NEET Chemistry: Equilibrium questions with solutions

Q1. A beaker contains a saturated solution of copper(l) chloride. Copper(l) chloride is slightly soluble salt with a solubility product of \( 1.2 \times 10^{-6} \) Which of the following salts when added to the solution would precipitate copper(l) chloride?

1. Sodium chloride
2. Potassium bromide
3. Silver(I) nitrate
4. Lead(II) acetate E. Magnesium iodide

Answer: Silver(I) nitrate

Silver(I) nitrate provides Ag+ ions, which react with Cl− to form AgCl, an insoluble precipitate. Removing chloride from solution shifts the equilibrium for CuCl(s) formation, causing more copper(I) chloride to precipitate.

Q2. The function of 'buffer solution' is:

1. to increase the pH value in chemical reaction
2. to decrease the pH value on chemical reaction
3. to increase or decrease the pH value of according the need
4. to keep the pH value constante in chemical reaction

Answer: to keep the pH value constante in chemical reaction

A buffer solution is designed to maintain a nearly constant pH by neutralizing small additions of acid or base. That is why its main function is to keep the pH stable in a chemical reaction.

Q3. When \( C a(O H)_{2} \) attains solubility equilibrium, the:

1. solution is saturated
2. \( p H \) will be less than 7
3. trial \( K_{s p} \) is less than the \( K_{s p} \)
4. concentrations of the ions are equal

Answer: solution is saturated

When Ca(OH)2 reaches solubility equilibrium, the rate of dissolving equals the rate of crystallizing, so no more solid can dissolve under those conditions. That means the solution is saturated.

Q4. Predict if there will be any precipitate by mixing \( 50 \mathrm{mL} \) of \( 0.01 \mathrm{M} \) NaCl and 50 mL of \( 0.01 \mathrm{M} A g N O_{3} \) solution. The solubility product of \( A g C l \) is \( 1.5 \times \) \( 10^{-10} \)

1. since ionic product is greater than solubility product no precipitate will be formed
2. since ionic product is lesser than solubility product. precipitation will occur
3. since ionic product is greater than solubility product. precipitation will occur.
4. since ionic product and solubility product are same, precipitation will not occur

Answer: since ionic product is greater than solubility product. precipitation will occur.

The ionic product of AgCl is calculated by multiplying the concentrations of Ag+ and Cl- ions, and if it is greater than the solubility product, precipitation will occur. [AI-generated key — verify before high-stakes use]

Q5. Calculate the ratio of \( H C O O^{-} \) and \( F^{-} \) in a mixture of \( 0.2 M \) HCOOH \( \left(K_{a}=\right. \) \( \left.2 \times 10^{-4}\right) \) and \( 0.1 M H F\left(K_{a}=6.6 \times\right. \) \( \left.10^{-4}\right) \)

1. 1: 6.6
2. 1: 3.3
3. 2: 3.3
4. 3.3 : 2

Answer: 1: 3.3

For each weak acid, the conjugate base formed is proportional to its dissociation, so [HCOO^-]/[HCOOH]  and [F^-]/[HF]  can be estimated from Ka and initial concentration. Comparing those two ratios gives the relative amounts of HCOO^-  and F^- , which simplifies to 1:3.3.

Q6. What is the [OH⁻] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂?

1. 0.40 M
2. 0.005 M
3. 0.12 M
4. 0.10 M

Answer: 0.12 M

Ba(OH)₂ provides 2 OH⁻ ions per molecule. Moles of OH⁻ = 30.0 mL × 0.10 M × 2 = 0.006 mol. Moles of H⁺ = 20.0 mL × 0.050 M = 0.001 mol. Excess OH⁻ = 0.006 - 0.001 = 0.005 mol. Total volume = 50.0 mL = 0.050 L. [OH⁻] = 0.005 mol / 0.050 L = 0.10 M.

Q7. In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?

1. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
2. 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
3. 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
4. 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen

Answer: 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

In the Haber process, the reaction is N₂ + 3H₂ → 2NH₃. Initially, 30 L of N₂ and 30 L of H₂ are taken. The limiting reagent is H₂, as it requires 3 times the amount of N₂. For 50% yield, 15 L of H₂ reacts with 5 L of N₂ to form 10 L of NH₃. Remaining gases are 25 L of N₂ and 15 L of H₂.

Q8. The equilibrium constant of the following are: N2 + 3H2 ⇌ 2NH3 K1 N2 + O2 ⇌ 2NO K2 H2 + 1/2 O2 ⇌ H2O K3 The equilibrium constant (K) of the reaction: 2NH3 + 5/2 O2 ⇌ 2NO + 3H2O, will be:

1. K2K3^3 / K1
2. K2K3 / K1
3. K2^2K3^3 / K1
4. K1K3^3 / K2

Answer: K2K3^3 / K1

To derive the equilibrium constant for the given reaction, we manipulate the given reactions and their equilibrium constants. The reverse of the first reaction gives 2NH3 ⇌ N2 + 3H2 with equilibrium constant 1/K1. Adding this to the second and third reactions (appropriately scaled) gives the desired reaction, and the equilibrium constant becomes K2K3^3 / K1.

Q9. If the value of an equilibrium constant for a particular reaction is 1.6 × 10^12, then at equilibrium the system will contain:

1. mostly reactants
2. mostly products
3. similar amounts of reactants and products
4. all reactants

Answer: mostly products

An equilibrium constant (K) value much greater than 1, such as 1.6 × 10^12, indicates that the reaction strongly favors the formation of products at equilibrium.

