chemistry : class11 Industrial Chemical Equilibria : Equilibrium in chemical processes : Equilibrium MCQ Questions and Answers
Practice important
Industrial Chemical Equilibria
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH < 0. Which combination of changes increases the equilibrium yield of ammonia while also increasing the rate of attainment of equilibrium?
A.
Increase pressure, decrease temperature, and add an iron catalyst
B.
Increase pressure, increase temperature, and add an iron catalyst
C.
Decrease pressure, decrease temperature, and remove ammonia continuously
D.
Increase pressure, decrease temperature, and add an inert gas at constant volume
Correct Answer:
Increase pressure, decrease temperature, and add an iron catalyst
Explanation:
The forward reaction is exothermic and produces fewer gas molecules, so lower temperature and higher pressure favor ammonia formation. The iron catalyst increases the rate of reaching equilibrium but does not change the equilibrium composition.
2. For the Contact process equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH < 0, which industrial strategy gives a high equilibrium conversion without using an impractically low temperature?
A.
Use a catalyst, moderately high pressure, and remove SO3 as it forms
B.
Use a catalyst, very low pressure, and continuously add SO3
C.
Use no catalyst, high temperature, and continuously add SO3
D.
Use a catalyst, constant-volume inert gas addition, and recycle SO3
Correct Answer:
Use a catalyst, moderately high pressure, and remove SO3 as it forms
Explanation:
The forward reaction is exothermic and changes three moles of gas into two, so pressure favors SO3. Removing SO3 shifts the equilibrium forward, while the catalyst permits a practical operating temperature by increasing reaction rate.
3. In the industrial synthesis of ammonia, the equilibrium constant Kp for N2(g) + 3H2(g) ⇌ 2NH3(g) is fixed at a given temperature. If an iron catalyst is introduced after equilibrium has been established, which quantity changes?
A.
Kp only
B.
The equilibrium partial pressure of ammonia only
C.
The time required to reach equilibrium only
D.
The equilibrium constant and the equilibrium partial pressures
Correct Answer:
The time required to reach equilibrium only
Explanation:
A catalyst lowers the activation energy of both forward and reverse reactions. Therefore, equilibrium is reached faster, but Kp and the equilibrium composition remain unchanged at the same temperature and pressure.
4. For the Haber equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), an inert gas is added while the total volume and temperature are kept constant. What is the immediate effect on the equilibrium?
A.
The equilibrium shifts toward ammonia because total pressure increases
B.
The equilibrium shifts toward nitrogen and hydrogen because ammonia is diluted
C.
There is no shift because the partial pressures of the reacting gases remain unchanged
D.
The equilibrium constant decreases because the inert gas increases total pressure
Correct Answer:
There is no shift because the partial pressures of the reacting gases remain unchanged
Explanation:
At constant volume and temperature, adding an inert gas increases total pressure but does not change the partial pressure of any reacting gas. The reaction quotient therefore remains unchanged, so the equilibrium does not shift.
5. In the Contact process reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), the reaction mixture is compressed at constant temperature. Which statement correctly describes the immediate and eventual effects?
A.
Qp becomes smaller than Kp, so the reaction proceeds forward
B.
Qp becomes larger than Kp, so the reaction proceeds backward
C.
Qp remains equal to Kp, so no reaction occurs
D.
Kp increases immediately, causing SO3 formation
Correct Answer:
Qp becomes smaller than Kp, so the reaction proceeds forward
Explanation:
On compression, all reacting partial pressures increase, but Qp is proportional to pSO3²/(pSO2²pO2). Because the denominator contains three total powers of pressure and the numerator contains two, Qp decreases, becoming less than Kp; the system then forms more SO3.
6. In the Ostwald process, ammonia is oxidized to nitric oxide over a platinum-rhodium catalyst. If the oxidation step is exothermic, which operating choice is most consistent with obtaining a high equilibrium conversion while maintaining a useful production rate?
A.
Very low temperature without a catalyst
B.
Moderately high temperature with a catalyst and rapid removal of products
C.
Very high temperature with continuous addition of nitric oxide
D.
Low pressure with continuous recycling of nitric oxide into the reactor
Correct Answer:
Moderately high temperature with a catalyst and rapid removal of products
Explanation:
Lower temperature favors an exothermic oxidation equilibrium, but an excessively low temperature gives an impractically slow rate. Industrial operation uses a catalyst and a compromise temperature, while product removal drives the equilibrium toward products.
7. For the equilibrium CO(g) + H2O(g) ⇌ CO2(g) + H2(g), used in hydrogen manufacture, which change increases the equilibrium conversion of carbon monoxide when the temperature is held constant?
A.
Add steam and remove hydrogen
B.
Add carbon dioxide and remove steam
C.
Add hydrogen and remove carbon monoxide
D.
Add an inert gas at constant volume
Correct Answer:
Add steam and remove hydrogen
Explanation:
Steam is a reactant and hydrogen is a product. Increasing the reactant concentration and continuously removing a product makes the reaction quotient smaller than K, causing the equilibrium to shift toward CO2 and H2.
8. For the exothermic reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), which temperature change causes the greatest decrease in the equilibrium constant Kp?
A.
A small increase in temperature
B.
A large increase in pressure
C.
Addition of a vanadium(V) oxide catalyst
D.
Removal of sulfur trioxide at constant temperature
Correct Answer:
A small increase in temperature
Explanation:
For an exothermic reaction, increasing temperature favors the reverse reaction and decreases Kp. Pressure changes, catalysts, and product removal can alter composition or reaction rate, but they do not change Kp unless temperature changes.
9. An ammonia plant removes NH3 from the reacting gas mixture by cooling and condensation after equilibrium. The condensed ammonia is separated, and unreacted N2 and H2 are recycled. What is the principal equilibrium advantage of this operation?
A.
It increases the value of Kp at the operating temperature
B.
It continuously lowers the ammonia partial pressure, driving further ammonia formation
C.
It makes the forward reaction activation energy zero
D.
It changes the stoichiometric ratio of nitrogen and hydrogen required by the reaction
Correct Answer:
It continuously lowers the ammonia partial pressure, driving further ammonia formation
Explanation:
Removing ammonia, a product, decreases its partial pressure. The reaction quotient becomes smaller than Kp, so additional nitrogen and hydrogen react to restore equilibrium. Recycling the unreacted gases improves overall plant conversion.
10. For N2(g) + 3H2(g) ⇌ 2NH3(g), a reaction vessel is expanded at constant temperature after the system reaches equilibrium. Which statement is correct?
A.
The equilibrium shifts toward NH3 because ammonia has the smaller stoichiometric coefficient
B.
The equilibrium shifts toward N2 and H2 because expansion favors the side with more gas molecules
C.
The equilibrium does not shift because Kp changes with volume
D.
The equilibrium shifts toward NH3 because its partial pressure decreases least
Correct Answer:
The equilibrium shifts toward N2 and H2 because expansion favors the side with more gas molecules
Explanation:
Expansion lowers pressure. The equilibrium shifts toward the side with more gaseous molecules, namely the reactant side with four moles compared with two moles of ammonia. Kp remains constant because temperature is unchanged.
Short Questions and Answers
Practice short questions and answers related to
Industrial Chemical Equilibria
.
Short questions are currently unavailable for this page.