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

Solutions

Chapter

6

Thermodynamics

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.

6. For the reaction PCl₅(g)  PCl₃(g) + Cl₂(g)

(1) H = E (2) H > E (3) H < E (4) Can’t predicted

Sol. Answer (2) 5 3 2 PCl (g)PCl (g) Cl (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 (2) E = r – s (3) E = r (4) E = 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

(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

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

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, n_g = 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 = – x₂cal

What is the heat of transition of graphite into diamond?

(1) x₁ + x₂ (2) x₂ – x₁ (3) x₁ – x₂ (4) x₁x₂ Sol. Answer (2) Required equation graphite diamond C C graphite 2 2

 

C O CO g ; H  x, cal  

 

2 diamond 2 2 CO C O g ; H  x cal

Adding, we get C_graphite_ C_diamond_   H x – x cal2 1 13. For the given reactions, A D ; H = x. Steps involved are

A B ; H₁ = x₁ B C ; H₂ = ? C D ; H₃ = x₃

(1) H₂ = x – (x₁ + x₃) (2) H₂ = x + x₁ + x₃ (3) H₂ = x₁ – x₃ – x (4) H₂ = (x₁ + x) – x₃ Sol. Answer (1)

Given reactions are

A H1 B H2 C H3 D H We know, H = H1 + H2 + H3 or, H₂ = H – H₁ – H₃ = x – (x₁ + x₃)

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 Sol. Answer (4) yellow 2

 

2 5

 

1 5 1 P O g P O s ; H 9.91kJ 2 2       red 2

 

2 5

 

2 5 1 P O g P O s ; H 8.78 kJ 2 2      Rearranging, we get yellow 2

 

2 5

 

5 P O g P O g ; H 9.91kJ 2       

 

2 5 red 2 5 P O P O g ; H 8.78 kJ 2     

Adding these two equations, we get

yellow  red





P P ; 9.91 8.78 kJ   1.13 kJ

15. If the heat of formation of NO₂ is ‘x’

[½ N_2(g) + O_2(g)  NO_2(g)] the heat of reaction N_2(g) + O_2(g)  2NO_(g) is y and the heat of reaction 2NO_(g) + O_2(g)  2NO_2(g) is z, then

(1) 2x + z = y (2) 2y + z = x (3) 2x – z = y (4) 2z + x = y Sol. Answer (3) Given:

 

 

 

2 2 2 1 2 N g O g NO g ; H x 2 ⎛ ⎞ _⎜     _⎟ ⎝ ⎠ = N g₂

 

2O₂ 2NO g ; H₂

 

 ₁2x Also,

 

2 2 2 2 N O 2NO g ; H y 2NO O 2NO ; H z         Adding, we get 2 2 2 2 N 2O 2NO ; H  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 (3) x = y

Sol. Answer (3)

2

HCl NaOH NaCl H 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,

2 4 2 4 2

H SO 2NaOHNa SO 2H O 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. H_f 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:

graphite 2

 

2 4

 

f

2C 2H g C H g ; H

Hf = Bond dissociation enthalpy of reactants – Bond dissociation enthalpy of products

= 12.5 = (171 × 2) + 2 × 104.3 – (4 × 99.3 + BE_{C = 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:

 

 

 

 

6 6 2 2 2 2C H l 15O g 12CO g 6 H O 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

Sol. Answer (4) rhombic 2

 

2

 

S O g SO g ; H –297.5 kJ  …(1) monoclinic 2

 

2

 

S O g SO g ; H –300 kJ  …(2) Subtracting (2) from (1), rhombic monoclinic





S S ; 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)

 

 

2 2 2 4 f H g  S 2O g H SO ; Given: 2 2 S O SO ; H = – 298.2 kJ 2 1 2 3 SO O SO 2   ; H = – 98.7 kJ 3 2 2 4 SO H OH SO ; H = – 130.2 kJ 2 1 2 2 H O H O 2   ; H = – 287.3 kJ

Additing all these, we get

2 2 2 4

H  S 2O H SO ; H' = – 814.4 kJ

 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 = – P_ex + V.

