Equilibrium

17 NEET questions on Equilibrium. A sample is shown below with full explanations; the rest are available free after signing up.

Equilibrium2026hard

Q1.The correct order of solubility of the given salts in water at 298 K is

  • 1.AgBr — Ksp at 298 K = 5.0 × 10⁻¹³
  • 2.Zn(OH)₂ — Ksp at 298 K = 1.0 × 10⁻¹⁵
  • 3.Hg₂Cl₂ — Ksp at 298 K = 1.3 × 10⁻¹⁸
  • AHg₂Cl₂ > Zn(OH)₂ > AgBr
  • BAgBr > Zn(OH)₂ > Hg₂Cl₂
  • CHg₂Cl₂ > AgBr > Zn(OH)₂
  • DZn(OH)₂ > AgBr > Hg₂Cl₂Answer

Explanation

Ksp values cannot be compared directly across salts of different stoichiometry — that is the whole trap. AgBr is AB type, so S = √Ksp ≈ 7.1 × 10⁻⁷. Zn(OH)₂ and Hg₂Cl₂ are AB₂ type with Ksp = 4S³, giving S ≈ 6.3 × 10⁻⁶ and 6.9 × 10⁻⁷. So Zn(OH)₂ is the most soluble despite having a smaller Ksp than AgBr.

Source: NEET 2026 re-exam (21 June), Code 50, Q52

Equilibrium2025medium

Q2.For the reaction A(g) ⇌ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given: R = 0.0831 L atm mol⁻¹ K⁻¹] K_P for the reaction at 1000 K is

  • A83.1
  • B2.077 × 10⁵
  • C0.033Answer
  • D0.021

Explanation

At equilibrium K_C is the ratio of the forward to the backward rate constant, so K_C = k_f/k_b = 1/2500 = 4 × 10⁻⁴. The two constants are related by K_P = K_C(RT)^Δn, where Δn is the change in the number of moles of gas. Here Δn = 2 − 1 = 1, so K_P = (1/2500) × 0.0831 × 1000 = 83.1/2500 = 0.033.

Source: NEET 2025 (4 May), Code 45, Q73

Equilibrium2025medium

Q3.Higher yield of NO in N₂(g) + O₂(g) ⇌ 2NO(g) can be obtained at [ΔH of the reaction = +180.7 kJ mol⁻¹]

  • A. Higher temperature
  • B. Lower temperature
  • C. Higher concentration of N₂
  • D. Higher concentration of O₂
  • AA, D only
  • BB, C only
  • CB, C, D only
  • DA, C, D onlyAnswer

Explanation

By Le Chatelier’s principle, the reaction is endothermic since ΔH is positive, so supplying heat drives the equilibrium forward and a higher temperature raises the yield — which rules B out and puts A in. Increasing the concentration of either reactant also shifts the equilibrium to the right, so both C and D help. Note that pressure has no effect here, because the number of gaseous moles is two on each side.

Source: NEET 2025 (4 May), Code 45, Q88

Practise the full bank, free

1286 questions with worked explanations, timed mocks and a syllabus map that shows where you are weak.

Create a free account

Questions are reformatted from NEET previous year papers with original explanations and topic tagging. NEET Prep is not affiliated with the NEET examining body.