Chapter-8 Reversible Reactions and Equilibrium — Online MCQ Test
CHEMISTRY · Grade 10 · IGCSE cambridge
Practice Chapter-8 Reversible Reactions and Equilibrium with a free chapter-wise online MCQ test.
This chapter covers: Reversible reactions dynamic equilibrium and the conditions used in industrial processes..
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Chapter-8 Reversible Reactions and Equilibrium — Important Questions & Answers
What is a reversible reaction?
- A. A reaction that can only proceed in one direction
- B. A reaction where reactants can form products and products can reform reactants
- C. A reaction that requires a catalyst to proceed
- D. A reaction that releases energy continuously
Answer: B. A reaction where reactants can form products and products can reform reactants
A reversible reaction is one where products can decompose back into reactants, represented by the double arrow (⇌) symbol.
A reversible reaction is one where products can decompose back into reactants, represented by the double arrow (⇌) symbol.
Which symbol is used to represent a reversible reaction?
- A. →
- B. ⇌
- C. ←
- D. ⇄
Answer: B. ⇌
The double-headed arrow (⇌) represents a reversible reaction where both forward and backward reactions occur simultaneously.
The double-headed arrow (⇌) represents a reversible reaction where both forward and backward reactions occur simultaneously.
Which factor does NOT directly affect the position of equilibrium?
- A. Temperature
- B. Pressure
- C. Concentration of reactants
- D. The presence of a catalyst
Answer: D. The presence of a catalyst
A catalyst speeds up both forward and backward reactions equally, so it does not shift the equilibrium position, only helps reach it faster.
A catalyst speeds up both forward and backward reactions equally, so it does not shift the equilibrium position, only helps reach it faster.
Consider the equilibrium: N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ/mol. Which change will increase the proportion of NO₂ at equilibrium?
- A. Decreasing temperature
- B. Increasing pressure
- C. Increasing temperature
- D. Using a catalyst
Answer: C. Increasing temperature
Since the forward reaction is endothermic (+ΔH), increasing temperature shifts equilibrium right, favoring NO₂ formation and increasing its proportion.
Since the forward reaction is endothermic (+ΔH), increasing temperature shifts equilibrium right, favoring NO₂ formation and increasing its proportion.
Why is a lower temperature NOT preferred in the industrial synthesis of ammonia via the Haber Process, despite the forward reaction being exothermic?
- A. Lower temperature increases the equilibrium constant more than high temperature
- B. Lower temperature makes the reaction rate extremely slow, reducing practical yield per unit time
- C. Lower temperature shifts equilibrium backward
- D. The catalyst only works at high temperatures
Answer: B. Lower temperature makes the reaction rate extremely slow, reducing practical yield per unit time
While lower temperature favors product formation (exothermic), it drastically slows the reaction rate. Industry uses ~450°C as a compromise between yield and practical production speed.
While lower temperature favors product formation (exothermic), it drastically slows the reaction rate. Industry uses ~450°C as a compromise between yield and practical production speed.