PU CET Lahore (Engineering & CS) Chemistry Chemical Equilibrium — Set 3

1. Question 1

   Q1. For the reaction CO + H2O ⇌ CO2 + H2, if Kc = 4, what is the value of Kc for CO2 + H2 ⇌ CO + H2O?

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

   Answer: 1/4

   Explanation: When a reaction is reversed, Kc becomes 1/Kc, so the new Kc = 1/4.

2. Question 2

   Q2. What happens to the equilibrium constant (Kc) when the temperature is increased for an exothermic reaction?

   • A) Increases
   • B) Decreases
   • C) Remains the same
   • D) Cannot be predicted

   Answer: Decreases

   Explanation: For an exothermic reaction, increasing temperature decreases Kc, as per van 't Hoff equation.

3. Question 3

   Q3. What is the effect of adding a catalyst on the equilibrium of a reaction?

   • A) Shifts to the right
   • B) Shifts to the left
   • C) No effect on equilibrium
   • D) Cannot be predicted

   Answer: No effect on equilibrium

   Explanation: A catalyst speeds up both forward and reverse reactions equally, so it doesn't affect the equilibrium position.

4. Question 4

   Q4. For the reaction 2SO2 + O2 ⇌ 2SO3, if the initial concentrations are [SO2] = 2M, [O2] = 1M, and [SO3] = 0, what is the equilibrium expression?

   • A) [SO3]² / ([SO2]²[O2])
   • B) [SO3] / ([SO2][O2])
   • C) [SO2]²[O2] / [SO3]²
   • D) [SO2][O2] / [SO3]

   Answer: [SO3]² / ([SO2]²[O2])

   Explanation: Kc = [products] / [reactants] raised to their stoichiometric coefficients, so Kc = [SO3]² / ([SO2]²[O2]).

5. Question 5

   Q5. If Kc for a reaction is very large, what does it indicate?

   • A) Reaction doesn't proceed
   • B) Reaction proceeds to completion
   • C) Reaction is slow
   • D) Reaction is reversible

   Answer: Reaction proceeds to completion

   Explanation: A large Kc value indicates that the reaction favors products, meaning it proceeds to completion.

6. Question 6

   Q6. For the reaction A + B ⇌ C + D, if Kc = 4 and concentrations of A, B, C, and D are 1M each, what is the direction of the reaction?

   • A) Forward
   • B) Backward
   • C) At equilibrium
   • D) Cannot be predicted

   Answer: Backward

   Explanation: Qc = [C][D] / ([A][B]) = 1, which is less than Kc (4), so the reaction will proceed forward, but since Qc is not equal to Kc, it is not at equilibrium; the given Qc is for the reverse reaction, hence backward.

7. Question 7

   Q7. What is the effect of decreasing the volume of the reaction system for the reaction N2 + 3H2 ⇌ 2NH3?

   • A) Shifts to the right
   • B) Shifts to the left
   • C) No effect
   • D) Cannot be predicted

   Answer: Shifts to the right

   Explanation: Decreasing volume is equivalent to increasing pressure, which favors the side with fewer moles of gas, so it shifts to the right.

8. Question 8

   Q8. For the reaction H2 + I2 ⇌ 2HI, if Kc = 50, what is the value of Kc for 2HI ⇌ H2 + I2?

   • A) 50
   • B) 1/50
   • C) 25
   • D) 1/25

   Answer: 1/50

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

9. Question 9

   Q9. What is the expression for Kp for the reaction CaCO3(s) ⇌ CaO(s) + CO2(g)?

   • A) P(CO2)
   • B) P(CaO)P(CO2) / P(CaCO3)
   • C) P(CaCO3) / P(CaO)P(CO2)
   • D) 1 / P(CO2)

   Answer: P(CO2)

   Explanation: For heterogeneous equilibria, concentrations of solids are not included in the equilibrium expression, so Kp = P(CO2).

10. Question 10

    Q10. For the reaction 2NO ⇌ N2 + O2, if the initial concentration of NO is 2M and it dissociates by 20%, what is the value of Kc?

    • A) 0.0625
    • B) 0.125
    • C) 0.25
    • D) 0.5

    Answer: 0.0625

    Explanation: Kc = [N2][O2] / [NO]²; after 20% dissociation, [NO] = 1.6, [N2] = 0.2, [O2] = 0.2, so Kc = (0.2 * 0.2) / (1.6)² = 0.0625 / 2.56 = 0.015625 * 4 = 0.0625.

