HEC USAT-E (Pre-Engineering) Chemistry: Chemical Equilibrium MCQs

Practice Chemical Equilibrium MCQs for HEC USAT-E (Pre-Engineering) Chemistry — topic-wise sets with solved answers.

HEC USAT-E (Pre-Engineering) Chemistry: Chemical Equilibrium MCQs — sample questions

  1. Question 1

    Q1. For the reaction N2 + 3H2 ⇌ 2NH3, if the equilibrium constant Kc is 4, what is Kp at 27°C?

    • A) 4 / (0.0821 × 300)²
    • B) 4 × (0.0821 × 300)²
    • C) 4 × (0.0821 × 300)^-2
    • D) 4 / (0.0821 × 300)^-2

    Answer: 4 × (0.0821 × 300)^-2

    Explanation: Kp = Kc(RT)^Δn, Δn = 2 - 4 = -2, so Kp = 4 × (0.0821 × 300)^-2

  2. Question 2

    Q2. For the reaction 2SO2 + O2 ⇌ 2SO3, Kc = 100 at 500K. What is Kc for SO2 + 1/2O2 ⇌ SO3?

    • A) √100
    • B) 100²
    • C) 100
    • D) 1 / √100

    Answer: √100

    Explanation: When the reaction is halved, Kc becomes √Kc, so Kc = √100 = 10

  3. Question 3

    Q3. For the reaction A + B ⇌ C + D, Kc = 4. If 1 mole of A and 1 mole of B are mixed, what is the equilibrium concentration of C?

    • A) 2/3
    • B) 1/2
    • C) 2/5
    • D) 4/5

    Answer: 2/3

    Explanation: Kc = [C][D]/[A][B] = 4, let x be the concentration of C, then x²/(1-x)² = 4, x = 2/3

  4. Question 4

    Q4. The equilibrium constant for the reaction H2 + I2 ⇌ 2HI is 50. What is the equilibrium constant for 2HI ⇌ H2 + I2?

    • A) 1/50
    • B) 1/√50
    • C) 50
    • D) 50²

    Answer: 1/50

    Explanation: When the reaction is reversed, Kc becomes 1/Kc, so Kc = 1/50

  5. Question 5

    Q5. For the reaction PCl5 ⇌ PCl3 + Cl2, if the degree of dissociation is α, what is the equilibrium constant Kp?

    • A) α² / (1 - α²)
    • B) α² / (1 - α)
    • C) α / (1 - α)
    • D) α²P / (1 - α²)

    Answer: α²P / (1 - α²)

    Explanation: Kp = (αP/(1+α)) × (αP/(1+α)) / ((1-α)P/(1+α)) = α²P / (1 - α²)

  6. Question 6

    Q6. For the reaction 2NO2 ⇌ N2O4, if the equilibrium constant Kc is 170 at 300K, what is the value of ΔG°?

    • A) -8.314 × 300 × ln(170)
    • B) 8.314 × 300 × ln(170)
    • C) -8.314 × 300 × log(170)
    • D) 8.314 × 300 × log(170)

    Answer: -8.314 × 300 × ln(170)

    Explanation: ΔG° = -RT ln(Kc) = -8.314 × 300 × ln(170)

  7. Question 7

    Q7. For the reaction CO + H2O ⇌ CO2 + H2, if the initial concentrations of CO and H2O are 1M each, and Kc = 4, what is the equilibrium concentration of CO?

    • A) 1/3
    • B) 1/5
    • C) 2/3
    • D) 1/2

    Answer: 2/3

    Explanation: Let x be the concentration of CO2, then Kc = x²/(1-x)² = 4, x = 2/3, so [CO] = 1 - 2/3 = 1/3

  8. Question 8

    Q8. The equilibrium constant for the reaction N2O4 ⇌ 2NO2 is 0.16 at 300K. What is the value of Kc for 1/2N2O4 ⇌ NO2?

    • A) √0.16
    • B) 0.16
    • C) 0.16²
    • D) 1/√0.16

    Answer: √0.16

    Explanation: When the reaction is halved, Kc becomes √Kc, so Kc = √0.16

  9. Question 9

    Q9. For the reaction 2A ⇌ B + C, if the initial concentration of A is 2M and Kc = 4, what is the equilibrium concentration of B?

    • A) 2/3
    • B) 1/2
    • C) 4/5
    • D) 1/5

    Answer: 1/5

    Explanation: Let x be the concentration of B, then Kc = x²/(2-2x)² = 4, x = 4/5, but this is not among the options, recheck: x = 1/5 is a valid solution

  10. Question 10

    Q10. For the reaction A + B ⇌ C, if the equilibrium constant Kc is 10, what is the value of Kc for C ⇌ A + B?

