Tertiary butyl bromide (C4H9Br) reacts with water in the following manner: C4H9Br + H2O = C4H9OH + HBr. The reaction follows the first order rate law rate = K(C4H9Br). If one begins with 1 gram of tertiary butyl bromide and K = 5.0 x 10-3 sec-1, how much tertiary butyl bromide will be left after 3 minutes? What is the half life of this reaction?

Answered: Tertiary butyl bromide (C4H9Br) reacts…

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter11: Chemical Kinetics: Rates Of Reactions

Section: Chapter Questions

Problem 121QRT: If you know some calculus, derive the integrated first-order rate law for the reaction by following...

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Tertiary butyl bromide (C₄H₉Br) reacts with water in the following manner: C₄H₉Br + H₂O = C₄H₉OH + HBr. The reaction follows the first order rate law rate = K(C₄H₉Br). If one begins with 1 gram of tertiary butyl bromide and K = 5.0 x 10^-3 sec^-1, how much tertiary butyl bromide will be left after 3 minutes? What is the half life of this reaction?

Expert Solution

Step 1

Mass of tertiary butyl bromide initially = 1 g

Time = 3 minutes

Rate constant k = 5.0 × 10^-3s^-1

We need to find mass of reactant after 3 min.

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