heat and does 4 kJ of work. The internal energy of the system (1) Decreases by 6 kJ

(2)

Increases by 6 kJ

(3) Decreases by 14 kJ

(4)

Increases by 14 kJ

Sol. Answer (2) As per Ist law, U = q + w q = +10 kJ (heat is absorbed by the system) w = –4 kJ (work is done by the system)  U = q + w = 10 – 4 kJ = +6 kJ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

45

10. In a reaction, all reactant and products are liquid, then (1) H > E

(2)

H < E

(3)

H = E

(4)

Can't predicted

Sol. Answer (3) ∵ All reactants and products are liquid, hence, ng = 0 where, ng signifies change in moles of gaseous substances.  H = U + ng RT or, H = U 11. Regarding the internal energy of the molecule, which of the following statement is correct? (1) Its absolute value can be successfully calculated (2) Its absolute value cannot be determined (3) It is the sum of vibrational and rotational energies (4) Both (1) & (3) Sol. Answer (2) The absolute value of internal energy cannot be determined since it is the sum total of all the energies at a molecular level. These energies cannot be determined and hence the absolute value of U cannot be determined. 12. Consider the following reaction : C (graphite) + O2(g)CO2(g) ; H = – x1 cal C (diamond) + O2(g) CO2(g) ; H = – x2 cal What is the heat of transition of graphite into diamond? (1) x1 + x2

(2)

x2 – x1

(3)

x1 – x2

(4)

x1 x 2

(4)

H2 = (x1 + x) – x3

Sol. Answer (2) Required equation

C graphite   C diamond C graphite   O2  CO2  g ; H   x, cal CO2  C diamond  O2  g ; H   x2 cal Adding, we get C graphite   C diamond  H  x2 – x1 cal

 D ; H = x. Steps involved are 13. For the given reactions, A 

 B ; H1 = x1 A   C ; H2 = ? B   D ; H3 = x3 C  (1) H2 = x – (x1 + x3)

(2)

H2 = x + x1 + x3

(3)

H2 = x1 – x3 – x

Sol. Answer (1) Given reactions are We know,

A

H1

B

H2

C

H3

D

H = H1 + H2 + H3 or, H2 = H – H1 – H3 = x – (x1 + x3)

H

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46

Thermodynamics

Solution of Assignment (Set-2)

14. The heats of combustion of yellow P and red P are –9.91kJ and –8.78kJ respectively. The heat of transition of yellow to red phosphorus is (1) –18.69 kJ

(2)

+1.13 kJ

(3)

+18.69 kJ

(4)

–1.13 kJ

(3)

2x – z = y

(4)

2z + x = y

(3)

x=

y 2

(4)

x y

Sol. Answer (4)

P yellow   P red 

5 1 O2  g  P2O5  s  ; H1   9.91kJ 2 2

5 1 O2  g  P2O5  s  ; H2   8.78 kJ 2 2

Rearranging, we get P yellow  

5 O2  g  P2O5  g ; H   9.91kJ 2

P2O5  Pred 

5 O2  g ; H   8.78 kJ 2

Adding these two equations, we get

P yellow   Pred ;  9.91  8.78  kJ   1.13 kJ 15. If the heat of formation of NO2 is ‘x’ [½ N2(g) + O2(g)  NO2(g)] the heat of reaction N2(g) + O2(g)  2NO(g) is y and the heat of reaction 2NO(g) + O2(g)  2NO2(g) is z, then (1) 2x + z = y

(2)

2y + z = x

Sol. Answer (3) Given: ⎛1 ⎞ 2  ⎜ N2  g   O2  g   NO2  g  ; H  x ⎟ 2 ⎝ ⎠

= N2  g  2O2  2NO2  g ; H1  2x Also,

N2  O2  2NO  g ; H  y 2NO  O2  2NO2 ; H  z Adding, we get N2  2O2  2NO2 ; H2  y  z

∵ H1 = H2

(∵ H is state function)

 2x = y + z 16. In the reactions HCl + NaOH  NaCl + H2O + x cal. H2SO4 + 2NaOH  Na2SO4 + 2H2O + y cal. (1) x = y

(2)

x = 2y

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Solution of Assignment (Set-2)

Thermodynamics

47

Sol. Answer (3) HCl  NaOH  NaCl  H2 O  x cal

where x cal is the heat released due to neutralisation of 1 g equivalent of acid by 1 g equivalent base. In the 2nd reaction,

H2SO4  2NaOH  Na2SO4  2H2O  y cal Now, 2 g equivalents of strong acid reacts with 2 g equivalent of strong base, thus releasing double of energy released in the 1st reaction.  y = 2x 17. Hf C2H4 = 12.5 kcal Heat of atomisation of C = 171 kcal Bond energy of H2 = 104.3 kcal Bond energy C – H = 99.3 kcal What is C = C bond energy? (1) 140.7 kcal

(2)

49 kcal

(3)

40 kcal

(4)

76 kcal

Sol. Answer (1) Given:

2C graphite   2H2  g  C2H4  g ; Hf Hf = Bond dissociation enthalpy of reactants – Bond dissociation enthalpy of products = 12.5 = (171 × 2) + 2 × 104.3 – (4 × 99.3 + BEC = C) 

BE(C = C) = 140.9 kcal

18. The difference between H and E for the reaction 2C6H6 (l) +15O2(g)  12CO2(g) + 6H2O (l) at 25°C in kJ is (1) –7.43 kJ

(2)

+3.72 kJ

(3)

–3.72 kJ

(4)

+7.43 kJ

Sol. Answer (1) Given reaction:

2C6H6  l  15O2  g  12CO2  g  6 H2O  l ng = 12 – 15 = – 3 mol H = U + ng RT or, H – U = – 3 × 8.314 Jk–1 mol–1 × 298 K × mol = –7.432 kJ 19. S (rhombic) + O2 (g) SO2 (g); H = –297.5 kJ S (monoclinic) + O2 (g) SO2(g); H = –300 kJ The above data can predict that (1) Rhombic sulphur is yellow in colour (2) Monoclinic sulphur has metallic lustre (3) Monoclinic sulphur is more stable (4) H(Transition) of S(R) to S(M) is endothermic process Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

48

Thermodynamics

Solution of Assignment (Set-2)

Sol. Answer (4)

Sr hombic   O2  g  SO2  g ; H  –297.5 kJ

…(1)

Smonoclinic   O2  g  SO2  g ; H  –300 kJ

…(2)

Subtracting (2) from (1),

Sr hombic   Smonoclinic  ; H   297.5  300  kJ   2.5 kJ  This transition is endothermic. 20. If S + O2 SO2; H = –298.2 kJ SO2 + 1/2O2 SO3; H = –98.7 kJ SO3 + H2O H2SO4; H = –130.2 kJ H2 + 1/2O2 H2O; H = –287.3 kJ then the enthalpy of formation of H2SO4 at 298 K is (1) –814.4 kJ

(2)

–650.3 kJ

(3)

–320.5 kJ

(4)

–433.5 kJ

Sol. Answer (1)

H2  g  S  2O2  g  H2SO4 ;  f Given: S  O2  SO2

SO2 

1 O2  SO3 2

SO3  H2O  H2SO 4

H2 

1 O2  H2O 2

;

H = – 298.2 kJ

;

H = – 98.7 kJ

;

H = – 130.2 kJ

;

H = – 287.3 kJ

;

H' = – 814.4 kJ

Additing all these, we get H2  S  2O2  H2 SO 4

 Hf = H' = – 814.4 kJ 21. The volume of a gas expands by 0.25 m3 at a constant pressure of 103N m–2. The work done is equal to (1) 2.5 erg

(2)

250 J

(3)

250 watt

(4)

250 newton

Sol. Answer (2) We know, work done, w = – Pex + V. Given, pressure is 103 N m–2 and, V = 0.25 m3  w = – 103 Nm–2 × 0.25 m3 = – 250 J So, work done by the gas is 250 J. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

49

22. When 1 g of anhydrous oxalic acid is burnt at 25°C, the amount of heat liberated is 2.835 kJ. H combustion is (oxalic acid : C2H2O4) (1) –255.15 kJ

(2)

–445.65 kJ

(3)

–295.24 kJ

(4)

–155.16 kJ

Sol. Answer (1) H2C2O 4  s  

 oxalic acid

1 O2  g   2CO2  g   H2O  l  2

We know, Hcombustion = Amount of heat liberated when 1 mole of substance reacts with oxygen. Mol. wt. of oxalic acid = 90 g mol–1 1 g oxalic acid liberates 2.835 kJ  90 g (1 mole) oxalic acid liberates 2.835 × 90 kJ mol–1 = 255.15 kJ  Heat involved = – 255.15 kJ 23. The heat of neutralization of LiOH and HCl at 25°C is 34.868 kJ mol–1. The heat of ionisation of LiOH will be (1) 44.674 kJ

(2)

22.232 kJ

(3)

32.684 kJ

(4)

96.464 kJ

Sol. Answer (2) Let heat of ionization be a of LiOH. 

