![Determine the units of the specific rate constant k given the rate law d[x]/dt = k[B]^(3/4).](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1778543275819-331f4b531d15e99c.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
Hey ThegirlGodishelping🥰💙🫠, good to see you again. To find the units of the specific rate constant k, we need to analyze the given rate equation and the units of the other terms. Step 1: Identify the units of each term in the rate equation. The rate equation is given as: (d[x])/(dt) = k[B]^3/4 The term (d[x])/(dt) represents the rate of reaction. Since [x] is concentration in M (Molarity) and t is time in seconds (s), the units of the rate are Ms. The term [B] represents concentration, so its unit is M. Therefore, the unit of [B]^3/4 is M^3/4. Step 2: Rearrange the rate equation to solve for k. From the equation (d[x])/(dt) = k[B]^3/4, we can express k as: k = (d[x])/(dt)[B]^3/4 Step 3: Substitute the units into the rearranged equation for k. Units of k = Units of (d[x])/(dt)Units of [B]^3/4 = M/sM^3/4 Step 4: Simplify the units. Units of k = M^1 · s^-1 · M^-3/4 Combine the terms with M: Units of k = M^(1 - 3/4) · s^-1 Units of k = M^(4/4 - 3/4) · s^-1 Units of k = M^1/4 · s^-1 The units of the specific rate constant k are M^1/4s^-1. Send me the next one 📸