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For a reaction involving A and B, the rate expression is
Rate = k[A]ⁿ[B]^m
If the concentration of A is made twice and the concentration of B is reduced to one-half, then the new rate compared with the original rate is given by:
- m + n
- n - m
- 2^(n - m)
- 1/2^(m + n)
Correct answer: 2^(n - m)
Solution
New rate / old rate = [2A]^n [B/2]^m / ([A]^n [B]^m) = 2^n * (1/2)^m = 2^(n-m). The factor by which the rate changes is 2^(n-m).
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