Hne

x

4  LiCl  H O LiOH  Li  OH  HCl  2

H = – 34.868 kJ  x – 57.1 kJ mol–1 = – 34.868 kJ mol–1 or, x = (57.1 – 34.868) kJ mol–1 = 22.232 kJ 24. Which compound will absorb the maximum amount of heat when dissolved in the same amount of water? (Integral heats of solution at 25°C in kcal/mol of each solute are given in brackets) (1) HCl (H = –17.74)

(2)

HNO3 (H = –7.85)

(3) NH4NO3 (H = +16.08)

(4)

NaCl (H = +1.02)

(3)

(q2 – q1)

Sol. Answer (3) Maximum heat is absorbed by NH4NO3  H = + 16.08 (maximum positive value) 25. HA + OH–H2O + A– + q1 kJ H+ + OH–  H2O + q2 kJ The enthalpy of dissociation of HA is (1) (q1 + q2)

(2)

(q1 – q2)

(4)

–(q1 + q2)

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50

Thermodynamics

Solution of Assignment (Set-2)

Sol. Answer (3) HA  H  A  ; H

Given: HA  OH  H2O  A   q1 kJ ; H  – q1 kJ

q2 kJ  H2O  H  OH ; H   q2 kJ Adding, we get

HA  H  A  ; H   q2 – q1  kJ 26. An athlete takes 100 g of glucose of energy equivalent to 1560 kJ. How much amount of energy is uptaken by 1 g molecule of glucose? (1) 15.6 kJ

(2)

2808 kJ

(3)

1560 kJ

(4)

28.08 kJ

Sol. Answer (2) 100 g

1560 kJ

180 g

1560  180 kJ 100

∵ wt. of 1 gram molecule = 180 g which gives

1560  180 kJ = 2808 kJ 100

27. C6H12 (l) + 9O2(g)  6H2O(l) + 6CO2(g);H298= – 936.9 kcal. Thus (1) –936.9 = E –(2×10–3 × 298 × 3) kcal

(2)

+936.9 = E +(2×10–3 × 298 × 3) kcal

(3) –936.9 = E –(2×10–3 × 298 × 2) kcal

(4)

–936.9 = E +(2×10–3 × 298 × 2) kcal

Sol. Answer (1) ng = 6 – 9 = – 3  H = U + ng RT or, U = H – ng RT = – 936.9 kcal – (– 3 R × 298 K) or, – 936.9 kcal = U + (– 3 × 2 cal k–1 mol–1 × 298 K) = U – (3 × 2 × 298 × 10–3 ) kcal U = E = change in internal energy. 28. For strong acid strong base neutralisation energy for 1 mole H2O formation is –57.1 kJ. If 0.25 mole of strong monoprotic acid is reacted with 0.5 mole of strong base then enthalpy of neutralisation is (1) –(0.25 × 57.1)

(2)

0.5 × 57.1

(3)

57.1

(4)

–(0.5 × 57.1)

Sol. Answer (1) 1 mole of strong monoprotich acid reacts with 1 mole of strong base to give – 57.1 kJ  0.25 mol of strong acid will react with only 0.25 mol of strong base (and not 0.5 mol)  Energy involved = – 57.1 kJ × 0.25 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

51

29. The heat of combustion of solid benzoic acid at constant volume is –321.3 kJ at 27°C. The heat of combustion at constant pressure is (1) –321.3 – 300R

(2)

–321.30 + 300R

(3)

–321.3 – 150R

(3)

x1 

(4)

–321.3 + 900R

(4)

x1 

Sol. Answer (3)

COOH



ng = 7 –

15 O2 g  7CO2 g  3H2O 2

15 1 =– 2 2

We know, H = U + ng RT H = qP U = qV  qP = qV + ng RT ⎛ 1 ⎞ = – 321.3 + ⎜   300 K  R ⎟ ⎝ 2 ⎠

= – 321.3 – 150 R

30.

1 H2 (g)  O2 (g)  H2O( ) 2

H2O(l)  H2O(g); H = x4 Given, EH–H = x1 EO=O = x2 EO–H = x3 HF of H2O vapour is (1) x1 

x2  x3  x 4 2

(2)

2x 3  x1 

x2  x4 2

x2  2x 3  x 4 2

x2  2x 3  x 4 2

Sol. Answer (4)

H2 g 

1 Hf  H2O g O2 g  2

H H2O () Now, for H2  g 

x4 1 O2  g  H2O    ; H 2

H = (B.D.E)reactants – (B.D.E)products = x1  Hf = H + x4 = x1 

1 x 2  2x3 2

1 x 2  2x3  x 4 2

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52

Thermodynamics

Solution of Assignment (Set-2)

31. A cylinder contains either ethylene or propylene. 12 ml of gas required 54 ml of oxygen for complete combustion. The gas is (1) Ethylene

(2)

Propylene

(3) 1 : 1 mixture of two gases

(4)

1 : 2 mixture

Sol. Answer (2) 12 ml of gas requires 54 ml of O2 or, 1 mole of gas requires

9 mole of O2 2

If the gas is C2H4 C2H4  3O2  2CO2  2H2O

If the gas is propylene C3H6 

9 O2  3CO2  3H2O 2

It is clear that propylene (1 mole) requires

9 moles of oxygen. As per the data, 12 ml of gas requires 54 2

ml of oxygen and hence 1 part of gas requires

9 parts of oxygen by moles. 2

 The gas is propylene 32. The specific heat of a gas is found to be 0.075 calories at constant volume and its formula wt is 40. The atomicity of the gas would be (1) One

(2)

Two

(3)

Three

(4)

Four

Sol. Answer (1) Specific heat = 0.075 calories  Molar specific heat capacity, CV = 0.075 × 40 = 3 cal mol–1 k–1  CP = CV + R = 3 cal mol–1 k–1 + 2 cal mol–1 k–1 = 5 cal mol–1 k–1 

CP 5     1.66 CV 3

 Monoatomic gas. 33. H(g) + O(g)  O – H(g); H for this reaction is (1) Heat of formation of O – H

(2)

Bond energy of O – H

(3) Heat of combustion of H2

(4)

Zero at all temperatures

Sol. Answer (2)

H  g  O  g  O – H  g ; H H = (B.D.E)reactants – (B.D.E)Products = O – Bond energy of O – H  H = – Bond energy of O – H Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

53

34. Energy required to dissociate 4 g of gaseous H2 into free gaseous atoms is 872 kJ at 25°C. The bond energy of H-H bond will be (1) 8.72 kJ

(2)

4.36 kJ

(3)

436 kJ

(4)

43.6 kJ

Sol. Answer (3)

 

H2 g 1mole (2g)

 2H  g ; H

H = (B.D.E)reactants – (B.D.E)Products = Bond energy of H2 Given that 4 g of H2 requires 872 kJ to dissociate  2 g of H2 requires 436 kJ  1 mole of H2 require 436 kJ  H = +436 kJ = Bond energy of H2 35. The dissociation energy of CH4 (g) is 360 kcal mol–1 and that of C2H6 (g) is 620 kcal mol–1. The C – C bond energy (1) 260 kcal mol–1

(2)

180 kcal mol–1

130 kcal mol–1

(3)

(4)

80 kcal mol–1

Sol. Answer (4)

H H

C H

H

H H H C–C–H H H

C(g) + 4 H (g) ; H7

2 C(g) + 6 H (g) ; H2

Given: H1 = 360 kcal mol–1 ; H2 = 620 kcal mol–1 Also, H1 = 4 × Bond energy of C – H = 360 kcal mol–1  Bond energy of C – H = 90 kcal mol–1 Now, H2 = 6 × Bond energy of C – H + Bond energy of C – C H2 = 6 × 90 kcal mol–1 + Bond energy of C – C ∵ H2 = 620 kcal mol–1  620 kcal mol–1 = 540 kcal mol–1 + Bond energy of C – C or, Bond energy of C – C = 80 kcal mol–1 36. The enthalpy of reaction, 2HCCH + 5O2  4CO2 + 2H2O If the bond energies of C–H, CC, O=O, C=O and O–H bonds are p, q, r, s, t respectively (1) [8s + 4t] – [4p + q + 5r]

(2)

[4p + 2q + 5r] – [8s + 4t]

(3) [4p + 2q + 5r + 8s + 4t]

(4)

[2p + q + 5r] – [8s + 4t]

Sol. Answer (2) Given: 2H – C  C – H + 5O = O  4O = C = O + 2

H

O

H

H is enthalpy of reaction. H = (B.D.E)reactants – (B.D.E)products



= 2  ECH  ECH   2 ECC  5 EOO

 –  4  ECO  ECO   2 EOH  EOH  

=  4ECH  2 ECC  5 EOO    8ECO  4EOH  = 4p + 2q + 5r – (8s + 4t) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

54

Thermodynamics

Solution of Assignment (Set-2)

37. Using bond energy data, calculate heat of formation of isoprene 5C(s) + 4H2(g)

H 2C = C – CH = CH 2 CH3

Given C–H, H–H, C–C, C = C and C(s)  C(g) respectively as 98.8 kcal, 104 kcal, 83 kcal, 147 kcal, 171 kcal (1) – 21 kcal

(2)

21 kcal

(3)

40 kcal

(4)

50 kcal

Sol. Answer (2) Given

5 C(s) + 4 H2(g)

H

C=C–C=C H

5x C(g) + 4 H2(g)

H

H

Hf

H

C H

H

H

H

 H = (B.D.E)reactants – (B.D.E)products = 4 EH – H – (2 EC = C + 8 EC – H + EC – C)  Hf = 5x + H (∵ H is a state function) x = 171 kcal  Hf = 5 × 171 kcal + [4 × 104 kcal – (2 × 147) – 8 × 98.8 kcal] – 2 × 83 kcal = 855 kcal + (416 kcal – 294 kcal – 790.4 kcal – 166 kcal) = 20.6 kcal  21 kcal 38. In a flask colourless N2O4 is in equilibrium with brown coloured NO2. At equilibrium when the flask is heated at 100°, the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, H for formation of NO2 is (1) Negative

(2)

Positive

(3)

Zero

(4)

Undefined

Sol. Answer (2)

  N2O4   2NO2 Upon heating, brown colour deepens, i.e. NO2 is formed.  The reaction is as follows

  N2O4  H   2NO2 The reaction is, hence, endothermic. 39. For which of these reactions will there be S positive? (1) H2O(g) H2O(l)

(2)

H2(g) + I2(g) 2HI(g)

(3) CaCO3(s) CaO(s) + CO2(g)

(4)

N2(g) + 3H2(g)2NH3(g)

Sol. Answer (3) For the 3rd reaction, CaCO3 (s)  CaO(s)  CO2 (g)

1 solid reactant gives 1 solid and 1 gaseous product and as a result increases disorder liness  S = positive Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

55

40. For stretched rubber, Entropy (1) Increases

(2)

First increases then decreases

(3) Decreases

(4)

First decreases then increases

(1) Ice

(2)

Liquid water

(3) Steam

(4)

Randomness is same in all

Sol. Answer (3) For stretched rubber, entropy decreases. ∵ Upon releasing , it regains its original shape  Spontaneous process, S = positive So, for the reverse process (stretching), S must have been negative. 41. The least random state of H2O is

Sol. Answer (1) Least random state of H2O would be its solid state, i.e. ice. 42. S for the reaction: MgCO3(s)  MgO(s) + CO2(g) (1) Zero

(2)

–ve

(3)

+ve

(4)



Sol. Answer (3)

MgCO3  s   MgO  s   CO2  g S = SCO  SMgO  SMgCO 2 3 ∵ MgO and MgCO3 are solids, so their entropy is almost same.  S = SCO

2

i.e., S is positive.