Given, pressure is 103 _{N m}–2

and, V = 0.25 m3

 w = – 103 _Nm–2 _{× 0.25 m}3 _{= – 250 J}

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)

 

2 2 4oxalic acid 2

 

2

 

2

 

1 H C O s O g 2CO g H O 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.

 x Hne4 2 LiOHLi OH HCl LiCl H O 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)

Sol. Answer (3)

Maximum heat is absorbed by NH₄NO₃

 H = + 16.08 (maximum positive value) 25. HA + OH–H₂O + A– + q₁ kJ

H+ _{+ OH}–_{ H}

2O + q2 kJ

The enthalpy of dissociation of HA is

Sol. Answer (3) HAH A ; H   Given: 2 1 1 HA OH  H O A q kJ ; H – q kJ  2 2 2 q kJ H O HOH ; H   q kJ Adding, we get



2 1



HAHA ; H    q – q 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 180 100  kJ ∵ wt. of 1 gram molecule = 180 g which gives 1560 180 100  kJ = 2808 kJ

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 – n_g 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

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 (4) –321.3 + 900R Sol. Answer (3) COOH

 

 

2 2 2 15 O g 7CO g 3H O 2    ng = 7 – 15 2 = – 1 2 We know, H = U + ng RT H = qP U = qV  qP = qV + ng RT = – 321.3 + ⎛⎜ ₂1 300 K R ⎞⎟ ⎝ ⎠ = – 321.3 – 150 R 30. H (g)₂ 1₂O (g)₂ H O( )₂  H₂O_(l)  H₂O_(g); H = x₄ Given, E_H–H = x₁ E_O=O = x₂ E_O–H = x₃ HF of H2O vapour is (1) ₁ 2 x₃ x₄ 2 x x    (2) _{3 1} 2 x₄ 2 x x x 2    (3) ₁ 2 2x₃ x₄ 2 x x    (4) ₁ 2 2x₃ x₄ 2 x x    Sol. Answer (4)

 

 

Hf

 

2 1 2 2 H g O g H O g 2    H O ( )_{2 } H x₄ Now, for H g₂

 

1O g₂

 

H O₂

 

; H 2    

H = (B.D.E)reactants – (B.D.E)products = 1 2 3

1 x x 2x 2   Hf = H + x4 = 1 2 3 4 1 x x 2x x 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

2 mole of O2

If the gas is C₂H₄

2 4 2 2 2

C H 3O 2CO 2H O

If the gas is propylene

3 6 9 2 2 2

C H O 3CO 3H O

2

  

It is clear that propylene (1 mole) requires 9

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

ml of oxygen and hence 1 part of gas requires 9

2 parts of oxygen by moles.

 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

 P

V

C _{5 1.66}

C    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

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)

 

 

2 1mole (2g) H g 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 (3) 130 kcal mol–1 (4) 80 kcal mol–1

Sol. Answer (4) C H H H H C(g) + 4 H (g) ; H 7 2 C(g) + 6 H (g) ; H 2 C – C – H H H H H H

Given: H₁ = 360 kcal mol–1 _; H

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

=



2 E



_{C H} E_{C H}



2 E_{C C} 5 E_{O O}



–



4 E



_{C O} E_{C O}



2 E



_{O H} E_{O H}





=



4E_{C H} 2 E_{C C} 5 E_{O O}

 

 8E_{C O} 4E_{O H}



37. Using bond energy data, calculate heat of formation of isoprene

H C = C – CH = CH_{2 2} CH3

5C(s) + 4H (g)₂

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 H (g)2 Hf _{C = C – C = C} H H H C(g) + 4 H (g)2 5x C H H H H H H

 H = (B.D.E)reactants – (B.D.E)products

= 4 E_{H – H} – (2 E_{C = C} + 8 E_{C – H} + E_{C – 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)

2 4 2

N O 2NO

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

2 4 2

N O   H_2NO

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,

3 2

CaCO (s)CaO(s) CO (g)

1 solid reactant gives 1 solid and 1 gaseous product and as a result increases disorder liness  S = positive

40. For stretched rubber, Entropy

(1) Increases (2) First increases then decreases

(3) Decreases (4) First decreases then increases

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

(1) Ice (2) Liquid water

(3) Steam (4) Randomness is same in all

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)

 

 

 

3 2

MgCO s MgO s CO g

S = S_CO2S_MgOS_MgCO3

∵ MgO and MgCO3 are solids, so their entropy is almost same.

