For 2NO plus 2H2, doubling [H2] doubles the rate while doubling [NO] quadruples it. What is the overall order?

1. Take the pair of experiments in which [NO] is held fixed and [H2] is doubled. 2. The rate ratio there is 2, so 2 = 2^n and the order in H2 is 1. 3. Now take the pair in which [H2] is held fixed and [NO] is doubled. 4. The rate ratio there is 4, so 4 = 2^m and the order in NO is 2. 5. The overall order is the sum of the two, that is 2 plus 1. 6. So the reaction is third order overall and the rate law is rate = k[NO]^2[H2]. 7. Answering second order comes from reading the coefficients of the equation instead of the data. _Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.6 Reaction Mechanisms_