43. The standard entropies of N2 (g), H2 (g) and NH3 (g) are 191.5, 130.5, 192.6 JK–1 mol–1. The value of Sº of formation of ammonia is (1) –98.9 JK–1 mol–1

(2)

Zero

(3)

+129.4 JK–1 mol–1

(4)

–29.4 JK–1 mol–1

Sol. Answer (1)

1 3 N2  H2  NH3 2 2 S° = SNH  SH  SN 3 2 2 ⎛3 ⎞ 1 191.5 JK 1 ⎟ = (192.6 JK–1 mol–1 × 1 mol) – ⎜  130.5 JK  2 2 ⎝ ⎠

= 192.6 JK–1 – 195.75 JK–1 – 95.75 JK–1 = – 98.9 JK–1 for 1 mole of NH3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

56

Thermodynamics

Solution of Assignment (Set-2)

44. What is the increase in entropy when 11.2 L of O2 are mixed with 11.2 L of H2 at STP? (1) 0.576 J/K

(2)

5.76 J/K

(3)

7.56 J/K

(4)

2.76 J/K

(4)

315 J

Sol. Answer (2)



Smix = – n R × 2.303 xH2 log H2  xO2 log O2



Total moles = (0.5 + 0.5) moles = 1 mole

xH2  xO2 

1 2

∵ moles of O2 =

11.2 = 0.5 22.4

moles of H2 =

11.2 = 0.5 22.4

S = – 1 mol × 8.314 JK–1 (0.5 log 0.5 + 0.5 log 0.5) = – 8.314 JK–1 (– log 2) = + 5.76 JK–1  45. Given S C2H6 = 225 J mol-1K–1,

 o SC = 220 J mol–1K–1, SH = 130 J mol–1K–1. Then S° for the process 2H4 2

C2H4 + H2  C2H6 is (1) +25 J

(2)

–125 J

(3)

135 J

Sol. Answer (2) For the reaction, C2H4  H2  C2H6

 S° = SC2H4  SC2H4  SH2 = (225 – 220 – 130) JK–1 46. For the melting of NaCl heat required is 7.26 kcal mol–1 and S increases by 6.73 cal mol–1k–1. The melting point of the salt is (1) 805.75°C

(2)

500 K

(3)

1.77 K

(4)

1.77°C

Sol. Answer (1) Let melting temperature = T  Sfusion =

Hfusion T

Hfusion  T = S fusion

=

7.26 cal mol1 6.73 cal mol1 k 1

= 1078.75 K

 T = 1078.75 K or (1078.75 – 273)°C = 805.75°C Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

57

47. The S for the reaction o

2H2(g) + O2(g)  2H2O(l) at 300 K when SHo 2 (g) = 126.6, SO2 (g) = 201.20,

SH 2O (l)= 68.0JK–1mol–1 respectively is (1) –318.4JK–1mol–1

(2)

318.4JK–1mol–1

(3)

31.84 JK–1mol–1

(4)

3.184 JK–1mol–1

Sol. Answer (1)

2H2  g  O2  g  2H2O   



 S = 2  SH2O – 2  SH2  SO2





1 1 1 1 = 2 × 68.0 JK–1 mol–1 – 2  126.6 J mol J K  201.20 J K mol



= [136 – (253.2 + 201.2)] J K–1 mol–1 = – 318.4 J K–1 mol–1 48. Which of the following is correct? H

S

Nature of reaction

(1) (–)

(+)

Spontaneous only at high temperature

(2) (+)

(–)

Nonspontaneous regardless of temperature

(3) (+)

(+)

Spontaneous only at low temperature

(4) (–)

(–)

Spontaneous at all temperatures

Sol. Answer (2) If H > 0 and S < 0  – TS would always be positive 

(∵ T is positive always)

  TS is always positive H    

Always Always positive positive

 G is always positive. 49. Entropy of vaporisation of water at 100°C, if molar heat of vaporisation is 9710 cal mol—1 will be (1) 20 cal mol–1 K–1

(2)

26.0 cal mol–1 K–1

(3)

24 cal mol–1 K–1

(4)

28.0 cal mol–1 K–1

Sol. Answer (2) Molar entropy of vaporisation of water,



Svap

m



 Hvap m T



9710 cal mol1 = 26.03 cal mol–1 K–1 373 K

50. A particular reaction at 27°C for which H > 0 and S > 0 is found to be non-spontaneous. The reaction may proceed spontaneously if (1) The temperature is decreased

(2)

The temperature is increased

(3) The temperature is kept constant

(4)

It is carried in open vessel at 27°C

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58

Thermodynamics

Solution of Assignment (Set-2)

Sol. Answer (2) A reaction is spontaneous when G < 0, We know, G =

  TS H    

Positive Negative

So, G < 0 only when H < TS So, the reaction would proceed only when temperature is high. 51. It is impossible for a reaction to take place if (1) H is +ve and S is +ve

(2)

H is –ve and S is +ve

(3) H is +ve and S is –ve

(4)

H is –ve and S is –ve

Sol. Answer (3) It is impossible for a reaction to occur when G > 0 which is possible only when H > 0 and when S < 0. 52. The standard free energy change G° is related to K (equilibrium constant) as (1) G° = –2.303 RT logK

(2)

G° = 2.303 RT logK

(3) G° = RT logK

(4)

G° = –RT logK

Sol. Answer (1) We know, G = G° + RT ln Q ; Q = Reaction quotient. At equilibrium, G = 0 ; Q = K  0 = G° + RT ln K or, G° = – RT ln K = – 2.303 RT log K 53. The sole criterion for the spontaneity of a process is (1) Tendency to acquire minimum energy (2) Tendency to acquire maximum randomness (3) Tendency to acquire minimum energy and maximum randomness (4) Tendency to acquire maximum stability Sol. Answer (4) The sole criterion for the spontaneity of a process is the tendency to acquire maximum stability. 54. For an endothermic reaction to be spontaneous (1) G = 0

(2)

G > 0

(3) G < 0

(4)

G may be +ve or –ve

Sol. Answer (3) Whatever the process be (endothermic or exothermic), G has to be negative for the process to be spontaneous. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

59

55. At 27°C the reaction, C6H6(l) + 15/2 O2(g) 6CO2(g) + 3H2O(l) proceeds spontaneously because the magnitude of (1) H = TS

(2)

H > TS

(3)

H < TS

(4)

H > 0 and TS < 0

(4)

Cp – Cv = 0

Sol. Answer (2) For the given reaction, S = negative

(∵ Lesser number of gaseous products are formed)

 –TS = positive But H = –ve Since, it is a combustion reaction and hence exothermic. ∵ G = H – TS ; So for G to be negative, |H| > |TS|

SECTION - B Objective Type Questions 1.

For two mole of an ideal gas (1) Cp, m – Cv, m = R

(2)

Cp – Cv =

R 2

(3)

Cv – Cp = –2R

Sol. Answer (1) For two, three or even thousand moles of an ideal gas, Cp, m – Cv, m = R ; where Cp, m is molar heat capacity at const. pressure while Cv, m is molar heat capacity at const. volume. 2.

When an ideal gas is compressed adiabatically and reversibly, the final temperature is (1) Higher than the initial temperature

(2)

Lower than the initial temperature

(3) The same as the initial temperature

(4)

Dependent on the rate of compression

Sol. Answer (1) When an ideal gas is compressed adiabatically and reversibly, Then q = 0  According to 1st law: U = q + w = w ∵ Gas is compressed  work done is positive i.e. w is positive in magnitude  U is positive  T is positive as well. 3.

S° will be highest for the reaction (1) Ca 

1 O2 (g)  CaO(s) 2

(3) C(s) + O2(g)  CO2(g)

(2)

CaCO3(s)  CaO(s) + CO2(g)

(4)

N2(g) + O2(g)  2NO(g)

Sol. Answer (2) S° would be highest for the reaction for which ng is most positive. In this case, CaCO3  s   CaO  s   CO2  g ng = +1 which is maximum for this case. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

60 4.

Thermodynamics

Solution of Assignment (Set-2)

In an irreversible process, the value of Ssystem + Ssurr is (1) +ve

(2)

–ve

(3)

Zero

(4)

All of these

Sol. Answer (1) For any process, whether reversible or irreversible, Suniverse > 0  Ssystem + Ssurroundings > 0. 5.

