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Equilibrium · Question 2

JEE Main 2026 (Online) 8th April Evening ShiftNumericalMedium📈 Topic in 19% papers+4
1

Solid carbon, CaO and CaCO3 are mixed and allowed to attain equilibrium at T K. CaCO3(s) CaO(s) + CO2(g) Kp1 = 0.08 atm C(s) + CO2(g) 2CO(g) Kp2 = 2 atm The partial pressure of CO is 10-1 atm

Answer

4

Solution

Both equilibria exist together in the same vessel with CO2 common to both.

From (1): Kp1 = pCO2 (solids do not appear) ⇒ pCO2 = 0.08 atm

From (2): Kp2 = pCO2pCO2

⇒ pCO2 = 2 0.08 = 0.16

⇒ pCO = 0.4 atm = 4 10-1 atm

Answer: 4

JEE Main 2026 (Online) 8th April Evening Shift

Keep practising — Equilibrium

2

One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. A(g) B(g) Kc for this reaction at 400 K is 4.0. The partial pressures (in atm) of He and B(g) are respectively (at equilibrium) (Assume He, A(g) and B(g) behave as ideal gases) (Given : R = 0.082 L atm K-1 mol-1)

  1. 3.28, 2.624
  2. 2.624, 3.28
  3. 3.28, 0.656
  4. 0.656, 6.56
JEE Main 2026 (Online) 6th April Morning Shift
3

Given is a concentrated solution of a weak electrolyte AxBy of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :

  1. [K cx+y-1xxyy ]x+y
  2. ( Kcx+y-1xxyy ) 1x+y
  3. ( cx+y-1xxyyK )x+y
  4. ( cx+y-1xxyyK ) 1x+y
JEE Main 2026 (Online) 6th April Evening Shift
4

In a closed flask at 600 K, one mole of X2Y4(g) attains equilibrium as given below : X2Y4(g) 2XY2(g) At equilibrium, 75% X2Y4(g) was dissociated and the total pressure is 1 atm. The magnitude of rG (in kJ mol-1) at this temperature is . (Nearest Integer) (Given : R = 8.3 J mol-1 K-1; ln 10 = 2.3, log 2 = 0.3, log 3 = 0.48, log 5 = 0.69, log 7 = 0.84)

JEE Main 2026 (Online) 6th April Evening Shift
5

M3A2 is a sparingly soluble salt of molar mass y g mol-1 and solubility x g L-1. The ratio of the molar concentration of the anion ( A3-) to the solubility product of the salt is

  1. 154 y4x4
  2. y5108x4
  3. 108 x5y5
  4. 1108 y4x4
JEE Main 2026 (Online) 5th April Morning Shift
6

Arrange the following resultant mixtures in increasing order of their pH values Choose the correct answer from the options given below :

  1. B < C < A
  2. C < A < B
  3. C < B < A
  4. A < C < B
JEE Main 2026 (Online) 5th April Morning Shift
7

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below A(g) B(g) + C(g) array|c|c| 1T ( K-1) & 10 Kp \ 0.05 & 3.5 \ 0.06 & 2.5 \ 0.07 & 1.5 \ array The magnitude of H R calculated from the above data is . (Nearest integer)

JEE Main 2026 (Online) 5th April Morning Shift
8

The reaction A(g) B(g) + C(g) was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a − x) at a total pressure of p. The equilibrium constant Kp of the reaction can be calculated from the expression :

  1. x2a2 + x2 p
  2. x2a2 - x2 p
  3. a + x2x2 p
  4. a2 - x2x2 p
JEE Main 2026 (Online) 5th April Evening Shift
9

At T(K), the equilibrium constant of A2(g) + B2(g) C(g) is 2.7 10-5. What is the equilibrium constant for 13A2(g) + 13B2(g) 13C(g) at the same temperature?

  1. (2.7 10-5)3
  2. 6 10-2
  3. 2.7 10-5
  4. 3 10-2
JEE Main 2026 (Online) 4th April Morning Shift
10

The pH of a solution obtained by mixing 5 mL of 0.1 M NH4OH solution with 250 mL of 0.1 M NH4Cl solution is 10-2. (Nearest integer) Given: pKb(NH4OH) = 4.74 2 = 0.30 3 = 0.48 5 = 0.70

JEE Main 2026 (Online) 4th April Morning Shift
11

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (pKa value of acetic acid is 4.75).

  1. 7.0
  2. 4.75
  3. 3.5
  4. 4.82
JEE Main 2026 (Online) 4th April Evening Shift

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