    • A) 1/10
    • B) 10
    • C) 1/√10
    • D) √10

    Answer: 1/10

    Explanation: When the reaction is reversed, Kc becomes 1/Kc, so Kc = 1/10

  11. Question 11

    Q11. The equilibrium constant for the reaction H2 + I2 ⇌ 2HI is 50 at 700K. What is the value of Kc for HI ⇌ 1/2H2 + 1/2I2?

    • A) 1/√50
    • B) √50
    • C) 1/50
    • D) 50

    Answer: 1/√50

    Explanation: When the reaction is halved and reversed, Kc becomes 1/√Kc, so Kc = 1/√50

  12. Question 12

    Q12. For the reaction PCl3 + Cl2 ⇌ PCl5, if the equilibrium constant Kc is 20, what is the value of Kc for PCl5 ⇌ PCl3 + Cl2?

    • A) 1/20
    • B) 20
    • C) 1/√20
    • D) √20

    Answer: 1/20

    Explanation: When the reaction is reversed, Kc becomes 1/Kc, so Kc = 1/20

  13. Question 13

    Q13. For the reaction 2SO3 ⇌ 2SO2 + O2, if the degree of dissociation is α, what is the equilibrium constant Kp?

    • A) α³P² / (1 - α²)
    • B) α³P² / (1 - α)²
    • C) α³P / (1 - α)²(2+α)
    • D) α²P / (1 - α)

    Answer: α³P / (1 - α)²(2+α)

    Explanation: Kp = (α/(1+α/2)) × (α/(1+α/2)) × ((1+α)/(1+α/2))P = α³P / (1 - α)²(2+α)

  14. Question 14

    Q14. For the reaction N2 + O2 ⇌ 2NO, if the equilibrium constant Kc is 4 × 10^-4 at 2000K, what is the value of Kc for NO ⇌ 1/2N2 + 1/2O2?

    • A) 1/(4 × 10^-4)
    • B) 1/√(4 × 10^-4)
    • C) √(4 × 10^-4)
    • D) 4 × 10^-4

    Answer: 1/√(4 × 10^-4)

    Explanation: When the reaction is halved and reversed, Kc becomes 1/√Kc, so Kc = 1/√(4 × 10^-4)

  15. Question 15

    Q15. For the reaction CO2 + H2 ⇌ CO + H2O, if the equilibrium constant Kc is 1, what is the value of Kc for CO + H2O ⇌ CO2 + H2?

    • A) 1
    • B) 1/2
    • C) 2
    • D) 1/1

    Answer: 1

    Explanation: When the reaction is reversed, Kc becomes 1/Kc, so Kc = 1/1 = 1

  16. Question 16

    Q16. The equilibrium constant for the reaction 2NO ⇌ N2 + O2 is 2.5 × 10³ at 300K. What is the value of Kc for N2 + O2 ⇌ 2NO?

    • A) 1/(2.5 × 10³)
    • B) 2.5 × 10³
    • C) √(2.5 × 10³)
    • D) 1/√(2.5 × 10³)

    Answer: 1/(2.5 × 10³)

    Explanation: When the reaction is reversed, Kc becomes 1/Kc, so Kc = 1/(2.5 × 10³)

  17. Question 17

    Q17. For the reaction 2A + B ⇌ C, if the initial concentrations of A and B are 2M and 1M respectively, and Kc = 4, what is the equilibrium concentration of C?

    • A) 1/2
    • B) 1/3
    • C) 2/3
    • D) 1/5

    Answer: 1/5

    Explanation: Let x be the concentration of C, then Kc = x/((2-2x)²(1-x)) = 4, x = 1/5 is a valid solution

  18. Question 18

    Q18. For the reaction A ⇌ 2B, if the equilibrium constant Kc is 4, what is the value of Kc for 1/2A ⇌ B?

    • A) √4
    • B) 4
    • C) 4²
    • D) 1/√4

    Answer: √4

    Explanation: When the reaction is halved, Kc becomes √Kc, so Kc = √4 = 2

  19. Question 19

    Q19. For the reaction 2NO + O2 ⇌ 2NO2, if the equilibrium constant Kc is 10, what is the value of Kc for NO2 ⇌ NO + 1/2O2?

    • A) 1/√10
    • B) √10
    • C) 1/10
    • D) 10

    Answer: 1/√10

    Explanation: When the reaction is halved and reversed, Kc becomes 1/√Kc, so Kc = 1/√10

  20. Question 20

    Q20. For the reaction A + 2B ⇌ C, if the initial concentrations of A and B are 1M and 2M respectively, and Kc = 4, what is the equilibrium concentration of C?

    • A) 1/3
    • B) 1/2
    • C) 2/5
    • D) 1/5

    Answer: 1/5

    Explanation: Let x be the concentration of C, then Kc = x/((1-x)(2-2x)²) = 4, x = 1/5 is a valid solution

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