Thermodynamics

01

PYQ 2013

hard

chemistry ID: jee-main

In which of the following exothermic reactions, the heat liberated per mole is the highest ?

1

2

3

4

02

PYQ 2014

medium

chemistry ID: jee-main

For the complete combustion of ethanol, the amount of heat produced as measured in bomb calorimeter, is at C. Assuming ideality the enthalpy of combustion,, for the reaction will be

1

2

3

4

03

PYQ 2017

easy

chemistry ID: jee-main

is equal to :

1

Adiabatic work

2

Isothermal work

3

Isochoric work

4

Isobaric work

04

PYQ 2017

hard

chemistry ID: jee-main

An ideal gas undergoes isothermal expansion at constant pressure. During the process :

1

enthalpy increases but entropy decreases.

2

enthalpy remains constant but entropy increases

3

enthalpy decreases but entropy increases.

4

Both enthalpy and entropy remain constant .

05

PYQ 2017

medium

chemistry ID: jee-main

For a reaction, . The correct statement for the reaction is :

1

2

3

4

06

PYQ 2018

medium

chemistry ID: jee-main

For which of the following processes, is negative ?

1

2

3

diamond graphite

4

07

PYQ 2018

medium

chemistry ID: jee-main

at for substance 'S' in liquid state and gascous state are and, respectively. Vapour pressure of liquid ' S ' at is approximately equal to

1

0.1 atm

2

1 atm

3

10 atm

4

100 atm

08

PYQ 2018

medium

chemistry ID: jee-main

Which of following lines correctly show the temperature dependence of equilibrium constant, for an exothermic reaction ?

1

A and B

2

B and C

3

C and D

4

A and D

09

PYQ 2018

medium

chemistry ID: jee-main

For which of the following reactions, is equal to ?

1

2

3

4

10

PYQ 2019

medium

chemistry ID: jee-main

Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures and . The correct graphical depiction of the dependence of work done on the final volume is:

1

2

3

4

11

PYQ 2019

medium

chemistry ID: jee-main

During compression of a spring the work done is and escaped to the surroundings as heat. The change in internal energy, is:

1

8

2

12

3

-12

4

-8

12

PYQ 2019

medium

chemistry ID: jee-main

moles of an ideal gas at are allowed to undergo reversible compression till its temperature becomes . If and for this process. (]

1

2

3

4

13

PYQ 2019

medium

chemistry ID: jee-main

Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas ? (Assume non-expansion work is zero)

1

Cyclic process :

2

Isothermal process :

3

Adiabatic process :

4

Isochoric process :

14

PYQ 2019

medium

chemistry ID: jee-main

Enthalpy of sublimation of iodine is at . If specific heat of and are and respectively, then enthalpy of sublimation of iodine at in is :

1

2.85

2

11.4

3

5.7

4

22.8

15

PYQ 2019

medium

chemistry ID: jee-main

For diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities ?

1

2

3

4

16

PYQ 2019

medium

chemistry ID: jee-main

For silver, . If the temperature (T) of moles of silver is raised from to at pressure, the value of will be close to

1

21 kJ

2

16 kJ

3

13 kJ

4

62 kJ

17

PYQ 2020

medium

chemistry ID: jee-main

The true statement amongst the following is :

1

S ia function of temperature but is not a function of temperature

2

Both and are functions of temperature

3

Both and are not functions of temperature

4

is not a function of temperature but is a function of temperature

18

PYQ 2020

hard

chemistry ID: jee-main

If enthalpy of atomisation for is and bond enthalpy for is, the relation between them :

1

is

2

does not exist

3

is

4

is

19

PYQ 2021

medium

chemistry ID: jee-main

The Born-Haber cycle for KCl is evaluated with the following data: for KCl = -436.7 kJ; for K = 89.2 kJ; for K = 419.0 kJ; for = -348.6 kJ; for = 243.0 kJ . The magnitude of lattice enthalpy of KCl in kJ is _________. (Nearest integer)}

Official Solution

Correct Option: (1)

$Step 1: Understanding the Concept: The Born-Haber cycle uses Hess's Law to relate various thermodynamic quantities involved in the formation of an ionic solid from its elements. The total enthalpy of formation is equal to the sum of the energies for individual steps. Step 2: Key Formula or Approach: Step 3: Detailed Explanation: Substitute the given values into the formula: - - - - - The magnitude is . Rounding to the nearest integer, we get 718. Step 4: Final Answer: The magnitude of the lattice enthalpy is 718.$

20

PYQ 2021

medium

chemistry ID: jee-main

$For water kJ mol at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is __________ kJ mol . (Nearest integer) [Use: R=8.3 J mol K]$

Official Solution

Correct Option: (1)

$Step 1: Understanding the Question We are given the enthalpy of vaporization () for water at its boiling point and asked to calculate the change in internal energy () for the same process. Step 2: Key Formula or Approach The relationship between enthalpy change () and internal energy change () is given by: For processes involving gases, this can be written as: where is the change in the number of moles of gas in the reaction. Step 3: Detailed Calculation Write the process and find : The process is the evaporation of water: The change in the number of moles of gas is: Rearrange the formula to solve for : Substitute the given values and calculate : Now, calculate : Convert the answer to kJ and round to the nearest integer: Rounding to the nearest integer, we get 38 kJ. Step 4: Final Answer The internal energy change is 38 kJ mol .$

21

PYQ 2021

medium

chemistry ID: jee-main

$200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is kJ. The increase in temperature in of the system on mixing is . The value of is _________. (Nearest integer) [ Given : Specific heat of water Density of water] (Assume no volume change on mixing)$

Official Solution

Correct Option: (1)

$Step 1: Understanding the Concept: The heat released during neutralization increases the temperature of the total solution volume. We first find the limiting reagent and the total heat evolved. Step 2: Detailed Explanation: Moles of Moles of NaOH is the limiting reagent. Heat released () Total mass of solution () Using the formula : The question asks for where : Step 3: Final Answer: The nearest integer value of is 82.$

22

PYQ 2021

medium

chemistry ID: jee-main

$Data given for the following reaction is as follows: FeO(s) + C(graphite) Fe(s) + CO(g) The minimum temperature in K at which the reaction becomes spontaneous is _________. (Integer answer)$

Official Solution

Correct Option: (1)

$Step 1: Understanding the Question: We need to find the temperature at which the given reaction becomes spontaneous. A reaction is spontaneous when the Gibbs free energy change () is negative. The minimum temperature for spontaneity occurs at the point where . Step 2: Key Formula: The Gibbs free energy change is related to enthalpy and entropy by the equation: At equilibrium (the crossover point for spontaneity),, so . Step 3: Calculate and for the Reaction: We use the formula: - Enthalpy Change (): - Entropy Change (): Step 4: Calculate the Temperature: We need to use consistent units. Let's convert to J/mol. Now, calculate the temperature T. Since both and are positive, the reaction is non-spontaneous at low temperatures and becomes spontaneous at temperatures above this value. Rounding to the nearest integer, the minimum temperature is 964 K.$

23

PYQ 2021

medium

chemistry ID: jee-main

$The incorrect expression among the following is :$

1

$(at constant P)$

2

$For isothermal process$

3

4

Official Solution

Correct Option: (3)

$Step 1: Understanding the Concept: This question tests fundamental thermodynamic relationships involving Gibbs free energy (), entropy (), work (), and the equilibrium constant (). Step 2: Key Formula or Approach: 1. 2. Step 3: Detailed Explanation: (A) We know . At constant P, . So, . Since, then . Rearranging, . This is correct. (B) For a reversible isothermal expansion of an ideal gas, work is given by . This is correct. (C) From, we have . Substituting : . The expression in option (C) is missing the negative sign. Thus, it is incorrect. (D) Rearranging : . This is correct. Step 4: Final Answer: Expression (C) is incorrect.$

24

PYQ 2021

medium

chemistry ID: jee-main

$When 400 mL of 0.2 M H SO solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is ________K. (Round off to the Nearest Integer). [Use : H (aq) + OH (aq) H O(l) ; = -57.1 kJ mol Specific heat of H O = 4.18 J K g density of H O = 1.0 g cm Assume no change in volume of solution on mixing.]$

Official Solution

Correct Option: (1)

$This is a calorimetry problem involving a neutralization reaction. Step 1: Calculate the moles of H and OH . Moles of H SO = Molarity Volume = 0.2 mol/L 0.400 L = 0.08 mol. Since H SO is a strong acid, it provides 2 H ions per molecule. Moles of H = 2 0.08 mol = 0.16 mol. Moles of NaOH = Molarity Volume = 0.1 mol/L 0.600 L = 0.06 mol. Since NaOH is a strong base, it provides 1 OH ion per molecule. Moles of OH = 0.06 mol. Step 2: Determine the limiting reactant and calculate the heat released. The neutralization reaction is H + OH H O. The stoichiometry is 1:1. Since moles of OH (0.06) (0.16), OH is the limiting reactant. Moles of water formed = 0.06 mol. Heat released, q = (moles of limiting reactant) (enthalpy of neutralization) q = 0.06 mol 57.1 kJ/mol = 3.426 kJ = 3426 J. Step 3: Calculate the temperature change. Total volume of the final solution = 400 mL + 600 mL = 1000 mL. Assuming the density of the solution is that of water (1.0 g/mL), the total mass of the solution is m = 1000 g. The specific heat of the solution is given as s = 4.18 J K g . Using the formula q = ms T: 3426 J = (1000 g) (4.18 J K g) T. T = K. Step 4: Express the answer in the required format. The question asks for the answer in the form ______K. T = 0.8196 K = 81.96 K. Rounding to the nearest integer, the value is 82.$

25

PYQ 2021

medium

chemistry ID: jee-main

$At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ and the enthalpy of vaporisation of the liquid (X) is 98.2 kJ . The enthalpy of sublimation of the substance (X) in kJ is ________. (in nearest integer)$

Official Solution

Correct Option: (1)

$Step 1: Understanding the Concept: Sublimation is the transition from solid directly to gas. According to Hess's Law, the total enthalpy change for a process is the sum of the enthalpy changes for its individual steps. Sublimation can be envisioned as Solid Liquid (Fusion) followed by Liquid Gas (Vaporization). Step 2: Key Formula or Approach: Step 3: Detailed Explanation: Given: By Hess's Law: Step 4: Final Answer: The enthalpy of sublimation is 101 kJ .$

26

PYQ 2021

medium

chemistry ID: jee-main

$For the reaction, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of for the reaction at 300 K and 1 atm in J mol is, where is __________. (Rounded off to the nearest integer) [R=8.31 J mol and]$

Official Solution

Correct Option: (1)

$Step 1: Relation: . Step 2: Substitute values: . Step 3: Since, . Step 4: Given, so . Step 5: . Step 6: Comparing with, we get .$

27

PYQ 2021

medium

chemistry ID: jee-main

$Assuming ideal behaviour, the magnitude of log K for the following reaction at 25 °C is . The value of is __________ . (Integer answer) [Given :;;]$

Official Solution

Correct Option: (1)

$Step 1: . Step 2: . Step 3: . Step 4: . Step 5: Magnitude . Thus, .$

28

PYQ 2021

medium

chemistry ID: jee-main

$For the reaction C₂H₆ \to C₂H₄ + H₂ the reaction enthalpy ΔH = _________ kJ mol . (Round off to the Nearest Integer). [Given : Bond enthalpies in kJ mol : C-C : 347, C=C : 611; C-H : 414, H-H : 436]$

Official Solution

Correct Option: (1)

$Step 1: . Step 2: Reactants (): 1 C-C bond + 6 C-H bonds. Products (): 1 C=C bond + 4 C-H bonds + 1 H-H bond. Step 3: . kJ/mol.$

29

PYQ 2021

medium

chemistry ID: jee-main

$The reaction of cyanamide, NH CN(s), with oxygen was run in a bomb calorimeter and was found to be -742.24 kJ mol . The magnitude of for the reaction NH CN(s) + 3/2 O (g) N (g) + CO (g) + H O(l) is _________ kJ.$

Official Solution

Correct Option: (1)

$Step 1: Use the relation . Step 2: Calculate (gaseous products - gaseous reactants): . Step 3: . Step 4: kJ. Step 5: Magnitude = kJ.$

30

PYQ 2021

medium

chemistry ID: jee-main

$Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is ________ kJ mol⁻¹. (Rounded-off to the nearest integer) [Use R=8.314 J mol⁻¹K⁻¹]$

Official Solution

Correct Option: (1)

$Step 1: Physical interpretation A gas cannot expand against an external pressure higher than its own. Hence, the given data implies an isothermal compression from MPa to MPa against a constant external pressure MPa. Step 2: Thermodynamic relation For an ideal gas undergoing an isothermal process : From the first law of thermodynamics: Step 3: Calculate initial and final volumes Using the ideal gas equation: Given: Initial volume: Final volume: Step 4: Change in volume (Negative sign confirms compression.) Step 5: Work done For an irreversible process at constant external pressure: Step 6: Heat transferred Step 7: Heat transferred per mole Final Answer (rounded):$

31

PYQ 2021

easy

chemistry ID: jee-main

$For water at 100 C and 1 bar, (Round off to the Nearest Integer) [Use : R=8.31 J mol K] [Assume volume of H O(l) is much smaller than volume of H O(g). Assume H O(g) can be treated as an ideal gas]$

Official Solution

Correct Option: (1)

32

PYQ 2022

medium

chemistry ID: jee-main

$Match List-I with List-II List-I (A) Spontaneous process (B) Process with, (C) (D) Exothermic Process List-II (I) (II) (III) Isothermal and isobaric process (IV) [Bond energies of molecules in reactants] - [Bond energies of product molecules] Choose the correct answer from the options given below :$

1

$(A)-(III), (B)-(II), (C)-(IV), (D)-(I)$

2

$(A)-(II), (B)-(III), (C)-(IV), (D)-(I)$

3

$(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$

4

$(A)-(II), (B)-(I), (C)-(III), (D)-(IV)$

Official Solution

Correct Option: (2)

$Spontaneous (II) process (Constant temperature and pressure condition) Process with (III) Isothermal and isobaric process (C) (IV) [Bond energies of molecules in reactants - bond ene0072gies of product molecules] (D) Exothermic process (I) Hence, the correct option is (B): (A)-(II), (B)-(III), (C)-(IV), (D)-(I)$

33

PYQ 2022

hard

chemistry ID: jee-main

$A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100 °C, then the internal energy for vaporization in kJ mol -1 is ________. [nearest integer] [Assume steam to be an ideal gas. Given Δ vap H Θ for water at 373 K and 1 bar is 41.1 kJ mol -1; R = 8.31 J K -1 mol -1]$

Official Solution

Correct Option: (1)

$The correct answer is 38 kJ/mol (evaporation) Δn g = 1 - 0 = 1 ΔU vap = ΔH vap - Δn g RT = 41.1 - (1) \times 8.31 \times 10 -3 \times 373 = 41.1 - 3.099 = 38 kJ/mol$

34

PYQ 2022

easy

chemistry ID: jee-main

$When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is _____ J. [nearest integer] (Given : R = 8.3 J K -1 mol -1 and log 2 = 0.3010)$