A closed flask contains a substance in all its three states, solids, liquids and vapour at its triple point. In this situation the average KE of the water molecule will be (1) Maximum in vapour state

(2)

Maximum in solid state

(3) Greater in the liquid than in vapour state

(4)

Same in all the three states

Sol. Answer (4) The triple point for water exists at a particular temperature. ∵ Temperature is same  K.E. of water is also same. 6.

In thermodynamics a process is called reversible when (1) System and surrounding change into each other (2) There is no boundary between system and surrounding (3) The surroundings are always in equilibrium with the system (4) The system changes into the surroundings spontaneously

Sol. Answer (3) A process is reversible only when the system and surroundings are always in equilibrium with each other. 7.

The molar heat capacity of water at constant pressure P is 75 J K–1 mol–1. When 1.0 kJ of heat is supplied to 1000 g of water, which is free to expand, the increase in temperature of water is (1) 1.2 K

(2)

2.4 K

(3)

4.8 K

(4)

0.24 K

Sol. Answer (4) Molar heat capacity of water is 75 J K–1 mol–1  To raise temperature by 1°C (or 1 K), heat required is 75 J for 1 mole of water i.e. 75 J for 18 g of H2O. So for 1 g of H2O, heat required is  Specific heat capacity =

75 J 18

75 J g–1 K–1 18

 q = mCT or, T =

=

q  mC

1000 J 75 1000 g  J K 1 g1 18

18 K = 0.24 K 75

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Solution of Assignment (Set-2)

8.

Thermodynamics

61

16 kg oxygen gas expands at STP (1 atm) isobarically to occupy double of its original volume. The work done during the process is nearly (1) 260 kcal

(2)

180 kcal

(3)

130 kcal

(4)

271 kcal

Sol. Answer (4) We know, work done, w = – Pex + V = – P V

(∵ pressure is constant)

= – P(2V – V) ; where V is initial volume = – PV = – nRT n=

16000 g 32 g mol1

(Considering ideal behaviour)  500 mol

R = 2 cal mol–1 K–1 T = 273 K  w = – 500 mol × 2 × 273 cal mol–1 K–1 × K = – 273 cal  work done by oxygen gas is 271 cal. 9.

The enthalpy and entropy change for a chemical reaction are –2.5 × 103 cal and 7.4 cal K–1 respectively. Predict the nature of reaction at 298 K is (1) Spontaneous

(2)

Reversible

(3)

Irreversible

(4)

Non-spontaneous

262.12 K

(4)

562.12 K

Sol. Answer (1) H = – 2.5 × 103 cal S = + 7.4 cal K–1 ∵ H < 0 and S > 0  G < 0  Process is spontaneous 10. The temperature at which the given reaction is at equilibrium Ag2O(s)  2Ag(s) +

1 O (g) 2 2

H = 30.5 kJ mol–1 and S 0.066 kJ mol–1 K–1 (1) 462.12 K

(2)

362.12 K

(3)

Sol. Answer (1) At equilibrium, G = 0  H = TS (at equilibrium)

T=

H 30.5 k J mol1   462.12 K S 0.066 k J mol1 K 1

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62

Thermodynamics

Solution of Assignment (Set-2)

11. 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 = 30.0 L atm. The change in enthalpy of the process in L atm is (1) 40.0

(2)

42.3

(3)

44.0

(4)

56.0

(4)

All of these

Sol. Answer (3) Enthalpy is given as, H = U + pV This can be rewritten as H = U + (pV) = U + (p2V2 – p1V1) = 30.0 L atm + (4.0 atm × 5.0 L – 2.0 atm × 2.0 L) = (30.0 + 14.0) L atm = 44.0 L atm 12. Which of the following can be zero for isothermal reversible expansion? (1) E

(2)

H

(3)

T

Sol. Answer (4) For an isothermal reversible expansion, T = 0  U or E = 0 Also, H = m CpT ∵ T = 0  H = 0 13. In an insulated container water is stirred with a rod to increase the temperature. Which of the following is true? (1) U = w  0, q = 0

(2)

U = w = q  0

(3) U = 0, w = q  0

(4)

w = 0, U = q  0

Sol. Answer (1) ∵ Container is isolated  q=0  According to first law, U = q + w or U = w  U and w are both same in magnitude ∵ T increases  U is positive  U + w = 2 U = positive and hence non zero. 14. Two atoms of hydrogen combine to form a molecule of hydrogen gas the energy of the H2 molecule is (1) Greater than that of separate atoms

(2)

Equal to that of separate atoms

(3) Lower than that of separate atoms

(4)

Sometimes lower and sometimes higher

Sol. Answer (3) 2 atoms of hydrogen forms bond to form H2 molecule.  Bond is formed  Attractive forces  Energy is released during the process. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

63

15. The temperature of 15 ml of a strong acid increases by 2°C when 15 ml of a strong base is added to it. If 5 ml of each are mixed, temperature should increase by (1) 0.6°C

(2)

0.3°C

(3)

2°C

(4)

6°C

Sol. Answer (3) Heat of neutralization, H depends upon number of gram equivalents of strong acid and strong base. Number of g equivalents depends upon volume of acid/base taken.  H  g equivalents of acid and base  H  volume of acid and base Also, H = qp = mCPT or, T =

H mCp

Cp is intensive variable m = d  m = dV V

and

 T =

H 1  V Cp  d

We know, H  V  H = KV (where K is proportionality constant)  T =

K  Change in temperature remains constant when all these conditions are same. Cp  d

16. The standard heat of formation of NO2(g) and N2O4(g) are 8.0 and 4.0 kcal mol–1 respectively. The heat of dimerisation of NO2 in kcal is (1) –12 kcal

(2)

12 kcal

(3)

4 kcal

(4)

16 kcal

Sol. Answer (1)

1 N2  g   O2  g  NO2  g  ; H  8.0 kcal mol1 2 Reversing and multiplying 2, 2NO2  N2  2O2 ; H  – 16 kcal

Also, N2  2O2  N2O4 ; H  4 kcal

Adding, we get 2NO2  N2O 4 ; H   12 kcal Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

64

Thermodynamics

Solution of Assignment (Set-2)

1 1 X2O(s)  X(s) + O2(g); H = 90 kJ. Then heat change during reaction of metal X with one mole of O2 2 4 to form oxide to maximum extent is

17. If

(1) 360 kJ

(2)

–360 kJ

(3)

–180 kJ

(4)

+180 kJ

Sol. Answer (2) Given, 

1 1  X2O  s   X  s   O2  g  ; H  90 kJ 2 4

X  s 

1 1 O2  g  X2O  s  ; H   90 kJ 4 2

Multiplying with 4, we get

4X  s   O2  g  2X2O  s  ; H   360 kJ 18. For a gaseous reaction : A(g) + 3B(g)  3C(g) + 3D(g) E is 17 kcal at 27°C. Assuming R = 2 cal K–1 mol–1 the value of H for the above reaction will be (1) 15.8 kcal

(2)

16.4 kcal

(3)

18.2 kcal

(4)

20.0 kcal

Sol. Answer (3) Given: A  g  3B  g  3C  g  3D  g ng = 6 – 4 = 2 We know, H = U + pV Considering ideal behaviour, we have H = U + ngRT = 17 kcal + 2 × 2 cal K–1 × 298 K = (17000 + 4 × 298) cal = 18192 kcal = 18.2 kcal 19. A mixture of 2 mole of CO and 1 mol of O2 is ignited. Correct relationship is (1) H = U (2) H > U (3) H < U (4) The relationship depends upon the capacity of vessel Sol. Answer (3)

2CO  g  O2  g  2CO2  g ng = 2 – 3 = –1 We know, H = U + ngRT H = U – 1 × RT  H = U – RT  H < U

⎡8  10  2 ⎤ ⎢ ⎥ ⎣⇒ 8  10 ⎦

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Solution of Assignment (Set-2)

Thermodynamics

65

20. Bond dissociation energy of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and Hf of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be (1) 800 kJ mol–1

(2)

200 kJ mol–1

(3)

300 kJ mol–1

(4)

400 kJ mol–1

(3)

32

(4)

64

Sol. Answer (1) B.D.E of XY, X2 and Y2 are in the ratio 1 : 1 : 0.5 Let B.D.E. of XY be x, X2 be x and Y2 be 0.5 x. Now,

X2  g  Y2  g  2XY  g ; H

 H = – 2 × (B.D.E)XY +

B.D.E

X2

 B.D.E. Y

2



= – 2x + (x + 0.5 x) = – 0.5x Also, Hf of XY is – 200 kJ mol–1 i.e.

