The rate law for A + B → products is rate = k[A]^m[B]^n. If doubling [A] (with B constant) doubles the rate, what is the order with respect to A?

• A. Zero order
• B. First order
• C. Second order
• D. Third order

Answer: (B)

The rate dependence on A is captured by the exponent m in the rate law. If you double [A] while keeping [B] constant, the rate should change by a factor of 2^m. Since the rate doubles, 2^m = 2, which gives m = 1. That means the reaction is first order with respect to A. The term involving B becomes a constant multiplier because [B] is unchanged, so it doesn’t affect this proportionality. If the dependence were zero order, the rate wouldn’t change when [A] changes; if it were second order, doubling [A] would quadruple the rate. The observed doubling selects the first-order dependence on A.