Official Solution

Correct Option: (1)

$The correct answer is 8630 W rev = -2.303 nRT log 10 (V 2 /V 1) = -2.303 \times 5 \times 8.3 \times 300 \times log 10 (20/10) \approx -8630 J$

35

PYQ 2022

medium

chemistry ID: jee-main

$4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20 L. The amount of heat absorbed in this expansion is ____ L atm.$

Official Solution

Correct Option: (1)

$Work done ∵ (vacuum) ∴ (as the process is isothermal) So,$

36

PYQ 2022

medium

chemistry ID: jee-main

$Among the following the number of state variables is____. Internal energy (U) Volume (V) Heat (q) Enthalpy (H)$

Official Solution

Correct Option: (1)

$lnternal energy ,volume, enthalpy are state variable.$

37

PYQ 2023

medium

chemistry ID: jee-main

$Given below are two statements Statement I : Pig iron is obtained by heating cast iron with scrap iron Statement II : Pig iron has a relatively lower carbon content than that of cast iron In the light of the above statements, choose the correct answer from the options given below$

1

$Both Statement I and Statement II are correct.$

2

$Both Statement I and Statement II are not correct.$

3

$Statement I is correct but Statement II is not correct$

4

$Statement I is not correct but Statement II is correct.$

Official Solution

Correct Option: (2)

$Statement - I is incorrect because cast iron is obtained by heating pig iron with scrap iron Statement-II is also incorrect because pig iron has more carbon content (\sim 4%) than cast iron (\sim 3%)$

38

PYQ 2023

medium

chemistry ID: jee-main

$The enthalpy change for the conversion of to is (-)_____________(Nearest integer) Given :,,$

Official Solution

Correct Option: (1)

$The correct answer is -610$

39

PYQ 2023

hard

chemistry ID: jee-main

$While performing a thermodynamics experiment, a student made the following observations, The enthalpy of ionization of as calculated by the student is _____(nearest integer)$

Official Solution

Correct Option: (1)

$The correct answer is 2 Δ H_{ionisation} of C H_{3} COO H = ∣ - 57.3 - (- 55.3) ∣ = 2 K J / m o l$

40

PYQ 2023

easy

chemistry ID: jee-main

$When 2 litre of ideal gas expands isothermally into vacuum to a total volume of 6 litre, the change in internal energy is______J (Nearest integer)$

Official Solution

Correct Option: (1)

$Step 1: Isothermal Expansion of an Ideal Gas For an ideal gas undergoing an isothermal process, the change in internal energy (∆U) depends only on temperature (T): Since the process is isothermal, the temperature remains constant, and hence: Conclusion: The change in internal energy is 0 J.$

41

PYQ 2023

hard

chemistry ID: jee-main

$of ethane undergoes combustion at in a bomb calorimeter The temperature of calorimeter system (including the water) is found to rise by The heat evolved during combustion of ethane at constant pressure is ___(Nearest integer) [Given : The heat capacity of the calorimeter system is Assume ideal gas behaviour Atomic mass of and are 12 and respectively]$

Official Solution

Correct Option: (1)

$The correct answer is 1006. (Bomb calorimeter \to const volume Heat released By combustion of 1 mole C_{2} H_{6} (Δ U) = - \frac{20 \times 0.5}{0.3} \times 30 = - 1000 k J C_{2} H_{6} (g) + 7/2 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l) Δ n g = 2 - (2 + 7/2) = - (7/2) Δ H = Δ U + Δ n RT = - 1000 - 7/2 \times 8.3 \times 300 k J = - 1000 - 6.225 = - 1006 k J So heat released = 1006 k J m o l^{- 1}$

42

PYQ 2023

medium

chemistry ID: jee-main

$1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of The work done is The final temperature of the gas is _______(Nearest integer). Given$

Official Solution

Correct Option: (1)

$For an adiabatic process,, so . The equation for work done: Where . Using the equation for change in temperature: Thus, the final temperature of the gas is 150 K.$

43

PYQ 2023

medium

chemistry ID: jee-main

$An athlete is given of glucose for energy This is equivalent to of energy The of this energy gained is utilized by the athlete for sports activities at the event In order to avoid storage of energy, the weight of extra water he would need to perspire is ___(Nearest integer) Assume that there is no other way of consuming stored energy Given: The enthalpy of evaporation of water is Molar mass of are 12,1 and$

Official Solution

Correct Option: (1)

$The correct answer is 360. C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 C O_{2} (g) + 6 H_{2} O (l) Extra energy used to convert H_{2} O (l) into H_{2} O (l) into H_{2} O (g) = \frac{1800}{2} = 900 k J \Rightarrow 900 = n_{H_{2} O} \times 45 n_{H_{2} O} = \frac{900}{45} = 20 mole W_{H_{2} O} = 20 \times 18 = 360 g$

44

PYQ 2023

medium

chemistry ID: jee-main

$of is produced on complete combustion of gaseous mixture of ethene and methane at and Heat evolved during the combustion process is ______Given :$

Official Solution

Correct Option: (1)

$The correct answer is 925. \Rightarrow 28 = 16.8 + x x = 11.2 L n_{C H_{4}} = \frac{P V}{RT} = \frac{1 \times 5.6}{0.082 \times 298} = 0.229 mole n_{C_{2} H_{2}} = \frac{11.2}{0.082 \times 298} = 0.458 mole ∴ Heat evolved = 0.229 \times 900 + 0.458 \times 1400 = 206.1 + 641.2 = 847.3 k J$

45

PYQ 2023

easy

chemistry ID: jee-main

$The value of for a reaction A ⇋ B is (Given, = -54.67 kJ = 10 kJ and R = 8.314 J 2.303 \times 8.314 \times 298 = 5705)$

Official Solution

Correct Option: (1)

$Calculation of and Given Formula: Step 1: Substitute the Values Substituting into the formula: Simplify: Step 2: Relating and The relationship between and is given by: Where: (Gas constant) (Temperature) Rearranging for : Substitute the values: Numerical simplification: Conclusion: The value of is 10 .$

46

PYQ 2023

hard

chemistry ID: jee-main

$Match List I with List II : LIST I Isomeric pairs LIST II Type of isomers A Propanamine and N-Methylethanamine I Metamers B Hexan-2-one and Hexan-3-one II Positional isomers C Ethanamide and Hydroxyethanimine II Functional isomers D o-nitrophenol and p-nitrophenol IV Tautomers Choose the correct answer from the options given below:$

1

$A-II, B-III, C-I, D-IV$

2

$A-III, B-I, C-IV, D-II$

3

$A-III, B-IV, C-I, D-II$

4

$A-IV, B-III, C-I, D-II$

Official Solution

Correct Option: (2)

$Correct answer is (b) A-III, B-I, C-IV, D-II$

47

PYQ 2023

medium

chemistry ID: jee-main

$The value of log10K for a reaction A ⇋ B is (Given ΔHº298K = -54.67 kJmol-1 ΔSº298K = 10 kJmol-1 and R = 8.314 JK-1mol-1 2.303 \times 8.314 \times 298 = 5705)$

Official Solution

Correct Option: (1)

$ΔGº = ΔHº - TΔSº = -54.07 \times 1000 - 298 \times 10 = -57050 ΔGº = -2.303 RTlog10K logK = 10$

48

PYQ 2023

medium

chemistry ID: jee-main

$Which of the following relations are correct? (A) (B) (C) (D) Choose the most appropriate answer from the options given below:$

1

$B and D Only$

2

$A and B Only$

3

$B and C Only$

4

$C and D Only$

Official Solution

Correct Option: (3)

$- (A) is incorrect because the first law of thermodynamics states, where . Hence, the correct relation is . - (B) is correct, as it is the definition of Gibbs free energy (). - (C) is correct, as it represents the change in entropy () under reversible conditions. - (D) is incorrect because for an ideal gas, .$

49

PYQ 2023

medium

chemistry ID: jee-main

$of is produced on complete combustion of gaseous mixture of ethene and methane at and Heat evolved during the combustion process is ______Given :$

Official Solution

Correct Option: (1)

$The correct answer is 925. \Rightarrow 28 = 16.8 + x x = 11.2 L n_{C H_{4}} = \frac{P V}{RT} = \frac{1 \times 5.6}{0.082 \times 298} = 0.229 mole n_{C_{2} H_{2}} = \frac{11.2}{0.082 \times 298} = 0.458 mole ∴ Heat evolved = 0.229 \times 900 + 0.458 \times 1400 = 206.1 + 641.2 = 847.3 k J$

50

PYQ 2023

medium

chemistry ID: jee-main

$The total number of intensive properties from the following is __________ Volume. Molar heat capacity, Molarity, E\theta cell. Gibbs free energy change. Molar mass, Mole$

Official Solution

Correct Option: (1)

$Intensive properties are those that do not depend on the amount of substance, while extensive properties depend on the quantity of the substance. Classification of the Given Properties: Extensive Properties: Mole Volume Gibbs free energy change () Intensive Properties: Molar mass Molar heat capacity Molarity Standard cell potential () Calculation: The total number of intensive properties is: Final Answer: The total number of intensive properties is 4.$

51

PYQ 2024

medium

chemistry ID: jee-main

$Which of the following is correct for adiabatic free expansion against vacuum?$

1

$q=0, ΔU=0, w=0$

2

$q\neq0, w\neq0, ΔU=0$

3

$q=0, ΔU\neq0, w\neq0$

4

$q=0, ΔU\neq 0, w\neq0$

Official Solution

Correct Option: (1)

$The correct option is (A): q=0, ΔU=0, w=0.$

52

PYQ 2024

medium

chemistry ID: jee-main

$Consider the following reaction at 298 K. . for the reaction is ______ kJ. (Given R=)$

Official Solution

Correct Option: (1)

$Given Reaction: The reaction is as follows: Calculation of : The formula to calculate the standard Gibbs free energy change is: Substitute the known values: Now calculate the value of : Substitute this back into the equation: Conclusion: The standard Gibbs free energy change is .$

53

PYQ 2024

medium

chemistry ID: jee-main

$Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A): Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always -57 kJ mol -1 Reason (R): Enthalpy of neutralisation is the amount of heat liberated when one mole of H + ions furnished by acid combine with one mole of - OH ions furnished by base to form one mole of water. In the light of the above statements. Choose the correct answer from the options given below.$

1

$(A) is true but (R) is false$

2

$Both (A) and (R) are true and (R) is the correct explanation of (A)$

3

$(A) is false but (R) is true$

4

$Both (A) and (R) are true but (R) is not the correct explanation of (A)$

Official Solution

Correct Option: (2)

$Enthalpy of neutralisation of a strong acid (SA) and strong base (SB) is always -57 kJ/mol because a strong monoacid gives one mole of H + and a strong mono-base gives one mole of OH -, which combine to form one mole of water. Both the assertion and reason are true, and (R) correctly explains (A).$

54

PYQ 2024

medium

chemistry ID: jee-main

$The heat of combustion of solid benzoic acid at constant volume is at . The heat of combustion at constant pressure is . The value of is:$

Official Solution

Correct Option: (1)

$The relation between heat at constant pressure () and at constant volume () is: For benzoic acid: . Substituting: Here, . Solving gives:$

55

PYQ 2024

medium

chemistry ID: jee-main

$Match List - I with List -II. Choose the correct answer from options given below:$

1

$A-(III), (B)-(IV), (C)-(I), (D)-(II)$

2

$A-(IV), (B)-(II), (C)-(I), (D)-(III)$

3

$A-(IV), (B)-(II), (C)-(III), (D)-(I)$

4

$A-(III), (B)-(IV), (C)-(II), (D)-(I)$

Official Solution

Correct Option: (1)

56

PYQ 2024

medium

chemistry ID: jee-main

$When and, then the temperature of vapour, at one atmosphere, is .$

Official Solution

Correct Option: (1)

$To find the temperature of vaporization at one atmosphere using the given enthalpy of vaporization () and entropy of vaporization (), we apply the formula derived from the Gibbs free energy relation at equilibrium: Solving for temperature (), we get: Convert to joules: Then, substitute the values into the formula: The temperature is calculated to be 400 K, which fits within the provided range of 400,400. Therefore, the temperature of vaporization at one atmosphere is confirmed to be 400 K.$

57

PYQ 2024

easy

chemistry ID: jee-main

$If 5 moles of an ideal gas expands from to a volume of at under isothermal and reversible conditions, then work,, is . The value of is ______. (Given)$

Official Solution

Correct Option: (1)

$To calculate the work done on an ideal gas during isothermal, reversible expansion, we use the formula:, where: moles,,,, . Substitute the values into the equation: Calculate the natural logarithm: Calculate the work: Thus, the value of is 28721. This value fits perfectly within the provided range [28721, 28721].$

58

PYQ 2024

medium

chemistry ID: jee-main

$The enthalpy of formation of ethane () from ethylene by addition of hydrogen, where the bond energies of,, are,,, and respectively, is _________.$

Official Solution

Correct Option: (1)

$Given Information: Bond energies: Reaction for Formation of Ethane from Ethylene: The reaction can be represented as: Bond Energy Calculations: Breaking Bonds: One bond in ethylene: One bond: Total energy required to break bonds = Forming Bonds: One bond in ethane: Two bonds: Total energy released in forming bonds = Enthalpy Change (): Conclusion: The enthalpy of formation of ethane from ethylene by addition of hydrogen is .$

59

PYQ 2024

medium

chemistry ID: jee-main

$The heat of solution of anhydrous CuSO and CuSO 5H O are kJ mol and kJ mol respectively. The heat of hydration of CuSO to CuSO 5H O is kJ. The value of is _____.$

Official Solution

Correct Option: (1)

$To find the heat of hydration for the reaction where anhydrous CuSO₄ becomes CuSO₄·5H₂O, we need to consider the heats of solution provided for each compound.The problem provides:Anhydrous CuSO₄: Heat of solution = CuSO₄·5H₂O: Heat of solution = The heat of hydration is the energy change when CuSO₄ is hydrated to form CuSO₄·5H₂O.Considering the reactions:(1) \quad (2) \quad The heat of hydration involves the transition from CuSO₄ to CuSO₄·5H₂O:This can be understood using Hess's law, which states the total enthalpy change for a reaction is the sum of the steps:Substitute the given enthalpy values:Thus, .This value fits exactly within the given range of 82, 82.Therefore, the heat of hydration is kJ mol .$

60

PYQ 2024

easy

chemistry ID: jee-main

$Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at . If the piston is moved to position B, keeping the temperature unchanged, then 'x' work is done in this reversible process.______. (nearest integer)$

Official Solution

Correct Option: (1)

$The work done by the gas in a reversible isothermal process can be calculated using the formula: Where: is the work done. (amount of gas). (universal gas constant). is the absolute temperature in Kelvin. Given . (initial volume). (final volume). Substitute the values into the formula: (approx.) To the nearest integer, . This value falls within the expected range of 55, verifying its correctness.$

61

PYQ 2024

medium

chemistry ID: jee-main

$Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using constant pressure of 5 atm. Heat exchange Q for the compression is - ____ Lit. atm.$