1 1 X2  g   Y2  g  XY  g ; Hf 2 2

 H = 2 × Hf = 2 × – 200 kJ mol–1 = – 400 kJ So, 0.5x = + 400 kJ  x = 800 kJ 21. Vapour density of a gas is 8. Its molecular mass will be (1) 8

(2)

16

Sol. Answer (2) We know, Vapour density =

Molecular mass of substance Molecular mass of H2

 Molecular mass of substance = 8 × 2 = 16 22. If x mole of ideal gas at 27°C expands isothermally and reversibly from a volume of y to 10y, then the work done is (1) w = x R 300 ln y

(2)

y w = – 300x R ln 10 y

(3) w = – 300x R ln 10

(4)

1 w = 100x R ln y

Sol. Answer (3)

V2 Work done, w = – nRT ln V 1 = – x × R × 300 ln

10 y y

= – x × R × 300 ln 10 = – 300x R ln 10 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

66

Thermodynamics

Solution of Assignment (Set-2)

23. Enthalpy of formation of NH3 is – X kJ and HH–H, HN–H are respectively Y kJ mol–1 andZ kJ mol–1. The value of HN  N is X 3

(3)

3Y + 6Z + X

(1) Reversible process

(2)

Cyclic process

(3) Cyclic reversible process

(4)

Isochoric process

(1) Y – 6Z +

(2)

– 3Y + 6Z – 2X

(4)

Y + 6X + Z

Sol. Answer (2)

1 3 N2  g  H2  g  NH3  g ; Hf  – x kJ 2 2 Hf = H = (B.D.E)reactants – (B.D.E)products =

1 3 HN2  HH2  3NH 2 2

=

HNN 3  XY  3XZ 2 2

We have, –X=

HNN 3Y  3 Z 2 2

3Y ⎛ ⎞  X ⎟  2 = 6Z – 3Y – 2X or, HNN = ⎜ 3Z  2 ⎝ ⎠

24. A system X undergoes following changes

X

(P1V1T1 )



W

(P2 V2T1 )



Z

(P3 V2T2 )



X

(P1V1T1 )

The overall process may be called as

Sol. Answer (2) ∵ The system returns to its original state, the whole process is called a cyclic process. 25. The heat of neutralisation for strong acid and strong base forming 2 moles of water is (1) – 2 × 57.1 kJ (2) – 57.1 kJ (3) 

57.1 kJ 2

(4) Strong acid and strong base will not undergo neutralisation Sol. Answer (1) When 1 mole of water is formed upon neutralization, – 57.1 kJ is released when 2 moles of water are formed, – 57.1 × 2 kJ of energy is released. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

67

26. The value of H° in kJ for the reaction will be CS2(l) + 4NOCl(g)  CCl4(l) + 2SO2 (g) + 2N2(g) if

Hf (CS2 )   x

Hf (NOCl)   y

Hf (CCl4 )   z

Hf (SO2 )   r

(1) x + 4y – z – 2r

(2)

r + z + 4y – x

(3) 2r + z + 4y + x

(4)

x + 4y + z – 2r

Sol. Answer (4) For the reaction,

CS2  l   4NOCl  g  CCl4  l   2SO2  g  2N2  g



 



    H° = HfCCl4  HfSO2  2  HfN2  2  HfCS2 (l)   HNOCl   4

= z + 2(–r) + 2(0) – (– x) – (– y) = x – 2r + x + y Hf of N2(g) is o since if is in its reference state. 27. The heat liberated on complete combustion of 1 mole of CH4 gas to CO2(g) and H2O(l) is 890 kJ. Calculate the heat evolved by 2.4 L of CH4 on complete combustion. (1) 95.3 kJ

(2)

8900 kJ

(3)

890 kJ

(4)

8.9 kJ

Sol. Answer (1)

CH4  g  2O2  g  CO2  g  2H2O  l  ; H H is 890 kJ for 1 mole CH4 We know, 22.4 L 2.4 L

means 1 mol means

1  2.4 mol = 0.107 mol 22.4

 H = 890 kJ × 0.107 = 95.23 kJ 28. The work done in an open vessel at 300 K, when 112 g iron reacts with dil HCl to give FeCl2, is nearly (1) 1.1 kcal

(2)

0.6 kcal

(3)

0.3 kcal

(4)

0.2 kcal

Sol. Answer (1)

2Fe  s   4HCl  2FeCl2  2H2  ng = 2 mol Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

68

Thermodynamics

Solution of Assignment (Set-2)

We know, w = – pV = – ngRT = – 2 mol × 2 cal mol–1 K–1 × 300 K = – 1.2 kcal So, work done = 1.2 kcal 29. Which statement is correct?

⎛ dH ⎞ ⎛ dE ⎞ ⎟ ⎜ ⎟ (1) ⎜ ⎝ dT ⎠P ⎝ dT ⎠ V

(2)

⎛ dH ⎞ ⎛ dE ⎞ ⎜ ⎟ ⎜ ⎟ R dT ⎝ ⎠P ⎝ dT ⎠ V

⎛ dE ⎞ ⎟ for ideal gas is zero (3) ⎜ ⎝ dV ⎠ T

(4)

All of these

Sol. Answer (3) For an ideal gas, dU or dE = 0 when T is constant 

⎛ dU ⎞ dU ⎜ dV ⎟ i.e., at constant volume is 0. ⎝ ⎠T dV

30. A schematic representation of enthalpy changes for the C( graphite )  The missing value is

Cgraphite + O2(g)

– 393.5 kJ CO2(g) (1) + 10.5 kJ

1 O (g)  CO (g) reaction, is given below. 2 2

(2)

– 11.05 kJ

?? CO(g) + ½O(g) – 283.0 kJ

(3)

– 110.5 kJ

(4)

– 10.5 J

Sol. Answer (3) Since enthalpy is state function, H + (– 283.0 kJ) = – 393.5 kJ or, H = – 110.5 kJ 31. Which of the following equations represent standard heat of formation of CH4? (1) C(diamond) + 2H2(g)  CH4(g)

(2)

C(graphite) + 2H2(g)  CH4(g)

(3) C(diamond) + 4H(g)  CH4(g)

(4)

C(graphite) + 4H(g)  CH4(g)

Sol. Answer (2) Standard heat of formation is enthalpy change when 1 mole of substance is formed from its constituents elements in their reference states. Reference state of carbon  graphite



C graphite   2H2  CH4  g

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Solution of Assignment (Set-2)

Thermodynamics

69

32. Different types of systems are given below

Surrounding

Surrounding

Work Heat System

Matter Energy System

A

B

The A and B systems respectively are (1) Open system, Closed system

(2)

Isolated system, Closed system

(3) Adiabatic system, Isolated system

(4)

Closed system, Isolated system

Sol. Answer (4) A  Both energy and matter is exchanged B  Neither energy nor matter is exchanged  A  open, B  isolated 33. Set of intensive properties is shown by (1) Mole fraction, standard electrode potential, heat capacity (2) Viscosity, refractive index, specific heat (3) Density, Gibbs free energy, internal energy (4) Number of moles, molarity, electrode potential Sol. Answer (2) Viscosity, refractive index, and specific heat do not depend upon mass and hence are intensive properties. 34. For the expansion occurring from initial to final stage in finite time, which is incorrect? (1) Equilibrium exist in initial and final stage (2) Work obtained is maximum (3) Driving force is much greater than the opposing force (4) Both (1) & (2) Sol. Answer (2) Work obtained is maximum in case of reversible process (i.e. process occurring in infinite time) Hence, work done infinite time is not maximum. 35. Calorific value of ethane, in kJ/g if for the reaction 2C2H6 + 7O2  4CO2 + 6H2O; H = –745.6 kcal (1) –12.4

(2)

–52

(3)

–24.8

(4)

–104

Sol. Answer (2) In the reaction, 2 moles of ethane yielded – 745.6 kcal i.e., 2 × molar mass of C2H6 = 60 g  60 g yields – 745.6 kcal  1 g yields –  1 g yields

75.6 kcal 60 g

745.6  4.2 kJ g1  –52.2 kJ g1 60

Hence, calorific value of ethane is – 52.2 kJ g–1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

70

Thermodynamics

Solution of Assignment (Set-2)

SECTION - C Previous Years Questions 1.

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? (1) Mg(s) 

1 O2 (g)   MgO(s) 2

(3) C (graphite) 

1 O2 (g)   CO(g) 2

(2)

1 1 1 C (graphite)  O2 (g)   CO2 (g) 2 2 2

(4)

CO(g) 

1 O2 (g)   CO2 (g) 2

Sol. Answer (3) For the reaction C graphite  

ng = 1 

1 O2  g  CO  g 2

1 1  2 2

 S is positive and hence G decreases with increase in temperature ∵ G = H – TS 2.

Standard enthalpy of vapourisation vapH for water at 100° C is 40.66 kJmol–1. The internal energy of vapourisation of water at 100°C (in kJmol–1) is (1) +43.76

(2)

+40.66

(3)

+37.56

(4)

–43.76

(Assume water vapour to behave like an ideal gas) Sol. Answer (3) We know, H = U + ngRT or, U = H – ngRT The reaction is

H2O  l   H2O  g  Dng = +1  U = 40.66 kJ mol–1 – 1 mol × 8.314 J mol–1 K–1 × 373 K = 40.66 – 3101 J = + 37.56 J 3.

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is (1) 5.260 cal/(mol K)

(2)

0.526 cal/(mol K)

(3) 10.52 cal/(mol K)

(4)

21.04 cal/(mol K)

Sol. Answer (1) Sfus 

Hfus 1.435  103 cal mol1  T 273 K

= 5.26 cal mol–1 K–1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

4.

Thermodynamics

71

Which of the following is correct option for free expansion of an ideal gas under adiabatic condition? (1) q = 0, T < 0, w  0

(2)

q = 0, T  0, w  0

(3) q  0, T = 0, w  0

(4)

q = 0, T = 0, w  0

Sol. Answer (4) Free expansion  Pext = 0  w=0  U = w + q = q

(∵ w = 0)

∵ Expansion is adiabatic  q=0 Hence U = 0  T = 0 5.

(∵ U is proportional to temperature)

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol–1 at 27°C, the entropy change for the process would be (1) 100 J mol–1 K–1

(2)

10 J mol–1 K–1

(3)

1.0 J mol–1 K–1

(4)

0.1 J mol–1 K–1

Sol. Answer (1) Svap 

6.

Hvap T



30000 J mol1  100 J mol1 K 1 300 K

Enthalpy change for the reaction, 4H(g)  2H2(g) is –869.6 kJ. The dissociation energy of H – H bond is (1) +217.4 kJ

(2)

–434.8 kJ

(3)

–869.6 kJ

(4)

+434.8 kJ

(3)

525 kJ/mol

(4)

– 175 kJ/mol

Sol. Answer (4) Given reaction:

4 H  g  2 H2  g  H = B.D.E. of reactants – B.D.E. of products = 0 – 2 × HH–H = – 2HH–H This is equal to – 869.6 kJ  – 2HH–H = – 869.6 kJ  HH–H = + 434.8 kJ 7.