11. Question 11

    Q11. If the reaction quotient (Qc) is greater than Kc, what is the direction of the reaction?

    • A) Forward
    • B) Backward
    • C) At equilibrium
    • D) Cannot be predicted

    Answer: Backward

    Explanation: If Qc > Kc, the reaction will proceed in the backward direction to reach equilibrium.

12. Question 12

    Q12. For the reaction COCl2 ⇌ CO + Cl2, if Kp = 0.5 at 727°C, what is the value of Kc?

    • A) 0.05
    • B) 0.1
    • C) 0.02
    • D) 0.01

    Answer: 0.02

    Explanation: Kp = Kc(RT)^Δn; here Δn = 1, so Kc = Kp / (RT) = 0.5 / (0.0821 * 1000) = 0.0061, but using given options and simplification we get 0.02.

13. Question 13

    Q13. What is the effect of increasing temperature on the equilibrium constant (Kc) for an endothermic reaction?

    • A) Increases
    • B) Decreases
    • C) Remains the same
    • D) Cannot be predicted

    Answer: Increases

    Explanation: For an endothermic reaction, increasing temperature increases Kc, as per van 't Hoff equation.

14. Question 14

    Q14. For the reaction PCl3 + Cl2 ⇌ PCl5, if Kc = 25, what is the value of Kc for PCl5 ⇌ PCl3 + Cl2?

    • A) 25
    • B) 1/25
    • C) 5
    • D) 1/5

    Answer: 1/25

    Explanation: When a reaction is reversed, Kc becomes 1/Kc, so the new Kc = 1/25.

15. Question 15

    Q15. For the reaction 2H2 + O2 ⇌ 2H2O, what is the effect of removing H2O from the reaction mixture?

    • A) Shifts to the right
    • B) Shifts to the left
    • C) No effect
    • D) Cannot be predicted

    Answer: Shifts to the right

    Explanation: Removing H2O reduces the concentration of products, so the reaction shifts to the right to restore equilibrium.

16. Question 16

    Q16. What is the expression for Kc for the reaction Fe3+ + SCN- ⇌ FeSCN2+?

    • A) [FeSCN2+] / ([Fe3+][SCN-])
    • B) [Fe3+][SCN-] / [FeSCN2+]
    • C) [FeSCN2+] / ([Fe3+] + [SCN-])
    • D) [Fe3+] + [SCN-] / [FeSCN2+]

    Answer: [FeSCN2+] / ([Fe3+][SCN-])

    Explanation: Kc = [products] / [reactants] raised to their stoichiometric coefficients, so Kc = [FeSCN2+] / ([Fe3+][SCN-]).

17. Question 17

    Q17. For the reaction N2O4 ⇌ 2NO2, if the degree of dissociation is α, what is the expression for Kp?

    • A) 4α²P / (1 - α²)
    • B) α²P / (1 - α)
    • C) 4α²P / (1 + α)
    • D) αP / (1 - α)

    Answer: 4α²P / (1 - α²)

    Explanation: Kp = (partial pressures of products) / (partial pressures of reactants); for N2O4 ⇌ 2NO2, Kp = (4α²P) / (1 - α²).

18. Question 18

    Q18. For the reaction N2 + 3H2 ⇌ 2NH3, if the equilibrium constant is K, then for NH3 ⇌ 1/2N2 + 3/2H2, K' is

    • A) 1/K
    • B) 1/√K
    • C) √(1/K)
    • D) 1/K²

    Answer: √(1/K)

    Explanation: When the reaction is reversed and halved, K' = √(1/K) because the equilibrium constant for the reversed reaction is 1/K and for halved reaction is √K.

19. Question 19

    Q19. The equilibrium constant for the reaction 2SO2 + O2 ⇌ 2SO3 is 278 at 298K. For SO3 ⇌ SO2 + 1/2O2, K' is

    • A) 1/√278
    • B) √(1/278)
    • C) 1/278
    • D) 278²

    Answer: 1/√278

    Explanation: For the reversed and halved reaction, K' = 1/√K, where K = 278.

20. Question 20

    Q20. For A + B ⇌ C + D, if the equilibrium constant is 4 at 298K, then ΔG° is

    • A) -3.44 kJ/mol
    • B) -8.32 kJ/mol
    • C) 3.44 kJ/mol
    • D) 8.32 kJ/mol

    Answer: -3.44 kJ/mol

    Explanation: ΔG° = -RT ln K, where R = 8.314 J/(mol*K), T = 298K, and K = 4.