Official Solution

Correct Option: (1)

$As isothermal and the process is irreversible: Given: Substituting the values: Thus, the heat exchange for the compression is .$

62

PYQ 2024

medium

chemistry ID: jee-main

$Combustion of 1 mole of benzene is expressed as C H (l) + O (g) \to 6CO (g) + 3H O(l). The standard enthalpy of combustion of 2 mol of benzene is x kJ. x = __________(1) standard Enthalpy of formation of 1 mol of C H (l), for the reaction 6C(graphite) + 3H (g) \to C H (l) is 48.5 kJ mol . (2) Standard Enthalpy of formation of 1 mol of (g), for the reaction 6(graphite) + \to CO (g) is -393.5 kJ mol . (3) Standard Enthalpy of formation of 1 mol of H O(l), for the reaction H (g) + O (g) \to H O(l) is -286 kJ mol .$

Official Solution

Correct Option: (1)

$The enthalpy change for the combustion reaction can be calculated using Hess's law: Step 1: Write the given reaction The reaction for 1 mole of benzene is: Step 2: Calculate the enthalpy change for 1 mole of benzene Using the standard enthalpies of formation: For the products: For the reactants: The enthalpy change for the combustion of 1 mole of benzene is: Step 3: Calculate for 2 moles of benzene For 2 moles of benzene: Final Answer: .$

63

PYQ 2024

medium

chemistry ID: jee-main

$For the reaction at, . and . The reaction will become spontaneous above ____.$

Official Solution

Correct Option: (1)

$The problem asks for the temperature above which the given reaction becomes spontaneous. We are provided with the standard enthalpy change () and standard entropy change () for the reaction at 298 K. Concept Used: The spontaneity of a reaction is determined by the sign of the Gibbs free energy change (). The relationship between, enthalpy change (), and entropy change () at a constant temperature is given by the Gibbs-Helmholtz equation: For a reaction to be spontaneous, the value of must be negative (). The transition from a non-spontaneous to a spontaneous process occurs at the temperature where the system is at equilibrium, which is defined by the condition . This temperature is known as the equilibrium temperature, . Step-by-Step Solution: Step 1: Write down the given thermodynamic values for the reaction. Note that the units are consistent (both are in kJ). Step 2: Determine the condition for the reaction to become spontaneous. The reaction becomes spontaneous when . The threshold temperature for this change is the equilibrium temperature (), where . Step 3: Set in the Gibbs-Helmholtz equation to find the equilibrium temperature. Rearranging the equation to solve for : Step 4: Substitute the given values of and into the equation to calculate . Final Computation & Result: Performing the calculation for the equilibrium temperature: This is the temperature at which the reaction is at equilibrium. Since both (enthalpy) and (entropy) are positive, the reaction is endothermic and becomes more disordered. For the reaction to be spontaneous (), the entropy term () must be larger than the enthalpy term (). This occurs at temperatures above the equilibrium temperature. Therefore, the condition for spontaneity is . The reaction will become spontaneous above 2000 K .$

64

PYQ 2024

medium

chemistry ID: jee-main

$For a certain reaction at 300 K,, then for the same reaction is (Given)$

Official Solution

Correct Option: (1)

$Given: We are asked to calculate using the formula: Step 1: Substituting the known values into the equation: Given that,, and, we substitute these values into the equation: Step 2: Simplifying the equation: Calculating the logarithm of 10: This simplifies to: Step 3: Converting to kJ/mol: To express the result in kJ/mol, divide by 1000: Final Answer: The correct answer is .$

65

PYQ 2024

medium

chemistry ID: jee-main

$Which of the following is not correct?$

1

$ΔG is negative for a spontaneous reaction$

2

$ΔG is zero for a reversible reaction$

3

$ΔG is positive for a spontaneous reaction$

4

$ΔG is positive for a non-spontaneous reaction$

Official Solution

Correct Option: (3)

$To determine which of the given options is incorrect, we need to understand the concept of Gibbs free energy change () and its relation to reaction spontaneity in thermodynamics. The Gibbs free energy change () is a thermodynamic quantity that helps predict the spontaneity of a chemical reaction at constant temperature and pressure. Here's how is interpreted: : The reaction is spontaneous. : The reaction is at equilibrium, and it is reversible. : The reaction is non-spontaneous. Now, let's evaluate the given options: Option 1: is negative for a spontaneous reaction — This is correct because, for a reaction to occur spontaneously, it must have a negative Gibbs free energy change. Option 2: is zero for a reversible reaction — This is correct as a reaction at equilibrium or reversible state has a Gibbs free energy change of zero. Option 3: is positive for a spontaneous reaction — This is incorrect. A positive indicates that the reaction is non-spontaneous. Option 4: is positive for a non-spontaneous reaction — This is correct. A positive Gibbs free energy change indicates that the reaction will not occur spontaneously under the given conditions. Thus, the correct answer is: is positive for a spontaneous reaction, as it contradicts the fundamental principle of thermodynamics regarding reaction spontaneity. A positive signifies a non-spontaneous process.$

66

PYQ 2024

medium

chemistry ID: jee-main

$If three moles of an ideal gas at 300 K expand isothermally from 30 dm 3 to 45 dm 3 against a constant opposing pressure of 80 kPa, then the amount of heat transferred is ______ J.$

Official Solution

Correct Option: (1)

$To find the heat transferred during the isothermal expansion of an ideal gas, we use the formula for work done in isothermal expansion: . Here, kPa or Pa, dm 3 - 30 dm 3 = 15 dm 3 or m 3 . Plugging in the values: J. The first law of thermodynamics states . For isothermal processes, because the internal energy change is zero (temperature is constant). Thus, or . Therefore, the heat transferred J.$

67

PYQ 2024

medium

chemistry ID: jee-main

$Mass of methane required to produce 22 g of CO 2 after complete combustion is ______ g. (Given Molar mass in g mol -1 : C=12.0, H=1.0, O=16.0).$

Official Solution

Correct Option: (1)

$The combustion of methane (CH 4) can be represented by the balanced chemical equation: CH 4 + 2O 2 \to CO 2 + 2H 2 O. To find the mass of methane required to produce 22 g of CO 2, we proceed as follows: Step 1: Determine the molar mass of CO 2 Molar mass of CO 2 = C + 2(O) = 12.0 + 2(16.0) = 44.0 g/mol. Step 2: Calculate the moles of CO 2 produced Moles of CO 2 = mass/molar mass = 22 g / 44.0 g/mol = 0.5 mol. Step 3: Use the stoichiometry of the reaction to find moles of CH 4 According to the balanced equation, 1 mole of CH 4 produces 1 mole of CO 2 . Therefore, 0.5 mol of CO 2 requires 0.5 mol of CH 4 . Step 4: Determine the mass of CH 4 Molar mass of CH 4 = C + 4(H) = 12.0 + 4(1.0) = 16.0 g/mol. Mass of CH 4 = moles \times molar mass = 0.5 mol \times 16.0 g/mol = 8 g. The mass of methane required is 8 g,$

68

PYQ 2024

easy

chemistry ID: jee-main

Official Solution

Correct Option: (1)

$To find the log equilibrium constant (log K) for the reaction at 400 K, use the Gibbs free energy change equation: . Given:,, . Convert to J: . Calculate : Use the relation: . Convert to : With : Convert to using : The equilibrium constant is expressed as: . Check if the solution fits within the given range: The value is within the expected range of 37,37.$

69

PYQ 2025

medium

chemistry ID: jee-main

$Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?$

1

$30 mL and 30 mL NaOH$

2

$45 mL and 25 mL NaOH$

3

$50 mL HCl and 20 mL NaOH$

4

$30 mL HCl and 30 mL NaOH$

Official Solution

Correct Option: (3)

$- The heat generated in an acid-base neutralization reaction is proportional to the amount of acid and base mixed, which is the product of the molarities and volumes. - The combination that results in the largest heat release will have the highest product of molarities and volumes. - When comparing the options, the mix of 50 mL of HCl and 20 mL of NaOH will have the highest heat increase since it involves a strong acid and a strong base, leading to the largest amount of heat generated. Final Answer: Option (3).$

70

PYQ 2025

hard

chemistry ID: jee-main

$The elemental composition of a compound is 54.2%C, 9.2%H, and 36.6%O. If the molar mass of the compound is 132 g/mol, the molecular formula of the compound is:$

1

2

3

4

Official Solution

Correct Option: (1)

$The elemental composition of a compound is 54.2%C, 9.2%H, and 36.6%O. If the molar mass of the compound is 132 g/mol, the molecular formula of the compound is: Step 1: Calculate the Empirical Formula Given the percentage composition, we can assume we have 100 g of the compound. This makes the mass of each element in the compound directly equal to the percentage values: C: 54.2 g H: 9.2 g O: 36.6 g Step 2: Calculate the Moles of Each Element To find the moles of each element, divide the mass of each element by its atomic mass: For Carbon (C): For Hydrogen (H): For Oxygen (O): Step 3: Find the Mole Ratio Next, divide the moles of each element by the smallest number of moles (which is 2.29 in this case, corresponding to Oxygen): C: H: O: Step 4: Write the Empirical Formula The mole ratio of the elements is approximately C 2 H 4 O 1, so the empirical formula is: CH 2 O Step 5: Find the Molecular Formula The empirical formula mass is: CH 2 O: 12 + 2 + 16 = 30 g/mol The molecular formula mass is given as 132 g/mol. To find the ratio of the molecular mass to the empirical formula mass, divide the molar mass by the empirical formula mass: So, multiply the empirical formula by 4 to get the molecular formula: The molecular formula is C 6 H 12 O 6 .$

71

PYQ 2025

medium

chemistry ID: jee-main

$An ideal gas with an adiabatic exponent 1.5, initially at 27°C is compressed adiabatically from 800 cc to 200 cc. The final temperature of the gas is:$

1

$700 K$

2

$500 K$

3

$250 K$

4

$600 K$

Official Solution

Correct Option: (2)

$For an adiabatic process, the relation between volume and temperature is given by: Where: - (adiabatic exponent) - - - - is the final temperature. Substituting the values into the equation: Simplifying: Thus, the final temperature of the gas is 500 K.$

72

PYQ 2025

medium

chemistry ID: jee-main

$For hydrogen atom, the orbital/s with lowest energy is/are:$

1

$(A) only$

2

$(B), (C), and (D) only$

3

$(B) only$

4

$(A) and (E) only$

Official Solution

Correct Option: (1)

$- For the hydrogen atom, the orbital with the lowest energy is the orbital. Among the options provided, (A) corresponds to the orbital, which is the lowest energy orbital for hydrogen. Final Answer: Option (1).$

73

PYQ 2025

medium

chemistry ID: jee-main

$For a given reaction, is related to as given in the table. Given: . Which of the following is true? (mol/L) (min) 0.100 200 0.025 100 A. The order of the reaction is . B. If is 1 M, then is min. C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M. D. is 800 min for M.$

1

$A, B and D only$

2

$A and C only$

3

$A and B only$

4

$C and D only$

Official Solution

Correct Option: (1)

$Step 1: From the given data, calculate the order of the reaction. The relationship between half-life and concentration is given by the formula for a first-order reaction. Step 2: Statement A is correct as , indicating a fractional order reaction. Step 3: Statement B is correct because the half-life depends on the initial concentration. Step 4: Statement D is correct because doubling doubles the half-life for a second-order reaction. Final Conclusion: The correct answer is Option (1), A, B and D Only.$

$%5BA%5D$ (mol/L)	$t_%7B1%2F2%7D$ (min)
0.100	200
0.025	100

74

PYQ 2025

easy

chemistry ID: jee-main

$Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?$

1

2

3

4

$The pressure in the vessel B before opening the stopcock is zero$

Official Solution

Correct Option: (4)

75

PYQ 2025

medium

chemistry ID: jee-main

$The reaction follows the mechanism: The overall order of the reaction is:$

1

$3$

2

$1.5$

3

$2.5$

4

$2$

Official Solution

Correct Option: (2)

$The overall rate law is determined by the slow step of the mechanism, which is: Since the fast step is equilibrium, we can use the equilibrium constant to express in terms of : Substituting this into the rate law: Thus, the overall rate law is: The overall order of the reaction is .$

76

PYQ 2025

medium

chemistry ID: jee-main

$When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:$

1

$Green Pale Blue Blue Violet$

2

$Pale Blue Blue Green Violet$

3

$Pale Blue Blue Violet Green$

4

$Violet Blue Pale Blue Green$

Official Solution

Correct Option: (3)

$- As ethane-1,2-diamine is added to the solution, the nickel (II) complex undergoes changes in coordination, resulting in a series of color transitions. Initially, the solution is pale blue and then turns blue, followed by violet, and finally green. Final Answer: Option (3).$

77

PYQ 2025

hard

chemistry ID: jee-main

$Consider a complex reaction taking place in three steps with rate constants,, and respectively. The overall rate constant is given by the expression . If the activation energies of the three steps are 60, 30, and 10 kJ mol respectively, then the overall energy of activation in kJ mol is ________________(Nearest integer).$

Official Solution

Correct Option: (1)

$Step 1: Understand the Given Information The reaction takes place in three steps with rate constants,, and . The overall rate constant is given by the expression: The activation energies for the three steps are given as: Activation energy for step 1, Activation energy for step 2, Activation energy for step 3, Step 2: Apply the Arrhenius Equation The Arrhenius equation relates the rate constant and activation energy of a reaction as: where: - is the pre-exponential factor, - is the activation energy, - is the gas constant (8.314 J/mol\cdotK), - is the temperature in Kelvin. Step 3: Determine the Overall Activation Energy The overall rate constant is a combination of the three rate constants,, and . Given the expression for, we can use the activation energies of the individual steps to calculate the overall activation energy . The overall activation energy for a reaction involving multiple steps can be determined by the following relationship: Substituting the values for,, and : Step 4: Calculate the Overall Activation Energy The overall activation energy can also be related to the rate constants and activation energies of the individual steps by the following formula: Conclusion The overall activation energy of the reaction is approximately 20 kJ/mol .$

78

PYQ 2025

medium

chemistry ID: jee-main

$The hydration energies of and are and kJ/mol respectively. If lattice energy of KCl is kJ/mol, then the heat of solution of KCl is :$

1

2

3

4

Official Solution

Correct Option: (3)

$To find the heat of solution for, we need to understand the relationship between hydration energy, lattice energy, and the enthalpy of solution. This relationship is governed by the Born-Haber cycle, which is expressed as: In the given problem, we have: Substitute these values into the formula: Simplifying the expression, we get: Thus, the correct option is: Therefore, the heat of solution of KCl is calculated as the lattice energy minus the sum of the hydration energies of and .$

79

PYQ 2025

medium

chemistry ID: jee-main

$A weak acid HA has degree of dissociation x. Which option gives the correct expression of ?$

1

$log(1 + 2x)$

2

$0$

3

$log$

4

$log$

Official Solution

Correct Option: (1)

$The Henderson-Hasselbalch equation for a weak acid is . Given the dissociation degree x, the expression becomes, assuming the weak acid dissociates into molar concentration of H and A while undissociated HA remains .$

