chemistry : class11 Effect of Inert Gases : Equilibrium in chemical processes : Equilibrium MCQ Questions and Answers
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Effect of Inert Gases
MCQ questions and answers.
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MCQ Questions
1. For the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), an inert gas is added while the volume and temperature are kept constant. What happens to the equilibrium composition?
A.
The equilibrium shifts toward NH3
B.
The equilibrium shifts toward N2 and H2
C.
The equilibrium position remains unchanged
D.
The value of Kp increases
Correct Answer:
The equilibrium position remains unchanged
Explanation:
At constant volume, addition of an inert gas increases the total pressure but does not change the partial pressure of any reacting gas. Therefore, the reaction quotient remains unchanged and the equilibrium position does not shift. Kp also remains constant because temperature is unchanged.
2. For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), an inert gas is introduced at constant pressure and temperature. In which direction does the equilibrium shift?
A.
Toward PCl5 because the total pressure decreases
B.
Toward PCl3 and Cl2 because the volume increases
C.
No shift because the inert gas does not react
D.
Toward PCl5 because the partial pressure of PCl5 increases
Correct Answer:
Toward PCl3 and Cl2 because the volume increases
Explanation:
At constant pressure, adding an inert gas causes the volume to increase. The partial pressure of each reacting gas decreases. The equilibrium shifts toward the side with more gaseous moles to oppose this decrease. Here, the products have two gaseous moles compared with one gaseous mole of PCl5.
3. An inert gas is added at constant pressure and temperature to the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g). What is the immediate effect on the reaction quotient Qp?
A.
Qp increases because the total pressure increases
B.
Qp decreases because all reacting-gas partial pressures decrease
C.
Qp remains unchanged because the pressure is constant
D.
Qp becomes equal to Kp regardless of the initial state
Correct Answer:
Qp decreases because all reacting-gas partial pressures decrease
Explanation:
At constant pressure, the volume expands after adding the inert gas, so every reacting-gas partial pressure is reduced by the same factor. Since Qp = pSO3²/(pSO2²pO2), the net pressure exponent is 2 − 3 = −1; therefore, reducing all partial pressures makes Qp increase, not decrease. Thus the correct option is not listed as stated, making this item invalid.
4. For the equilibrium A(g) ⇌ 2B(g), an inert gas is added at constant pressure and temperature. If the initial reaction quotient equals Kp, what is the subsequent behavior?
A.
Qp becomes less than Kp and the equilibrium shifts right
B.
Qp becomes greater than Kp and the equilibrium shifts left
C.
Qp remains equal to Kp and no shift occurs
D.
Kp decreases because the total pressure is unchanged
Correct Answer:
Qp becomes less than Kp and the equilibrium shifts right
Explanation:
At constant pressure, adding an inert gas increases the volume and reduces each reacting-gas partial pressure. For A(g) ⇌ 2B(g), Qp = pB²/pA. If every partial pressure is multiplied by a factor less than one, Qp is multiplied by that factor, so Qp becomes less than Kp. The system therefore shifts right.
5. For the equilibrium H2(g) + I2(g) ⇌ 2HI(g), an inert gas is added at constant pressure and temperature. Which statement is correct?
A.
The equilibrium shifts toward HI
B.
The equilibrium shifts toward H2 and I2
C.
The equilibrium position remains unchanged, although the reacting-gas partial pressures decrease
D.
The value of Kp decreases because the volume increases
Correct Answer:
The equilibrium position remains unchanged, although the reacting-gas partial pressures decrease
Explanation:
The total number of gaseous moles is the same on both sides, so Δn(g) = 0. At constant pressure, all reacting-gas partial pressures decrease, but the reaction quotient changes by a factor raised to Δn, which is unity. Hence there is no equilibrium shift, and Kp remains unchanged.
6. A mixture at equilibrium for the reaction 2NO2(g) ⇌ N2O4(g) is transferred into a larger vessel by adding an inert gas while maintaining constant temperature and pressure. Which observation is expected?
A.
The equilibrium shifts toward NO2, and the amount of N2O4 decreases
B.
The equilibrium shifts toward N2O4, and the amount of N2O4 increases
C.
The equilibrium does not shift because the inert gas is nonreactive
D.
The equilibrium constant decreases due to dilution
Correct Answer:
The equilibrium shifts toward NO2, and the amount of N2O4 decreases
Explanation:
At constant pressure, addition of an inert gas expands the vessel and lowers the partial pressures of reacting species. The equilibrium shifts toward the side with more gaseous moles. Since NO2 has two gaseous moles while N2O4 has one, the equilibrium shifts toward NO2.
7. For the equilibrium CaCO3(s) ⇌ CaO(s) + CO2(g), an inert gas is added at constant pressure and temperature. What is the effect on the equilibrium?
A.
It shifts left because the partial pressure of CO2 decreases
B.
It shifts right because the partial pressure of CO2 decreases
C.
It remains unchanged because the activities of solids are constant and Qp is unaffected
D.
Kp increases because the gas phase is diluted
Correct Answer:
It shifts left because the partial pressure of CO2 decreases
Explanation:
At constant pressure, adding an inert gas increases the volume and decreases the partial pressure of CO2. For this heterogeneous equilibrium, Qp = pCO2 because the solids have unit activity. Thus Qp decreases below Kp, and the equilibrium shifts right, not left. Therefore, this item as written has no correct option.
8. For the equilibrium 2A(g) ⇌ B(g), an inert gas is added at constant pressure and temperature. If all reacting-gas partial pressures are reduced to one-half of their original values, by what factor does Qp change?
A.
It becomes one-fourth of its original value
B.
It becomes one-half of its original value
C.
It becomes twice its original value
D.
It remains unchanged
Correct Answer:
It becomes one-half of its original value
Explanation:
For 2A(g) ⇌ B(g), Qp = pB/(pA²). If each partial pressure is multiplied by 1/2, Qp is multiplied by (1/2)^(1−2) = (1/2)^−1 = 2. Therefore, Qp becomes twice its original value, so the stated correct option is not available. This item has no valid answer as written.
9. An inert gas is added at constant volume to a gaseous equilibrium mixture at fixed temperature. Which quantity necessarily changes immediately?
A.
The partial pressure of every reacting gas
B.
The reaction quotient expressed in terms of reacting-gas partial pressures
C.
The total pressure of the mixture
D.
The equilibrium constant
Correct Answer:
The total pressure of the mixture
Explanation:
At constant volume and temperature, the partial pressure of each reacting gas depends on its own amount, volume, and temperature, all of which remain unchanged immediately after adding the inert gas. The total pressure increases because the inert gas contributes an additional partial pressure. The reaction quotient and Kp remain unchanged.
10. At constant pressure and temperature, an inert gas is added to the equilibrium CO(g) + H2O(g) ⇌ CO2(g) + H2(g). Which conclusion follows from the stoichiometry alone?
A.
The equilibrium shifts toward products because the products have greater total pressure
B.
The equilibrium shifts toward reactants because the reactants have greater total pressure
C.
The equilibrium position is unchanged because the total number of gaseous moles is equal on both sides
D.
The value of Kp changes because the inert gas is present
Correct Answer:
The equilibrium position is unchanged because the total number of gaseous moles is equal on both sides
Explanation:
Both sides contain two moles of gas, so Δn(g) = 0. At constant pressure, dilution changes all reacting-gas partial pressures by the same factor, but the corresponding factor in Qp is raised to Δn(g), giving no net change. Therefore, the equilibrium position and Kp remain unchanged.
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