Consider the following process H(kJ/mol) 1 A B 2

+ 150

3B  2C + D

– 125

E + A  2D

+ 350

For B + D  E + 2C, H will be (1) – 325 kJ/mol

(2)

325 kJ/mol

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72

Thermodynamics

Solution of Assignment (Set-2)

Sol. Answer (4) Given:

1 A  B ; H = + 150 kJ mol–1 2 

A  2B ; H = + 300 kJ mol–1

…(1)

Also given: 3B  2C  D ; H = – 125 kJ mol–1

…(2)

E  A  2D ; H = + 350 kJ mol–1



2D  E  A ;H = – 350 kJ mol–1

…(3)

Adding (1), (2) and (3), we get B  D  E  2C ; H = – 175 kJ

8.

Which reaction, with the following values of H, S, at 400 K is spontaneous and endothermic? (1) H = –48 kJ; S = + 135 J/K

(2)

H = –48 kJ; S = – 135 J/K

(3) H = +48 kJ; S = + 135 J/K

(4)

H = +48 kJ; S = – 135 J/K

Sol. Answer (3)

∵ G = H – TS = + 48 kJ – 400 × 135 J K–1

(∵ Reaction has to be endothermic)

= (+ 48000 – 54000) J = – 6000 J  G < 0 9.

Which of the following are not state functions? (I) q + w

(II)

q

(III)

w

(IV) H – TS

(1) (II) and (III)

(2)

(I) and (IV)

(3)

(II), (III) and (IV)

(4)

(I), (II) and (III)

Sol. Answer (1) q (heat) and w (work done) depend on path and hence are path functions. 10. Three thermochemical equations are given below: (i) C(graphite) + O2(g)  CO2(g); rH° = x kJ mol–1 (ii) C(graphite) +

(iii) CO (g) +

1 O (g)  CO(g); rH° = y kJ mol–1 2 2

1 O (g)  CO2(g); rH° = z kJ mol–1 2 2

Based on the above equations, find out which of the relationship given below is correct: (1) x = y – z

(2)

z=x+y

(3)

x=y+z

(4)

y = 2z – x

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Solution of Assignment (Set-2)

Thermodynamics

73

Sol. Answer (3) Subtracting equation (ii) from (i), we get CO 

1 O2  CO2 ; H  x  y kJ 2

Also, H = z kJ mol–1  x–y=z or, x = (y + z) 11. Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJmol–1 respectively. Enthalpy of formation of HCl is (1) 245 kJmol–1

(2)

93 kJmol–1

(3)

–245 kJmol–1

(4)

–93 kJmol–1

Sol. Answer (4) H2  Cl2  2HCl ; H

 H = HH–H + HCl–Cl – 2HH–Cl = (434 + 242 – 2 × 431) kJ = – 186 kJ So, for 2 moles of HCl, H = – 186 kJ  for 1 mole of HCl, H = – 93 kJ 12. For the gas phase reaction, PCl5 (g)  PCl3(g) + Cl2 (g), which of the following conditions is correct? (1) H > 0 and S < 0

(2)

H = 0 and S < 0

(3) H > 0 and S > 0

(4)

H < 0 and S < 0

Sol. Answer (3)

PCl5  g  PCl3  g  Cl2  g ng = 2 – 1 = + 1 S = positive And since it is a decomposition reaction  Bonds need to be broken  Endothermic reaction (H = positive) 13. Following reaction occurring in an automobile 2C8H18(g) + 25O2(g)  16CO2(g) + 18H2O (g). The sign of H, S and G would be (1) –, +, +

(2)

+, +, –

(3)

+, –, +

(4)

–, +, –

Sol. Answer (4) For the reaction, ng = 18 + 16 – 25 – 2 = + 7  S = positive It is a combustion reaction, H = negative  G is negative, H is negative and S is positive. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

74

Thermodynamics

Solution of Assignment (Set-2)

14. When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 litre of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ (Hcomb (CH4) = 890 kJ mol–1, Hcomb (C3H8) = 2220 kJ mol–1) is (1) 32

(2)

38

(3)

317

(4)

477

Sol. Answer (3) Let volume of methane be x L  volume of propane be (5 – x) L For methane,

CH4 g x Volume reacted:



2 O2 g  CO2 g  2H2O l  2x

For propane,

Volume reacted:

C3H8  5O2  2CO2  4H2O 5  x  5 5  x 

We know, 16 L of oxygen is consumed.  2x + 5 (5 – x) = 16 or, 2x + 25 – 5x = 16 or, 25 – 3x = 16  x=3 So, volume of methane combusted = 3 L 3 mol 22.4

 Moles of methane combusted =

Energy due to methane combusted =

3  890 kJ 22.4

Volume of propane combusted = 2 L  Moles of propane combusted =

2 mol 22.4

 Energy due to propane combusted =

2  2220 kJ 22.4

 Total energy released 2 ⎛ 3 ⎞  890   2220 ⎟ kJ = ⎜ 22.4 ⎝ 22.4 ⎠

=

1  3  890  4440  kJ 22.4

=

1  2670  4440  kJ = 317.41 kJ 22.4

15. If enthalpies of formation for C2H4(g), CO2(g) and H2O(l) at 25°C and 1 atm pressure are 52, –394 and –286 kJ/mol respectively, then enthalpy of combustion of C2H4(g) will be (1) + 14.2 kJ/mol

(2)

+ 1412 kJ/mol

(3)

– 141.2 kJ/mol

(4)

– 1412 kJ/mol

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Solution of Assignment (Set-2)

Thermodynamics

75

Sol. Answer (4) C2H4  3 O2  2 CO2  2 H2 O ; Hr

Hr = Enthalpy of combustion Now,

2 C  s   2 H2  g  C2H4  g ; H1



2  C  s   O2  g   CO2  g  ; H2



…(1)

1 ⎛ ⎞ 2  ⎜ H2  O2  g   H2O  g ; H3 ⎟ 2 ⎝ ⎠

…(2)

Also, we have,

C2H4  2 C  s   2 H2  g ; – H1

…(3)

Adding (1), (2), (3), we get C2H4  3 O2  2 CO2  2 H2O ; Hr  – H1  2 H2  2 H3

Hr = [– 52 + 2 × (– 394) + 2 × (– 286)] kJ/mol = – 1412 kJ./mol 16. For a reaction to occur spontaneously (1) H must be negative

(2)

S must be negative

(3) (H – TS) must be negative

(4)

(H + TS) must be negative

(3)

y – 2x

Sol. Answer (3) For a reaction to occur spontaneously, G should be negative  H – TS should be negative 17. Given that C + O2  CO2, H° = –x kJ 2CO + O2  2CO2, H° = –y kJ What is heat of formation of CO? (1)

y  2x 2

(2)

2x – y

(4)

2x  y 2

Sol. Answer (1)

C  s 

1 O2  CO ; Hr 2

C  O2  CO2 ; H   x kJ

CO2  CO 

1 y O2 ; H  kJ 2 2

Adding, we get

C

1 y y  2x O2  CO : Hr   x  2 2 2

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76

Thermodynamics

Solution of Assignment (Set-2)

18. Identify the correct statement regarding entropy (1) At absolute zero of temperature, the entropy of all crystalline substances is taken to be zero (2) At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve (3) At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero (4) At 0°C, the entropy of a perfectly crystalline substance is taken to be zero Sol. Answer (3) 3rd law of thermodynamics states that at 0 K, entropy of a perfectly crystalline substance is zero. 19. One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres. The E for this process is (R = 2 cal. Mol–1K–1) (1) 1381.1 cal

(2)

Zero

(3)

163.7 cal

(4)

9 L atm

Sol. Answer (2)  The process is isothermal, T = 0 and, U  T  U = 0 20. In the reaction : S + 3/2 O2  SO3 + 2x kcal and SO2 + 1/2 O2  SO3 + y kcal, the heat of formation of SO2 is (1) (2x + y)

(2)

(x – y)

(3)

(x + y)

(4)

(y – 2x)

(4)

0.977 J/mol·K

(4)

H = E – 2RT

Sol. Answer (4) S  O2  SO2 : H

Given : S  and,

3 O2  SO3 : H1   2x kcal 2

SO3  SO2 

1 O2 : H2  y kcal 2

Adding, S  O2  SO2 : H  H1  H2  y  2x kcal 21. At 27°C latent heat of fusion of a compound is 2930 J/mol. Entropy change is (1) 9.77 J/mol·K

(2)

10.77 J/mol·K

(3)

9.07 J/mol·K

Sol. Answer (1) Sfusion 

Hfusion 2930 J mol1   9.77 J mol1 K 1 T 300 K

22. For the reaction C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l), which one is true? (1) H = E – RT

(2)

H = E + RT

(3)

H = E + 2RT

Sol. Answer (1) We know, H = U + pV For an ideal gas, H = U + ngRT For the given reaction, ng = 2 – 3 = – 1  H = U + (–1) RT = U – RT Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

77

23. Change in enthalpy for reaction, 2H2O2 (l )  2H2O (l ) +O2 (g) if heat of formation of H2O2(l ) and H2O(l ) are –188 and –286 kJ/mol respectively, is (1) –196 kJ/mol

(2)

+196 kJ/mol

(3)

+948 kJ/mol

(4)

–948 kJ/mol

Sol. Answer (1) 2H2O2  2H2O  O2 ; Hr

Given: H2  O2  H2O2 ; H1   188 kJ mol1





2  H2O2  H2  O2 ; – H1  188 kJ mol1



1 ⎛ ⎞ and 2  ⎜ H2  O2  H2O ; H2  – 286 kJ mol1 ⎟ 2 ⎝ ⎠

Adding, we get

2H2O2  2H2O  O2 ; Hr  2 2  2  – 1  = [– 286 × 2 + 2(188)] kJ = (– 572 + 376) kJ = – 196 kJ 24. When 1 mol of 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? (1) q = U = –500 J, w = 0

(2)

q = U = 500 J, w = 0

(3) q = w = 500 J, U = 0

(4)

U = 0, q = w = –500 J

Sol. Answer (2) ∵ Volume is constant  V = 0  – pex + V = 0  w=0 As per 1st law of thermodynamics, U = q + w = q = +500 J

1 O  CH3OH is negative. If enthalpy of combustion of CH4 and CH3OH are x and y 2 2 respectively. Then which relation is correct?

