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

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

• A. 2
• B. 0.5
• C. 1
• D. 0

Answer: (C)

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.