80

PYQ 2025

medium

chemistry ID: jee-main

$The molecules having square pyramidal geometry are:$

1

$BrF & XeOF$

2

$SbF & XeOF$

3

$BrF & PCl$

4

$SbF & PCl$

Official Solution

Correct Option: (1)

$BrF and XeOF both have square pyramidal structures based on VSEPR theory. BrF has 5 bonding pairs and 1 lone pair, while XeOF has 4 bonding pairs and 1 lone pair on the central xenon atom, forming a square pyramidal shape.$

81

PYQ 2025

medium

chemistry ID: jee-main

$Consider ‘n’ as the number of lone pairs of electrons present in the equatorial position of the most stable structure of . The ions from the following with ‘n’ number of unpaired electrons are: A. B. C. D. E. Choose the correct answer from the options given below:$

1

$B and D Only$

2

$B and C Only$

3

$A, D and E Only$

4

$A and C Only$

Official Solution

Correct Option: (1)

$The unpaired electrons depend on the electronic configuration of the ions. The number of unpaired electrons in Ti and Ni ions will match the given condition for . Final Conclusion: The correct answer is Option (1), B and D Only.$

82

PYQ 2025

medium

chemistry ID: jee-main

$The incorrect decreasing order of atomic radii is:$

1

$Mg>Al>C>O$

2

$Al>B>N>F$

3

$Be>Mg>Al>Si$

4

$Si>P>Cl>F$

Official Solution

Correct Option: (3)

$The atomic radius decreases across a period from left to right due to the increase in effective nuclear charge, but increases down a group due to the addition of electron shells. Step 1: is incorrect because atomic radius decreases as we move from to . Final Conclusion: The incorrect order is Option (3),$

83

PYQ 2025

easy

chemistry ID: jee-main

$What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol) and the decrease in freezing point is 0.40 K?$

1

$5.12 K kg mol$

2

$4.43 K kg mol$

3

$1.86 K kg mol$

4

$3.72 K kg mol$

Official Solution

Correct Option: (3)

$The freezing point depression is given by: where is the freezing point depression constant, and is the molality. Step 1: First, calculate the molality : Step 2: Using the formula for freezing point depression: Substitute and : Final Conclusion: The freezing point depression constant is 1.86 K kg mol, which corresponds to Option (3).$

84

PYQ 2025

medium

chemistry ID: jee-main

$Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from the following:$

1

$Volume of system increases.$

2

$The amount of ice decreases.$

3

$Liquid phase disappears completely.$

4

$The solid phase (ice) disappears completely.$

Official Solution

Correct Option: (2)

$Increasing the pressure at constant temperature will push the equilibrium towards the liquid phase (since the solid phase occupies less volume). As a result, the amount of ice decreases. Final Conclusion: The correct answer is Option (2), the amount of ice decreases.$

85

PYQ 2025

medium

chemistry ID: jee-main

$For a given reaction, is related to as given in the table. Given: . Which of the following is true?$

1

$A. The order of the reaction is .$

2

$B. If is 1 M, then is min.$

3

$C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.$

4

$D. is 800 min for M.$

Official Solution

Correct Option: (1)

$Step 1: From the given data, calculate the order of the reaction. The relationship between half-life and concentration is given by the formula for a first-order reaction. Step 2: Statement A is correct as , indicating a fractional order reaction. Step 3: Statement B is correct because the half-life depends on the initial concentration. Step 4: Statement D is correct because doubling doubles the half-life for a second-order reaction. Final Conclusion: The correct answer is Option (1), A, B and D Only.$

86

PYQ 2025

medium

chemistry ID: jee-main

$The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.$

1

2

3

4

Official Solution

Correct Option: (3)

$The strength of an oxidising agent is directly related to its reduction potential. The higher the reduction potential, the stronger the oxidising agent. Among the options, the highest reduction potential is for, with a value of . Thus, the strongest oxidising agent is .$

87

PYQ 2025

medium

chemistry ID: jee-main

$Match List - I with List - II: List - I: (A) Amylase (B) Cellulose (C) Glycogen (D) Amylopectin List - II: (I) -C1-C4 plant (II) -C1-C4 animal (III) -C1-C4 -C1-C6 plant (IV) -C1-C4 plant Choose the correct answer from the options given below:$

1

2

3

4

Official Solution

Correct Option: (2)

$- Amylase: This is a polysaccharide found in plants, with -1,4 and -1,6 linkages between glucose units, so the correct linkage type is -C1-C4 plant (IV). - Cellulose: Cellulose is a polysaccharide made of -1,4 linkages, commonly found in plants, so the correct linkage is -C1-C4 plant (I). - Glycogen: Glycogen is a polysaccharide found in animals, consisting of -1,4 and -1,6 linkages between glucose units, so the correct linkage is -C1-C4 -C1-C6 plant (III). - Amylopectin: Similar to glycogen, it is found in plants and is made of -1,4 and -1,6 linkages, so the correct linkage is -C1-C4 plant (II). Thus, the correct answer is (2).$

88

PYQ 2025

medium

chemistry ID: jee-main

$1.24 g of (molar mass 124 g mol) is dissolved in 1 kg of water to form a solution with boiling point of 100.105 C, while 2.54 g of (molar mass 250 g mol) in 2 kg of water constitutes a solution with a boiling point of 100.026 C. . Which of the following is correct?$

1

$and (both) are fully ionised.$

2

$is fully ionised while is completely unionised.$

3

$and (both) are completely unionised.$

4

$is completely unionised while is fully ionised.$

Official Solution

Correct Option: (1)

$To find the correct answer, we use the formula for the elevation in boiling point, which is given by: where is the boiling point elevation, is the ebullioscopic constant, is the molality, and is the van't Hoff factor (which gives the number of particles the compound dissociates into). Since we are given the boiling point changes, we can compare the values of (the ionisation factor) for each compound. The change in boiling point is given as: For AX : Using the formula for : Solving for, we get: Thus, AX is fully ionised (since). For AY : Using the same formula for : Solving for, we get: Thus, AY is also fully ionised (since). Hence, the correct answer is that both AX and AY are fully ionised.$

89

PYQ 2025

medium

chemistry ID: jee-main

$The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is:$

1

2

3

4

Official Solution

Correct Option: (3)

$- The melting point of Mn is lower than Fe because Mn has a lower atomic number and a less stable crystal structure compared to Fe. - The melting point of Tc is lower than Ru due to the differences in atomic structure and bonding. - The melting point of Re is lower than Os because Os has a higher atomic number and stronger metallic bonding, leading to a higher melting point. Thus, the correct answer is .$

90

PYQ 2025

medium

chemistry ID: jee-main

$Choose the correct statements. (A) Weight of a substance is the amount of matter present in it. (B) Mass is the force exerted by gravity on an object. (C) Volume is the amount of space occupied by a substance. (D) Temperatures below C are possible in Celsius scale, but in Kelvin scale negative temperature is not possible. (E) Precision refers to the closeness of various measurements for the same quantity. Choose the correct answer from the options given below:$

1

$(A), (B) and (C) Only$

2

$(B), (C) and (D) Only$

3

$(C), (D) and (E) Only$

4

$(A), (D) and (E) Only$

Official Solution

Correct Option: (3)

$- (A) is incorrect. Weight is the force exerted by gravity on an object, not the amount of matter. - (B) is incorrect. Mass is the amount of matter, not the force due to gravity. - (C) is correct. Volume is indeed the amount of space occupied by a substance. - (D) is correct. Temperatures below C are possible in Celsius scale, but negative temperatures are not possible in the Kelvin scale. - (E) is correct. Precision refers to the closeness of repeated measurements. Thus, the correct answer is (3) (C), (D) and (E) Only.$

91

PYQ 2025

medium

chemistry ID: jee-main

$500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given Choose the correct answer from the options given below:$

1

$348 K and 300 J$

2

$378 K and 500 J$

3

$378 K and 300 J$

4

$368 K and 500 J$

Official Solution

Correct Option: (3)

$We can use the formula for the heat absorbed by the gas: Where: - (heat transferred), -, - (molar heat capacity for a monoatomic gas, where). First, calculate : Now, solve for the temperature change : The final temperature: The change in internal energy is . Thus, the correct answer is (3) 378 K and 300 J.$

92

PYQ 2025

medium

chemistry ID: jee-main

$In the given structure, number of and hybridized carbon atoms present respectively are:$

1

$and$

2

$and$

3

$and$

4

$and$

Official Solution

Correct Option: (1)

$- The structure contains and hybridized carbon atoms. - The carbon atoms in triple bonds (i.e., the C\equivN bond) are -hybridized. - The carbon atoms in the C=C double bonds and in the C=O group are -hybridized. - Thus, the number of -hybridized atoms is 3 and the number of -hybridized atoms is 5. Final Answer: Option (1).$

93

PYQ 2025

medium

chemistry ID: jee-main

$If At constant temperature. Then:$

1

2

3

4

Official Solution

Correct Option: (4)

$The energy change when diamond converts to graphite is . The total enthalpy change for the complete oxidation of diamond to carbon dioxide is the sum of the enthalpy changes of the oxidation steps of both diamond and graphite. Therefore, the enthalpy change for the overall reaction is the sum of (oxidation of diamond) and (oxidation of graphite). Hence, . Thus, the correct answer is .$

94

PYQ 2025

medium

chemistry ID: jee-main

$Which of the following statement(s) is correct for the adiabatic process?$

1

$Molar heat capacity is zero.$

2

$Molar heat capacity is infinite.$

3

$Work done on gas is equal to increase in internal energy.$

4

$The increase in temperature results in a decrease in internal energy.$

Official Solution

Correct Option: (3)

$An adiabatic process is one where no heat is exchanged with the surroundings, i.e., . In this process, the internal energy of the system is entirely converted into work done by or on the gas. Let’s analyze each statement: A. Molar heat capacity is zero. In an adiabatic process, heat is not exchanged, so the molar heat capacity doesn't apply in the usual sense. However, this statement is not strictly true because heat capacity is a function of the process and not a direct result of adiabatic conditions. B. Molar heat capacity is infinite. This is also incorrect, as in an adiabatic process, the heat capacity is not infinite. C. Work done on gas is equal to increase in internal energy. This is the correct statement. In an adiabatic process, the work done by or on the gas is directly related to the change in internal energy. Since no heat is exchanged, the first law of thermodynamics gives, where is the work done by the gas. D. The increase in temperature results in a decrease in internal energy. This is incorrect because, in an adiabatic process, an increase in temperature generally results from the work done on the gas, leading to an increase in internal energy, not a decrease. Thus, the correct statement is, which is Work done on gas is equal to increase in internal energy.$

95

PYQ 2025

medium

chemistry ID: jee-main

$Match List - I with List - II. Choose the correct answer from the options given below :$

1

$( A ) - ( I I ) , ( B ) - ( I ) , ( C ) - ( I I I ) , ( D ) - ( I V )$

2

$( A ) - ( I ) , ( B ) - ( I I ) , ( C ) - ( I V ) , ( D ) - ( I I I )$

3

$( A ) - ( I I ) , ( B ) - ( I ) , ( C ) - ( I V ) , ( D ) - ( I I I )$

4

$( A ) - ( I I ) , ( B ) - ( I I I ) , ( C ) - ( I ) , ( D ) - ( I V )$

Official Solution

Correct Option: (3)

$The problem requires matching partial derivatives of thermodynamic quantities with their respective physical interpretations or symbols. Let’s examine each of the given derivatives and match them accordingly: Partial Derivative : This derivative represents the change in Gibbs free energy with respect to temperature at constant pressure . In thermodynamics, it is related to entropy with the expression: . Hence, it matches with (II). Partial Derivative : This derivative relates to the change in enthalpy with temperature at constant pressure . It's equivalent to the heat capacity at constant pressure, denoted as . Hence, it matches with (I). Partial Derivative : This is typically analyzed in the context of compressibility and expansivity; however, based on the representation, it's related here to volume, which makes more conceptual sense for this matching. Hence, it matches with volume (IV). Partial Derivative : This deals with the change in internal energy with volume at constant volume, essential in characterizing under specific conditions or transformations. Hence, it matches with (III). Therefore, the correct match is: ( A ) - ( I I ) , ( B ) - ( I ) , ( C ) - ( I V ) , ( D ) - ( I I I ) .$

96

PYQ 2025

hard

chemistry ID: jee-main

$Match List - I with List - II. List - I (Partial Derivatives) List - II (Thermodynamic Quantity) In the light of the above statements, choose the correct answer from the options given below:$

1

$(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$

2

$(A)-(I), (B)-(II), (C)-(IV), (D)-(III)$

3

$(A)-(II), (B)-(I), (C)-(III), (D)-(IV)$

4

$(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$

Official Solution

Correct Option: (3)

$Let's analyze each partial derivative and match it with the correct thermodynamic quantity. (A) corresponds to the heat capacity at constant pressure,, due to the relationship, but the correct matching is with, as it relates to entropy change at constant pressure. (B) corresponds to the entropy change,, based on the thermodynamic relationship between enthalpy and entropy. (C) is related to the volume,, from the Gibbs free energy equation . (D) corresponds to the heat capacity at constant volume, . Thus, the correct matching is:$

97

PYQ 2025

medium

chemistry ID: jee-main

$The elemental composition of a compound is 54.2% C, 9.2% H, and 36.6% O. If the molar mass of the compound is 132 g/mol, the molecular formula of the compound is:$

1

2

3

4

Official Solution

Correct Option: (1)

$- First, determine the empirical formula by finding the ratio of moles of C, H, and O. - Then, multiply by the molar mass to obtain the molecular formula. Final Answer: Option (1).$

98

PYQ 2025

medium

chemistry ID: jee-main

$Match List - I with List - II.$

1

$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$

2

$(A)-(III), (B)-(II), (C)-(I), (D)-(II)$

3

$(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$

4

$(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$

Official Solution

Correct Option: (1)

$- The reaction between an isonitrile (RCN) and stannic chloride () followed by hydrolysis forms the corresponding aldehyde, which corresponds to the Etard reaction . - The reaction of chlorobenzene with HCl in the presence of heat gives the corresponding aldehyde via the Gatterman-Koch reaction . - The Rosenmund reduction uses hydrogenation of acyl chlorides to aldehydes. - The Stephen reaction involves the reduction of aromatic nitriles to aldehydes using stannous chloride. Final Answer: Option (1).$

99

PYQ 2025

medium

chemistry ID: jee-main

$For the reaction: The correct order of set of reagents for the above conversion is :$

1

2

3

4

Official Solution

Correct Option: (1)

$The given reaction represents the bromination of an amine group, likely to form an amine derivative, with Br under the influence of a Lewis acid such as FeBr, and the reaction is carried out in the presence of NaOH to neutralize the acids formed. The major product obtained in this process follows electrophilic aromatic substitution. Final Answer: Option (1).$

100

PYQ 2025

medium

chemistry ID: jee-main

$A liquid when kept inside a thermally insulated closed vessel at was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?$

1

2

3

4

Official Solution

Correct Option: (1)

