Solution

WS

JEE main - Chemistry

Section A
1.
(c) -2.27 kJ
Explanation:
-2.27 kJ

2.
(c) q = 0, ΔT = 0, w = 0
Explanation:
For free expansion of gas w = 0
Also for adiabatic process q = 0; Thus ΔT = 0 because
Δu = q + w =0

3. (a) 0.385 J g-1 K-1

Explanation:
Heat capacity of copper
Heat absorbed 3.85 J
= =
Rise in temperature 2 K

Specific heat capacity of copper

Heat capacity
= 0.385 J g-1 K-1
3.85 J
= =
Mass (ing) 2 K×5 g

4.
(c) -114 kJ
Explanation:
In the reaction, NaOH(aq)+ HCl(aq) ⟶ NaCl(aq)+H2​O(l), one equivalent of hydrogen ions is neutralized with one equivalents
of hydroxide ions. The enthalpy change is -57 kJ.
In the reaction Ba(OH)2​+H2​SO4​(aq) ⟶ BaSO4​(s)+2H2​O(l), two equivalents of hydrogen ions are neutralized with 2
equivalents of hydroxide ions. The enthalpy of neutralization is 2 × (−57kJ)=−114 kJ

5. (a) 62 kJ
Explanation:
According to Kirchoff s relation,
T2
ΔH = n ∫ Cp dT ‘
T1

where, ΔH = Change in enthalpy.

Cp = Heat capacity at constant pressure.
Given, n = 3 moles, T1 = 300K, T2 = 1000 K, Cp = 23 + 0.01 T
On susbstituting the given values in Eq. (i), we get
1000 100
ΔH = 3 ∫ (23 + 0.01T )dT = 3 ∫ 23dT + 0.01T dT
300 300

2 1000
0.01T
= 3[23T + ]
2
300

0.01 2 2
= 3 [23(1000 − 300) + (1000 − 300 )]
2

= 3 [16100 + 4550] = 61950 J ≈ 62 kJ

6. (a) (Cv - R) In 2
Explanation:
The expression of entropy is given as
T2 V2
Δ S = 2.303 Cvlog T1
+ 2.303 R log V1

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 We have T2 = 2T1 and V1 = 2V2
So we get,
ΔS = 2.303 Cvlog + 2.303 R log
2 1

1 2

Or we can write it as
ΔS = (Cv - R) In 2

7.
V2
(c) w = nRT ln (isothermal reversible expansion of an ideal gas)
V
1

Explanation:
For isothermal reversible expansion.
V2
w = -nRT ln V1

8.
(c) -602 kJ/mol
Explanation:
Heat of decomposition, ΔE = ms ΔT
= 1 × 1.23 × 6.12 = 7.5276 kJ
Molar heat of decomposition for NH4NO3
= 7.5276 × 80 = 602.2 kJ/mol

9.
(b) ΔU = 0, ΔStotal ≠ 0
Explanation:
For spontaneous process, ΔStotal > 0 and since irreversible is always spontaneous
∴ Δ Stotal > 0
For isothermal process, ΔT = 0. So, ΔU = 0

10.
(c) ΔG = -ve
Explanation:
Precipitation of BaCrO4 takes place spontaneously.

11.
(b) 373.4 K
Explanation:
ΔG = ΔH - TΔS

ΔG = 0, ∴ Δ H = TΔS
3
ΔH 40.63 × 10
T= ΔS
=
108.8
= 373.4 K

12.
(b) ΔSA > ΔSB
Explanation:
ΔSA > ΔSB

13.
(d) Heat supplied to the system
Explanation:
Heat supplied to the system is not a state function as during a process its value depends on the path followed. The value of free
energy (Gibbs energy), internal energy, and entropy depends on the state and not on the path followed to get that state, hence

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 these are state functions.

14. (a) There is an increase in entropy.

Explanation:
The mixing of non-reacting gases increases randomness and so increases entropy.
15.
(b) x < y
Explanation:
x<y

16.
(d) i > ii > iv > iii
Explanation:
As stability of cation increases rate of hydrohalogenation increases.

17. (a)

Explanation:
For geometrical isomerism, the carbon atoms involved in the double bond must have different substituents attached to it.

In option , one of the carbon atoms with a double bond has two same groups (CH3) attached to it.

Hence, it does not show geometrical isomerism.