Q10. A 20 litre container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value, will be: (Given that: SrCO3(s) ⇌ SrO(s) + CO2(g), Kp = 1.6 atm)

1. 20 litre
2. 4 litre
3. 2 litre
4. 5 litre

Answer: 4 litre

The equilibrium constant Kp = 1.6 atm determines the maximum pressure of CO2. Using the ideal gas law, the maximum volume is calculated when the pressure of CO2 equals Kp. At P = 1.6 atm, V = nRT/P = (0.4 × 20)/1.6 = 4 L.

Q11. If the equilibrium constant for N2(g) + O2(g) ⇌ 2NO(g) is K, the equilibrium constant for 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) will be:

1. K^2
2. 2K
3. K
4. √K

Answer: √K

When the stoichiometric coefficients of a reaction are halved, the equilibrium constant is raised to the power of 1/2 (square root). Thus, the equilibrium constant for the given reaction is √K.

Q12. The values of KP1 and KP2 for the reactions X ⇌ Y + Z ...(a) and A ⇌ 2B ...(b) are in the ratio of 9:1. If degree of dissociation of X and A be equal, then total pressure at equilibrium for (a) and (b) are in the ratio:

1. 3:1
2. 1:9
3. 36:1
4. 1:1

Answer: 3:1

The equilibrium constant KP is proportional to the total pressure raised to the power of the change in the number of moles (Δn). For reaction (a), Δn = 1, and for reaction (b), Δn = 1. Since the degree of dissociation is equal and KP1/KP2 = 9/1, the total pressure ratio is √9:√1 = 3:1.

Q13. Conjugate acid of NH₂⁻ is:

1. NH₄⁺
2. NH₃
3. NH₂
4. NH

Answer: NH₃

The conjugate acid of a base is formed by adding a proton (H⁺) to it. Adding H⁺ to NH₂⁻ gives NH₃.

Q14. The pH value of a 10 M solution of HCl is:

1. less than 0
2. equal to 0
3. equal to 1
4. equal to 2

Answer: less than 0

The pH is calculated as -log[H+]. For a 10 M HCl solution, [H+] = 10 M, so pH = -log(10) = -1, which is less than 0.

Q15. The value of equilibrium constant of the reaction HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g) is 8.0. The equilibrium constant of the reaction H2(g) + I2(g) ⇌ 2HI(g) will be:

1. 1/16
2. 1/64
3. 16
4. 1/8

Answer: 16

The equilibrium constant for the reverse reaction is the reciprocal of the given constant, and since the stoichiometry is doubled, the new constant is the square of the reciprocal: (1/8)^2 = 16.

Q16. The following equilibrium constants are given: N2 + 3H2 ⇌ 2NH3; K1 N2 + O2 ⇌ 2NO; K2 H2 + 1/2 O2 ⇌ H2O; K3 The equilibrium constant for the oxidation of 2 moles NH3 by oxygen to give NO is:

1. K2 × K3 / K1
2. K2² × K3 / K1
3. K1 × K2 / K3
4. K2 × K3³ / K1

Answer: K2² × K3 / K1

The reaction for the oxidation of 2 moles of NH3 to NO can be derived by combining the given reactions. Using the equilibrium constant relationships, the correct expression is K2² × K3 / K1.

Q17. Given reaction are X ⇌ Y + Z and 4 ⇌ 2B. Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then KP1/KP2 = ?

1. 9
2. 1
3. KP1 = 9
4. KP2 = 1

Answer: 9

For reaction (i), X ⇌ Y + Z, Δn = 1 (1+1-1). For reaction (ii), 4 ⇌ 2B, Δn = -2 (2-4). The ratio of KP1/KP2 depends on the relationship between Δn and pressure terms. Simplifying, KP1/KP2 = 9.

Q18. Given: Equilibrium constant (K1) for the reaction: HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g); K1 = 8. To find equilibrium constant for the following reaction: H2(g) + I2(g) ⇌ 2HI(g); K2 = ?

1. 1/64
2. 64
3. 8^2 = 64
4. K2 = 1/64

Answer: 8^2 = 64

The given reaction is the reverse of the original reaction and its stoichiometry is doubled. The equilibrium constant for the reverse reaction is the reciprocal of K1, and doubling the stoichiometry squares the constant. Thus, K2 = (1/K1)^2 = (1/8)^2 = 64.

Q19. Given, N2 + 3H2 ⇌ 2NH3; K1. N2 + O2 ⇌ 2NO; K2. H2 + 1/2 O2 ⇌ H2O; K3. We have to calculate 4NH3 + 5O2 ⇌ 4NO + 6H2O; K = ?

1. K = [NO]^2[H2O]^3 / [NH3]^2[O2]^5/2
2. K1 = [NH3]^2 / [N2][H2]^3
3. K2 = [NO]^2 / [N2][O2]
4. K3 = [H2O]^3 / [H2][O2]^3/2

Answer: K = [NO]^2[H2O]^3 / [NH3]^2[O2]^5/2

The equilibrium constant for the given reaction can be derived by combining the equilibrium constants K1, K2, and K3 appropriately. The resulting expression matches option A.

Q20. What is the pH of a solution with molarity (M) = 10M of HCl?

1. pH = log[H+] = log[10] = 1
2. Kw at 25°C = 1 × 10−14
3. Kw = [H+][OH−] = 10−14
4. pH = log[H+] = log[55 × 10−14]1/2

Answer: pH = log[H+] = log[10] = 1

HCl is a strong acid and dissociates completely in water. For a 10M HCl solution, [H+] = 10M, so pH = -log[H+] = -log(10) = 1.