 S = S_CO2 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) 2 2 3 1_N 3_{H NH} 2  2  S° = 3 2 2 NH H N S    S S = (192.6 JK–1 _mol–1 _{× 1 mol) –} 3 130.5 JK 1 191.5JK 1 2 2   ⎛ _{ } ⎞ ⎜ ⎟ ⎝ ⎠ = 192.6 JK–1 _{– 195.75 JK}–1 _{– 95.75 JK}–1 = – 98.9 JK–1 _{for 1 mole of NH} 3

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

Sol. Answer (2)

Smix = – n R × 2.303



xH2 log H2 xO2 log O2



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

2 2 H O 1 x x 2   ∵ moles of O2 = 11.2 22.4 = 0.5 moles of H₂ = 11.2 22.4 = 0.5

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,  4 2H C

S = 220 J mol–1K–1, SoH₂ = 130 J mol–1K–1. Then S° for the process

C₂H₄ + H₂  C₂H₆ is

(1) +25 J (2) –125 J (3) 135 J (4) 315 J

Sol. Answer (2) For the reaction,

2 4 2 2 6

C H H C H

 S° = SC H4₂  S C H_{2 4}  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 = fusion H T   T = fusion fusion H S   = 1 1 1 7.26 cal mol 6.73 cal mol k    = 1078.75 K  T = 1078.75 K or (1078.75 – 273)°C = 805.75°C

47. The S for the reaction 2H₂(g) + O₂(g)  2H₂O(l) at 300 K when 2 o H S (g) = 126.6, 2 O So (g) = 201.20,  O H2 S (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)

 

 

 

2 2 2

2H g O g 2H O 

 S = 2 S _{H O2} – 2 S



 _H2 S_O2



= 2 × 68.0 JK–1 _mol–1 _–



2 126.6 J mol J K 1 1201.20 J K mol1 1



= [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)

    _H T S is always positive 

Always

positive Alwayspositive

 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,



 

vap m



1 vap m H _{9710 cal mol} S T 373 K      = 26.03 cal mol–1 _K–1

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

Sol. Answer (2)

A reaction is spontaneous when G < 0, We know,

G =    H _T S_

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

55. At 27°C the reaction,

C₆H_6(l) + 15/2 O_2(g)6CO_2(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

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) C_{p, m} – C_{v, m} = R (2) C_p – C_v = R

2 (3) Cv – Cp = –2R (4) Cp – Cv = 0

Sol. Answer (1)

For two, three or even thousand moles of an ideal gas,

C_{p, m} – C_{v, m} = R ; where C_{p, m} is molar heat capacity at const. pressure while C_{v, 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) 2 1 Ca O (g) CaO(s) 2   (2) CaCO₃(s)  CaO(s) + CO₂(g) (3) C(s) + O₂(g)  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, CaCO s₃

 

CaO s

 

CO g₂

 

ng = +1 which is maximum for this case.

4. 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 75

18 J

 Specific heat capacity = 75₁₈ J g–1 _K–1

 q = mCT or, T = 1 1 q 1000 J 75 mC _{1000 g J K g} 18     = 18 K 75 = 0.24 K

8. 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 = – P_ex + V

= – P V (∵ pressure is constant)

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

= – nRT (Considering ideal behaviour)

n = 16000 g₁ 500 mol

32 g 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

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

2O2(g)

H = 30.5 kJ mol–1 _{and S 0.066 kJ mol}–1 _K–1

(1) 462.12 K (2) 362.12 K (3) 262.12 K (4) 562.12 K Sol. Answer (1) At equilibrium, G = 0  H = TS (at equilibrium) T = 1 1 1 H 30.5 k J mol _{462.12 K} S _{0.066 k J mol K}       

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

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 (4) All of these

Sol. Answer (4)

For an isothermal reversible expansion, T = 0

 U or E = 0 Also, H = m C_p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

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 = p H mC  C_p is intensive variable and m V = d  m = dV  T = p H 1 V C d    We know, H  V

 H = KV (where K is proportionality constant)  T =

p

K

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

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 2 2 2 1 N g O g NO g ; H 8.0 kcal mol 2     

Reversing and multiplying 2,

2 2 2 2NO N 2O ; H – 16 kcal  Also, 2 2 2 4 N 2O N O ; H 4 kcal  Adding, we get 2 2 4 2NO N O ; H  12 kcal

17. If 1

2X2O(s)  X(s) +

1

4O2(g); H = 90 kJ. Then heat change during reaction of metal X with one mole of O2

to form oxide to maximum extent is

(1) 360 kJ (2) –360 kJ (3) –180 kJ (4) +180 kJ Sol. Answer (2) Given, 1₂X O s₂

 