18.
(c) 120°
Explanation:
120°

19. (a) CH3MgBr, C2H5Br

Explanation:

20.
(d)

Explanation:

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 ⊕ N a, N H3 (l)
H

⟶ −→
− −−−−−−→

21.
(b) C2H2
Explanation:
− + 1
H− C ≡ C − H + NaNH 2 → H − C ≡ C Na + H2
2
( C2 H2 ) Sodium ethynide

22.

(c)

Explanation:

Section B
23. 925
Explanation:
Let, Volume of C2H4 is x litre
C2 H4 + 3O2 → 2CO2 + 2H2 O

Initial x −

Final − 2x

CH4 + 2O2 → CO2 + 2H2 O

Initial (16.8 − x) −

Final − (16.8 − x)

Total volume of CO2 = 2x + 16.8 - x

⇒ 28 = 16.8 + x
x = 11.2 L
Volume of C2H4 = 4.2 L, Volume of CH4
= (16.8 - 11.2) L = 5.6 L
n 5.6 n 11.2
CH4 = = 0.25 mol; C2 H4 = = 0.5 mol
22.4 22.4

Heat evolved = 0.25 × (-900) + 0.5 (-1400) = -925 kJ

∴

24. -14.6
Explanation:

i. 2Cu(s) + 1

2
O2(g) → Cu2O(s) : ΔGo = -78 kJ/mol

ii. H2(g) + 1

2
O2(g) → H2O(g), ΔGo = -78 kJ/mol
(i)-(ii) then
2Cu(s) + H2O(g) → Cu2O(s) + H2(g)

Δ Go = -78 + 178 = 100 kJ/mol = 105 J/mol

Now for the above reaction
PH

ΔG = ΔG
∘
+ RT ln( 2

PH O
)
2

To prevent the above reaction: ΔG ≥ 0

PH

Δ Go + RT ln( 2

PH O
) ≥ 0
2

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 PH

105 + 8 × 1250 ln( 2

PH O
) ≥ 0
2

104 (ln P - ln P ) ≥ -105

H2 H2

ln P ≥ -10 + ln P
H2 H2 O

Now,
PH
2O
=X H2 O × Ptotal = 0.01 × 1 = 10-2
ln P H2 ≥ -10 - 2 ln 10
ln P H2 ≥ -10 - 2 × 2.3 (Given ln 10 = 2.3)
ln P H2 ≥ -10 - 4.6

ln P H2 ≥ -14.6

∴ Minimum ln P = -14.6 H2

25. -2.76
Explanation:
For given isothermal reversible expansion,
V2 = 10 V1
∴ For isothermal reversible expansion of 1 mol of an ideal gas if final volume (V2) is 10 times the initial volume (V1), then log10
V2
(
V1
) =1
∴ w = -2303 RT ≈ -(19.15 × T) Joule ≈ -(4.6 × T) cal
For one mole of gas,
w = -(4.6 × T) cal
∴ For 2 moles of gas,

w = -(2 × 4.6 × T) cal

= -(2 × 4.6 × 300) cal
= -2760 cal = -2.76 cal
26. 935
Explanation:
SnO2(S) + C(S) ⟶ Sn(S) + CO2(g)

Δr Ho = [-394] - [-581] = 187 kJ/mole = 187 × 103 J/mol

Δr S0 = [52 + 210] - [56 + 6]
= 200 Jk-1 mol-1
o 3
Δr H
T= o
=
187×10

200
= 935 K
Δr S

27. -2.7
Explanation:
Δ U = 2.1 kcal = 2.1 × 103 cal
Δ ng = 2
Δ H = ΔU + ΔngRT

= 2.1 × 103 + 2 × 2 × 300

= 2100 + 1200
= 3300 cal
Δ G = ΔH - TΔS
= 3300 - 300 × 20
= 3300 - 6000
= -2700 cals
= -2.7 kcal
28. 117
Explanation:
Molar mass of NH3 = 17 g/mol
moles of NH3 given = 17

17
= 1 mole
For 1 mole enthalpy change = 23.4 kJ
5 moles enthalpy change = 2.3 × 5 = 117 kJ

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29. 2480.3
Explanation:
According to Gibbs-Helmholtz equation,
△r G° = △rH° - T△rS°
For a reaction to be feasible (spontaneous)
△r Go < 0