25. Enthalpy of CH4 +

(1) x > y

(2)

x < y

(3)

x = y

(4)

xy

Sol. Answer (1)

CH4 

1 O2  CH3 OH ; H  0 2

Given: CH4  2 O2  CO2  2 H2O ; H1   x

CH3 OH 

…(1)

3 O2  CO2  2 H2O ; Hr  y 2

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78

Thermodynamics



Solution of Assignment (Set-2)

CO2  2H2O  CH 3 OH 

3 O2 ; H'   y 2

…(2)

Adding (1) and (2), we get CH4 

1 O2  CH3 OH ; H  x  y 2

We know, H < 0  x–y|y|

26. Unit of entropy is (1) JK–1 mol–1

(2)

J mol–1

(3)

J–1 K–1 mol–1

(4)

JK mol–1

Sol. Answer (1) Unit of molar entropy is J K–1 Unit of entropy is J K–1 Entropy is represented in terms of molar entropy to make it an intensive variable. 27. In a closed insulated container a liquid is stirred with a paddle to increase the temperature which of the following is true? (1) E = W  0, q = 0

(2)

E = W = q  0

(3) E = 0, W = q  0

(4)

W = 0, E = q  0

Sol. Answer (1) Since container is insulated, q=0 According to 1st Law, U = q + W = W  0 [work is done] 28. 2 mole of ideal gas at 27°C temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change (R = 2 cal/mol K) (1) 92.1

(2)

0

(3)

4

(4)

9.2

Sol. Answer (4) We know,

V S = 2.303 nR log 2 V1 Given: V2 = 20 L ; V1 = 2 L n = 2 mol., R = 2 cal mol–1 K–1  S = 2.303 × 2 mol × 2 cal mol–1 K–1 × log

20 = 9.2 cal K–1 2

29. Heat of combustion for C(s), H 2 (g) and CH 4 (g) are –94, –68 and –213 kcal/mol, then H for C(s) + 2H2(g)  CH4(g) is (1) –17 kcal

(2)

–111 kcal

(3)

–170 kcal

(4)

–85 kcal

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Solution of Assignment (Set-2)

Thermodynamics

79

Sol. Answer (1) C  2H2  CH4 ; H

Given; C  O2  CO2 ; H1

…(1)

1 ⎛ ⎞ 2  ⎜ H2  O2  H2O ; H2 ⎟ 2 ⎝ ⎠

…(2)

CH4  2 O2  CO2  2 H2O ; H3



CO2  2 H2O  CH4  2 O2

…(3)

Adding (1), (2) and (3), we get C  2 H2  CH4 ; H  H  2H2 – H3

= [– 94 + 2 (– 68) – (– 213)] kcal = – 17 kcal 30. For the reaction C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) at constant temperature, H – E is (1) + RT

(2)

– 3RT

(3)

+ 3RT

(4)

– RT

Sol. Answer (2) For the reaction, ng = 3 – (5 + 1) = – 3  H = U + ngRT  H – U = –3RT 31. What is the entropy change (in JK–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) (1) 20.13

(2)

2.013

(3)

2.198

(4)

21.98

Sol. Answer (4) Sfusion 

Hfusion 6.0 kJ mol1   21.98 J mol1 K 1 T 273 K

  32. For which one of the following equations is Hreact equal to Hf for the product? (1) N2(g) + O3(g)  N2O3(g)

(2)

CH4(g) + 2Cl2(g)  CH2Cl2(l) + 2HCl(g)

(3) Xe(g) + 2F2(g)  XeF4(g)

(4)

2CO(g) + O2(g)  2CO2(g)

Sol. Answer (3) In the 3rd reaction, All the reactants are in their referance states and 1 mole of product is formed. Hence, Hf = Hreaction Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

80

Thermodynamics

Solution of Assignment (Set-2)

33. The molar heat capacity of water at constant pressure, C, is 75 J K–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 (1) 1.2 K

(2)

2.4 K

(3)

4.8 K

(4)

6.6 K

Sol. Answer (2) Given: Cp, m = 75 J K–1 mol–1 We know, qp = Cp , m × n T or, T =

qp Cp,m  n



1000 J 75 J K

1

mol

1



100 mol = 2.4 K 18

34. Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are –382.64 kJ mol–1 and –145.6 J mol–1, respectively. Standard Gibb’s energy change for the same reaction at 298 K is (1) –221.1 kJ mol–1

(2)

–339.3 kJ mol–1

(3)

–439.3 kJ mol–1

(4)

–523.2 kJ mol–1

Sol. Answer (2) G = H – TS ⎡ ⎛ 145.6 ⎞ ⎤ = ⎢ 382.64  298 K ⎜ 1000 ⎟ ⎥ kJ mol–1 ⎝ ⎠⎦ ⎣

= – 339.3 kJ mol–1 35. Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is (1) Ssystem + Ssurroundings > 0

(2)

Ssystem – Ssurroundings > 0

(3) Ssystem > 0 only

(4)

Ssurroundings > 0 only

Sol. Answer (1) For a spontaneous process, Suniverse > 0 or, Ssystem + Ssurrounding > 0 36. 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 (1 L atm = 101.32 J) (1) –6 J

(2)

–608 J

(3)

+304 J

(4)

–304 J

Sol. Answer (2) We know, w = – pext V = – 3 atm (6 L – 4 L ) = – 3 × 2 L atm = – 6 × 101.32 J = – 607.92 J 37. Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? (1) Exothermic and increasing disorder

(2)

Exothermic and decreasing disorder

(3) Endothermic and increasing disorder

(4)

Endothermic and decreasing disorder

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Solution of Assignment (Set-2)

Thermodynamics

81

Sol. Answer (1) A chemical reaction is certainly spontaneous if H < 0 and S > 0 for which G < 0 at all temperature. 38. A reaction occurs spontaneously if (1) TS < H and both H and S are +ve

(2)

T S > H and H is +ve and S is –ve

(3) T S > H and both H and S are +ve

(4)

T S = H and both H and S are +ve

Sol. Answer (3) A reaction is spontaneous if G < 0.  H – TS < 0 or, TS > H Also, reaction is spontaneous when, TS > H and H and S are +ve. Reaction is never spontaneous if H is +ve and is -ve 39. The absolute enthalpy of neutralisation of the reaction MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O(l) will be (1) –57.33 kJ mol–1

(2)

Greater than –57.33 kJ mol–1

(3) Less than – 57.33 kJ mol–1

(4)

57.33 kJ mol–1

Sol. Answer (3) For this reaction, MgO  s   2 HCl  aq  MgCl2  H2O

(Week base)

∵ MgO is weak, so extra energy is required for its complete dissociation Hence, Hn < 57.1 kJ mol–1 40. The enthalpy and entropy change for the reaction : Br2(l) + Cl2(g)  2BrCl(g) are 30 kJ mol–1 and 105 JK–1 mol–1 respectively. The temperature at which the reaction will be in equilibrium is (1) 300 K

(2)

285.7 K

(3)

273 K

(4)

450 K

Sol. Answer (2) At equilibrium, G = 0  H = TS

or, T =

30000 J mol1 H  285.7 K = S 105 J K 1 mo1

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82

Thermodynamics

Solution of Assignment (Set-2)

41. 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 (1) –358.5 kJ mol–1

(2)

–508.9 kJ mol–1

(3)

–208.1 kJ mol–1

(4)

–269.9 kJ mol–1

Sol. Answer (3) Enthalpy of hydrogenation of cyclohexene = – 119.5 kJ mol–1  Expected enthalpy of hydrogenation of benzene = (– 119.5 × 3) kJ mol–1

non-resonated benzene Eres resonated benzene

Energy E

cyclohexane  E + Eres = Expected heat of hydrogenation or, E = (– 119.5 × 3 + 150.4) kJ mol–1 = – 208.1 kJ mol–1 42. Consider the following reactions (i) H+(aq) + OH–(aq)  H2O(l), H = –X1 kJ mol–1 (ii) H2(g)+

1 O (g)  H2O(l), H =–X2 kJ mol–1 2 2

(iii) CO2(g) + H2(g)  CO(g) + H2O, H = –X3, kJ mol–1 (iv) C2H2(g) +

5 O 2 (g)  2CO2(g) + H2O(l), H = +X4 kJ mol–1 2

Enthalpy of formation of H2O(l) is (1) +X3 kJ mol–1

(2)

–X4 kJ mol–1

(3)

+X1 kJ mol–1

(4)

–X2 kJ mol–1

(3)

–501 J

(4)

–731 J

Sol. Answer (4)

H2 g 

1 O g  H2O l  ; H  – X2 kJ mol1 2 2

Elements in reference states

1 mole of substance is formed

 Hf = – X2 kJ mol–1 43. 2Zn + O2  2ZnO;

G° = –616 J

2Zn + S2  2ZnS;