X s

 

 1₄ O g ; H 90 kJ₂

 

   X s

 

1O₂

 

g 1 X O s ; H₂

 

90 kJ 4 2     

Multiplying with 4, we get

 

2

 

2

 

4X s O g 2X O 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)

 

2

 

2

 

2CO g O g 2CO g ng = 2 – 3 = –1 We know, H = U + ngRT H = U – 1 × RT  H = U – RT  H < U 8 10 2 8 10   ⎡ ⎤ ⎢_{⇒ } ⎥ ⎣ ⎦

20. Bond dissociation energy of XY, X₂ and Y₂ (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and H_f 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

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, X g₂

 

Y g₂

 

2XY g ; H

 



 H = – 2 × (B.D.E)XY +





B.D.E



X₂ 



B.D.E.



Y₂



= – 2x + (x + 0.5 x) = – 0.5x Also, Hf of XY is – 200 kJ mol –1 i.e. 1 X g₂

 

1 Y g₂

 

XY g ; H

 

_f 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 (3) 32 (4) 64

Sol. Answer (2) We know,

Vapour density =

2

Molecular mass of substance Molecular mass of H

 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) w = – 300x R ln ₁₀y_y (3) w = – 300x R ln 10 (4) w = 100x R ln _y1 Sol. Answer (3) Work done, w = – nRT ln 2 1 V V = – x × R × 300 ln 10 y_y = – x × R × 300 ln 10 = – 300x R ln 10

23. Enthalpy of formation of NH₃ is – X kJ and H_H–H, H_N–H are respectively Y kJ mol–1 andZ kJ mol–1. The value of H_{N  N} is (1) Y – 6Z + 3 X (2) – 3Y + 6Z – 2X (3) 3Y + 6Z + X (4) Y + 6X + Z Sol. Answer (2)

 

 

 

2 2 3 f 1_{N g} 3_{H g NH g ; H – x kJ} 2  2   

Hf = H = (B.D.E)reactants – (B.D.E)products

= 1 H_N2 3 H_H2 3 _{N H} 2   2     = HN N 3 XY 3XZ 2 2  _{ } We have, – X = HN N 3Y 3 Z 2 2  _{ } or, H_{N N} = ⎛⎜3Z3Y₂ X⎟⎞2 ⎝ ⎠ = 6Z – 3Y – 2X

24. A system X undergoes following changes

) T V P ( ) T V P ( ) T V P ( ) T V P ( ₁X₁₁  ₂W_{2 1} ₃Z_{2 2}  ₁X₁₁ The overall process may be called as

(1) Reversible process (2) Cyclic process

(3) Cyclic reversible process (4) Isochoric process

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) 2 1 . 57  kJ

(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.

26. The value of H° in kJ for the reaction will be CS₂(l) + 4NOCl(g)  CCl₄(l) + 2SO₂ (g) + 2N₂(g) if f 2 H (CS ) x     H (NOCl)_f   y f 4 H (CCl ) z     H (SO )_f ₂   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,

 

 

 

 

 

2 4 2 2 CS l 4NOCl g CCl l 2SO g 2N g H° = Hf_CCl4  



Hf_SO2

  

  2 Hf_N2   2 Hf_CS2(l)  



HNOCl



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)

 

 

 

 

4 2 2 2 CH g 2O g CO g 2H O l ; H H is 890 kJ for 1 mole CH4

We know, 22.4 L means 1 mol

2.4 L means 1 2.4

22.4 mol = 0.107 mol

 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)

 

2 2

2Fe s 4HCl2FeCl 2H

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?

(1) V P dT dE dT dH _⎟ ⎠ ⎞ ⎜ ⎝ ⎛  ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ (2) R dT dE dT dH V P  ⎟ ⎠ ⎞ ⎜ ⎝ ⎛  ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ (3) T dV dE ⎟ ⎠ ⎞ ⎜ ⎝

⎛ _{for ideal gas is zero (4) All of these}

Sol. Answer (3) For an ideal gas,

dU or dE = 0 when T is constant  T dU dV ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ i.e., dU dV at constant volume is 0.