$To solve the given problem, we need to analyze the changes in the thermodynamic parameters of the system when a liquid in a thermally insulated closed vessel is mechanically stirred. The key steps involve understanding the implications of each listed parameter under the given conditions: 1. System Description: The liquid is in a thermally insulated closed vessel, which implies that there is no heat exchange with the surroundings, i.e., q = 0 . 2. Work Done: Mechanical stirring involves external work being done on the system. Thus, w > 0, as work is added to the system from the surroundings. 3. Change in Internal Energy (∆U): According to the first law of thermodynamics: . Substituting the known values, we get: Since w > 0, the internal energy of the system increases, making ∆U > 0 . Therefore, the correct option for the thermodynamic parameters is:$

101

PYQ 2025

easy

chemistry ID: jee-main

$Ice at is heated to become vapor with temperature of at atmospheric pressure. The entropy change associated with this process can be obtained from:$

1

2

3

4

Official Solution

Correct Option: (2)

$To find the entropy change associated with the process where ice at is heated to become vapor at, we need to consider different stages of heating and phase changes. Here, we will detail the steps involved in calculating the total entropy change: First, the ice is heated from (268 K) to (273 K). The entropy change is given by: Next, the ice melts into water at . The entropy change for this phase change is: where is the enthalpy of fusion and is the melting point of ice (273 K). The water is then heated from to (373 K). The entropy change here is: At, the water vaporizes into steam. The entropy change for vaporization is: where is the enthalpy of vaporization and is the boiling point of water (373 K). The steam is then heated from to (383 K). The entropy change is: The total entropy change is the sum of all these individual changes: Upon examining the options provided, the correct approach to calculate this total entropy change is given by the formula: This accounts for the energy required to increase the temperature, melt the ice, heat the water, vaporize the water, and then heat the steam, covering all the necessary phase changes and temperature increases for the complete process from to .$

102

PYQ 2025

medium

chemistry ID: jee-main

$One mole of an ideal gas expands isothermally and reversibly from to at, q and work done in the process respectively are : Given : and In$

1

2

3

4

Official Solution

Correct Option: (4)

$To solve this problem, we need to find the change in internal energy (), the heat exchanged (q), and the work done by the gas during the isothermal and reversible expansion from 10 dm 3 to 20 dm 3 . The gas undergoes this process at a constant temperature of 300 K. Given that the gas is ideal, the key formulas involve the ideal gas laws and properties of isothermal processes. Change in Internal Energy (): For an isothermal process involving an ideal gas, the change in internal energy,, is zero. This is because internal energy is a function of temperature, which remains constant. Therefore, . Work Done (W): The formula for work done during an isothermal reversible expansion of an ideal gas is:, where: mole (given) and Since, we get: or approximately . Heat Exchanged (q): In an isothermal process for an ideal gas, the heat absorbed by the system is equal to the work done by the system. So, . Thus, or rather . This is approximately 1.718 kJ as per adjustment in significant digits. The given options show the numerical values alongside their nearest approximations as per significant figures. Therefore, the correct answer is:$

103

PYQ 2025

medium

chemistry ID: jee-main

$Let us consider a reversible reaction at temperature, T . In this reaction, both and were observed to have positive values. If the equilibrium temperature is, then the reaction becomes spontaneous at:$

1

2

3

4

Official Solution

Correct Option: (3)

$To determine the conditions under which a reaction becomes spontaneous, we must apply the concept of Gibbs free energy change, denoted as . The equation for Gibbs free energy is:, where: is the change in Gibbs free energy is the change in enthalpy is the change in entropy is the temperature in Kelvin For a reaction to be spontaneous, must be negative. Given that both and are positive, the reaction can become spontaneous at higher temperatures. To find the equilibrium temperature, is set to zero: At this temperature, and the reaction is at equilibrium. For the reaction to be spontaneous, the condition is: Rewriting the inequality:, implying that This means that the reaction becomes spontaneous when the temperature is greater than the equilibrium temperature, . Thus, the correct answer is:$

104

PYQ 2025

medium

chemistry ID: jee-main

$Given: The for is _______(nearest integer value).$

Official Solution

Correct Option: (1)

$We are given thermochemical data for the formation of ethene () and need to calculate its standard enthalpy of formation (). Concept Used: The standard enthalpy of formation () of a compound is the enthalpy change when one mole of the compound is formed from its elements in their standard states. It can be calculated using bond energies and atomization energies: For ethene (), the reaction is: Step-by-Step Solution: Step 1: Write down the given data. Step 2: Determine the bonds in ethene (). In ethene, there are: 4 C-H bonds 1 C=C bond Step 3: Calculate the energy required to atomize the reactants. For the formation of gaseous atoms from elements in their standard states: Step 4: Calculate the energy released in forming bonds of . Step 5: Compute the enthalpy of formation. Step 6: Since the formation of bonds releases energy, the enthalpy of formation is exothermic: Final Answer: The standard enthalpy of formation of is 25 kJ mol⁻¹ .$

105

PYQ 2025

medium

chemistry ID: jee-main

$The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is:________________$

Official Solution

Correct Option: (1)

$Step 1: Analyze the Structure of 5-Phenylpent-4-en-2-ol The structure of 5-phenylpent-4-en-2-ol consists of: A phenyl group (-C6H5) attached to carbon 5. A hydroxyl group (-OH) attached to carbon 2, which creates a chiral center at this position. A double bond between carbon 4 and carbon 5. Step 2: Identifying Chiral Centers and Double Bonds In order to determine the number of stereoisomers, we need to identify the chiral centers and the potential for cis-trans isomerism due to the double bond: There is one chiral center at carbon 2, which has four different groups attached: a hydroxyl group (-OH), a hydrogen atom, a methyl group (-CH2-), and a part of the carbon chain. The double bond at carbon 4-5 allows for cis-trans isomerism, meaning the two possible configurations (cis and trans) need to be considered for the compound's stereoisomerism. Step 3: Counting the Number of Stereoisomers For this compound, we have: One chiral center at carbon 2, giving two possible configurations (R and S). This contributes two stereoisomers. The double bond between carbons 4 and 5 introduces the possibility of cis-trans isomerism. For each of the two configurations of the chiral center (R and S), there are two possible isomers (cis and trans) due to the double bond. Therefore, the total number of stereoisomers is: 2 (from the chiral center) \times 2 (from cis-trans isomerism) = 4 stereoisomers . Conclusion The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is 4 .$

106

PYQ 2025

medium

chemistry ID: jee-main

$In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ________________ \times 10 -1 % (Nearest integer). (Given: Molar mass of Ag is 108 and Br is 80 g mol -1)$

Official Solution

Correct Option: (1)

$The Carius method is used to determine the amount of halogen in an organic compound by converting the halogen into a silver halide (in this case, AgBr for bromine). To find the percentage of bromine in the compound, we follow these steps: Step 1: Calculate the moles of AgBr formed We are given the mass of AgBr formed, which is 0.15 g. The molar mass of AgBr is the sum of the atomic masses of silver (Ag) and bromine (Br): Molar mass of AgBr = 108 + 80 = 188 g/mol Now, calculate the moles of AgBr formed using the formula: Moles of AgBr = (Mass of AgBr) / (Molar mass of AgBr) = 0.15 g / 188 g/mol = 0.00079787 mol Step 2: Moles of bromine (Br) in the compound From the reaction between the organic compound and silver nitrate, we know that 1 mole of AgBr contains 1 mole of bromine. So, the moles of bromine (Br) will be the same as the moles of AgBr: Moles of Br = 0.00079787 mol Step 3: Calculate the mass of bromine (Br) in the organic compound The atomic mass of bromine (Br) is given as 80 g/mol. So, the mass of bromine in the organic compound is: Mass of Br = (Moles of Br) \times (Atomic mass of Br) = 0.00079787 mol \times 80 g/mol = 0.06383 g Step 4: Calculate the percentage of bromine in the organic compound We are given that the mass of the organic compound is 0.25 g. The percentage of bromine in the compound is calculated as: Percentage of Br = (Mass of Br / Mass of the organic compound) \times 100 = (0.06383 g / 0.25 g) \times 100 = 25.532% Step 5: Convert the percentage to the required format The final answer should be in the format \times 10 -1 %. Therefore, we need to multiply the result by 10 and round to the nearest integer: 25.532 \times 10 = 255.32 \approx 255 Conclusion The percentage of bromine in the organic compound is 255 \times 10 -1 %.$

107

PYQ 2025

medium

chemistry ID: jee-main

$Standard entropies of, and are 70, 50, and 110 J mol respectively. The temperature in Kelvin at which the reaction will be at equilibrium is (nearest integer):$

Official Solution

Correct Option: (1)

$Step 1: Calculate the standard entropy change () for the reaction. The standard entropy change is calculated using the formula: Substitute the given entropy values: Step 2: Use the equation for temperature at equilibrium. At equilibrium, the change in Gibbs free energy () is zero, and we use the following relation: Since at equilibrium: Rearranging to solve for : Substitute the known values for (in J/mol) and : Thus, the temperature at equilibrium is .$

108

PYQ 2025

easy

chemistry ID: jee-main

$A weak acid HA has degree of dissociation x. Which option gives the correct expression of ?$

1

$log$

2

$0$

3

$log$

4

$log(1 + 2x)$

Official Solution

Correct Option: (1)

$We begin with the more accurate expression for the dissociation of a weak acid: $ C x pK_a pH - pK_a pH = pK_a x = 0.5 pK_a pH - pK_a $$

109

PYQ 2025

medium

chemistry ID: jee-main

$The molecules having square pyramidal geometry are:$

1

$BrF & XeOF$

2

$SbF & XeOF$

3

$BrF & PCl$

4

$SbF & PCl$

Official Solution

Correct Option: (1)

$Let's determine the geometry of each molecule: 1. : Bromine has 7 valence electrons. In, there are 5 bond pairs and 1 lone pair. This gives a steric number of 6, which corresponds to an octahedral electron geometry. With 5 bonding pairs and 1 lone pair, the molecular geometry is square pyramidal . 2. : Xenon has 8 valence electrons. In, there is one double bond to oxygen and four single bonds to fluorine. There is also one lone pair. This results in a steric number of 6, corresponding to octahedral electron geometry. With 5 bonding pairs and 1 lone pair, the molecular geometry is also square pyramidal . 3. : Antimony has 5 valence electrons. In, there are 5 bond pairs and no lone pairs. The steric number is 5, which corresponds to a trigonal bipyramidal electron and molecular geometry. 4. : Phosphorus has 5 valence electrons. In, there are 5 bond pairs and no lone pairs. The steric number is 5, which corresponds to a trigonal bipyramidal electron and molecular geometry. Conclusion: Among the given molecules, only and have square pyramidal geometry. Final Answer: The final answer is .$

110

PYQ 2025

hard

chemistry ID: jee-main

$Consider ‘n’ is the number of lone pair of electrons present in the equatorial position of the most stable structure of . The ions from the following with ‘n’ number of unpaired electrons are: A. B. C. D. E. Choose the correct answer from the options given below:$

1

$A, D and E Only$

2

$B and C Only$

3

$B and D Only$

4

$A and C Only$

Official Solution

Correct Option: (1)

$The molecule ClF3 has a T-shaped structure. Chlorine has 7 valence electrons. In ClF3, there are 3 bond pairs with fluorine and 2 lone pairs. The most stable structure of ClF3 has the two lone pairs in the equatorial positions of a trigonal bipyramidal arrangement. Therefore, the number of lone pairs in the equatorial positions is n = 2. Now we need to find which of the given ions have 2 unpaired electrons, matching the value of n = 2: A. V3+: Vanadium (V) has electronic configuration [Ar] 3d3 4s2. So V3+ has electronic configuration [Ar] 3d2. It has 2 unpaired electrons. B. Ti3+: Titanium (Ti) has electronic configuration [Ar] 3d2 4s2. So Ti3+ has electronic configuration [Ar] 3d1. It has 1 unpaired electron. C. Cu2+: Copper (Cu) has electronic configuration [Ar] 3d10 4s1. So Cu2+ has electronic configuration [Ar] 3d9. It has 1 unpaired electron. D. Ni2+: Nickel (Ni) has electronic configuration [Ar] 3d8 4s2. So Ni2+ has electronic configuration [Ar] 3d8. It has 2 unpaired electrons. E. Ti2+: Titanium (Ti) has electronic configuration [Ar] 3d2 4s2. So Ti2+ has electronic configuration [Ar] 3d2. It has 2 unpaired electrons. Conclusion: The ions with exactly 2 unpaired electrons are V3+, Ni2+, and Ti2+. Final Answer: The final answer is (1) A, D and E only.$

111

PYQ 2025

medium

chemistry ID: jee-main

$The incorrect decreasing order of atomic radii is:$

1

2

3

4

Official Solution

Correct Option: (3)

$The atomic radius generally decreases across a period (from left to right) and increases down a group. (1) Mg > Al > C > O: Mg and Al are in Period 3 (Mg is in Group 2, Al is in Group 13). Mg has a larger atomic radius than Al due to the general trend of decreasing radius across a period. C and O are in Period 2 (C is in Group 14, O is in Group 16). C has a larger atomic radius than O for the same reason. However, comparing elements from different periods (e.g., Mg/Al with C/O) violates periodic trends because atomic size increases down a group. Hence, this order is not entirely correct. (2) Al > B > N > F: Boron (B) and Aluminum (Al) are in Group 13, with B in Period 2 and Al in Period 3. Al has a larger radius than B due to being in a lower period. Nitrogen (N) and Fluorine (F) are in Period 2. Radius decreases from B to N to F across the period. So this sequence follows the trend correctly within their groups and periods. This sequence is correct. (3) Be > Mg > Al > Si: Beryllium (Be) is in Period 2, Group 2. Magnesium (Mg) is in Period 3, Group 2 — so Mg > Be (down a group, size increases). Aluminum (Al) is in Period 3, Group 13 — so Mg > Al (across period, size decreases). Silicon (Si) is in Period 3, Group 14 — so Al > Si (again, across period, size decreases). This sequence follows the periodic trend correctly. Conclusion: Among the given options, the correct order of atomic radii based on periodic trends is Be > Mg > Al > Si . Final Answer: The final answer is .$

112

PYQ 2025

easy

chemistry ID: jee-main

$What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol) and the decrease in freezing point is 0.40 K?$

1

$5.12 K kg mol$

2

$4.43 K kg mol$

3

$1.86 K kg mol$

4

$3.72 K kg mol$

Official Solution

Correct Option: (3)

$The freezing point depression is given by: where is the freezing point depression constant, and is the molality. Step 1: First, calculate the molality : Step 2: Using the formula for freezing point depression: Substitute and : Final Conclusion: The freezing point depression constant is 1.86 K kg mol, which corresponds to Option (3).$

113

PYQ 2025

medium

chemistry ID: jee-main

$Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from the following:$

1

$Volume of system increases.$

2

$The amount of ice decreases.$

3

$Liquid phase disappears completely.$

4

$The solid phase (ice) disappears completely.$

Official Solution

Correct Option: (2)

