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JEE Main 35 Year PYQs- Equilibrium - Free MCQ Test with solutions [Updated]


MCQ Practice Test & Solutions: Test: JEE Main 35 Year PYQs- Equilibrium (52 Questions)

You can prepare effectively for JEE Chapter-wise Tests for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: JEE Main 35 Year PYQs- Equilibrium". These 52 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 100 minutes
  • - Number of Questions: 52

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Test: JEE Main 35 Year PYQs- Equilibrium - Question 1
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PARAGRAPH 1

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)

Q. Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt.2 is

Detailed Solution: Question 1

Let the heat capacity of insulated beaker be C.
Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g

⇒ ± Total heat capacity = (C + 200 × 4.2) J/K Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1

⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ = 5.7 × 1000 J

⇒ 5.7 × 1000 = (C + 200 × 4.2) × Δ´T.
5.7 × 1000 = (C + 200 + 4.2) × 5.7

⇒ (C + 200 × 4.2) = 1000
In second experiment,

Total mass of aqueous content = 200 g

⇒ Total heat capacity = (C + 200 × 4.2) = 1000

⇒ Heat released = 1000 × 5.6 = 5600 J.
Overall, only 0.1 mol of CH3COOH undergo neutralization.

⇒ Δ Hneutralization of CH3COOH = -5600/0.1

= – 56000 J/mol
= – 56 KJ/mol.

⇒ Δ Hneutralization of CH3COOH = 57 – 56 = 1 KJ/mol

Test: JEE Main 35 Year PYQs- Equilibrium - Question 2
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PARAGRAPH 1

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)

Q. The pH of the solution after Expt. 2 is

Detailed Solution: Question 2

Final solution contain 0.1 mole of CH3COOH2 and CH3COONa each.
Hence it is a buffer solution.

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Test: JEE Main 35 Year PYQs- Equilibrium - Question 3
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PARAGRAPH 2

Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:

The standard reaction Gibbs energy, ΔrG°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.

(Given R = 0.083 L bar K–1 mol–1)

Q. The equilibrium constant Kp for this reaction at 298 K, in terms of βequilibrium, is 

Detailed Solution: Question 3




Test: JEE Main 35 Year PYQs- Equilibrium - Question 4
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PARAGRAPH 2

Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:

The standard reaction Gibbs energy, ΔrG°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.

(Given R = 0.083 L bar K–1 mol–1)

Q. The INCORRECT statement among the following, for this reaction, is

Detailed Solution: Question 4

(A) Correct statement.
As on decrease in pressure reaction move indirection where no. of gaseous molecules increase.
(B) Correct statement At the start of reaction QP < KP so dissociation of X2 take place spontaneousely.
(C) Incorrect statement as

Test: JEE Main 35 Year PYQs- Equilibrium - Question 5
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Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 5 choices (a), (b), (c) and  (d) out of which ONLY ONE is correct. 

Q. 

Statement -1 The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature.

Statement -2  When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change.

Detailed Solution: Question 5

The statement-1 is clearly wrong in context to LeChateliers principle, which states that “increase in temperature shifts the equilibrium in the forward direction of those reactions which proceed with absorption of heat (endothermic reactions), and in the backward direction of those reactions which proceed with the evolution of heat (exothermic reactions).” Statement -2 is clearly true again according to Lechatelier principle.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 6
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Read the following statement and explanation and answer as per the options given below :

Statement -1  HNO3 is a stronger acid than HNO2

Statement -2 In HNO3 there are two nitrogen-to-oxygen bonds whereas in HNO2 there is only one.

Detailed Solution: Question 6

TIPS/Formulae :

Among oxyacids, the acidic character increases with increase in oxidation state of the central atom.
O.S. of  N in HNO3 = + 5
O.S.of  N in HNO2  = + 3

thus HNO3 stronger  than HNO2. Hence assertion is correct.

The assertion is true but the reason is wrong as can be clearly seen from the above structures.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 7
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Read the following statement and explanation and answer as per the options given below :

Statement -1 For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero.

Statement -2 At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

Detailed Solution: Question 7

We know that for every chemical reaction at equilibrium, Gibb's free energy (ΔG = 0) is zero.
However standard Gibb's free energy (ΔG°) may or may not be zero. Thus statement 1 is False.
For a spontaneous reaction, at constant temperature and pressure, the reaction proceeds in the direction in which ΔG is < 0 i.e. in the direction of decreasing Gibb's energy (G) so the statement 2 is True.
Thus the only such option is (d) which is correct answer.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 8
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1 M NaCl and 1 M HCl are present in an aqueous solution.

The solution is

Detailed Solution: Question 8

NOTE :  A buffer is a solution of weak acid and its salt with strong base and vice versa.
HCl is strong acid and NaCl is its salt with strong base. pH is less than 7 due to HCl.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 9
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Species acting as both Bronsted acid and base is

Detailed Solution: Question 9

(HSO4)– can accept and donate a proton
(HSO4)– + H+ → H2SO4  (acting as base)
(HSO4)– – H+ → SO42–.   (acting as acid)

Test: JEE Main 35 Year PYQs- Equilibrium - Question 10
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Let the solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is

Detailed Solution: Question 10

Test: JEE Main 35 Year PYQs- Equilibrium - Question 11
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Change in volume of the system does not alter  which of the following equilibria?

