![If the rate law for a reaction is Rate = k[B]^(3/2), by what factor does the reaction rate increase if the concentration of B is doubled?](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1778543672250-20a2e9b74ddb21d5.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
Here's how to determine the increase in reaction rate: Step 1: Write down the initial rate law. The given experimental rate law is: Rate_1 = k[B]^3/2 where [B] is the initial concentration of B. Step 2: Write down the new rate law when the concentration of B is doubled. If the concentration of B is doubled, the new concentration is 2[B]. The new rate, Rate_2, will be: Rate_2 = k(2[B])^3/2 Step 3: Simplify the new rate expression. Rate_2 = k · 2^3/2 · [B]^3/2 Step 4: Calculate the ratio of the new rate to the initial rate to find the factor of increase. Rate_2Rate_1 = k · 2^3/2 · [B]^3/2k[B]^3/2 Cancel out k and [B]^3/2: Rate_2Rate_1 = 2^3/2 Step 5: Evaluate the numerical value. 2^3/2 = 2^1 · 2^1/2 = 2sqrt(2) The reaction rate will increase by a factor of 2sqrt(2). The reaction rate will increase by a factor of 2sqrt(2). That's 2 down. 3 left today — send the next one.