In a rate law rate = k[A]^m[B]^n, what is the overall order?

• A. m
• B. m+n
• C. n
• D. m-n

Answer: (B)

The rate depends on each reactant’s concentration raised to its own exponent, and the total or overall order is the sum of those exponents. Since rate ∝ [A]^m [B]^n, how fast the reaction goes changes with A and B as if you’re multiplying the effects: doubling A changes the rate by 2^m, doubling B changes it by 2^n, and doubling both changes it by 2^(m+n). So the overall order is m + n. It’s also worth remembering that m and n come from experiment and describe how the mechanism responds to each reactant, not necessarily the stoichiometric coefficients. For example, if m = 1 and n = 2, the rate law is rate ∝ [A][B]^2 and the overall order is 3.