Detailed Solution: Question 11

In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 12
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For the reaction CO (g) + (1/2) O2 (g) = CO2 (g), Kp / Kc is

Detailed Solution: Question 12

Kp = Kc (RT)Δn;

Test: JEE Main 35 Year PYQs- Equilibrium - Question 13
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Which one of the following statements is not true? 

Detailed Solution: Question 13

pH of an acidic solution should be less than 7. The reason is that from H2O. [H+] = 10–7M  which cannot be neglected in comparison to 10–8M. The pH can be calculated as.
from acid, [H+]  = 10–8M.

from H2O, [H+] = 10–7M
∴ Total [H+] = 10–8 + 10–7 = 10–8 (1 + 10) = 11× 10–8
∴ pH = – log [H+] = –log 11×10–8 = –[log11 + 8 log 10]            
= –[1.0414 – 8] = 6.9586

Test: JEE Main 35 Year PYQs- Equilibrium - Question 14
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The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10–5 mol L–1. Its solubility product number will be

Detailed Solution: Question 14

Test: JEE Main 35 Year PYQs- Equilibrium - Question 15
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For the reaction equilibrium

the concentrations of N2O4 and NO2 at equilibrium are 4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of Kc for the reaction is

Detailed Solution: Question 15

Test: JEE Main 35 Year PYQs- Equilibrium - Question 16
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Consider the reaction equilibrium

On the basis of Le Chatelier’s principle, the condition favourable for the forward reaction is

Detailed Solution: Question 16

Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles.
∴ High pressure will be required.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 17
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When rain is accompanied by a thunderstorm, the collected rain water will have a  pH value

Detailed Solution: Question 17

The rain water after thunderstorm contains dissolved acid and therefore the pH is less than rain water without thunderstorm.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 18
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The conjugate base of 

Detailed Solution: Question 18

NOTE : Conjugate acid-base differ by H+

Test: JEE Main 35 Year PYQs- Equilibrium - Question 19
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What is the equilibrium expression for the reaction 

Detailed Solution: Question 19


 The solids have con centration unity

Test: JEE Main 35 Year PYQs- Equilibrium - Question 20
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For the reaction,  is equal to

Detailed Solution: Question 20


Test: JEE Main 35 Year PYQs- Equilibrium - Question 21
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The equilibrium constant for the reaction  at temperature T is 4×10–4.

The value of Kc for the reaction

at the same temperature is

Detailed Solution: Question 21


Test: JEE Main 35 Year PYQs- Equilibrium - Question 22
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The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’ is given in term of Ksp by the relation :

Detailed Solution: Question 22


Test: JEE Main 35 Year PYQs- Equilibrium - Question 23
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If α is the degree of dissociation of Na2SO4 , the Vant Hoff’s factor (i) used for calculating the molecular mass is

Detailed Solution: Question 23

Test: JEE Main 35 Year PYQs- Equilibrium - Question 24
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The solubility product of a salt having general formula MX2 , in water is : 4 × 10 -12. The concentration of M2+ ions in the aqueous solution of the salt is

Detailed Solution: Question 24

Where  s is the solubility of MX2 then Ksp = 4s3; s × (2s)2 = 4×10–12 = 4s3; s = 1 × 10–4
∴ [M++] = s = 1[M++] = 10 × 10– 4

Test: JEE Main 35 Year PYQs- Equilibrium - Question 25
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The exothermic formation of CIF3 is represented by the equation :

Which of the following will increase the quantity of CIF3 in an equilibrium mixture of Cl2 , F2 and ClF3 ?

Detailed Solution: Question 25

The reaction given is an exothermic reaction thus accordingly to Lechatalier’s principle lowering of temperature, addition of F2 and  or Cl2 favour the for ward direction and hence the production of ClF3.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 26
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For the reaction  

(Kc = 1.8 x 10-6 at 184°C) (R = 0.0831 kJ/ (mol. K))

When Kp and Kc are compared at 184°C, it is found that

Detailed Solution: Question 26

For the reaction:- 

Given Kc = 1.8 × 10–6   at 184ºC
R = 0.0831 kj/mol. k
Kp = 1.8 × 10–6  × 0.0831 × 457 = 6.836 × 10–6
[∵ 184°C = (273 + 184) = 457 k, Δn = (2 + 1, –1) = 1]
Hence it is clear that  Kp >  Kc

Test: JEE Main 35 Year PYQs- Equilibrium - Question 27
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Hydrogenion concentration in mol/L in a solution of pH = 5.4 will be :

Detailed Solution: Question 27

On solving,  [H+] = 3.98 × 10–6

Test: JEE Main 35 Year PYQs- Equilibrium - Question 28
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What is the conjugate base of OH - ?

Detailed Solution: Question 28

Conjugate acid-base pair differ by only one proton.

Conjugate base of OH– is O2–

Test: JEE Main 35 Year PYQs- Equilibrium - Question 29
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An amount of solid NH4 HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The equilibrium constant for NH4HS decomposition at this temperature is

Detailed Solution: Question 29

Then 0.5 + x + x = 2x + 0.5 = 0.84  (given) ⇒ x = 0.17 atm.

Test: JEE Main 35 Year PYQs- Equilibrium - Question 30
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Phosphorus pentachloride dissociates as follows, in a closed reaction vessel

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be

Detailed Solution: Question 30

Total moles after dissociation 1 – x + x + x = 1 + x

= mole fraction of PCl3 × Total pressure

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