△r Ho - T△rSo < 0

Given, △rHo = + 491.1 KJ mol-1

△r So = 198 JK-1 mol-1

∴ 491.1 × 103 -T × 198 < 0

3

T> 491.1×10

198
= 2480.3 K
∴ Above 2480.3 K reaction will become spontaneous.
30. 27.2
Explanation:
Δ H = ΔU + ΔnRT
or
Δ U = ΔH - ΔnRT
= 29 - (6 - 3) × 2.0 × 10-3 × 300
= 29 - 1.8 = 27.2 kcal
31. 8630
Explanation:
n = 5 mol; T= 300 K; V1 - 10L; V2 = 20L
Work done in isothermal condition.
V2
wrev = −nRT In V1

20
= −5 × 8.3 × 300 ln = −8630.38 J
10

32. 60
Explanation:
q = 50 J; w = +10J
∴ ΔU = q + W = 50 + 10

∴ ΔU = 60J

33. 360
Explanation:
C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O(g)
50% energy used to convert H2O(l) into H2O(g) = 1800

2
= 900 kJ
The energy required to evaporate 1 mole of water = 45 kJ
900
⇒ 900 = n × 45 ⇒ n
H2 O = = 20 mole
H2 O
45

W
H2 O
= 20 × 18 = 360 g
34. -13538
Explanation:
From ΔHo = ΔUo + ΔngRT

Δ Ho = -20 × 1000 - 1 × 8.314 J/mol. K × 298K

= -22477.57 J
Δ Go = ΔHo - TΔSo = -22477.57 - (298 × -30)
= -13538 J
35. 98.7
Explanation:
For isothermal irreversible expansion,
w = -pex × ΔV = -2.0 × (1.0 - 0.5)
= -1.0 L. atm

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 = -1.0 × 101.3 = -101.3 J
For first law of thermodynamics,
ΔU = q + w

= 200 - 101.3 = +98.7 J

36. 0.88
Explanation:
Heat evolved by 100 watt bulb in 15 minutes
= W × t( sec )

4
= 100 × 60 × 15 = 9 × 10 J

Volume of room = volume of air = 5 × 4 × 3 = 60m 3

6 −1
= 60 × 10 mL

∴ Mass of air in room

6 −6
= 60 × 10 × 1.22 × 10 kg

= 73.2 kg
Since, heat given by bulb = Heat taken by (roof + wall) + Heat taken by air
or 9 × 10 = 50 × 10 × ΔT + 73.2 × 10 × 0.71 × ΔT
4 3 3

∴ ΔT = 0.88K

37. 3
Explanation:
State Variable is an independent variable of a state function.
Internal energy, volume and enthalpy are state variable.
38. 62
Explanation:
According to Kirchoff s relation,
T2
ΔH = n ∫ Cp dT ‘
T1

where, ΔH = Change in enthalpy.

Cp = Heat capacity at constant pressure.
Given, n = 3 moles, T1 = 300K, T2 = 1000 K, Cp = 23 + 0.01 T
On susbstituting the given values in Eq. (i), we get
1000 100
ΔH = 3 ∫ (23 + 0.01T )dT = 3 ∫ 23dT + 0.01T dT
300 300

2 1000
0.01T
= 3[23T + ]
2
300

0.01 2 2
= 3 [23(1000 − 300) + (1000 − 300 )]
2

= 3 [16100 + 4550] = 61950 J ≈ 62 kJ

39. 4
Explanation:
Extensive ⇒ Mole, Volume, Gibbs free energy.
Intensive ⇒ Molar mass, Molar heat capacity, Molarity, E cell. ⊖

40. 610
Explanation:
1

2
Cl2(g) → Cl(g) → Cl−(g) → Cl-(aq)
ΔH
reaction =
1

2
× 240 + (-350) + (-380) = -610
41. 1070
Explanation:
△H
△S =
Tmp

30.4×1000
28.4 =
Tmp

Tmp = 1070.422 K
42. -192.5
Explanation:
C(s) + O2(g) → CO2(g);

ΔC H0 [Cgraphite] = -286.0 kJ/mol ...(i)

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 H2(g) + 1

2
O2(g) → H2O(g);

ΔC H0[H2(g)] = -393.5 kJ/mol ...(ii)

C2H6(g) + 7

2
O2(g) → 2CO2(g) + 3H2O(g);

ΔC H0[C2H6(g)] = -1560 kJ/mol ...(iii)

The reaction of formation of ethane is
2C(s) + 3H2(g) → C2H6(g)

