![A reaction rate depends on the concentration of reactant B according to the rate law: Rate = k[B]^(3/2). If the concentration of B is doubled, by what factor does the rate increase?](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1778543778205-39f3d5bfa9dd3391.png&w=3840&q=75)
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
✅ Correct. Your selected answer is correct. The initial rate law is Rate_1 = k[B]^3/2. When the concentration of B is doubled, the new rate is Rate_2 = k(2[B])^3/2 = k · 2^3/2 · [B]^3/2. The ratio of the new rate to the initial rate is Rate_2Rate_1 = 2^3/2. Calculating the value: 2^3/2 = 2sqrt(2) ≈ 2 × 1.414 = 2.828. This value is approximately 2.8, which matches your selection. 3 done, 2 left today. You're making progress.