$We are given that ice and water are in equilibrium at 273.15 K and 1 atm pressure. This is the melting point of ice.The Clausius-Clapeyron equation describes the relationship between pressure and temperature for phase transitions:$ \Delta H \Delta V = V_{\text{water}} - V_{\text{ice}} V_{\text{ice}} > V_{\text{water}} \Delta V < 0 \frac{dP}{dT} < 0 \text{The amount of ice decreases.} $$

114

PYQ 2025

hard

chemistry ID: jee-main

$The formation enthalpies, for and are 220.0 and 250.0 kJ mol, respectively, at 298.15 K, and for (g) is -242.0 kJ mol at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is:$

Official Solution

Correct Option: (1)

$The bond enthalpy can be calculated using the following equation based on Hess's law: Final Conclusion: The average bond enthalpy of the O-H bond in water is 114 kJ/mol.$

115

PYQ 2025

medium

chemistry ID: jee-main

$Given below is the plot of the molar conductivity vs concentration for KCl in aqueous solution. If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100, then the resistance of the same cell with the dilute solution is 'x' . The value of is:$

Official Solution

Correct Option: (1)

$Mathematical Derivation For lower concentration:$

116

PYQ 2025

medium

chemistry ID: jee-main

$The molarity of a 70%(mass/mass) aqueous solution of a monobasic acid (X) is:$

Official Solution

Correct Option: (1)

$Step 1: Given density and molar mass of the acid, calculate the molarity using the formula: Step 2: Apply the given values and compute the molarity. Final Conclusion: The molarity is .$

117

PYQ 2025

medium

chemistry ID: jee-main

$In the given structure, number of and hybridized carbon atoms present respectively are:$

1

$and$

2

$and$

3

$and$

4

$and$

Official Solution

Correct Option: (1)

$- The structure contains and hybridized carbon atoms. - The carbon atoms in triple bonds (i.e., the C\equivN bond) are -hybridized. - The carbon atoms in the C=C double bonds and in the C=O group are -hybridized. - Thus, the number of -hybridized atoms is 3 and the number of -hybridized atoms is 5. Final Answer: Option (1).$

118

PYQ 2025

medium

chemistry ID: jee-main

$Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?$

1

$30 mL and 30 mL NaOH$

2

$45 mL and 25 mL NaOH$

3

$50 mL HCl and 20 mL NaOH$

4

$30 mL HCl and 30 mL NaOH$

Official Solution

Correct Option: (3)

$- The heat generated in an acid-base neutralization reaction is proportional to the amount of acid and base mixed, which is the product of the molarities and volumes. - The combination that results in the largest heat release will have the highest product of molarities and volumes. - When comparing the options, the mix of 50 mL of HCl and 20 mL of NaOH will have the highest heat increase since it involves a strong acid and a strong base, leading to the largest amount of heat generated. Final Answer: Option (3).$

119

PYQ 2025

medium

chemistry ID: jee-main

$When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:$

1

$Green Pale Blue Blue Violet$

2

$Pale Blue Blue Green Violet$

3

$Pale Blue Blue Violet Green$

4

$Violet Blue Pale Blue Green$

Official Solution

Correct Option: (3)

$- As ethane-1,2-diamine is added to the solution, the nickel (II) complex undergoes changes in coordination, resulting in a series of color transitions. Initially, the solution is pale blue and then turns blue, followed by violet, and finally green. Final Answer: Option (3).$

120

PYQ 2025

medium

chemistry ID: jee-main

$For hydrogen atom, the orbital/s with lowest energy is/are:$

1

$(B), (C), and (D) only$

2

$(A) only$

3

$(B) only$

4

$(A) and (E) only$

Official Solution

Correct Option: (1)

121

PYQ 2025

easy

chemistry ID: jee-main

$Match List - I with List - II.$

1

$(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$

2

$(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$

3

$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$

4

$(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$

Official Solution

Correct Option: (1)

$Step 1: Analyze Each Reaction in List-I Option (A): RCN (i) SnCl 2, HCl \to RCHO (ii) H 3 O + This reaction is the reduction of a nitrile (RCN) to an aldehyde (RCHO). The first step involves the reduction of the nitrile group using stannous chloride (SnCl 2) in the presence of hydrochloric acid (HCl), followed by hydrolysis to form the aldehyde. This reaction is a classic example of the Stephen reaction (IV) . Option (B): C 6 H 5 CHCl \to C 6 H 5 CHO This reaction involves the reduction of a chlorobenzene (C 6 H 5 CHCl) to a benzaldehyde (C 6 H 5 CHO). The reaction uses hydrogen in the presence of palladium (Pd) and barium sulfate (BaSO 4) as a catalyst. This reaction is known as the Rosenmund reduction (III) . Option (C): C 6 H 5 CH 3 (i) CrO 3, CS 2 \to (ii) H 2 O This reaction involves the oxidation of toluene (C 6 H 5 CH 3) to benzaldehyde. The first step uses chromium trioxide (CrO 3) as an oxidizing agent, followed by hydrolysis to form the aldehyde. This reaction is known as the Etard reaction (I) . Option (D): CO, HCl \to C 6 H 5 CHO This reaction involves the formation of benzaldehyde from chlorobenzene (C 6 H 5 Cl) in the presence of carbon monoxide (CO) and hydrochloric acid (HCl), with the catalyst being anhydrous AlCl 3 or CuCl. This reaction is known as the Gatterman-Koch reaction (II) . Step 2: Match the Reactions with Their Names Based on the analysis of the reactions above, we can now match the reactions with their corresponding names from List-II: (A) RCN (i) SnCl 2, HCl \to RCHO (ii) H 3 O + \to (IV) Stephen reaction (B) C 6 H 5 CHCl \to C 6 H 5 CHO \to (III) Rosenmund reduction (C) C 6 H 5 CH 3 (i) CrO 3, CS 2 \to (i) H 2 O \to (I) Etard reaction (D) CO, HCl \to C 6 H 5 CHO \to (II) Gatterman-Koch reaction Step 3: Conclusion The correct matching of List-I with List-II is: (A) \to (IV) Stephen reaction (B) \to (III) Rosenmund reduction (C) \to (I) Etard reaction (D) \to (II) Gatterman-Koch reaction This matches the correct answer.$

122

PYQ 2025

medium

chemistry ID: jee-main

$For the reaction: The correct order of set of reagents for the above conversion is :$

1

2

3

4

Official Solution

Correct Option: (1)

$The given reaction represents the bromination of an amine group, likely to form an amine derivative, with Br under the influence of a Lewis acid such as FeBr, and the reaction is carried out in the presence of NaOH to neutralize the acids formed. The major product obtained in this process follows electrophilic aromatic substitution. Final Answer: Option (1).$

123

PYQ 2025

medium

chemistry ID: jee-main

$500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given .$

1

$378 K and 300 J$

2

$378 K and 500 J$

3

$348 K and 300 J$

4

$368 K and 500 J$

Official Solution

Correct Option: (3)

$The problem involves calculating the final temperature and change in internal energy of Argon gas when energy is transferred as heat. Here's how we solve it: Identify the number of moles mol, initial temperature K, and heat transferred J. The heat capacity at constant volume for monoatomic gases like Argon, . Given, we have . The change in internal energy () for a monoatomic ideal gas is: . The heat added at constant volume is: . Therefore, . Calculate final temperature: K. Correction: K. Check the internal energy change through: . Correction: Use, hence, . Thus, the final temperature and change in internal energy is and respectively.$

124

PYQ 2025

medium

chemistry ID: jee-main

$The correct statement amongst the following is :$

1

$The term 'standard state' implies that the temperature is$

2

$The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K$

3

$is zero for O(g)$

4

$is zero for$

Official Solution

Correct Option: (4)

$The question asks us to determine the correct statement regarding the standard state and enthalpy change of formation. Let's consider each option and evaluate the correctness based on standard chemistry principles: The term 'standard state' implies that the temperature is : ^{}\right\wing This statement is incorrect. The term 'standard state' refers to a reference state for thermodynamic calculations. While it is often assumed to be 25°C (298 K) for simplicity, the standard state itself is not defined by a specific temperature. The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K: This statement is partially incorrect. The standard state for a gas is defined as the pure gas at a pressure of 1 bar. However, it does not specify a fixed temperature like 273 K; instead, standard enthalpies are commonly tabulated at 298 K, not 273 K. is zero for O(g): This statement is incorrect. refers to the standard enthalpy change of formation. For oxygen in its diatomic gaseous form ( ), is zero because it is the reference state. However, for monatomic oxygen ( ), it is not zero due to the energy required to dissociate the diatomic molecule. is zero for : This statement is correct. denotes the standard enthalpy change of formation at 500 K. Since is the reference state for oxygen, its standard enthalpy change of formation remains zero at any temperature including 500 K. Based on these evaluations, the correct statement is: is zero for . This is consistent with the fact that is the natural standard state for elemental oxygen.$

125

PYQ 2025

easy

chemistry ID: jee-main

$The formation enthalpies, for and are 220.0 and 250.0 kJ mol, respectively, at 298.15 K, and for (g) is -242.0 kJ mol at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is:$

Official Solution

Correct Option: (1)

$Given reactions and enthalpies: The reaction for and to form is given by: From the given data: 2 (for bonds) 250 kJ/mol (for bonds) (for the formation of) The bond energy formula is: Substituting the values: Solving for :$

126

PYQ 2025

medium

chemistry ID: jee-main

$Consider the given data: $ $ Choose the correct statement:$

1

$Dissolution of gas in water is an endothermic process$

2

$The heat of solution depends on the amount of solvent$

3

$The heat of dilution for the HCl (HCl.10H O to HCl.40H O) is 3.78 kJ/mol$

4

$The heat of formation of HCl solution is represented by both (a) and (b)$

Official Solution

Correct Option: (2)

$The question involves understanding the thermodynamics of dissolution and dilution of hydrochloric acid (HCl) gas in water, represented by two reactions with their enthalpy changes (). The process of dissolving a gas in water is generally exothermic, meaning it releases heat. That is why the enthalpy change () is negative in both reactions given in the question. The first option, "Dissolution of gas in water is an endothermic process," is incorrect because dissolution is exothermic (negative). The second option, "The heat of solution depends on the amount of solvent," suggests that the amount of heat evolved or absorbed can vary with different amounts of solvent. This statement matches our data, as the heat of solution does change between the two cases (and). Thus, this option is correct. The third option involves calculating the heat of dilution from to . The heat of dilution can be determined by calculating the for the change in the number of water molecules: Thus, this option is correct as well, since it calculates the heat of dilution correctly. The fourth option claims both expressions represent heat of formation, but generally, they represent the heat of solution rather than formation. Therefore, it does not accurately describe the scenario provided. Hence, the correct statement is: "The heat of solution depends on the amount of solvent," highlighting the impact of the volume of solvent on the enthalpy change during solution formation.$

127

PYQ 2025

medium

chemistry ID: jee-main

$Given below are two statements : Statement I : When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II : At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below:$

1

$Statement I is true but Statement II is false$

2

$Statement I is false but Statement II is true$

3

$Both Statement I and Statement II are true$

4

$Both Statement I and Statement II are false$

Official Solution

Correct Option: (3)

$To solve this question, let's examine each statement individually and understand the underlying principles of each. Statement I: When a system containing ice in equilibrium with water is heated, heat is absorbed by the system, and there is no change in the temperature of the system until the whole ice gets melted. Explanation: This describes the process of melting, where the temperature remains constant as heat is absorbed. This heat is absorbed as latent heat of fusion, which is used to change the ice to water without increasing the temperature until all ice has melted. Conclusion: Statement I is true. Statement II: At the melting point of ice, there is an absorption of heat to overcome intermolecular forces of attraction within the molecules of water in ice, and the kinetic energy of molecules is not increased at the melting point. Explanation: At the melting point, the absorbed heat breaks the solid lattice (overcoming intermolecular forces) rather than increasing the kinetic energy, which explains why temperature remains constant during this phase transition. Conclusion: Statement II is true. Thus, both statements correctly describe the thermodynamics involved in the melting of ice: Correct Answer: Both Statement I and Statement II are true This choice is consistent with the principles of phase transition and heat absorption without temperature change (latent heat) during the melting process.$

128

PYQ 2025

medium

chemistry ID: jee-main

$A perfect gas (0.1 mol) having R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is ____ J (nearest integer) [Given : R = 0.082 L atm K]$

Official Solution

Correct Option: (1)

$The problem asks for the calculation of the total work done (w) when a perfect gas undergoes a series of reversible transformations as depicted in the given P-V diagram, from an initial point 1 to a final point 4. Concept Used: The work done (w) in a reversible thermodynamic process is calculated by the integral of pressure with respect to volume. The convention used in chemistry is: For a reversible process, . The total work done for a process consisting of multiple steps is the sum of the work done in each individual step: . For an isochoric process (constant volume),, so the work done is . For an isobaric process (constant pressure), the work done is . The final answer needs to be converted from L\cdotatm to Joules (J) using the conversion factor . Step-by-Step Solution: Step 1: Analyze and calculate the work done for the path from point 1 to point 2 (). This is an isochoric process, as the volume remains constant at . For any isochoric process, the change in volume is zero. Step 2: Analyze and calculate the work done for the path from point 2 to point 3 (). This is an isobaric process, as the pressure remains constant at . The volume changes from to . First, calculate the change in volume : Convert the volume from to Liters (L), since : Now, calculate the work done in L\cdotatm: Step 3: Analyze and calculate the work done for the path from point 3 to point 4 (). This is also an isochoric process, with the volume constant at . Step 4: Calculate the total work done () for the entire transformation from 1 to 4. The total work is the sum of the work done in each step. Step 5: Convert the total work from L\cdotatm to Joules (J). Using the conversion factor : Step 6: Round the result to the nearest integer as requested by the question. The question asks for the value to be filled in the blank of `(-)___ J`. Therefore, the value is the magnitude of the work done. The total work done (w) is 304 J.$

129

PYQ 2025

medium

chemistry ID: jee-main

$A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K . As a result of combustion, the temperature of the calorimeter increased by 5 K. The magnitude of the heat of combustion at constant volume is ___$

Official Solution

Correct Option: (1)

$Mass of n-octane = 1.14 g Heat capacity of the bomb calorimeter (C) = 5 kJ K Increase in temperature () = 5 K The heat evolved during the combustion of n-octane at constant volume () is absorbed by the calorimeter, causing the temperature increase. Magnitude of heat evolved = This is the heat evolved from the combustion of 1.14 g of n-octane. We need to find the heat of combustion per mole of n-octane. The molecular formula of n-octane is C H . The molar mass of n-octane is: Number of moles of n-octane burnt = The heat evolved from the combustion of 0.01 mol of n-octane is 25 kJ. The heat of combustion per mole of n-octane () at constant volume is: The magnitude of the heat of combustion at constant volume is 2500 kJ mol . The nearest integer is 2500.$

130

PYQ 2025

medium

chemistry ID: jee-main

$Given below is the plot of the molar conductivity vs concentration for KCl in aqueous solution. If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100, then the resistance of the same cell with the dilute solution is 'x' . The value of is:$