30. A schematic representation of enthalpy changes for the C_(graphite)₂1O₂(g)CO(g)reaction, is given below. The missing value is

Cgraphite+ O (g)₂ CO (g)₂ CO(g) + ½O(g) – 393.5 kJ – 283.0 kJ ?? (1) + 10.5 kJ (2) – 11.05 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

32. Different types of systems are given below Work Heat System Surrounding Matter Energy System Surrounding 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 – 75.6 kcal_{60 g}  1 g yields 745.6 4.2kJ g 1 –52.2 kJ g 1 60     _

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) 1O (g)₂ MgO(s) 2   (2) 1C (graphite) 1O (g)₂ 1CO (g)₂ 2 2 2 (3) 2 1 C (graphite) O (g) CO(g) 2   (4) CO(g) 1O (g)₂ CO (g)₂ 2   Sol. Answer (3) For the reaction

graphite 2

 

 

1 C O g CO g 2   ng = 1 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 – n_gRT The reaction is

 

 

2 2 H O l H O 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)

3 1

fus

fus H 1.435 10 cal mol

S T 273 K       = 5.26 cal mol–1 _K–1

4. 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 (∵ U is proportional to temperature)

5. 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) 1 vap 1 1 vap H _{30000 J mol} S 100 J mol K T 300 K        

6. Enthalpy change for the reaction, 4H_(g) 2H_2(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

Sol. Answer (4) Given reaction:

 

2

 

4 H g 2 H 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

Sol. Answer (4) Given: 1 A B 2  ; H = + 150 kJ mol–1  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) + 1

2O2(g)  CO(g); rH° = y kJ mol–1

(iii) CO (g) + 1

2O2(g)  CO2(g); rH° = z kJ mol –1

Based on the above equations, find out which of the relationship given below is correct:

Sol. Answer (3)

Subtracting equation (ii) from (i), we get

2 2 1 CO O CO ; 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) 2 2 H Cl 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)

 

 

 

5 3 2 PCl g PCl g Cl 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

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 (H_comb (CH₄) = 890 kJ mol–1_{, H}

comb (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,

 

 

 

 

4 2 2 2 CH g 2 O g CO g 2H O x 2x    l Volume reacted: For propane,



C H5 x3 8

 

5 5 x5O2



2CO2 4H O2      Volume reacted:

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

 Moles of methane combusted = _22.43 mol

Energy due to methane combusted = 3 890

22.4 kJ

Volume of propane combusted = 2 L

 Moles of propane combusted = 2

22.4 mol

 Energy due to propane combusted = 2 2220

22.4 kJ

 Total energy released

= ⎛⎜_22.43 890 _22.42 2220⎞⎟ ⎝ ⎠kJ = 1 3 890 4440 kJ





22.4   = 1 2670 4440 kJ





22.4  = 317.41 kJ

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

Sol. Answer (4) 2 4 2 2 2 r C H 3 O 2 CO 2 H O ; H Hr = Enthalpy of combustion Now,

 

2

 

2 4

 

1 2 C s 2 H g C H g ; H

 

 

 



2 2 2



2 C s O g CO g ; H …(1)

 

 

2 1 2 2 3 2 H O g H O g ; H 2 ⎛ ⎞ _⎜    _⎟ ⎝ ⎠ …(2) Also, we have,

 

 

2 4 2 1 C H 2 C s 2 H g ; – H …(3) Adding (1), (2), (3), we get 2 4 2 2 2 r 1 2 3 C H 3 O 2 CO 2 H O ; H  – H  2 H  2 H 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

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) 2 x 2 y  (2) 2x – y (3) y – 2x (4) 2 y x 2  Sol. Answer (1)

 

1 2 r C s O CO ; H 2    2 2 C O CO ; H   x kJ 2 1 2 y CO CO O ; H kJ 2 2      Adding, we get 2 r 1 y y 2x C O CO : H x 2 2 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) Sol. Answer (4) 2 2 S O SO : H Given : S 3 O₂ SO : H_{3 1} 2x kcal 2      and, SO₃ SO₂ 1O₂ 2   : H ₂ y kcal Adding, _{S O 2 SO : H2   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 (4) 0.977 J/mol·K

Sol. Answer (1) 1 1 1 fusion fusion H 2930 J mol S 9.77 J mol K 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 (4) H = E – 2RT

Sol. Answer (1) We know,

H = U + pV For an ideal gas,

H = U + ngRT

For the given reaction, n_g = 2 – 3 = – 1  H = U + (–1) RT

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) 2 2 2 2 r 2H O 2H O O ; H  Given: 1 2 2 2 2 1 H O H O ; H  188 kJ mol  2



H O2 2 H2 O ; – H2  1188 kJ mol1



1 2 1 2 2 2 and 2 H O H O ; H – 286 kJ mol 2  ⎛ ⎞ _⎜     _⎟ ⎝ ⎠ Adding, we get





2 2 2 2 r 2 1 2H O 2H O O ; H    2 2 – = [– 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 25. Enthalpy of CH₄ +

2 1

O₂  CH₃OH is negative. If enthalpy of combustion of CH₄ and CH₃OH are x and y respectively. Then which relation is correct?

