For rate = k[NO][O3], a trial with [NO] = 1.00 x 10^-6 M and [O3] = 3.00 x 10^-6 M gives a rate of 6.60 x 10^-5 mol/L/s. What is k?

1. Rearrange the rate law to k = rate divided by the product of the two concentrations. 2. The product of the concentrations is 1.00 x 10^-6 mol/L times 3.00 x 10^-6 mol/L. 3. That product is 3.00 x 10^-12 mol^2/L^2. 4. Divide 6.60 x 10^-5 mol/L/s by 3.00 x 10^-12 mol^2/L^2. 5. The result is 2.20 x 10^7 L/mol/s. 6. The value 2.20 x 10^-7 L/mol/s comes from multiplying by the concentrations instead of dividing. 7. The value 6.60 x 10^7 L/mol/s keeps the rate figure of 6.60 and only fixes the power of ten. _Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.3 Rate Laws_