Official Solution

Correct Option: (1)

$Step 1: From the graph, we see that as the concentration decreases, the molar conductivity decreases linearly. Step 2: Based on the relation, and assuming molar conductivity is inversely proportional to concentration, the value of for dilute solution can be estimated. Final Conclusion: The resistance of the cell with the dilute solution is .$

131

PYQ 2025

medium

chemistry ID: jee-main

$The formation enthalpies, for and are 220.0 and 250.0 kJ mol, respectively, at 298.15 K, and for (g) is -242.0 kJ mol at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is:$

Official Solution

Correct Option: (1)

$The bond enthalpy can be calculated using the following equation based on Hess's law: Final Conclusion: The average bond enthalpy of the O-H bond in water is 114 kJ/mol.$

132

PYQ 2025

hard

chemistry ID: jee-main

$The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is:$

Official Solution

Correct Option: (1)

$Step 1: Given density and molar mass of the acid, calculate the molarity using the formula: Step 2: Apply the given values and compute the molarity. Final Conclusion: The molarity is .$

133

PYQ 2025

medium

chemistry ID: jee-main

$Consider a complex reaction taking place in three steps with rate constants,, and respectively. The overall rate constant is given by the expression . If the activation energies of the three steps are 60, 30, and 10 kJ mol respectively, then the overall energy of activation in kJ mol is (Nearest integer).$

Official Solution

Correct Option: (1)

$The expression for the overall rate constant is given as: Now, from the Arrhenius equation: where is the activation energy, is the pre-exponential factor, is the gas constant, and is the temperature. Taking the natural logarithm of the rate constants for each step: We need to determine the overall activation energy, which can be computed by adding the activation energies of the individual steps weighted by their respective rate constants. From the given activation energies,, and kJ mol, the overall activation energy for the reaction is found to be approximately: However, by adjusting for the relative weightings of the rate constants, we find the overall activation energy to be closer to:$

134

PYQ 2025

medium

chemistry ID: jee-main

$The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is:$

Official Solution

Correct Option: (1)

$Step 1: The given compound is -phenylpent-4-en-2-ol. This is a compound with one chiral center at and one at, which means it can form stereoisomers. Step 2: Since the structure is of an alkene (with a double bond) at the position 4, and the hydroxyl group () at position 2, the molecule can form cis and trans stereoisomers depending on the orientation of the substituents about the double bond. Step 3: There are two chiral centers in the molecule. For each chiral center, there are two possibilities (R or S), leading to a total of stereoisomers. However, the molecule also has a plane of symmetry due to the phenyl group, which reduces the total number of stereoisomers. Step 4: Hence, the molecule has only 2 stereoisomers.$

135

PYQ 2025

medium

chemistry ID: jee-main

$In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is times 10^{-1}% (Nearest integer). (Given: Molar mass of Ag is 108 and Br is 80 g mol^{-1}$

Official Solution

Correct Option: (1)

$Using the Carius method, the amount of bromine can be calculated as follows: We are given that 0.25 g of the organic compound yields 0.15 g of silver bromide (AgBr). The moles of AgBr are calculated using: Next, we calculate the mass of bromine in the organic compound: Now, calculate the percentage of bromine: Rounding this to the nearest integer, the final answer is:$

136

PYQ 2025

medium

chemistry ID: jee-main

$The molar mass of the water insoluble product formed from the fusion of chromite ore with in presence of is ....... g mol .$

Official Solution

Correct Option: (1)

$The reaction for the fusion of chromite ore with sodium carbonate () in the presence of oxygen () is given by: The molar mass of the water-insoluble product is the sum of the molar masses of the components involved. The product formed here is chromium oxide,, with a molar mass of: Thus, the molar mass of the water-insoluble product is .$

137

PYQ 2025

medium

chemistry ID: jee-main

$The sum of sigma () and pi () bonds in Hex-1,3-dien-5-yne is:$

Official Solution

Correct Option: (1)

$Hex-1,3-dien-5-yne has the following structure: - 1 single bond between carbon atoms (sigma bonds). - 2 double bonds (each having one sigma and one pi bond). - 1 triple bond (with one sigma and two pi bonds). - Single bonds between carbon and hydrogen atoms. The total number of bonds is: - carbon-hydrogen single bonds. - carbon-carbon sigma bonds (from the backbone of the chain). - 2 pi bonds from the double bonds. - 2 pi bonds from the triple bond. Thus, the total number of and bonds is: - -bonds: - -bonds: So, the sum of sigma and pi bonds is .$

138

PYQ 2025

medium

chemistry ID: jee-main

$Given below are some nitrogen containing compounds: Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ....... mg of HCl. (Given Molar mass in g mol : C: 12, H: 1, O: 16, Cl: 35.5.)$

Official Solution

Correct Option: (1)

$In order to solve this, we need to identify the most basic compound from the given list. The basicity is typically determined by the ability of the compound to donate an electron pair to form a salt with HCl. In this case, primary amines are generally the most basic. Aniline () is a primary amine and will react with HCl to form an ammonium salt. Methyleneamine () is another primary amine and more basic than aniline. Acetanilide () is an amide derivative of aniline and will be less basic due to the electron-withdrawing nature of the acetyl group. Nitroaniline () is less basic due to the strong electron-withdrawing group (-NO2) attached to the aromatic ring. Thus, methyleneamine () is the most basic compound. Now we calculate how much HCl will be consumed by 1.0 g of methyleneamine. Molar mass of methyleneamine: The number of moles of methyleneamine in 1.0 g: Since methyleneamine is a primary amine, it will react with 1 mole of HCl per mole of amine group: Thus, 1.0 g of the most basic compound will consume 1.26 g of HCl. Since the question asks for the answer in mg, we convert: Correct Answer: The most basic compound, methyleneamine, will consume 1260 mg of HCl.$

139

PYQ 2026

hard

chemistry ID: jee-main

$A solution contains 0.25 moles of non-volatile solute dissolved in one mole of solvent. Then calculate the % vapor pressure of solution relative to the vapor pressure of pure solvent.$

1

$20$

2

$40$

3

$60$

4

$80$

Official Solution

Correct Option: (4)

$The vapor pressure of a solution is affected by the amount of solute present. Raoult's Law for the vapor pressure of a solution is given by: Where: - is the vapor pressure of the solution. - is the vapor pressure of the pure solvent. - is the mole fraction of the solvent in the solution. The mole fraction of the solvent is given by: Given: - Moles of solute - Moles of solvent Thus, the mole fraction of the solvent is: The vapor pressure of the solution is then: The percentage of vapor pressure of the solution relative to the vapor pressure of the pure solvent is: Step 1: Conclusion. Thus, the % vapor pressure of the solution relative to the vapor pressure of pure solvent is . Therefore, the correct answer is option (D). Final Answer: 80$

140

PYQ 2026

medium

chemistry ID: jee-main

$Match List-I with List-II. Choose the correct answer.$

1

$A-II, B-III, C-I, D-IV$

2

$A-II, B-I, C-IV, D-III$

3

$A-II, B-I, C-III, D-IV$

4

$A-I, B-II, C-III, D-IV$

Official Solution

Correct Option: (3)

$Concept: Thermodynamic expressions depend on the type of process. Step 1: {Isothermal reversible expansion} Heat absorbed equals work done: Thus Step 2: {Free expansion} Occurs in vacuum: Thus Step 3: {Irreversible compression} Work: Thus Step 4: {Cyclic reversible process} Thus$

141

PYQ 2026

medium

chemistry ID: jee-main

$Calculate for the following reaction (in kJ/mol): Given data:$

Official Solution

Correct Option: (1)

$Step 1: Write the formula for calculating . Step 2: Apply the given values for . Step 3: Substitute the values for each compound. Step 4: Calculate the result.$

142

PYQ 2026

medium

chemistry ID: jee-main

$Initial pressure and volume of monoatomic gas is and . It is expanded adiabatically to times of initial volume. Find magnitude of change in internal energy.$

1

2

3

4

Official Solution

Correct Option: (3)

$Concept: For an adiabatic process For a monoatomic gas Internal energy change: Using ideal gas relation this becomes Step 1: Use adiabatic relation. Given Step 2: Calculate change in internal energy. Step 3: Simplify the expression. Since magnitude is required,$

143

PYQ 2026

medium

chemistry ID: jee-main

$Arrange the following isothermal processes in order of the magnitude of the work (p - V) involved between states 1 and 2. A. Expansion in single stage B. Expansion in multi stages C. Compression in single stage D. Compression in multi stages Choose the correct option.$

1

2

3

4

Official Solution

Correct Option: (3)

$In thermodynamics, the work done during a p-V process is represented by the area under the curve on a P-V diagram. For an isothermal process between two fixed states (1 and 2), we compare expansion and compression work.Step 1: Compare Expansion Work ( vs ).For expansion (from volume to where ), the maximum work is obtained during a reversible process. In an irreversible expansion, work done depends on the external pressure. A multi-stage expansion ( ) approximates the reversible path more closely than a single-stage expansion ( ). Thus, the magnitude of work done in multi-stages is greater than in a single stage:$ w_C w_D V_2 V_1 w_C w_D P_{ext} \ge P_{internal} P_{ext} \le P_{internal} $$

144

PYQ 2026

medium

chemistry ID: jee-main

$An ideal gas is placed in a container at and another ideal gas is placed in a different container at are mixed at final pressure of and final volume of . Calculate the final temperature.$

1

2

3

4

Official Solution

Correct Option: (2)

$textbf{Step 1: Use Ideal Gas Law.} The ideal gas law states: where is the pressure, is the volume, is the number of moles, and is the temperature. Step 2: Combine the Ideal Gas Laws for both gases. Let the number of moles of gas in the first container be and in the second container be . The ideal gas law for each gas before mixing is: After mixing, the total pressure is and the total volume is . Using the ideal gas law for the final state, we have: where is the final temperature. Step 3: Solve for the final temperature. From the equations above, we get: Substitute these values into the equation for the final temperature: Simplifying this gives:$

145

PYQ 2026

medium

chemistry ID: jee-main

$Consider the following data for the reaction at 600 K. The (in kJ) for the reaction is :}$

1

2

3

4

Official Solution

Correct Option: (2)

$Step 1: Understanding the Concept: Gibbs Free Energy change for a reaction can be calculated using the relation . We first need to calculate the standard enthalpy and entropy change for the reaction using the provided formation data. : Key Formula or Approach: Step 2: Detailed Explanation: 1. Calculation of : Reaction: . 2. Calculation of : . 3. Calculation of : Convert to kJ units: . . . Step 3: Final Answer: The for the reaction at 600 K is kJ .$

146

PYQ 2026

medium

chemistry ID: jee-main

$Which of the following statements are not correct? A. For water, magnitude of is more than the magnitude of . B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point. C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers. D. The dimerised form of benzoic acid in benzene is C₆H₅-C-OH \dots O=C-C₆H₅$

1

$A and B only$

2

$A and D only$

3

$A, B and D only$

4

$A, C and D only$

Official Solution

Correct Option: (3)

$Step 1: Understanding the Concept: Colligative properties include,, and . Constants and are solvent-specific. Step 2: Detailed Explanation: Statement A: For water, and . Thus, . Statement A is Incorrect. Statement B: Since and, and, the magnitude of depression in freezing point is larger than the elevation in boiling point. Statement B is Incorrect. Statement C: This is Correct. Osmotic pressure is measured at room temperature and the values are larger and easier to measure for macromolecules. Statement D: Benzoic acid dimers form via two hydrogen bonds involving both oxygen atoms of the carboxyl group. The structure shown in D is incomplete/incorrectly represented as a simple chain. Statement D is Incorrect. Step 3: Final Answer: Statements A, B, and D are not correct.$

147

PYQ 2026

medium

chemistry ID: jee-main

$For given four processes compare magnitude of work:$

1

2

3

4

Official Solution

Correct Option: (4)

$Step 1: Work in thermodynamic processes. In thermodynamics, the work done during an expansion or compression process is related to the pressure-volume relationship. The work done in an isothermal process depends on whether the process is single-step or multi-step, and also whether the process involves expansion or compression. Step 2: Comparison of processes. - (I) Isothermal multistep expansion (W): The work done in a multi-step expansion process is generally larger than in a single-step process, due to the gradual increase in volume. - (II) Isothermal single-step expansion (W): In a single-step expansion, the work is smaller compared to a multistep expansion. - (III) Isothermal single-step compression (W): In a single-step compression, the work done is smaller compared to a multistep compression. - (IV) Isothermal multi-step compression (W): The work done in a multi-step compression process is larger than in a single-step compression process. Step 3: Conclusion. Thus, the correct comparison of magnitudes of work is given by option (4), where . Final Answer: (4)$

148

PYQ 2026

medium

chemistry ID: jee-main

$Statement-1: Heat capacity at constant volume is always greater than heat capacity at constant pressure. Statement-2: At constant volume as work done is zero, heat given to the chaotic motion is reflected by increase in temperature.$

1

$Statement-1 and statement-2 both are correct.$

2

$Statement-1 is correct but statement-2 is incorrect.$

3

$Statement-1 is incorrect but statement-2 is correct.$

4

$Both statement-1 and statement-2 are incorrect.$

Official Solution

Correct Option: (3)

$Step 1: Analyze Statement-1. The heat capacity at constant volume () and constant pressure () are related by the equation: where is the gas constant. As is positive, it is clear that: Thus, Statement-1 is incorrect because heat capacity at constant volume is always less than or equal to heat capacity at constant pressure, not greater. Step 2: Analyze Statement-2. At constant volume, there is no work done, as . The heat added to the system at constant volume only increases the internal energy of the system, which results in an increase in temperature. Therefore, Statement-2 is correct. Step 3: Conclusion. Since Statement-1 is incorrect and Statement-2 is correct, the correct answer is (C). (C) Statement-1 is incorrect but statement-2 is correct.$

149

PYQ 2026

medium

chemistry ID: jee-main

$For the reversible reaction at 300 K, and equilibrium constant, then calculate the magnitude of in Joule/K-mole. [Given: log 2 = 0.3,,, log 3 = 0.47]$

Official Solution

Correct Option: (1)

$Step 1: Use the relationship between,, and . For a reaction at equilibrium, we know the following relationship: At equilibrium, . So, the equation becomes: Step 2: Substitute the known values. Given,, and, we can write: Step 3: Calculate . We can break into two parts: Step 4: Substitute into the equation and solve for . Now, substitute into the equation: Solving for : Step 5: State the final answer. The magnitude of is:$

150

PYQ 2026

medium

chemistry ID: jee-main

$1 mole of He and 1 mole A are taken in a 10 L rigid container at 400 K, and equilibrium is established. Calculate the partial pressure of He and B at equilibrium if .$

1

$3.28 atm, 2.62 atm$

2

$2.6 atm, 3.28 atm$

3

$2.6 atm, 2.6 atm$

4

$3.28 atm, 3.28 atm$

Official Solution

Correct Option: (1)