G° = –293 J

S2 + 2O2  2SO2;

G° = –408 J

G° for the following reaction 2ZnS + 3O2  2ZnO + 2SO2 is (1) –1462 J

(2)

–1317 J

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Solution of Assignment (Set-2)

Thermodynamics

83

Sol. Answer (4) Given: 2 Zn  O2  ZnO ; G  – 616 J 2 ZnS  2 Zn  S2 ; G   293 J

S2  2 O2  2 SO2 ; G   408 J Adding, we get

2 ZnS  3 O2  2 ZnO  2 SO2 ; G   616  293  408  J   1462 J 44. 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, H

H

H

H

C=C+H–HH–C–C–H H

H

H

H

will be (1) 553.0 kJ mol–1

(2)

1523.6 kJ mol–1

(3)

–243.6 kJ mol–1

(4)

–120.0 kJ mol–1

Sol. Answer (4) Given reaction :

H H

C=C

H H

H H C H

+H–H

C

H H ; Hr H

Hr = (B.D.E)reactants – (B.D.E)products = HC = C + 4 HC – H + HH – H – Hc – C – (HC – H) × 6 = (606.10 + 4 × 410.5 + 431.37 – 336.49 – 6 × 410.5) kJ mol–1 = – 120.0 kJ mol–1 45. The values of H and S for the reaction, C(graphite)  CO2 (g)  2CO(g) are 170 kJ and 170 JK–1, respectively. This reaction will be spontaneous at (1) 510 K

(2)

710 K

(3)

910 K

(4)

1110 K

Sol. Answer (4) Reaction will be spontaneous at higher temperature when G < 0 At 1110 K H – TS G = (17000 J – 1110 × 170 J) = – 171.7 kJ i.e., G = – ve Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

84

Thermodynamics

Solution of Assignment (Set-2)

46. Standard entropies of X 2, Y 2 and XY 3 are 60, 40 and 50 JK –1 mol –1 respectively. For the reaction 1 3   X  Y   XY3 , H  30kJ to be at equilibrium, the temperature should be 2 2 2 2

(1) 500 K

(2)

750 K

(3)

1000 K

(4)

1250 K

Sol. Answer (2) For the reaction to be at equilibrium, G = 0  H = TS Now,

3 ⎛1 ⎞ S =  ⎜ S X2  S Y2 ⎟  S XY3 2 2 ⎝ ⎠ 60 3 ⎛ ⎞   40 ⎟ J K 1 = ⎜ 50  2 2 ⎝ ⎠

= – 40 J K–1  T=

H 30000 J   750 K S 40

47. Match Column-I (Equations) with Column-II (Type of process) and select the correct option Column-I

Column-II

Equations

Type of processes

a. Kp > Q

(i) Non-spontaneous

b. G° < RT ln Q

(ii) Equilibrium

c. Kp = Q

(iii) Spontaneous and endothermic

d.

T

H S

(iv) Spontaneous

(1) a(i), b(ii), c(iii), d(iv)

(2)

a(iii), b(iv), c(ii), d(i)

(3) a(iv), b(i), c(ii), d(iii)

(4)

a(ii), b(i), c(iv), d(iii)

Sol. Answer (3) 48. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be (1) Infinite

(2)

3 joules

(3)

9 joules

(4)

Zero

Sol. Answer (4) For a spontaneous process, G = H– TS < 0 or, H < TS or,

H T S

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Solution of Assignment (Set-2)

Thermodynamics

85

For a non-spontaneous process, G > 0 or, G° + RT ln Q > 0 [∵ G = G° + RT ln Q] ∵ RT ln Q > 0 So, G° has to be greater in magnitude than RT ln Q for the reaction to be spontaneous. ∵ G° < RT ln Q Non spontaneous kp = Q  G = 0 i.e. process is at equilibrium. When kp > Q.  Reaction will move in forward direction  Spontaneous. 49. For vaporization of water at 1 atmospheric pressure, the values of H and S are 40.63 kJ mol–1 and 108.8 JK–1 mol–1, respectively. The temperature when Gibbs energy change (G) for this transformation will be zero, is (1) 273.4 K

(2)

393.4 K

(3)

373.4 K

(4)

293.4 K

(3)

–10.3 kJ

(4)

+6.2 kJ

Sol. Answer (3) ∵ Pext = 0  w=0 50. The following two reactions are known : Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g); H = – 26.8 kJ FeO(s) + CO(g)  Fe(s) + CO2(g); H = – 16.5 kJ The value of H for the following reaction Fe2O3(s) + CO(g)  2FeO(s) + CO2(g) is (1) +10.3 kJ

(2)

– 43.3 kJ

Sol. Answer (4) Fe2O3  CO  2 FeO  CO2 ;  r

Given: Fe2O3  3CO  2 Fe  3CO2 ; 1

…(1)

FeO  CO  Fe  CO2 ;  2

Multiplying by (2), and reversing, we get

2 Fe  2 CO2  2 FeO  2CO ;   2  2

…(2)

Adding 1 and 2, we get Fe2O3  CO  2 FeO  CO2 ; Hr  H1 – 2 H2

= [– 26.8 – (2 × – 16.5)] kJ = + 6.2 kJ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

86

Thermodynamics

Solution of Assignment (Set-2)

SECTION - D Assertion - Reason Type Questions 1.

A : Cdiamond  Cgraphite H and U are same for this reaction. R : Entropy increases during the conversion of diamond to graphite.

Sol. Answer (2) For the reaction, ng = 0  H = U Also, graphite is more disordered because of sp2 hybridization and  electrons.  S is +ve 2.

A : Specific heat is an intensive property. R : Heat capacity is an extensive property.

Sol. Answer (2) q Specific heat C = m T q Extensive   Intensive m Extensive

Heat capacity, S =

q T

∵ q is extensive  S is also extensive 3.

A : All reactions which are exothermic are spontaneous. R : All reactions in which entropy increases are spontaneous.

Sol. Answer (4) Not all exothermic reactions are spontaneous. Neither those reactions in which entropy increases. Only those reactions are spontaneous for which G < 0. 4.

A : Enthalpy of neutralisation of 1 equivalent each of HCl and H2SO4 with NaOH is same. R : Enthalpy of neutralisation is always the heat evolved when 1 mole acid is neutralised by a base.

Sol. Answer (3) 1 eq of strong acid and strong base gives out – 57.1 kJ g eq–1 Enthalpy of neutralisation is heat evolved when 1 g equivalent of acid reacts with 1 g equivalent of base. 5.

A : q and w are path function. R : q + w is a state function.

Sol. Answer (2) q and w are path functions, while, q + w = U is a state function and do not depend upon path. 6.

A : Dissolution of sugar in water proceed via increases in entropy. R : Entropy decreases when egg is hard boiled.

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Thermodynamics

87

Sol. Answer (3) Dissolution of sugar in water results in increase in entropy

C12H22O11  H2O  C6H12O6  C6H12O6 Glucos e

fructose

When egg is boiled, bonds are broken and egg attains a more disordered state  S = +ve Also, egg is boiled  spontaneous process. So, G < 0 at 100°C ∵ H is +ve, S must be +ve for G to be –ve 7.

A : For an isolated system G = –TStotal. R : For an isolated system q = 0.

Sol. Answer (2) For an isolated system, H = 0  G = – TStotal 8.

A : Combustion is an exothermic process. R : Combustion is a spontaneous process.

Sol. Answer (2) Combustion is exothermic and also spontaneous. Since G < 0. 9.

A : Total enthalpy change of a multistep process is sum of H1 + H2 + H3 + ..... R : When heat is absorbed by the system, the sign of q is taken to be negative.

Sol. Answer (3) ∵ H is a state function, So for multistep process, H = H1 + H2 + H3 + …… Also, when heat is absorbed by the system, q is +ve. 10. A : Bond energy is equal to enthalpy of formation with negative sign. R : Bond energy is energy required to dissociate 1 mole single bond. Sol. Answer (3) Bond energy is the energy required to dissociate 1 mole of bonds (can be single, double or triple)  Reason is false. Now, Hr = (B.E.)reactants – (B.E.)products Lets take an example, of H2O

H2 

1 O2  H2O ; Hf 2

Hf = (B.E.)reactants – (B.E.)products Hf = – (Hf)reactants + (Hf)products Comparing, (Hf) = – B.E. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

88

Thermodynamics

Solution of Assignment (Set-2)

11. A : H is positive for endothermic reactions. R : If total enthalpies of reactants and products are HR & HP respectively then for an endothermic reaction HR < HP. Sol. Answer (1) If HR and HP are total enthalpies of reactants and products respectively, then for an endothermic reaction HR < HP or, HP – HR < 0  H < 0 12. A : The energy of the universe is constant, whereas the entropy of the universe is continuously increasing. R : For spontaneous process S > 0. Sol. Answer (2) For universe, energy is constant while S > 0 Also, for a spontaneous process, Suniverse > 0 13. A : A non-spontaneous process becomes spontaneous when coupled with a suitable spontaneous reaction. R : The overall free energy of coupled spontaneous reaction is negative. Sol. Answer (1) Overall free energy change for coupled spontaneous reaction is –ve. Also, A non spontaneous reaction becomes spontaneous if coupled with another reaction so that Goverall < 0. 14. A : An ideal crystal has more entropy than a real crystal. R : An ideal crystal has more disorder. Sol. Answer (4) An ideal crystal has lesser entropy (more ordered) than a real crystal. 15. A : Work done in an irreversible isothermal process at constant volume is zero. R : Work is assigned negative sign during expansion and is assigned positive sign during compression. Sol. Answer (2) Work done at constant volume = 0 ∵ w = – pext V Also, according to 10 PAC convention, expansion  w = –ve compression  w = +ve.

  

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