∴ Δf H0 of C2H6(g) = 2 × Δ 0H[Cgraphite] + 3 + Δ H0[H2(g)] - Δ H0[C2H6(g)]

C C C

= 2 × (-286.0) + 3(-393.5) - (-1560)

= -192.5 kJ/mol
43. 8
Explanation:
In the given system, during the compression of the spring, the work done is 10 kJ and 2 kJ of heat is escaped to the surroundings.
So, q = - 2 kJ and W = 10 kJ According to the first law of thermodynamics,
ΔU = q + W = -2 kJ + 10 kJ

ΔU = 8 kJ
The change in internal energy, ΔU (in kJ) is 8 kJ.
44. -8
Explanation:
Given, = K ∴ K = atm L
P

V
(initial condition)
1

2
−1

V2 V2 V2 2 2
2 V V
V 2 1
W = − ∫ P ΔV = − ∫ K V ΔV = −K ∫ = −K [ − ]
2 2 2
V1 V1 V1

1 36 4
= − [ − ] = −8 L − atm
2 2 2

45. 2
Explanation:
ΔH
neutaralization = -57.3 kJ/mol
In case of acetic acid
ΔH = ΔH + ΔH − 55.3 = ΔH − 57.3
ioni neutralization ioni
ΔHion
i
= 2 kJ/mol
46. 309.16
Explanation:
Given S(s) + 3F2(g) → SF6(g); ΔH = -1100 kJ ...(i)
S(s) → S(g); ΔH = 275kJ ...(ii)
1

2
F2(g) → F(g); ΔH = 80 kJ ...(iii)
To get SF6(g) → S(g) + 6F(g); we can proceed as (ii) + 6 × (iii) - (i)
∴ SF6(g) → S(g) + 6F(g); ΔH = 1855 kJ
Thus, average bond energy for S-F bond
= = 309.16 kJ
1855

47. 6
Explanation:
∘ B⋅E
−ΔH = +S. E⋅ + I ⋅ E⋅1 + − EGE1 − U
f 2

154
−617 = +S. E. +I . E.1 + − 328 − 1047
2

S.E. + I.E. = 681

(6 + 8 + 1 = 15; 1 + 5 = 6)
48. 1411
Explanation:
C2 H4 ( g) + 3O2 ( g) → 2CO2 ( g) + 2H2 O(ℓ)
−1
ΔU = −1406 kJ mol , T = 300 K
ΔH = ΔU + Δng RT

Δng = Δnp − ΔnR = 2 - 4 = -2

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 ΔH = −1406 + (−2) × 8.3 × 300 = -1406 - 4.98
−1
= −1410.98 kJ mol ≈ −1411 kJ mol

ΔG = ΔH − TΔS, at equilibrium ΔG = 0
∴ ΔH = TΔS = -1411 kJ mol-1
49. 1006
Explanation:
In Bomb calorimeter, the heat is released at constant volume.
By combustion of 1 mole
20×0.5
C2 H6 (ΔU) = − × 30 = −1000 kJ
0.3
7
C2 H6 ( g) + O2 ( g) → 2CO2 ( g) + 3H2 O(l)
2
7 7
Δng = 2 − (2 + ) = −( )
2 2
7
= −1000 −
2
× 8.3 × 300kJ = -1000 - 6.225 = -1006 kJ
So heat released = 1006 kJ mol-1
50. 620
Explanation:
1 → 2 ⇒ Isobaric process

2 → 3 ⇒ Isochoric process

3 → 1 ⇒ Isothermal process

W = W1→2 + W2→3 + W3→1

V2
= (−P (V2 − V1 ) + 0 + [− P1 V ln( )])
1
V1

20
= [−1 × (40 − 20) + 0 + [−1 × 20 ln( )]]
40

= -250 + 20 In 2 = -20 + 20 × 2.3 × 0.3

= -6.2 L bar (1 bar = 100 J) ⇒ |W| = 6.2 L bar = 620 J
51. 101
Explanation:
ΔH
sub
= ΔH
fus.
+ ΔH
vap.
= 2.8 + 98.2 = 101 kJ/mol
52. 200
Explanation:
Moles of coal = 2.4

12
= 0.2

q = CvdT
q = 20 × (300 - 298) = 40kJ
0.2 moles evolved = 40 kJ
1 moles evolved = = 200 kJ/mol
40

0.2

ΔH = -200 kJ/mol

53. 3
Explanation:
There are 3 asymmetric carbon atoms in the given compound:

54. 2
Explanation:
Although the compound has two chiral carbons (indicated by stars), it does not has four optically active isomers as expected. It is
due to its existence in cis-form only.