$Step 1: Understanding the reaction. The given reaction is: We are given: - 1 mole of and 1 mole of are taken in a 10 L rigid container. - The equilibrium constant,, is given for the reaction. - The temperature is 400 K. At equilibrium, we need to calculate the partial pressures of and . We are also told that the volume of the container is 10 L. Step 2: Set up the ICE table (Initial, Change, Equilibrium). Since helium is inert and does not participate in the reaction, its number of moles remains unchanged at 1 mole. The expression for is: At equilibrium, the concentration of and will be: Substitute these into the equilibrium constant expression: Step 3: Solve for . Solve the equation: Multiplying both sides by : Step 4: Calculate the partial pressures. Now, substitute into the equilibrium expressions for and : The partial pressure is given by . Using the ideal gas law: For : For : Step 5: Final Answer. Thus, the partial pressure of is 3.28 atm and the partial pressure of is 2.62 atm. Therefore, the correct answer is: Final Answer: 3.28 atm, 2.62 atm.$

151

PYQ 2026

medium

chemistry ID: jee-main

$If at infinite dilution, molar conductivities of BaCl, HCl and H SO are,, S cm mol respectively, find solubility product of BaSO . Given (Specific conductance) = S cm$

1

2

3

4

Official Solution

Correct Option: (1)

$Step 1: Molar conductivity at infinite dilution. At infinite dilution, the molar conductivities of the ionic compounds are related to their respective ions. The solubility product of BaSO can be derived from the molar conductivities at infinite dilution of BaCl, HCl, and H SO . Step 2: Derivation of the solubility product. For the dissociation of BaSO, the ions produced are Ba and SO . The relationship between the molar conductivity of the salt and the dissociation constants gives the solubility product. The given equation for the specific conductance allows us to compute the solubility product. Step 3: Comparison with the options. - (1) is the correct expression for the solubility product as derived from the given data and equations. - Other options are incorrect as they do not match the derived formula for the solubility product. Step 4: Conclusion. Thus, the correct answer is option (A). Final Answer: (A)$

152

PYQ 2026

easy

chemistry ID: jee-main

$For process, work done by the gas is 10 J and heat absorbed in the process is 2 J. For the reverse process, heat evolved is 6 J. Find the work done for the reverse process.$

1

$Work done on the gas is 14 J$

2

$Work done by the gas is 2 J$

3

$Work done on the gas is 20 J$

4

$Work done by the gas is 12 J$

Official Solution

Correct Option: (1)

$Step 1: Understanding the first process . For the process, the work done by the gas is 10 J and the heat absorbed is 2 J. According to the first law of thermodynamics: where: - is the change in internal energy, - is the heat absorbed, - is the work done by the gas. For the process : Step 2: Understanding the reverse process . In the reverse process, heat is evolved (i.e., heat is released) and the amount of heat evolved is given as 6 J. The change in internal energy for the reverse process will be the same as for the forward process (since internal energy depends only on the initial and final states, not the path), so: For the reverse process, we know that heat is released, so (negative because heat is evolved). Using the first law of thermodynamics for the reverse process: Solving for : Thus, the work done on the gas is: Final Answer: Work done on the gas is 14 J.$

153

PYQ 2026

hard

chemistry ID: jee-main

$Match the LIST-I with LIST-II. Choose the correct answer from the options given below:$

1

$A-III, B-II, C-IV, D-I$

2

$A-II, B-I, C-III, D-IV$

3

$A-I, B-II, C-III, D-IV$

4

$A-II, B-III, C-I, D-IV$

Official Solution

Correct Option: (2)

$Step 1: Process A (Reversible isothermal work). Thus, A II. Step 2: Process B (Irreversible isothermal work). Thus, B I. Step 3: Process C (Change in internal energy). Thus, C III. Step 4: Process D (Change in enthalpy). Thus, D IV. Step 5: Final matching. Final Answer:$

154

PYQ 2026

medium

chemistry ID: jee-main

$Angular momentum of the electron in a hydrogen atom is then find total energy of electron (in eV/atom)$

1

$-1.51$

2

$-122.4$

3

$-40.8$

4

$-4.53$

Official Solution

Correct Option: (1)

$Based on Bohr's model of the atom, the angular momentum of an electron in a stationary orbit is quantized, and the energy of the electron depends on the principal quantum number (). 1. Determine the orbit number (n): According to Bohr's postulate, Angular Momentum = . Given: Comparing both, we get . 2. Calculate Total Energy: The formula for total energy of an electron in the orbit of a hydrogen-like atom is: For Hydrogen, . For : Rounding to two decimal places, the energy is -1.51 eV/atom. Option (1) is correct.$

155

PYQ 2026

medium

chemistry ID: jee-main

$For reaction, . Find the transition temperature (in) of the above reaction at 1 bar.$

Official Solution

Correct Option: (1)

$The transition temperature is the temperature at which the system is at equilibrium between two states, which implies . 1. Set equilibrium condition: At transition temperature, . 2. Solve for T: 3. Convert to Celsius: The transition temperature is 727 .$

156

PYQ 2026

medium

chemistry ID: jee-main

$For a reversible adiabatic process involving ideal gas if initial pressure and volume are 8 bar and respectively and final pressure is 1 bar. Calculate |work done| (in Kilo Joule) []$

Official Solution

Correct Option: (1)

$In a reversible adiabatic process for an ideal gas, the relationship between pressure and volume follows, where . 1. Find : 2. Calculate final volume (): Raising both sides to the power : 3. Calculate work done (W): For an adiabatic process: 4. Convert to Kilo Joules: Absolute value .$

157

PYQ 2026

easy

chemistry ID: jee-main

$500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given:$

1

$378 K and 500 J$

2

$368 K and 500 J$

3

$348 K and 300 J$

4

$378 K and 300 J$

Official Solution

Correct Option: (3)

$Concept: Argon is a monoatomic ideal gas. At constant pressure, heat supplied is given by: Change in internal energy is: For monoatomic gas: Step 1: Given data: Step 2: Calculate : Step 3: Temperature rise: Step 4: Final temperature: Step 5: Calculate : Step 6: Change in internal energy:$

158

PYQ 2026

medium

chemistry ID: jee-main

$For the following change: H O(l) H O(g) 5 C 100 C Select the correct answer$

1

$q = +ve, w = +ve, U = +ve$

2

$q = -ve, w = -ve, U = +ve$

3

$q = +ve, w = -ve, U = +ve$

4

$q = -ve, w = -ve, U = -ve$

Official Solution

Correct Option: (3)

$Step 1: Understanding the process. This process represents the phase transition of water from liquid to gas, known as evaporation. The process occurs at constant temperature, so the temperature change is zero. However, there is heat added to the system for the phase change to occur. Step 2: Determine the heat flow (q). Since water is converting from liquid to gas, energy must be supplied to the system, meaning that heat is absorbed. Hence, q = +ve. Step 3: Work (w). During the transition from liquid to gas, the volume of water increases as it turns into vapor. This expansion against the atmospheric pressure means that work is done by the system. Hence, w = -ve (work is done by the system on the surroundings). Step 4: Change in internal energy (U). The increase in heat results in an increase in the internal energy of the system, so U = +ve. Step 5: Conclusion. The correct answer is (3) q = +ve, w = -ve, U = +ve.$

159

PYQ 2026

medium

chemistry ID: jee-main

$For an ideal gas undergo isothermal reversible process from 0.5Mpa, 20dm to 0.2Mpa at 600K. Calculate correct option. [Given]$

1

$w = -3.9 kJ, U= 0, q = 3.9 kJ$

2

$w = -9.1 kJ, U= 0, q = 9.1 kJ$

3

$w = +9.1 kJ, U= 0, q = -9.1 kJ$

4

$w = +3.9 kJ, U= 0, q = -3.9 kJ$

Official Solution

Correct Option: (2)

$Ideal gas, Isothermal . Expansion (Pressure decreases from 0.5 to 0.2 MPa) Work done by gas . This eliminates options (3) and (4). Work Formula: . . . . Log term: . . . Calculation: . .$

160

PYQ 2026

medium

chemistry ID: jee-main

$Dissociation of a gas takes place according to the following chemical reaction. At equilibrium, the total pressure is at . The standard Gibbs energy of formation of the involved substances is given below: Substance (kJ mol) The degree of dissociation of is given by where________ (Nearest integer). [Given:,, . Assume degree of dissociation is not negligible.]$

Official Solution

Correct Option: (1)

$Step 1: Calculate standard Gibbs energy change of reaction. Step 2: Calculate equilibrium constant . Step 3: Write expression for in terms of degree of dissociation. Let degree of dissociation . Initial moles: Equilibrium moles: Total moles: Partial pressures: Step 4: Substitute . Step 5: Express in the required form. Final Answer:$

Substance	$%5CDelta%20G_f%5E%5Ccirc$ (kJ mol $%5E%7B-1%7D$ )
$A_2$	$-100.00$
$A$	$-50.832$

161

PYQ 2026

medium

chemistry ID: jee-main

$Select the correct match between List-I and List-II: List-I: (I) Isothermal reversible (1 mole ideal gas, T = 300K, 2dm to 20dm) calculate (II) Isothermal irreversible [3KPa, 1m to 3m] calculate (III) 1 mole gas undergoes constant pressure process in which change in temperature is 400K, C = 5R/2, will be (IV) 1 mole ideal gas having C = 3R/2 and = 320K, calculate List-II: (A) 8.32 (B) 6 (C) 4 (D) 5.74$

1

$I-A: II-B: III-D: IV-C$

2

$I-D: II-B: III-A: IV-C$

3

$I-B: II-A: III-C: IV-D$

4

$I-A: II-B: III-C: IV-D$

Official Solution

Correct Option: (2)

$Step 1: Analyzing each case. (I) Isothermal reversible: For isothermal reversible expansion, the work done is calculated as: Substituting the values: Thus, the work done is 8.32 kJ. Therefore, the correct match is I-A. (II) Isothermal irreversible: For isothermal irreversible expansion, the work done is calculated as: Substituting the values: Thus, the work done is 6 kJ. Therefore, the correct match is II-B. (III) Constant pressure process: For a constant pressure process, . Given K and, the enthalpy change will be: Thus, . Therefore, the correct match is III-A. (IV) Ideal gas with Cv: For an ideal gas, the change in internal energy is given by: Substituting the values: Thus, . Therefore, the correct match is IV-C. Step 2: Conclusion. The correct matches are I-D, II-B, III-A, IV-C, which corresponds to option (2).$

162

PYQ 2026

medium

chemistry ID: jee-main

$For Balanced chemical reaction + 6HCl 2AlCl + 3H (g)} \)$

1

$With excess of Al, volume of H gas produced per mole of HCl reacted will be 33.6 L at 1 atm & 273 K.$

2

$With excess of Al, volume of H gas produced per mole of HCl reacted will be 11.2 L at 1 atm & 273 K.$

3

$With excess of HCl, moles of AlCl produced per mole of Al reacted are 2.$

4

$At given P and T, 12 L HCl produce 6 L H gas.$

Official Solution

Correct Option: (2)

$Step 1: Analyzing the reaction. The balanced chemical equation shows that 2 moles of Al react with 6 moles of HCl to produce 3 moles of H . Therefore, the moles of H produced per mole of HCl reacted are: Step 2: Volume of H gas produced. Using the molar volume of an ideal gas (22.4 L at 1 atm and 273 K), the volume of H gas produced per mole of HCl reacted will be: Step 3: Conclusion. Thus, the correct answer is (2), as 11.2 L of H gas are produced per mole of HCl reacted at 1 atm and 273 K.$

163

PYQ 2026

medium

chemistry ID: jee-main

$For the reaction At 300 K and 1 atm pressure, the degree of dissociation of A gas at equilibrium is . Find .$

Official Solution

Correct Option: (1)

$Step 1: Using the formula for Gibbs Free Energy. For the reaction: Where is the equilibrium constant. Using the given data: Step 2: Calculating K. Substituting into the Gibbs free energy equation: Thus, Step 3: Determining the degree of dissociation. For the dissociation reaction: The equilibrium constant can also be expressed as: Let the initial concentration of A be 1 mole (for simplicity), and the concentration of A at equilibrium be . The degree of dissociation is given by: Substituting the value of : Solving for, we get: Step 4: Conclusion. Thus, the degree of dissociation is, or . The correct value of is approximately, which corresponds to the correct answer (58).$

164

PYQ 2026

hard

chemistry ID: jee-main

$If C(diamond) C(graphite) + C(diamond) + C(graphite) + at constant temperature, then$

1

2

3

4

Official Solution

Correct Option: (1)

$Concept: Enthalpy is a state function. According to Hess’s law, the enthalpy change of a reaction depends only on the initial and final states, not on the path followed. Step 1: Write the given reactions with their enthalpy changes: Step 2: Reverse reaction (iii): Step 3: Add reactions (ii) and reversed (iii): Step 4: Net enthalpy change:$

165

PYQ 2026

medium

chemistry ID: jee-main

$For an ideal gas, volume is made 8 times and temperature is decreased 4 times, and heat exchanged during the process is zero (q = 0), select the correct gas.$

1

$CH$

2

$He$

3

$CO$

4

$NH$

Official Solution

Correct Option: (2)

$Step 1: Using the First Law of Thermodynamics. For an ideal gas, the heat exchanged is related to the change in internal energy and work done on the system. If, then the process is adiabatic. For an ideal gas, the internal energy depends only on temperature. Step 2: Analyzing the gases. - For helium (He), which is a noble gas, the internal energy is only a function of temperature, and the temperature decrease leads to a decrease in internal energy without any heat exchange. - For other gases like CH, CO, and NH, their internal energies depend on both temperature and volume (since they have more complex interactions). Step 3: Conclusion. Thus, the correct gas is He, as it is an ideal gas and exhibits the behavior described in the problem. Hence, option (2) is the correct answer.$

166

PYQ 2026

medium

chemistry ID: jee-main

$The plot of versus is a straight line. The intercept and slope of this line are respectively given by (Where is the equilibrium constant.)$

1

2

3

4

Official Solution

Correct Option: (1)

$Concept: The temperature dependence of the equilibrium constant is given by the van’t Hoff equation : Step 1: Convert Natural Logarithm to Base-10 Logarithm Using: Step 2: Compare with Straight Line Equation General straight line form: Here: Thus, Final Conclusion:$

LIST I Isomeric pairs		LIST II Type of isomers
A	Propanamine and N-Methylethanamine	I	Metamers
B	Hexan-2-one and Hexan-3-one	II	Positional isomers
C	Ethanamide and Hydroxyethanimine	II	Functional isomers
D	o-nitrophenol and p-nitrophenol	IV	Tautomers

High-Yield Trend

Chapter Questions 166 MCQs

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Calculation of and

Given Formula:

Step 1: Substitute the Values

Step 2: Relating and

Conclusion:

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Classification of the Given Properties:

Calculation:

Final Answer:

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Concept Used:

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Final Computation & Result:

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Chapter Questions
166 MCQs

Calculation of $%5CDelta%20G%5E%5Ccirc$ and $%5Clog%20K$

Step 2: Relating $%5CDelta%20G%5E%5Ccirc$ and $%5Clog%20K$