For a reaction with rate = k[A]^m[B]^n, if doubling [A] while keeping [B] constant doubles the rate, what is m?

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Study for the IMAT Chemistry Exam. Access flashcards and multiple choice questions; hints and explanations included for each question. Prepare effectively for your exam.

Multiple Choice

For a reaction with rate = k[A]^m[B]^n, if doubling [A] while keeping [B] constant doubles the rate, what is m?

Explanation:
When the rate law is rate = k[A]^m[B]^n, changing a reactant’s concentration changes the rate by the concentration raised to its exponent. Doubling [A] multiplies the rate by 2^m (since [B] is held constant). Since the rate doubles, we have 2^m = 2, which gives m = 1. That means the reaction is first order in A. If the exponent were 0.5, doubling [A] would multiply the rate by √2, not 2; if it were 2, doubling would multiply the rate by 4.

When the rate law is rate = k[A]^m[B]^n, changing a reactant’s concentration changes the rate by the concentration raised to its exponent. Doubling [A] multiplies the rate by 2^m (since [B] is held constant). Since the rate doubles, we have 2^m = 2, which gives m = 1. That means the reaction is first order in A. If the exponent were 0.5, doubling [A] would multiply the rate by √2, not 2; if it were 2, doubling would multiply the rate by 4.

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