Question:
For an elementary reaction, \(2A+B\rightarrow A_{2}B\), if \(r=k[A]^{2}[B]\)
If the volume of a vessel is quickly reduced to half of its original volume, then the rate of reaction will be
For an elementary reaction, \(2A+B\rightarrow A_{2}B\), if \(r=k[A]^{2}[B]\)
If the volume of a vessel is quickly reduced to half of its original volume, then the rate of reaction will be
If the volume of a vessel is quickly reduced to half of its original volume, then the rate of reaction will be
Show Hint
Halving the volume doubles every concentration; the rate depends on concentration squared for A and first power for B.
Updated On: Nov 12, 2026
- remain unchanged.
- decreased by half.
- increased by eight times.
- increased by four times.
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Concentration is moles per volume. Halving the volume doubles every concentration.
Step 2: New rate
Original: \(r = k[A]^{2}[B]\).
New: \(r' = k(2[A])^{2}(2[B]) = 8\,k[A]^{2}[B]\).
Step 3: Result
\(r' = 8r\), so the rate increases eight times.
Final Answer:
The rate of reaction increases eight times. \[ \boxed{\text{(C) }\text{increased by eight times}} \]
Concentration is moles per volume. Halving the volume doubles every concentration.
Step 2: New rate
Original: \(r = k[A]^{2}[B]\).
New: \(r' = k(2[A])^{2}(2[B]) = 8\,k[A]^{2}[B]\).
Step 3: Result
\(r' = 8r\), so the rate increases eight times.
Final Answer:
The rate of reaction increases eight times. \[ \boxed{\text{(C) }\text{increased by eight times}} \]
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