Equilibrium

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Equilibrium — solved practice questions

8 JEE Chemistry questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   For the reaction 2SO2(g) + O2(g) <=> 2SO3(g), the relationship between Kp and Kc is:

   • AKp = Kc(RT)
   • BKp = Kc(RT)^-1
   • CKp = Kc
   • DKp = Kc(RT)^2
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   Correct answer: (B) Kp = Kc(RT)^-1

   Kp = Kc(RT)^(delta n), where delta n = 2 − 3 = -1 for gaseous species. Therefore Kp = Kc(RT)^-1 = Kc/(RT).

2. Q2easy

   The pH of a 0.001 M NaOH solution at 25 °C is:

   • A3
   • B7
   • C11
   • D12
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   Correct answer: (C) 11

   NaOH is a strong base, so [OH-] = 0.001 M, giving pOH = 3. Then pH = 14 − 3 = 11.

3. Q3medium

   For the equilibrium N2O4(g) <=> 2NO2(g), at equilibrium the concentrations are [NO2] = 0.02 M and [N2O4] = 0.1 M. The value of Kc is:

   • A0.004
   • B0.04
   • C0.2
   • D0.0004
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   Correct answer: (A) 0.004

   Kc = [NO2]^2 / [N2O4] = (0.02)^2 / 0.1 = 0.0004 / 0.1 = 0.004.

4. Q4medium

   According to Le Chatelier's principle, for the exothermic reaction N2(g) + 3H2(g) <=> 2NH3(g), the yield of NH3 is increased by:

   • AIncreasing temperature and decreasing pressure
   • BIncreasing temperature and increasing pressure
   • CDecreasing temperature and decreasing pressure
   • DDecreasing temperature and increasing pressure
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   Correct answer: (D) Decreasing temperature and increasing pressure

   Since the reaction is exothermic and proceeds with a decrease in moles of gas, high pressure and low temperature shift the equilibrium toward NH3, increasing its yield.

5. Q5medium

   The solubility product (Ksp) of AgCl is 1.6 x 10^-10 at 25 °C. Its molar solubility in pure water is:

   • A1.6 x 10^-10 mol/L
   • B1.26 x 10^-5 mol/L
   • C1.6 x 10^-5 mol/L
   • D4.0 x 10^-5 mol/L
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   Correct answer: (B) 1.26 x 10^-5 mol/L

   For AgCl <=> Ag+ + Cl-, Ksp = s^2, so s = sqrt(1.6 x 10^-10) = 1.26 x 10^-5 mol/L.

6. Q6medium

   The pH of a 0.1 M acetic acid solution (Ka = 1.8 x 10^-5) is approximately:

   • A1.0
   • B3.5
   • C2.87
   • D4.74
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   Correct answer: (C) 2.87

   For a weak acid, [H+] = sqrt(Ka x C) = sqrt(1.8 x 10^-5 x 0.1) = 1.34 x 10^-3 M. So pH = -log(1.34 x 10^-3) ≈ 2.87.

7. Q7hard

   A buffer solution contains 0.1 M acetic acid and 0.05 M sodium acetate (pKa of acetic acid = 4.74). The pH of the buffer is approximately:

   • A4.44
   • B5.04
   • C4.74
   • D5.44
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   Correct answer: (A) 4.44

   By the Henderson-Hasselbalch equation, pH = pKa + log([salt]/[acid]) = 4.74 + log(0.05/0.1) = 4.74 + log(0.5) = 4.74 − 0.30 = 4.44.

8. Q8medium

   For the decomposition CaCO3(s) <=> CaO(s) + CO2(g), the equilibrium constant Kp is equal to:

   • Ap(CaO) x p(CO2)
   • Bp(CO2) / p(CaCO3)
   • C1 / p(CO2)
   • Dp(CO2)
   Show answer & solution

   Correct answer: (D) p(CO2)

   Pure solids do not appear in the equilibrium expression. Only the gaseous CO2 contributes, so Kp = p(CO2), the partial pressure of carbon dioxide.

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