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 The above-shown transformation does not exist due to restricted rotation about the bridgehead carbons, hence only cis-form and
its mirror image exist.
55. 1
Explanation:
Possible structures of acyclic neutral compound:
a. HC ≡ C − CH = C − NH 2
sp sp 2 2
sp sp

b. CH 2
˙
= CH − C ≡ C − NH 2
2 2 sp sp
sp sp

∴ No sp3 hybridized carbon in any of the compound.

c. CH 3 − CH = CHp H − C ≡ N
3 2 2
sp sp
sp

∴ One sp3 hybridized carbon

56. 6.0
Explanation:
Dimethylcyclopentane has following isomers:

57. 7.0
Explanation:

including this structure there are 7

Hyperconjugation structure (Number of Hyperconjugation structure = no. α - hydrogens)
58. 5.0
Explanation:

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 CH3 CH CH3 CH CH3 CH
3 3 3

| | + | + | | |
H

C H3 −C − C H −C −C H3 ⟶ C H3 −C − C H −C −C H3 ⟶ (A) C H 3 −C −CH2 −C −C H3 + (B)

| | | − H2 O | | | | +

OH H OH H
H3 C H3 C H3 C

CH3 CH
3

| ∗ |

C H3 −C − C H −C −C H3
+ |

CH3

59. 7
Explanation:
As given in the figure stereocentres are visible, i.e

Hence, the total number of stereoisomers = 23 = 8

But out of these, the following one is optically inactive due to symmetry

Hence, the total number of optically active stereoisomers = 7

60. 9
Explanation:
hv
CH3 - CH3 + Br2 (Excess) ⟶

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61. 60
Explanation:
Staggered form is produced when the rearrangement of atoms or group takes place by an angle of 60°. 1 ,1,1 trichloroethane (CCl3
- CH3) in eclipsed form on rotation by 60° gives staggered form.

62. 3
Explanation:

Total numbers of isomer = 3

63. 1
Explanation:
Mg2C3 + 4H2O → 2Mg(OH)2 + C H 3 − C ≡ CH
(One mole)

64. 8
Explanation:
Number of monohalogen derivatives are equal to the number of chemically different Hydrogen atom present in the compound.
Possible number of structure for C5H12 are

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65. 2
Explanation:
The molecule cannot show geometrical isomerism, so only its mirror image will be the other stereoisomer.

66. 3
Explanation:

A Fischer projection always represents an eclipsed conformation that is least stable. Now, converting this to stable staggered
conformation by keeping one carbon atom fixed and the other by rotating, we get

120
∘
clockwise
−−−−−−−→

Hence, only 3 different staggered conformers are possible.

67. 70.0
Explanation:

Hence molar mass of hydrocarbon (X) is 70 g/mol.

68. 4.0
Explanation:

Total Possible Isomeric product = 1 + 3 = 4

69. 2.0
Explanation:

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70. 3.0
Explanation:
−3
17×10
Moles of hydrocarbon = 136
= 1.25 × 10
−4

−3
8.4×10
Moles of H2 gas = 22.4
= 0.375 × 10
−3 −4
= 3.75 × 10

Hydrogen molecule used for 1 molecule of hydrocarbon is 3.

−4
3.75×10
= = 3
−4
1.25×10

71. 4.0
Explanation:
10 mg
Moles of X = 80
= 0.125 m mol
8.4 mg
Moles consumed of H 2
=
22.4
0.375 m mol.
n
H2 0.375
= = 3
nX 0.125

So, the compound X have 3 double bond.

Ozonolysis of the compound yield formaldehyde and dialdehyde.
The compound is
H

H2 C = CH − CH = CH − C = CH2

Molecular mass = (12 × 6) + 1 × 8 = 72 + 8 = 80 amu

Ozonolysis form:

72. 2.0
Explanation:

Number of sp2 hybridised carbon atoms = (6 and 7)2

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73. 10.0
Explanation:
H

|σ
σ
σ σ σ σ σ
H C C C C ≡ C H
|σ |σ |σ

H H H

Numbers of σ bonds = 10
74. 7.0
Explanation:

In benzaldehyde, total number of sp2 carbons are 7.

75. 13
Explanation:

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