Rate Law And Rate Constant
For a reaction A + B → Products, the rate law is: rate = k[A]²[B]. If the concentration of A is doubled and B is halved, the new rate compared to the original rate is:
Select the correct option:
Solution
2 times the original rate
Original rate = k[A]²[B]. When [A] → 2[A] and [B] → [B]/2: New rate = k[2A]²[B/2] = k × 4[A]² × [B]/2 = 2k[A]²[B] = 2 × original rate. The quadrupling effect due to doubling of [A] is offset by halving of [B], resulting in a net doubling of the rate.
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About This Question
- Subject
- chemistry
- Chapter
- chemical kinetics
- Topic
- rate law and rate constant
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2 times the original rate
Original rate = k[A]²[B]. When [A] → 2[A] and [B] → [B]/2: New rate = k[2A]²[B/2] = k × 4[A]² × [B]/2 = 2k[A]²[B] = 2 × original rate. The quadrupling effect due to doubling of [A] is offset by halving of [B], resulting in a net doubling of the rate.
This medium difficulty chemistry question is from the chapter chemical kinetics, covering the topic of rate law and rate constant. It appeared in the 2025 exam.
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