(1) x > y (2) x < y (3) x = y (4) x  y Sol. Answer (1) 4 1 2 3 CH O CH OH ; H 0 2     Given: 4 2 2 2 1 CH 2 O CO 2 H O ; H  x …(1) 3 3 2 2 2 r CH OH O CO 2 H O ; H y 2     

 CO22H O2 CH OH3 3₂O ; H'2   y …(2) Adding (1) and (2), we get

4 1 2 3 CH O CH OH ; H x y 2      We know, H < 0  x – y < 0 ∵ | x | > | 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, S = 2.303 nR 2 1 V log V 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 _× log20

2 = 9.2 cal K–1

29. Heat of combustion for C(s), H₂(g) and CH₄(g) are –94, –68 and –213 kcal/mol, then H for

C(s) + 2H₂(g)  CH₄(g) is

Sol. Answer (1) 2 4 C 2H CH ; H Given; 2 2 1 C O CO ; H …(1) 2 1 2 2 2 2 H O H O ; H 2 ⎛ ⎞ _⎜    _⎟ ⎝ ⎠ …(2) 4 2 2 2 3 CH 2 O CO 2 H O ; H  CO₂2 H O₂ CH₄2 O₂ …(3) Adding (1), (2) and (3), we get

2 4 2 3

C 2 H CH ; H    H 2 H – H

= [– 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) 1 1 1 fusion fusion H 6.0 kJ mol S 21.98 J mol K T 273 K        

32. For which one of the following equations is  H_react equal to  H_f 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, H_f = H_reaction

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: C_{p, m} = 75 J K–1 _mol–1 We know, q_p = C_{p , m} × n T or, T = p _{1 1} p,m q _{1000 J 100} C _n _{75 J K mol}   18 mol = 2.4 K

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 = ⎡⎢382.64 298 K ⎛⎜₁₀₀₀145.6⎞⎟⎤⎥ ⎝ ⎠ ⎣ ⎦ 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) S_system + S_surroundings > 0 (2) S_system – S_surroundings > 0

(3) S_system > 0 only (4) S_surroundings > 0 only Sol. Answer (1)

For a spontaneous process, Suniverse > 0

or, S_system + S_surrounding > 0

36. The work done during the expansion of a gas from a volume of 4 dm3 _{to 6 dm}3 _{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 = – p_ext 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

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,

 

 

2 2

(Week base)

MgO s 2 HCl aq MgCl H O

∵ MgO is weak, so extra energy is required for its complete dissociation Hence, H_n < 57.1 kJ mol–1

40. The enthalpy and entropy change for the reaction : Br_2(l) + Cl_2(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 = H S   = 1 1 1 30000 J mol _{285.7 K} 105 J K mo    

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

Energy E Eres non-resonated benzene resonated benzene 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)+

2 1

O₂(g)  H₂O(l), H =–X₂ kJ mol–1

(iii) CO₂(g) + H₂(g)  CO(g) + H₂O, H = –X₃, kJ mol–1 (iv) C₂H₂(g) + O₂

2

5 _{(g)  2CO}

2(g) + H2O(l), H = +X4 kJ mol–1

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

Sol. Answer (4)

 

 



1 2 1 2 2 2 H g O g H O l ; H – X kJ mol 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

Sol. Answer (4) Given: 2 2 Zn O ZnO ; G  – 616 J 2 2 ZnS2 Zn S ; G    293 J 2 2 2 S 2 O 2 SO ; G   408 J Adding, we get





2 2 2 ZnS 3 O 2 ZnO 2 SO ; 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,

C = C + H – H H – C – C – H H H H 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 H H C = C + H – H H_H C C H ; H r H H H

Hr = (B.D.E)reactants – (B.D.E)products

= H_{C = C} + 4 H_{C – H} + H_{H – H} – H_{c – C} – (H_{C – 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